supercritical solvents · 2018-08-30 · learning outcomes 803 chapter summary and key terms...

800 chaptEr 18 chemistry of the Environment

shortcomings, the two-step process is replaced by a one-step process in which toluene is reacted with methanol at 425 °C over a special catalyst:

CH3OH+−H2, −H2O

Basecatalyst

CH3

Toluene Methanol

CH CH2

Styrene

The one-step process saves money both because toluene and methanol are less expen-sive than benzene and ethylene, and because the reaction requires less energy input. Additional benefits are that the methanol could be produced from biomass and that benzene is replaced by less-toxic toluene. The hydrogen formed in the reaction can be recycled as a source of energy. (This example demonstrates how finding the right cata-lyst is often key in discovering a new process.)

Let’s consider some other examples in which green chemistry can operate to improve environmental quality.

Supercritical SolventsA major area of concern in chemical processes is the use of volatile organic compounds as solvents. Generally, the solvent in which a reaction is run is not consumed in the reaction, and there are unavoidable releases of solvent into the atmosphere even in the most carefully controlled processes. Further, the solvent may be toxic or may decom-pose to some extent during the reaction, thus creating waste products.

The use of supercritical fluids represents a way to replace conventional solvents. Recall that a supercritical fluid is an unusual state of matter that has properties of both a gas and a liquid. (Section 11.4) Water and carbon dioxide are the two most popu-lar choices as supercritical fluid solvents. One recently developed industrial process, for example, replaces chlorofluorocarbon solvents with liquid or supercritical CO2 in the pro-duction of polytetrafluoroethylene 13CF2CF24n, sold as Teflon®2. Though CO2 is a green-house gas, no new CO2 need be manufactured for use as a supercritical fluid solvent.

As a further example, para-xylene is oxidized to form terephthalic acid, which is used to make polyethylene terephthalate (PET) plastic and polyester fiber (Section 12.8, Table 12.5):

CH3 CH3190 °C, 20 atm

Catalyst3 O2+ 2 H2O+CHO OH

O

para-Xylene

C

O

Terephthalic acid

This commercial process requires pressurization and a relatively high temperature. Oxygen is the oxidizing agent, and acetic acid 1CH3COOH2 is the solvent. An alter-native route employs supercritical water as the solvent and hydrogen peroxide as the oxidant. This alternative process has several potential advantages, most particularly the elimination of acetic acid as solvent.

give it some ThoughtWe noted earlier that increasing carbon dioxide levels contribute to global climate change, which seems like a bad thing, but now we are saying that using carbon dioxide in industrial processes is a good thing for the environment. Explain this seeming contradiction.

Greener Reagents and ProcessesLet us examine two more examples of green chemistry in action.

Hydroquinone, HO-C6H4 -OH, is a common intermediate used to make poly-mers. The standard industrial route to hydroquinone, used until recently, yields many by-products that are treated as waste:

sEction 18.5 Green chemistry 801

NH2

2 24 MnO2 5 H2SO4

Hydroquinone

++ (NH4)2SO4 4 MnSO4

Fe, HCl Waste

+ 4 H2O+

O

O

OH

OH

+

FeCl22 +

Using the principles of green chemistry, researchers have improved this process. The new process for hydroquinone production uses a new starting material. Two of the by-products of the new reaction (shown in green) can be isolated and used to make the new starting material.

CH3

CH3

CH2

CH3

HO

HO

C OH

OH

HO CCatalyst

By-products recycledto make starting material

H2O2

+

+

CH3

CH3

CO

OH

The new process is an example of “atom economy,” a phrase that means that a high percentage of the atoms from the starting materials end up in the product.

give it some ThoughtWhere might there be room to make changes in this process that would make hydroquinone production even greener?

Another example of atom economy is a reaction in which, at room temperature and in the presence of a copper(I) catalyst, an organic azide and an alkyne form one product molecule:

R1

R1

NN+

N−Azide Alkyne

Cu(I)+ CHC R2

R2

C

N

HC N

N


This reaction is informally called a click reaction. The yield—actual, not just theoretical—is close to 100%, and there are no by-products. Depending on the type of azide and type of alkyne we start with, this very efficient click reaction can be used to create any number of valuable product molecules.

give it some ThoughtWhat are the hybridizations of the two alkyne C atoms before and after the click reaction?

(a) Acid rain is no threat to lakes in areas where the rock is limestone (calcium carbonate), which can neutralize the acid. Where the rock is granite, however, no neutralization occurs. How does limestone neutralize acid? (b) Acidic water can be treated with basic substances to increase the pH, although such a procedure is usually only a temporary cure. Calculate the mini-mum mass of lime, CaO, needed to adjust the pH of a small lake 1V = 4 * 109 L2 from 5.0 to 6.5. Why might more lime be needed?

soLuTionAnalyze We need to remember what a neutralization reaction is and calculate the amount of a substance needed to effect a certain change in pH.Plan For (a), we need to think about how acid can react with calcium carbonate, a reaction that evidently does not happen with acid and granite. For (b), we need to think about what reaction between an acid and CaO is possible and do stoichiometric calculations. From the proposed change in pH, we can calculate the change in proton concentration needed and then figure out how much CaO is needed.Solve

(a) The carbonate ion, which is the anion of a weak acid, is basic (Sections 16.2 and 16.7) and so reacts with H+1aq2. If the concentration of H+1aq2 is low, the major product is the bicarbonate ion, HCO3

-. If the concentration of H+1aq2 is high, H2CO3 forms and decom-poses to CO2 and H2O. (Section 4.3)

(b) The initial and final concentrations of H+1aq2 in the lake are obtained from their pH values:

3H+4initial = 10-5.0 = 1 * 10-5 M and 3H+4final = 10-6.5 = 3 * 10-7 M

Using the lake volume, we can calculate the number of moles of H+1aq2 at both pH values:

11 * 10-5 mol>L214.0 * 109 L2 = 4 * 104 mol 13 * 10-7 mol>L214.0 * 109 L2 = 1 * 103 mol

Hence, the change in the amount of H+1aq2 is 4 * 104 mol - 1 * 103 mol ≈ 4 * 104 mol.Let’s assume that all the acid in the lake is completely ionized, so that only the

free H+1aq2 contributing to the pH needs to be neutralized. We need to neutralize at least that much acid, although there may be a great deal more than that amount in the lake.

The oxide ion of CaO is very basic. (Section 16.5) In the neutralization reaction, 1 mol of O2- reacts with 2 mol of H+ to form H2O. Thus, 4 * 104 mol of H+ requires

14 * 104 mol H+2a 1 mol CaO2 mol H+ b a 56.1 g CaO

1 mol CaOb = 1 * 106 g CaO

This is slightly more than a ton of CaO. That would not be very costly because CaO is inex-pensive, selling for less than $100 per ton when purchased in large quantities. This amount of CaO is the minimum amount needed, however, because there are likely to be weak acids in the water that must also be neutralized.

This liming procedure has been used to bring the pH of some small lakes into the range necessary for fish to live. The lake in our example would be about a half mile long and a half mile wide and have an average depth of 20 ft.

saMpLE inTEGraTivE ExErCisE putting Concepts Together

Learning outcomes 803

Chapter summary and Key TermsearTh’s aTmosphere (secTion 18.1) In this section we exam-ined the physical and chemical properties of Earth’s atmosphere. The complex temperature variations in the atmosphere give rise to four regions, each with characteristic properties. The lowest of these regions, the troposphere, extends from Earth’s surface up to an altitude of about 12 km. Above the troposphere, in order of increasing alti-tude, are the stratosphere, mesosphere, and thermosphere. In the upper reaches of the atmosphere, only the simplest chemical species can sur-vive the bombardment of highly energetic particles and radiation from the Sun. The average molecular weight of the atmosphere at high eleva-tions is lower than that at Earth’s surface because the lightest atoms and molecules diffuse upward and also because of photodissociation, which is the breaking of bonds in molecules because of the absorption of light. Absorption of radiation may also lead to the formation of ions via photoionization.

human acTiViTies anD earTh’s aTmosphere (secTion 18.2) Ozone is produced in the upper atmosphere from the reaction of atomic oxygen with O2. Ozone is itself decomposed by absorp-tion of a photon or by reaction with an active species such as Cl. chlorofluorocarbons can undergo photodissociation in the strato-sphere, introducing atomic chlorine, which is capable of catalytically destroying ozone. A marked reduction in the ozone level in the upper atmosphere would have serious adverse consequences because the ozone layer filters out certain wavelengths of harmful ultraviolet light that are not removed by any other atmospheric component. In the tro-posphere the chemistry of trace atmospheric components is of major importance. Many of these minor components are pollutants. Sulfur dioxide is one of the more noxious and prevalent examples. It is oxi-dized in air to form sulfur trioxide, which, upon dissolving in water, forms sulfuric acid. The oxides of sulfur are major contributors to acid rain. One method of preventing the escape of SO2 from industrial oper-ations is to react it with CaO to form calcium sulfite CaSO3.

photochemical smog is a complex mixture in which both nitrogen oxides and ozone play important roles. Smog components are gener-ated mainly in automobile engines, and smog control consists largely of controlling auto emissions.

Carbon dioxide and water vapor are the major components of the atmosphere that strongly absorb infrared radiation. CO2 and H2O are therefore critical in maintaining Earth’s surface temperature. The concentrations of CO2 and other so-called greenhouse gases in the atmosphere are thus important in determining worldwide climate.

Because of the extensive combustion of fossil fuels (coal, oil, and nat-ural gas), the concentration of carbon dioxide in the atmosphere is steadily increasing.

earTh’s WaTer (secTion 18.3) Earth’s water is largely in the oceans and seas; only a small fraction is freshwater. Seawater contains about 3.5% by mass of dissolved salts and is described as having a salinity (grams of dry salts per 1 kg seawater) of 35. Seawater’s density and salinity vary with depth. Because most of the world’s water is in the oceans, humans may eventually need to recover freshwater from seawater. The global water cycle involves continuous phase changes of water.

human acTiViTies anD WaTer qualiTy (secTion 18.4) Fresh-water contains many dissolved substances including dissolved oxy-gen, which is necessary for fish and other aquatic life. Substances that are decomposed by bacteria are said to be biodegradable. Because the oxidation of biodegradable substances by aerobic bacteria consumes dissolved oxygen, these substances are called oxygen-demanding wastes. The presence of an excess amount of oxygen-demanding wastes in water can sufficiently deplete the dissolved oxygen to kill fish and produce offensive odors. Plant nutrients can contribute to the problem by stimulating the growth of plants that become oxygen-demanding wastes when they die. Desalination is the removal of dissolved salts from seawater or brackish water to make it fit for human consumption. Desalination may be accomplished by distillation or by reverse osmosis.

The water available from freshwater sources may require treat-ment before it can be used domestically. The several steps generally used in municipal water treatment include coarse filtration, sedimen-tation, sand filtration, aeration, sterilization, and sometimes water softening.

Water supplies may be impacted by the practice of fracking, in which water laden with sand and a variety of chemicals is pumped at high pressure into rock formations to release natural gas and other petroleum materials.

green chemisTry (secTion 18.5) The green chemistry initiative promotes the design and application of chemical products and pro-cesses that are compatible with human health and that preserve the environment. The areas in which the principles of green chemistry can operate to improve environmental quality include choices of sol-vents and reagents for chemical reactions, development of alternative processes, and improvements in existing systems and practices.

Learning outcomes after studying this chapter, you should be able to:

• Describe the regions of Earth’s atmosphere in terms of how tem-perature varies with altitude. (Section 18.1)

• Describe the composition of the atmosphere in terms of the major components in dry air at sea level. (Section 18.1)

• Calculate concentrations of gases in parts per million (ppm). (Section 18.1)

• Describe the processes of photodissociation and photoionization and their role in the upper atmosphere. (Section 18.1)

• Use bond energies and ionization energies to calculate the minimum frequency or maximum wavelength needed to cause photodissocia-tion or photoionization. (Section 18.1)

• Explain how ozone in the upper atmosphere functions to filter short wavelength solar radiation. (Section 18.1)

• Explain how chlorofluorocarbons (CFCs) cause depletion of the ozone layer. (Section 18.2)

• Describe the origins and behavior of sulfur oxides and nitrogen oxides as air pollutants, including the generation of acid rain and photochemical smog. (Section 18.2)

• Describe how water and carbon dioxide cause an increase in atmo-spheric temperature near Earth’s surface. (Section 18.2)

• Describe the global water cycle. (Section 18.3)

• Explain what is meant by the salinity of water and describe the pro-cess of reverse osmosis as a means of desalination. (Section 18.4)

• List the major cations, anions, and gases present in natural waters and describe the relationship between dissolved oxygen and water quality. (Section 18.4)


visualizing Concepts

18.1 At 273 K and 1 atm pressure, 1 mol of an ideal gas occupies 22.4 L. (Section 10.4) (a) Looking at Figure 18.1 predict whether a 1 mol sample of the atmosphere in the middle of the strato-sphere would occupy a greater or smaller volume than 22.4 L (b) Looking at Figure 18.1, we see that the temperature is lower at 85 km altitude than at 50 km. Does this mean that one mole of an ideal gas would occupy less volume at 85 km than at 50 km? Explain. (c) In which parts of the atmosphere would you expect gases to behave most ideally (ignoring any photochemical reac-tions)? [Section 18.1]

18.2 Molecules in the upper atmosphere tend to contain double and triple bonds rather than single bonds. Suggest an expla-nation. [Section 18.1]

18.3 The figure shows the three lowest regions of Earth’s atmo-sphere. (a) Name each and indicate the approximate eleva-tions at which the boundaries occur. (b) In which region is ozone a pollutant? In which region does it filter UV solar radiation? (c) In which region is infrared radiation from Earth’s surface most strongly reflected back? (d) An aurora borealis is due to excitation of atoms and molecules in the atmosphere 55–95 km above Earth’s surface. Which regions in the figure are involved in an aurora borealis? (e) Com-pare the changes in relative concentrations of water vapor and carbon dioxide with increasing elevation in these three regions.

A

B

C

18.4 You are working with an artist who has been commissioned to make a sculpture for a big city in the eastern United States. The artist is wondering what material to use to make her sculpture because she has heard that acid rain in the eastern United States might destroy it over time. You take samples of granite, marble, bronze, and other materials, and place them

outdoors for a long time in the big city. You periodically examine the appearance and measure the mass of the sam-ples. (a) What observations would lead you to conclude that one or more of the materials are well-suited for the sculpture? (b) What chemical process (or processes) is (are) the most likely responsible for any observed changes in the materials? [Section 18.2]

18.5 Where does the energy come from to evaporate the estimated 425,000 km3 of water that annually leaves the oceans, as illus-trated here? [Section 18.3]

Earth’satmosphere

Worldocean

Water transport to the atmosphere

18.6 Describe the properties that most clearly distinguish among salt water, freshwater, and groundwater. [Section 18.3]

18.7 Describe what changes occur when atmospheric CO2 inter-acts with the world ocean as illustrated here. [Section 18.3]

CO2 (g)

CO2 (aq) Worldocean

• List the main steps involved in treating water for domestic uses. (Section 18.4)

• Describe the process of fracking and name its potential adverse environmental effects. (Section 18.4)

• Describe the main goals of green chemistry. (Section 18.5)

• Compare reactions and decide which reaction is greener. (Section 18.5)

Exercises

Exercises 805

18.14 From the data in Table 18.1, calculate the partial pressures of carbon dioxide and argon when the total atmospheric pres-sure is 1.05 bar.

18.15 The average concentration of carbon monoxide in air in an Ohio city in 2006 was 3.5 ppm. Calculate the number of CO molecules in 1.0 L of this air at a pressure of 759 torr and a temperature of 22 °C.

18.16 (a) From the data in Table 18.1, what is the concentration of neon in the atmosphere in ppm? (b) What is the concentra-tion of neon in the atmosphere in molecules per liter, assum-ing an atmospheric pressure of 730 torr and a temperature of 296 K?

18.17 The dissociation energy of a carbon–bromine bond is typi-cally about 210 kJ>mol. (a) What is the maximum wave-length of photons that can cause C ¬ Br bond dissociation? (b) Which kind of electromagnetic radiation—ultraviolet, vis-ible, or infrared—does the wavelength you calculated in part (a) correspond to?

18.18 In CF3Cl the C ¬ Cl bond-dissociation energy is 339 kJ>mol. In CCl4 the C ¬ Cl bond-dissociation energy is 293 kJ>mol. What is the range of wavelengths of photons that can cause C ¬ Cl bond rupture in one molecule but not in the other?

18.19 (a) Distinguish between photodissociation and photoioniza-tion. (b) Use the energy requirements of these two processes to explain why photodissociation of oxygen is more impor-tant than photoionization of oxygen at altitudes below about 90 km.

18.20 Why is the photodissociation of N2 in the atmosphere rela-tively unimportant compared with the photodissociation of O2?

human activities and Earth’s atmosphere (section 18.2)

18.21 Do the reactions involved in ozone depletion involve changes in oxidation state of the O atoms? Explain.

18.22 Which of the following reactions in the stratosphere cause an increase in temperature there?(a) O1g2 + O21g2 ¡ O3*1g2(b) O3*1g2 + M1g2 ¡ O31g2 + M*1g2(c) O21g2 + hn ¡ 2 O1g2(d) O1g2 + N21g2 ¡ NO1g2 + N1g2(e) All of the above

18.23 (a) What is the difference between chlorofluorocarbons and hydrofluorocarbons? (b) Why are hydrofluorocarbons poten-tially less harmful to the ozone layer than CFCs?

18.24 Draw the Lewis structure for the chlorofluorocarbon CFC-11, CFCl3. What chemical characteristics of this substance allow it to effectively deplete stratospheric ozone?

18.25 (a) Why is the fluorine present in chlorofluorocarbons not a major contributor to depletion of the ozone layer? (b) What are the chemical forms in which chlorine exists in the strato-sphere following cleavage of the carbon–chlorine bond?

18.26 Would you expect the substance CFBr3 to be effective in depleting the ozone layer, assuming that it is present in the stratosphere? Explain.

18.8 The following picture represents an ion-exchange column, in which water containing “hard” ions, such as Ca2+, is added to the top of the column, and water containing “soft” ions, such as Na+, comes out the bottom. Explain what is happening in the column. [Section 18.4]

Ion-exchangeresin

Soft water comesout the bottom

Add hard waterto top of column

18.9 From study of Figure 18.22 describe the various ways in which operation of a fracking well site could lead to environ-mental contamination.

18.10 One mystery in environmental science is the imbalance in the “carbon dioxide budget.” Considering only human activities, scientists have estimated that 1.6 billion metric tons of CO2 is added to the atmosphere every year because of deforestation (plants use CO2, and fewer plants will leave more CO2 in the atmosphere). Another 5.5 billion tons per year is put into the atmosphere because of burning fossil fuels. It is further estimated (again, considering only human activities) that the atmosphere actually takes up about 3.3 billion tons of this CO2 per year, while the oceans take up 2 billion tons per year, leav-ing about 1.8 billion tons of CO2 per year unaccounted for. This “missing” CO2 is assumed to be taken up by the “land.” What do you think might be happening? [Sections 18.1–18.3]

Earth’s atmosphere (section 18.1)

18.11 (a) What is the primary basis for the division of the atmo-sphere into different regions? (b) Name the regions of the at-mosphere, indicating the altitude interval for each one.

18.12 (a) How are the boundaries between the regions of the atmo-sphere determined? (b) Explain why the stratosphere, which is about 35 km thick, has a smaller total mass than the tropo-sphere, which is about 12 km thick.

18.13 Air pollution in the Mexico City metropolitan area is among the worst in the world. The concentration of ozone in Mexico City has been measured at 441 ppb (0.441 ppm). Mexico City sits at an altitude of 7400 feet, which means its atmospheric pressure is only 0.67 atm. (a) Calculate the partial pressure of ozone at 441 ppb if the atmospheric pressure is 0.67 atm. (b) How many ozone molecules are in 1.0 L of air in Mexico City? Assume T = 25 °C.


many grams of water could be evaporated from a 1.00 square meter patch of ocean over a 12-h day. (b) The specific heat capacity of liquid water is 4.184 J>g °C. If the initial surface temperature of a 1.00 square meter patch of ocean is 26 °C, what is its final temperature after being in sunlight for 12 h, assum-ing no phase changes and assuming that sunlight penetrates uniformly to depth of 10.0 cm?

[18.38] The enthalpy of fusion of water is 6.01 kJ>mol. Sunlight striking Earth’s surface supplies 168 W per square meter 11 W = 1 watt = 1 J>s2. (a) Assuming that melting of ice is due only to energy input from the Sun, calculate how many grams of ice could be melted from a 1.00 square meter patch of ice over a 12-h day. (b) The specific heat capacity of ice is 2.032 J>g °C. If the initial temperature of a 1.00 square meter patch of ice is -5.0 °C, what is its final temperature after being in sunlight for 12 h, assuming no phase changes and assuming that sunlight penetrates uniformly to a depth of 1.00 cm?

18.39 A first-stage recovery of magnesium from seawater is precipi-tation of Mg1OH22 with CaO:

Mg2+1aq2 + CaO1s2 + H2O1l2 ¡ Mg1OH221s2 + Ca2+1aq2 What mass of CaO, in grams, is needed to precipitate 1000 lb

of Mg1OH22? 18.40 Gold is found in seawater at very low levels, about 0.05 ppb

by mass. Assuming that gold is worth about $1300 per troy ounce, how many liters of seawater would you have to pro-cess to obtain $1,000,000 worth of gold? Assume the density of seawater is 1.03 g>mL and that your gold recovery process is 50% efficient.

18.41 (a) What is groundwater? (b) What is an aquifer? 18.42 The Ogallala aquifer described in the Closer Look box in Sec-

tion 18.3, provides 82% of the drinking water for the people who live in the region, although more than 75% of the water that is pumped from it is for irrigation. Irrigation withdraw-als are approximately 18 billion gallons per day. (a) Assum-ing that 2% of the rainfall that falls on an area of 600,000 km2 recharges the aquifer, what average annual rainfall would be required replace the water removed for irrigation? (b) What process or processes accounts for the presence of arsenic in well water?

human activities and Water Quality (section 18.4)

18.43 Suppose that one wishes to use reverse osmosis to reduce the salt content of brackish water containing 0.22 M total salt concentration to a value of 0.01 M, thus rendering it usable for human consumption. What is the minimum pressure that needs to be applied in the permeators (Figure 18.19) to achieve this goal, assuming that the operation occurs at 298 K? (Hint: Refer to Section 13.5.)

18.44 Assume that a portable reverse-osmosis apparatus operates on seawater, whose concentrations of constituent ions are listed in Table 18.5, and that the desalinated water output has an effective molarity of about 0.02 M. What minimum pressure must be applied by hand pumping at 297 K to cause reverse osmosis to occur? (Hint: Refer to Section 13.5.)

18.45 List the common products formed when an organic ma-terial containing the elements carbon, hydrogen, oxygen,

18.27 For each of the following gases, make a list of known or pos-sible naturally occurring sources: (a) CH4, (b) SO2, (c) NO.

18.28 Why is rainwater naturally acidic, even in the absence of pol-luting gases such as SO2?

18.29 (a) Write a chemical equation that describes the attack of acid rain on limestone, CaCO3. (b) If a limestone sculpture were treated to form a surface layer of calcium sulfate, would this help to slow down the effects of acid rain? Explain.

18.30 The first stage in corrosion of iron upon exposure to air is oxidation to Fe2+. (a) Write a balanced chemical equation to show the reaction of iron with oxygen and protons from acid rain. (b) Would you expect the same sort of reaction to occur with a silver surface? Explain.

18.31 Alcohol-based fuels for automobiles lead to the production of formaldehyde (CH2O) in exhaust gases. Formaldehyde un-dergoes photodissociation, which contributes to photochemi-cal smog:

CH2O + hn ¡ CHO + H

The maximum wavelength of light that can cause this reaction is 335 nm. (a) In what part of the electromagnetic spectrum is light with this wavelength found? (b) What is the maximum strength of a bond, in kJ>mol, that can be broken by absorp-tion of a photon of 335-nm light? (c) Compare your answer from part (b) to the appropriate value from Table 8.4. What do you conclude about C ¬ H bond energy in formaldehyde? (d) Write out the formaldehyde photodissociation reaction, showing Lewis-dot structures.

18.32 An important reaction in the formation of photochemical smog is the photodissociation of NO2:

NO2 + hn ¡ NO1g2 + O1g2 The maximum wavelength of light that can cause this reaction

is 420 nm. (a) In what part of the electromagnetic spectrum is light with this wavelength found? (b) What is the maximum strength of a bond, in kJ>mol, that can be broken by absorp-tion of a photon of 420-nm light? (c) Write out the photodis-sociation reaction showing Lewis-dot structures.

18.33 Explain why an increasing concentration of CO2 in the atmo-sphere affects the quantity of energy leaving Earth but does not affect the quantity of energy entering from the Sun.

18.34 (a) With respect to absorption of radiant energy, what distin-guishes a greenhouse gas from a non-greenhouse gas? (b) CH4 is a greenhouse gas, but N2 is not. How might the molecular struc-ture of CH4 explain why it is a greenhouse gas?

Earth’s Water (section 18.3)

18.35 What is the molarity of Na+ in a solution of NaCl whose salin-ity is 5.6 if the solution has a density of 1.03 g>mol?

18.36 Phosphorus is present in seawater to the extent of 0.07 ppm by mass. Assuming that the phosphorus is present as dihy-drogenphosphate, H2PO4

-, calculate the corresponding molar concentration of phosphate in seawater.

18.37 The enthalpy of evaporation of water is 40.67 kJ>mol. Sun-light striking Earth’s surface supplies 168 W per square meter 11 W = 1 watt = 1 J>s2. (a) Assuming that evaporation of water is due only to energy input from the Sun, calculate how

Exercises 807

equations for the oxidation of Fe2+ to Fe3+ by dissolved ox-ygen and for the formation of Fe1OH231s2 by reaction of Fe3+1aq2 with HCO3

-1aq2. 18.54 What properties make a substance a good coagulant for water

purification? 18.55 (a) What are trihalomethanes (THMs)? (b) Draw the Lewis

structures of two example THMs. 18.56 (a) Suppose that tests of a municipal water system reveal the

presence of bromate ion, BrO3-. What are the likely origins of

this ion? (b) Is bromate ion an oxidizing or reducing agent? Write a chemical equation for the reaction of bromate ion with hyponitrite ion.

Green Chemistry (section 18.5)

18.57 One of the principles of green chemistry is that it is bet-ter to use as few steps as possible in making new chemicals. In what ways does following this rule advance the goals of green chemistry? How does this principle relate to energy efficiency?

18.58 Discuss how catalysts can make processes more energy efficient.

18.59 A reaction for converting ketones to lactones, called the Baeyer–Villiger reaction,

+H2C CH2

CH2

CH2H2C

OOHCl

C

O

+

OCl

OH

O

3-Chloroperbenzoic acid

3-Chlorobenzoic acid

Ketone

H2C

CH2

CH2

H2C

H2C

C

O

O

Lactone

is used in the manufacture of plastics and pharmaceuticals. 3-Chloroperbenzoic acid is shock-sensitive, however, and prone to explode. Also, 3-chlorobenzoic acid is a waste prod-uct. An alternative process being developed uses hydrogen peroxide and a catalyst consisting of tin deposited within a solid support. The catalyst is readily recovered from the re-action mixture. (a) What would you expect to be the other product of oxidation of the ketone to lactone by hydro-gen peroxide? (b) What principles of green chemistry are addressed by use of the proposed process?

18.60 The hydrogenation reaction shown here was performed with an iridium catalyst, both in supercritical CO2 (scCO22 and in the chlorinated solvent CH2Cl2. The kinetic data for the reac-tion in both solvents are plotted in the graph. In what respects is the use of scCO2 a good example of a green chemical reaction?

sulfur, and nitrogen decomposes (a) under aerobic conditions, (b) under anaerobic conditions.

18.46 (a) Explain why the concentration of dissolved oxygen in freshwater is an important indicator of the quality of the water. (b) Find graphical data in the text that show variations of gas solubility with temperature, and estimate to two signifi-cant figures the percent solubility of O2 in water at 30 °C as compared with 20 °C. How do these data relate to the quality of natural waters?

18.47 The organic anion

SO3H3C C

CH3

H

(CH2)9

is found in most detergents. Assume that the anion undergoes aerobic decomposition in the following manner:

2 C18H29SO3-1aq2 + 51 O21aq2 ¡

36 CO21aq2 + 28 H2O1l2 + 2 H+1aq2 + 2 SO42-1aq2

What is the total mass of O2 required to biodegrade 10.0 g of this substance?

18.48 The average daily mass of O2 taken up by sewage discharged in the United States is 59 g per person. How many liters of water at 9 ppm O2 are 50 % depleted of oxygen in 1 day by a population of 1,200,000 people?

18.49 Magnesium ions are removed in water treatment by the ad-dition of slaked lime, Ca1OH22. Write a balanced chemical equation to describe what occurs in this process.

18.50 (a) Which of the following ionic species could be responsible for hardness in a water supply: Ca2+, K+, Mg2+, Fe2+, Na+? (b) What properties of an ion determine whether it will contribute to water hardness?

18.51 In the lime soda process at one time used in large scale mu-nicipal water softening, calcium hydroxide prepared from lime and sodium carbonate are added to precipitate Ca2+ as CaCO31s2 and Mg2+ as Mg1OH22 1s2:

Ca2+1aq2 + CO32-1aq2 ¡ CaCO31s2

Mg 2+1aq2 + 2OH-1aq2 ¡ MgOH2 (aq2 How many moles of Ca1OH22 and Na2CO3 should be added

to soften 1200 L of water in which

3Ca2+4 = 5.0 * 10-4 M and3Mg2+4 = 7.0 * 10-4 M?

18.52 The concentration of Ca2+ in a particular water supply is 5.7 * 10-3 M. The concentration of bicarbonate ion, HCO3

-, in the same water is 1.7 * 10-3 M. What masses of Ca1OH22 and Na2CO3 must be added to 5.0 * 107 L of this water to reduce the level of Ca2+ to 20% of its original level?

18.53 Ferrous sulfate 1FeSO42 is often used as a coagulant in water purification. The iron(II) salt is dissolved in the water to be purified, then oxidized to the iron(III) state by dissolved oxygen, at which time gelatinous Fe1OH23 forms, assuming the pH is above approximately 6. Write balanced chemical


18.63 A friend of yours has seen each of the following items in newspaper articles and would like an explanation: (a) acid rain, (b) greenhouse gas, (c) photochemical smog, (d) ozone depletion. Give a brief explanation of each term and identify one or two of the chemicals associated with each.

18.64 Suppose that on another planet the atmosphere consists of 17% Kr, 38% CH4, and 45% O2. What is the average molar mass at the surface? What is the average molar mass at an alti-tude at which all the O2 is photodissociated?

18.65 If an average O3 molecule “lives” only 100–200 seconds in the stratosphere before undergoing dissociation, how can O3 offer any protection from ultraviolet radiation?

18.66 Show how Equations 18.7 and 18.9 can be added to give Equation 18.10.

18.67 What properties of CFCs make them ideal for various com-mercial applications but also make them a long-term problem in the stratosphere?

18.68 Halons are fluorocarbons that contain bromine, such as CBrF3. They are used extensively as foaming agents for fight-ing fires. Like CFCs, halons are very unreactive and ultimately can diffuse into the stratosphere. (a) Based on the data in Table 8.4, would you expect photodissociation of Br atoms to occur in the stratosphere? (b) Propose a mechanism by which the presence of halons in the stratosphere could lead to the depletion of stratospheric ozone.

18.69 (a) What is the difference between a CFC and an HFC? (b) It is estimated that the lifetime for HFCs in the stratosphere is 2–7 years. Why is this number significant? (c)Why have HFCs been used to replace CFCs? (d) What is the major disadvan-tage of HFCs as replacements for CFCs?

18.70 Explain, using Le Châtelier’s principle, why the equilibrium constant for the formation of NO from N2 and O2 increases with increasing temperature, whereas the equilibrium con-stant for the formation of NO2 from NO and O2 decreases with increasing temperature.

18.71 Natural gas consists primarily of methane, CH41g2. (a) Write a balanced chemical equation for the complete combustion of

methane to produce CO21g2 as the only carbon-containing product. (b) Write a balanced chemical equation for the in-complete combustion of methane to produce CO(g) as the only carbon-containing product. (c) At 25 °C and 1.0 atm pressure, what is the minimum quantity of dry air needed to combust 1.0 L of CH41g2 completely to CO21g2?

18.72 It was estimated that the eruption of the Mount Pinatubo vol-cano resulted in the injection of 20 million metric tons of SO2 into the atmosphere. Most of this SO2 underwent oxidation to SO3, which reacts with atmospheric water to form an aero-sol. (a) Write chemical equations for the processes leading to formation of the aerosol. (b) The aerosols caused a 0.590.6 °C drop in surface temperature in the northern hemisphere. What is the mechanism by which this occurs? (c) The sulfate aerosols, as they are called, also cause loss of ozone from the stratosphere. How might this occur?

18.73 One of the possible consequences of climate change is an in-crease in the temperature of ocean water. The oceans serve as a “sink” for CO2 by dissolving large amounts of it. (a) The figure below shows the solubility of CO2 in water as a func-tion of temperature. Does CO2 behave more or less similarly to other gases in this respect?

10 20 30 40 50 600Water temperature (°C)

Solu

bilit

y (g

gas

per

kg

wat

er)

3.5

4

3

2.5

1.5

2

1

0.5

0

CO2

18.61 In the following three instances which choice is greener in each situation? Explain. (a) Benzene as a solvent or water as a solvent. (b) The reaction temperature is 500 K, or 1000 K. (c) Sodium chloride as a by-product or chloroform 1CHCl32 as a by-product.

18.62 In the following three instances which choice is greener in a chemical process? Explain. (a) A reaction that can be run at 350 K for 12 h without a catalyst or one that can be run at 300 K for 1 h with a reusable catalyst. (b) A reagent for the reaction that can be obtained from corn husks or one that is obtained from petroleum. (c) A process that produces no by-products or one in which the by-products are recycled for another process.

0 2 4 6

Ir-catalyst

CH2Cl2

scCO2

scCO2

Chiral Ir-cat.

8 10 231 3 5 7 9 22 24

100

80

60

40

20

0

Con

vers

ion

[%]

Time [h]

IrP BN

O

F3C CF3

F3C

H3C Ph

F3CF3C

CF3

CF3

CF3

+ −

NH3C H

PhPh NPhH

additional Exercises

integrative Exercises 809

(d) OH + CH4 ¡ H2O + CH3

CH3 + O2 ¡ OOCH3

OOCH3 + NO ¡ OCH3 + NO2

(e) The concentration of hydroxyl radicals in the troposphere is approximately 2 * 106 radicals per cm3. This estimate is based on a method called long path absorption spec-troscopy (LPAS), similar in principle to the Beer’s law measurement discussed in the Closer Look essay on p. 582, except that the length of the light path in the LPAS mea-surement is 20 km. Why must the path length be so large?

(f) The reactions shown in (d) also illustrate a second char-acteristic of the hydroxyl radical: its ability to cleanse the atmosphere of certain pollutants. Which of the reactions in (d) illustrate this?

18.84 An impurity in water has an extinction coefficient of 3.45 * 103 M-1 cm-1 at 280 nm, its absorption maximum (A Closer Look, p. 582). Below 50 ppb, the impurity is not a prob-lem for human health. Given that most spectrometers cannot detect absorbances less than 0.0001 with good reliability, is mea-suring the absorbance of a water sample at 280 nm a good way to detect concentrations of the impurity above the 50-ppb threshold?

18.85 The concentration of H2O in the stratosphere is about 5 ppm. It undergoes photodissociation according to:

H2O1g2 ¡ H1g2 + OH1g2(a) Write out the Lewis-dot structures for both products and

reactant.(b) Using Table 8.4, calculate the wavelength required to

cause this dissociation.(c) The hydroxyl radicals, OH, can react with ozone, giving

the following reactions:OH 1g2 + O31g2 ¡ HO21g2 + O21g2HO21g2 + O1g2 ¡ OH1g2 + O21g2

What overall reaction results from these two elementary reactions? What is the catalyst in the overall reaction? Explain.

18.80 The estimated average concentration of NO2 in air in the United States in 2006 was 0.016 ppm. (a) Calculate the partial pressure of the NO2 in a sample of this air when the atmo-spheric pressure is 755 torr (99.1 kPa). (b) How many mol-ecules of NO2 are present under these conditions at 20 °C in a room that measures 15 * 14 * 8 ft?

[18.81] In 1986 an electrical power plant in Taylorsville, Georgia, burned 8,376,726 tons of coal, a national record at that time. (a) Assuming that the coal was 83% carbon and 2.5% sulfur and that combustion was complete, calculate the number of tons of carbon dioxide and sulfur dioxide produced by the plant during the year. (b) If 55% of the SO2 could be removed by reaction with powdered CaO to form CaSO3, how many tons of CaSO3 would be produced?

18.82 The water supply for a midwestern city contains the fol-lowing impurities: coarse sand, finely divided particulates, nitrate ion, trihalomethanes, dissolved phosphorus in the form of phosphates, potentially harmful bacterial strains, dissolved organic substances. Which of the following pro-cesses or agents, if any, is effective in removing each of these impurities: coarse sand filtration, activated carbon filtra-tion, aeration, ozonization, precipitation with aluminum hydroxide?

[18.83] The hydroxyl radical, OH, is formed at low altitudes via the reaction of excited oxygen atoms with water:

O*1g2 + H2O1g2 ¡ 2 OH1g2

(a) Write the Lewis structure for the hydroxyl radical. (Hint: It has one unpaired electron.)

Once produced, the hydroxyl radical is very reactive. Explain the significance of each of the following reac-tions or series of reactions with respect to pollution in the troposphere:

(b) OH + NO2 ¡ HNO3

(c) OH + CO + O2 ¡ CO2 + OOH OOH + NO ¡ OH + NO2

(d) Nitrogen dioxide dissolves in water to form nitric acid and nitric oxide.

18.77 (a) Explain why Mg1OH22 precipitates when CO32- ion is

added to a solution containing Mg2+. (b) Will Mg1OH22 pre-cipitate when 4.0 g of Na2CO3 is added to 1.00 L of a solution containing 125 ppm of Mg2+?

[18.78] It has been pointed out that there may be increased amounts of NO in the troposphere as compared with the past because of massive use of nitrogen-containing compounds in fertilizers. Assuming that NO can eventually diffuse into the stratosphere, how might it affect the conditions of life on Earth? Using the index to this text, look up the chemistry of nitrogen oxides. What chemical pathways might NO in the troposphere follow?

[18.79] As of the writing of this text, EPA standards limit atmo-spheric ozone levels in urban environments to 84 ppb. How many moles of ozone would there be in the air above Los Angeles County (area about 4000 square miles; consider a height of 100 m above the ground) if ozone was at this concentration?

(b) What are the implications of this figure for the problem of climate change?

18.74 The rate of solar energy striking Earth averages 168 watts per square meter. The rate of energy radiated from Earth’s surface averages 390 watts per square meter. Comparing these num-bers, one might expect that the planet would cool quickly, yet it does not. Why not?

18.75 The solar power striking Earth every day averages 168 watts per square meter. The peak electrical power usage in New York City is 12,000 MW. Considering that present technol-ogy for solar energy conversion is about 10% efficient, from how many square meters of land must sunlight be collected in order to provide this peak power? (For comparison, the total area of New York city is 830 km2.)

18.76 Write balanced chemical equations for each of the follow-ing reactions: (a) The nitric oxide molecule undergoes pho-todissociation in the upper atmosphere. (b) The nitric oxide molecule undergoes photoionization in the upper atmosphere. (c) Nitric oxide undergoes oxidation by ozone in the stratosphere.

integrative Exercises


tropospheric concentrations of OH and CF3CH2F are 8.1 * 105 and 6.3 * 108 molecules>cm3, respectively, what is the rate of reaction at this temperature in M>s?

18.92 The Henry’s law constant for CO2 in water at 25 °C is 3.1 * 10-2 M atm-1. (a) What is the solubility of CO2 in water at this temperature if the solution is in contact with air at nor-mal atmospheric pressure? (b) Assume that all of this CO2 is in the form of H2CO3 produced by the reaction between CO2 and H2O:

CO21aq2 + H2O1l2 ¡ H2CO31aq2

What is the pH of this solution?

18.93 The precipitation of Al1OH23 1Ksp = 1.3 * 10-332 is some-times used to purify water. (a) Estimate the pH at which precipitation of Al1OH23 will begin if 5.0 lb of Al21SO423 is added to 2000 gal of water. (b) Approximately how many pounds of CaO must be added to the water to achieve this pH?

[18.94] The valuable polymer polyurethane is made by a condensa-tion reaction of alcohols (ROH) with compounds that contain an isocyanate group (RNCO). Two reactions that can gener-ate a urethane monomer are shown here:

RNH2 CO2

Cl ClC

O

+ R N C O

R N C O R N

H

C

O

OR′R′OH

2 H2O+

+

RNH2

(i)

(ii) + R N C O

R N C O R N

H

C

O

OR′R′OH

2 HCl+

+

(a) Which process, i or ii, is greener? Explain.(b) What are the hybridization and geometry of the car-

bon atoms in each C-containing compound in each reaction?

(c) If you wanted to promote the formation of the isocyanate intermediate in each reaction, what could you do, using Le Châtelier’s principle?

18.95 The pH of a particular raindrop is 5.6. (a) Assuming the major species in the raindrop are H2CO31aq2, HCO3

-1aq2, and CO3 2-1aq2, calculate the concentrations of these species in the raindrop, assuming the total carbonate concentra-tion is 1.0 * 10-5 M. The appropriate Ka values are given in Table 16.3. (b) What experiments could you do to test the hypothesis that the rain also contains sulfur-containing species that contribute to its pH? Assume you have a large sample of rain to test.

18.86 Bioremediation is the process by which bacteria repair their environment in response, for example, to an oil spill. The efficiency of bacteria for “eating” hydrocarbons depends on the amount of oxygen in the system, pH, temperature, and many other factors. In a certain oil spill, hydrocarbons from the oil disappeared with a first-order rate constant of 2 * 10-6 s-1. At that rate, how many days would it take for the hydrocarbons to decrease to 10% of their initial value?

18.87 The standard enthalpies of formation of ClO and ClO2 are 101 and 102 kJ>mol, respectively. Using these data and the thermodynamic data in Appendix C, calculate the over-all enthalpy change for each step in the following catalytic cycle:

ClO1g2 + O31g2 ¡ ClO21g2 + O21g2ClO21g2 + O1g2 ¡ ClO1g2 + O21g2

What is the enthalpy change for the overall reaction that results from these two steps?

18.88 The main reason that distillation is a costly method for puri-fying water is the high energy required to heat and vapor-ize water. (a) Using the density, specific heat, and heat of vaporization of water from Appendix B, calculate the amount of energy required to vaporize 1.00 gal of water beginning with water at 20 °C. (b) If the energy is provided by electricity costing $0.085>kWh, calculate its cost. (c) If distilled water sells in a grocery store for $1.26 per gal, what percentage of the sales price is represented by the cost of the energy?

[18.89] A reaction that contributes to the depletion of ozone in the stratosphere is the direct reaction of oxygen atoms with ozone:

O1g2 + O31g2 ¡ 2 O21g2 At 298 K the rate constant for this reaction is 4.8 * 105 M-1 s-1.

(a) Based on the units of the rate constant, write the likely rate law for this reaction. (b) Would you expect this reaction to occur via a single elementary process? Explain why or why not (c) Use ∆Hf° values from Appendix C to estimate the enthalpy change for this reaction. Would this reaction raise or lower the temperature of the stratosphere?

18.90 The following data were collected for the destruction of O3 by H 1O3 + H ¡ O2 + OH2 at very low concentrations:

Trial 3o3 4 1M 2 [h] (M)initial rate 1M ,s 2

1 5.17 * 10-33 3.22 * 10-26 1.88 * 10-14

2 2.59 * 10-33 3.25 * 10-26 9.44 * 10-15

3 5.19 * 10-33 6.46 * 10-26 3.77 * 10-14

(a) Write the rate law for the reaction.(b) Calculate the rate constant.

18.91 The degradation of CF3CH2F (an HFC) by OH radicals in the troposphere is first order in each reactant and has a rate constant of k = 1.6 * 108 M-1s-1 at 4 °C. If the

Design an Experiment 811

design an ExperimentConsiderable fracking of petroleum/gas wells (see Closer Look box in Section 18.4) has occurred in recent years in a particular rural area. The residents have complained that the water in the residential wells serving their domestic water needs has become contaminated with chemicals as-sociated with the fracking operations. The well operators respond that the chemicals about which complaints are lodged occur naturally, and are not the result of well-drilling activities.

Describe experiments that you could conduct on the waters from residential wells to help determine whether and to what extent well contaminants are due to fracking operations. Among the chemicals that might be expected to be employed in fracking operations are hydrochloric acid, sodium chloride, ethylene glycol, borate salts, water-soluble gelling agents such as guar gum, citric acid, methanol, and other alcohols such as isopropanol, and methane. Assume that you have available the techniques to make measurements of the concentrations of these sub-stances in the residential wells. What experiments would you conduct and what analyses of the results would you carry out in an attempt to settle the question of whether fracking operations have led to contamination of the well water? Would simply measuring the concentrations of some or all of these substances in the well waters be sufficient to settle the issue?

19 Chemical ThermodynamicsThe amazing organization of living systems, from complex molecular structures such as the nucleosome, to cells, to tissues, and finally to whole plants and animals, is an unending source of wonder and delight to the chemists, biochemists, physicists, and biologists who study them. Energy must be spent, somehow, to keep all of these organized systems in good working order. But we have not yet learned enough about energy to understand how the underlying chemical and physical processes of life are governed.

19.3 The Molecular InTerpreTaTIon of enTropy and The ThIrd law of TherModynaMIcs On the molecular level, we learn that the entropy of a system is related to the number of accessible microstates. The entropy of the system increases as the randomness of the system increases. The third law of thermodynamics states that, at 0 K, the entropy of a perfect crystalline solid is zero.

19.4 enTropy changes In cheMIcal reacTIons Using tabulated standard molar entropies, we can calculate the standard entropy changes for systems undergoing reaction.

19.1 sponTaneous processes We see that changes that occur in nature have a directional character. They move spontaneously in one direction but not in the reverse direction.

19.2 enTropy and The second law of TherModynaMIcs We discuss entropy, a thermodynamic state function that is important in determining whether a process is spontaneous. The second law of thermodynamics tells us that in any spontaneous process, the entropy of the universe (system plus surroundings) increases.

WhaT’s ahead

▶ THE NUCLEOSOME Inside the nucleus of a living cell, DNA (the outer double-helical ribbon) surrounds eight protein molecules (the colored spiral ribbons). This overall DNA/protein structure, called the nucleosome, is the basic unit of chromosomes in the nuclei of our cells. These structures are highly ordered, yet also must be unraveled in order for gene expression to take place. Both packaging and unpackaging of DNA in the nucleosome involve changes in the energy of the system.

Understanding biochemical processes such as DNA replication, photosynthesis, or metabolism requires us to ask increasingly more sophisticated questions, and antici-pate increasingly sophisticated answers to them. Fortunately, the general laws that gov-ern chemical reactions—from those that we perform in the laboratory to those that occur in biology—can help us understand the most complicated of processes. Two of the most important questions chemists ask when designing and using chemical reac-tions are “How fast is the reaction?” and “How far does it proceed?” The first question is addressed by chemical kinetics, which we discussed in Chapter 14. The second ques-tion involves the equilibrium constant, the focus of Chapter 15. Let’s briefly review how these concepts are related.

In Chapter 14 we learned that the rate of any chemical reaction is controlled largely by a factor related to energy, namely, the activation energy of the reaction. (Section 14.5) In general, the lower the activation energy, the faster a reaction proceeds. In Chapter 15 we saw that chemical equilibrium is reached when a given reaction and its reverse reaction occur at the same rate. (Section 15.1)

Because reaction rates are closely tied to energy, it is logical that equilibrium also depends in some way on energy. In this chapter we explore the connection between

19.7 free energy and The equIlIbrIuM consTanT Finally, we consider how the standard free-energy change for a chemical reaction can be used to calculate the equilibrium constant for the reaction.

19.5 gIbbs free energy We encounter another thermodynamic state function, free energy (or Gibbs free energy), a measure of how far removed a system is from equilibrium. The change in free energy measures the maximum amount of useful work obtainable from a process and tells us the direction in which a chemical reaction is spontaneous.

19.6 free energy and TeMperaTure We consider how the relationship among free-energy change, enthalpy change, and entropy change provides insight into how temperature affects the spontaneity of a process.

814 chapTer 19 chemical Thermodynamics

energy and the extent of a reaction. Doing so requires a deeper look at chemical thermo-dynamics, the area of chemistry that deals with energy relationships. We first encoun-tered thermodynamics in Chapter 5, where we discussed the nature of energy, the first law of thermodynamics, and the concept of enthalpy. Recall that the enthalpy change for any system is the heat transferred between the system and its surroundings during a constant-pressure process. (Equation 5.10)

In the “Strategies in Chemistry” box in Section 5.4, we pointed out that the enthalpy change that takes place during a reaction is an important guide as to whether the reac-tion is likely to proceed. Now we will see that reactions involve not only changes in enthalpy but also changes in entropy—another important thermodynamic quantity. Recall that entropy is related to the degree of randomness in a system. (Section 13.1) Our discussion of entropy will lead us to the second law of thermodynamics, which pro-vides insight into why physical and chemical changes tend to favor one direction over another. We drop a brick, for example, and it falls to the ground. We do not expect the brick to spontaneously rise from the ground to our outstretched hand. We light a candle, and it burns down. We do not expect a half-consumed candle to regenerate itself spontaneously, even if we have captured all the gases produced when the candle burned. Thermodynamics helps us understand the significance of this directional character of processes, regardless of whether they are exothermic or endothermic.

19.1 | spontaneous ProcessesThe first law of thermodynamics states that energy is conserved. (Section 5.2) In other words, energy is neither created nor destroyed in any process, whether that pro-cess is a brick falling, a candle burning, or an ice cube melting. Energy can be trans-ferred between a system and the surroundings and can be converted from one form to another, but the total energy of the universe remains constant. In Chapter 5, we ex-pressed this law mathematically as ∆E = q + w, where ∆E is the change in the inter-nal energy of a system, q is the heat absorbed (or released) by the system from (or to) the surroundings, and w is the work done on the system by the surroundings, or on the surroundings by the system. (Equation 5.5) Remember that q 7 0 means that the system is gaining heat from the surroundings, and w 7 0 means that the system is gaining work from the surroundings (i.e., the surroundings are doing work on the system).

The first law helps us balance the books, so to speak, on the heat transferred between a system and its surroundings and the work done by or on a system. However, because energy is conserved, we cannot simply use the value of ∆E to tell us whether a process is favored to occur because anything we do to lower the energy of the system raises the energy of the surroundings, and vice versa. Nevertheless, experience tells us that certain processes always occur, even though, so far as we can observe in our study of chemical and physical processes, energy is conserved. Water placed in a freezer turns into ice, for instance, and if you touch a hot object, heat is transferred to your hand. The first law guarantees that energy is conserved in these processes, but it says noth-ing about the preferred direction for the process. Furthermore, the situations we have just mentioned occur without any outside intervention. We say they are spontaneous. A spontaneous process is one that proceeds on its own without any outside assistance.

A spontaneous process occurs in one direction only, and the reverse of any sponta-neous process is always nonspontaneous. Drop an egg above a hard surface, for example, and it breaks on impact (◀ Figure 19.1). Now, imagine seeing a video clip in which a broken egg rises from the floor, reassembles itself, and ends up in someone’s hand. You would conclude that the video is running in reverse because you know that broken eggs simply do not magically rise and reassemble themselves! An egg falling and breaking is spontaneous. The reverse process is nonspontaneous, even though energy is conserved in both processes. (Strategies in Chemistry Box, Section 5.4)

We have touched upon several spontaneous and nonspontaneous processes in our study of chemistry thus far. For example, a gas spontaneously expands into a vacuum (▶ Figure 19.2), but the reverse process, in which the gas moves back entirely into one

Spontaneous Notspontaneous

▲ Figure 19.1 A spontaneous process!

Go FiGureDoes the potential energy of the eggs change during this process?

secTiOn 19.1 spontaneous processes 815

of the flasks, does not happen. In other words, expansion of the gas is spontaneous, but the reverse process is nonspontaneous. In general, processes that are spontaneous in one direction are nonspontaneous in the opposite direction.

Experimental conditions, such as temperature and pressure, are often important in determining whether a process is spontaneous. We are all familiar with situations in which a forward process is spontaneous at one temperature but the reverse process is spontane-ous at a different temperature. Consider, for example, ice melting. At atmospheric pres-sure, when the temperature of the surroundings is above 0 °C, ice melts spontaneously, and the reverse process—liquid water turning into ice—is not spontaneous. However, when the temperature of the surroundings is below 0 °C, the opposite is true—liquid water turns to ice spontaneously, but the reverse process is not spontaneous (▼ Figure 19.3).

What happens at T = 0 °C, the normal melting point of water, when the flask of Figure 19.3 contains both water and ice? At the normal melting point of a substance, the solid and liquid phases are in equilibrium. (Section 11.6) At this temperature, the two phases are interconverting at the same rate and there is no preferred direction for the process.

Just because a process is spontaneous does not necessarily mean that it will occur at an observable rate. A chemical reaction is spontaneous if it occurs on its own accord, regardless of its speed. A spontaneous reaction can be very fast, as in the case of acid–base neutralization, or very slow, as in the rusting of iron. Thermodynamics tells us the direction and extent of a reaction but nothing about the rate; rate is the domain of kinetics rather than of thermodynamics.

Give It Some ThoughtIf a process is nonspontaneous, does that mean the process cannot occur under any circumstances?

0.5 atm0.5 atm

1 atm 0 atm

Evacuated �ask0 atm

Gas at1 atm

Closed stopcock

A B

A B

A B

This process is spontaneous

This process is not spontaneous

When stopcock opens, gas expands to occupy both �asks

All gas molecules move back into �ask A

▲ Figure 19.2 Expansion of a gas into an evacuated space is a spontaneous process. The reverse process—gas molecules initially distributed evenly in two flasks all moving into one flask—is not spontaneous.

Go FiGureIf flask B were smaller than flask A, would the final pressure after the stopcock is opened be greater than, equal to, or less than 0.5 atm?

Spontaneous for T > 0 °C

Spontaneous for T < 0 °C

▲ Figure 19.3 Spontaneity can depend on temperature. At T 7 0 °C, ice melts spontaneously to liquid water. At T 6 0 °C, the reverse process, water freezing to ice, is spontaneous. At T = 0 °C the two states are in equilibrium.

Go FiGureIn which direction is this process exothermic?

soluTionAnalyze We are asked to judge whether each process is spontane-ous in the direction indicated, in the reverse direction, or in neither direction.

samPle exerCise 19.1 identifying spontaneous Processes

Predict whether each process is spontaneous as described, spontaneous in the reverse direction, or at equilibrium: (a) Water at 40 °C gets hotter when a piece of metal heated to 150 °C is added. (b) Water at room temperature decomposes into H21g2 and O21g2. (c) Benzene vapor, C6H61g2, at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 °C.

Plan We need to think about whether each process is consistent with our experience about the natural direction of events or whether we expect the reverse process to occur.


Seeking a Criterion for SpontaneityA marble rolling down an incline or a brick falling from your hand loses potential en-ergy. The loss of some form of energy is a common feature of spontaneous change in mechanical systems. In the 1870s Marcellin Bertholet (1827–1907) suggested that the direction of spontaneous changes in chemical systems is determined by the loss of en-ergy. He proposed that all spontaneous chemical and physical changes are exothermic. It takes only a few moments, however, to find exceptions to this generalization. For example, the melting of ice at room temperature is spontaneous and endothermic. Sim-ilarly, many spontaneous dissolution processes, such as the dissolving of NH4NO3, are endothermic, as we discovered in Section 13.1. We conclude that although the majority of spontaneous reactions are exothermic, there are spontaneous endothermic ones as well. Clearly, some other factor must be at work in determining the natural direction of processes.

To understand why certain processes are spontaneous, we need to consider more closely the ways in which the state of a system can change. Recall from Section 5.2 that quantities such as temperature, internal energy, and enthalpy are state functions, properties that define a state and do not depend on how we reach that state. The heat transferred between a system and its surroundings, q, and the work done by or on the system, w, are not state functions—their values depend on the specific path taken between states. One key to understanding spontaneity is understanding differences in the paths between states.

Reversible and Irreversible ProcessesIn 1824, a 28-year-old French engineer named Sadi Carnot (1796–1832) published an analysis of the factors that determine how efficiently a steam engine can convert heat to work. Carnot considered what an ideal engine, one with the highest possible effi-ciency, would be like. He observed that it is impossible to convert the energy content of a fuel completely to work because a significant amount of heat is always lost to the surroundings. Carnot’s analysis gave insight into how to build better, more efficient engines, and it was one of the earliest studies in what has developed into the discipline of thermodynamics.

An ideal engine operates under an ideal set of conditions in which all the pro-cesses are reversible. A reversible process is a specific way in which a system changes its state. In a reversible process, the change occurs in such a way that the system and surroundings can be restored to their original states by exactly reversing the change. In other words, we can restore the system to its original condition with no net change to either the system or its surroundings. An irreversible process is one that cannot simply be reversed to restore the system and its surroundings to their original states. What Carnot discovered is that the amount of work we can extract from any process depends on the manner in which the process is carried out. He concluded that a revers-ible change produces the maximum amount of work that can be done by a system on its surroundings.

Solve

(a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hot-ter object to the colder one. (Section 5.1) Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere between the initial temperatures of the metal and the water. (b) Experience tells us that this process is not spontaneous— we certainly have never seen hydrogen and oxygen gases spontane-ously bubbling up out of water! Rather, the reverse process—the reac-tion of H2 and O2 to form H2O ¬ is spontaneous. (c) The normal boiling point is the temperature at which a vapor at 1 atm is in equilib-rium with its liquid. Thus, this is an equilibrium situation. If the tem-perature were below 80.1 °C, condensation would be spontaneous.

Practice exercise 1The process of iron being oxidized to make iron(III) oxide (rust) is spontaneous. Which of these statements about this process is/are true? (a) The reduction of iron(III) oxide to iron is also spontane-ous. (b) Because the process is spontaneous, the oxidation of iron must be fast. (c) The oxidation of iron is endothermic. (d) Equilib-rium is achieved in a closed system when the rate of iron oxidation is equal to the rate of iron(III) oxide reduction. (e) The energy of the universe is decreased when iron is oxidized to rust.

Practice exercise 2At 1 atm pressure, CO21s2 sublimes at -78 °C. Is this process spontaneous at -100 °C and 1 atm pressure?

secTiOn 19.1 spontaneous processes 817

Give It Some ThoughtSuppose you have a system made up of water only, with the container and everything beyond being the surroundings. Consider a process in which the water is first evaporated and then condensed back into its original container. Is this two-step process necessarily reversible?

Let’s next examine some aspects of reversible and irreversible processes, first with respect to the transfer of heat. When two objects at different temperatures are in con-tact, heat flows spontaneously from the hotter object to the colder one. Because it is impossible to make heat flow in the opposite direction, from colder object to hotter one, the flow of heat is an irreversible process. Given these facts, can we imagine any conditions under which heat transfer can be made reversible?

To answer this question, we must consider temperature differences that are infini-tesimally small, as opposed to the discrete temperature differences with which we are most familiar. For example, consider a system and its surroundings at essentially the same temperature, with just an infinitesimal temperature difference dT, between them (▲ Figure 19.4). If the surroundings are at temperature T and the system is at the infinitesimally higher temperature T + dT, then an infinitesimal amount of heat flows from system to surroundings. We can reverse the direction of heat flow by making an infinitesimal change of temperature in the opposite direction, lowering the system tem-perature to T - dT. Now the direction of heat flow is from surroundings to system. Reversible processes are those that reverse direction whenever an infinitesimal change is made in some property of the system.*

Now let’s consider another example, the expansion of an ideal gas at constant tem-perature (referred to as an isothermal process). In the cylinder-piston arrangement of Figure 19.5, when the partition is removed, the gas expands spontaneously to fill the evacuated space. Can we determine whether this particular isothermal expansion is reversible or irreversible? Because the gas expands into a vacuum with no external

*For a process to be truly reversible, the amounts of heat must be infinitesimally small and the transfer of heat must occur infinitely slowly; thus, no process that we can observe is truly reversible. The notion of infinitesi-mal amounts is related to the infinitesimals that you may have studied in a calculus course.

(a) (b)

Surroundings

System

Surroundings

System

Surroundings at temperature T Surroundings at temperature T

System at higher temperature T + δT

Small increment of heat transferred from system to surroundings

System at lower temperature T – δT

Small increment of heat transferred to system from surroundings

▲ Figure 19.4 Reversible flow of heat. Heat can flow reversibly between a system and its surroundings only if the two have an infinitesimally small difference in temperature dT. (a) Increasing the temperature of the system by dT causes heat to flow from the hotter system to the colder surroundings. (b) Decreasing the temperature of the system by dT causes heat to flow from the hotter surroundings to the colder system.

Go FiGureIf the flow of heat into or out of the system is to be reversible, what must be true of dT?


pressure, it does no P9V work on the surroundings. (Section 5.3) Thus, for the expansion, w = 0. We can use the piston to compress the gas back to its original state, but doing so requires that the surroundings do work on the system, meaning that w 7 0 for the compression. In other words, the path that restores the system to its original state requires a different value of w (and, by the first law, a different value of q) than the path by which the system was first changed. The fact that the same path cannot be followed to restore the system to its original state indicates that the process is irreversible.

What might a reversible isothermal expansion of an ideal gas be? This process will occur only if initially, when the gas is confined to half the cylinder, the external pressure acting on the piston exactly balances the pressure exerted by the gas on the piston. If the external pressure is reduced infinitely slowly, the piston will move outward, allow-ing the pressure of the confined gas to readjust to maintain the pressure balance. This infinitely slow process in which the external pressure and internal pressure are always in equilibrium is reversible. If we reverse the process and compress the gas in the same infinitely slow manner, we can return the gas to its original volume. The complete cycle of expansion and compression in this hypothetical process, moreover, is accomplished without any net change to the surroundings.

Because real processes can at best only approximate the infinitely slow change associated with reversible processes, all real processes are irreversible. Further, the reverse of any spontaneous process is a nonspontaneous process. A nonspontaneous process can occur only if the surroundings do work on the system. Thus, any sponta-neous process is irreversible. Even if we return the system to the original condition, the surroundings will have changed.

19.2 | entropy and the second law of Thermodynamics

Knowing that any spontaneous process is irreversible, can we make predictions about the spontaneity of an unfamiliar process? Understanding spontaneity requires us to ex-amine the thermodynamic quantity called entropy, which was first mentioned in Sec-tion 13.1. In general, entropy is associated either with the extent of randomness in a system or with the extent to which energy is distributed among the various motions of the molecules of the system. In this section we consider how entropy changes are related to heat transfer and temperature. Our analysis will bring us to a profound state-ment about spontaneity known as the second law of thermodynamics.

The Relationship between Entropy and HeatThe entropy, S, of a system is a state function just like internal energy, E, and enthalpy, H. As with these other quantities, the value of S is a characteristic of the state of a sys-tem. (Section 5.2) Thus, the change in entropy, ∆S, in a system depends only on

▲ Figure 19.5 An irreversible process. Initially an ideal gas is confined to the right half of a cylinder. When the partition is removed, the gas spontaneously expands to fill the whole cylinder. No work is done by the system during this expansion. Using the piston to compress the gas back to its original state requires the surroundings to do work on the system.

Piston

Vacuum Gas

Movable partition

Irreversible expansion of gasWork done by system = 0

CompressionWork done on system > 0

Work

If the partition is removed, the gas spontaneously �lls the evacuated space

Surroundings do work on the system to move the piston and thus compress the gas.

secTiOn 19.2 entropy and the second Law of Thermodynamics 819

the initial and final states of the system and not on the path taken from one state to the other:

∆S = Sfinal - Sinitial [19.1]

For the special case of an isothermal process, ∆S is equal to the heat that would be transferred if the process were reversible, qrev, divided by the absolute temperature at which the process occurs:

∆S =qrev

T 1constant T2 [19.2]

Although many possible paths can take the system from one state to another, only one path is associated with a reversible process. Thus, the value of q

rev is uniquely defined

for any two states of the system. Because S is a state function, we can use Equation 19.2 to calculate ∆S for any isothermal process between states, not just the reversible one.

Give It Some ThoughtHow can S be a state function when ∆S depends on q, which is not a state function?

∆S for Phase ChangesThe melting of a substance at its melting point and the vaporization of a substance at its boiling point are isothermal processes. (Section 11.4) Consider the melting of ice. At 1 atm pressure, ice and liquid water are in equilibrium at 0 °C. Imagine melt-ing 1 mol of ice at 0 °C, 1 atm to form 1 mol of liquid water at 0 °C, 1 atm. We can achieve this change by adding heat to the system from the surroundings: q = ∆Hfusion, where ∆Hfusion is the heat of melting. Imagine adding the heat infinitely slowly, raising the temperature of the surroundings only infinitesimally above 0 °C. When we make the change in this fashion, the process is reversible because we can reverse it by infinitely slowly removing the same amount of heat, ∆Hfusion, from the system, using immediate surroundings that are infinitesimally below 0 °C. Thus, qrev = ∆Hfusion for the melting of ice at T = 0 °C = 273 K.

The enthalpy of fusion for H2O is ∆Hfusion = 6.01 kJ>mol (a positive value because melting is an endothermic process). Thus, we can use Equation 19.2 to calculate ∆Sfusion for melting 1 mol of ice at 273 K:

∆Sfusion =qrev

T=

∆Hfusion

T=

11 mol216.01 * 103 J>mol2273 K

= 22.0 J>K

Notice that (1) we must use the absolute temperature in Equation 19.2, and (2) the units for ∆S, J>K, are energy divided by absolute temperature, as we expect from Equation 19.2.

soluTionAnalyze We first recognize that freezing is an exothermic process, which means heat is transferred from the system to the surround-ings and q 6 0. Because freezing is the reverse of melting, the enthalpy change that accompanies the freezing of 1 mol of Hg is - ∆Hfusion = -2.29 kJ>mol.Plan We can use - ∆Hfusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg. Then we use this value of q as qrev in Equation 19.2 to determine ∆S for the system.

samPle exerCise 19.2 Calculating ∆S for a Phase Change

Elemental mercury is a silver liquid at room temperature. Its normal freezing point is -38.9 °C, and its molar enthalpy of fusion is ∆Hfusion = 2.29 kJ>mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point?

Solve For q we have

q = 150.0 g Hg2a 1 mol Hg200.59 g Hg

b a -2.29 kJ1 mol Hg

b a 1000 J1 kJ

b = -571 J

Before using Equation 19.2, we must first convert the given Celsius temperature to kelvins:

-38.9 °C = 1-38.9 + 273.152 K = 234.3 K


The Second Law of ThermodynamicsThe key idea of the first law of thermodynamics is that energy is conserved in any pro-cess. (Section 5.2) Is entropy also conserved in a spontaneous process in the same way that energy is conserved?

Let’s try to answer this question by calculating the entropy change of a system and the entropy change of its surroundings when our system is 1 mol of ice (a piece roughly the size of an ice cube) melting in the palm of your hand, which is part of the surround-ings. The process is not reversible because the system and surroundings are at differ-ent temperatures. Nevertheless, because ∆S is a state function, its value is the same regardless of whether the process is reversible or irreversible. We calculated the entropy change of the system just before Sample Exercise 19.2:

∆Ssys =qrev

T=

11 mol216.01 * 103 J>mol2273 K

= 22.0 J>K

We can now calculate ∆Ssys:

∆Ssys =qrev

T=

-571 J234.3 K

= -2.44 J>K

Check The entropy change is negative because our qrev value is nega-tive, which it must be because heat flows out of the system in this exothermic process.Comment This procedure can be used to calculate ∆S for other isothermal phase changes, such as the vaporization of a liquid at its boiling point.

Practice exercise 1Do all exothermic phase changes have a negative value for the entropy change of the system? (a) Yes, because the heat

transferred from the system has a negative sign. (b) Yes, because the temperature decreases during the phase transition. (c) No, because the entropy change depends on the sign of the heat transferred to or from the system. (d) No, because the heat trans-ferred to the system has a positive sign. (e) More than one of the previous answers is correct.

Practice exercise 2The normal boiling point of ethanol, C2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is 38.56 kJ>mol. What is the change in entropy in the system when 68.3 g of C2H5OH1g2 at 1 atm condenses to liquid at the normal boiling point?

a Closer look

The Entropy Change When a Gas Expands Isothermally

In general, the entropy of any system increases as the system be-comes more random or more spread out. Thus, we expect the spon-taneous expansion of a gas to result in an increase in entropy. To see how this entropy increase can be calculated, consider the ex-pansion of an ideal gas that is initially constrained by a piston, as in the rightmost part of Figure 19.5. Imagine that we allow the gas to undergo a reversible isothermal expansion by infinitesimally de-creasing the external pressure on the piston. The work done on the surroundings by the reversible expansion of the system against the piston can be calculated with the aid of calculus (we do not show the derivation):

wrev = -nRT ln V2

V1

In this equation, n is the number of moles of gas, R is the ideal-gas constant (Section 10.4), T is the absolute temperature, V1 is the initial volume, and V2 is the final volume. Notice that if V2 7 V1, as it must be in our expansion, then wrev 6 0, meaning that the expanding gas does work on the surroundings.

One characteristic of an ideal gas is that its internal energy de-pends only on temperature, not on pressure. Thus, when an ideal gas expands isothermally, ∆E = 0. Because ∆E = qrev + wrev = 0, we

see that qrev = -wrev = nRT ln1V2>V12. Then, using Equation 19.2, we can calculate the entropy change in the system:

∆Ssys =qrev

T=

nRT ln V2

V1

T= nR ln

V2

V1 [19.3]

Let’ s calculate the entropy change for 1.00 L of an ideal gas at 1.00 atm pressure, 0 °C temperature, expanding to 2.00 L. From the ideal-gas equation, we can calculate the number of moles in 1.00 L of an ideal gas at 1.00 atm and 0 °C as we did in Chapter 10:

n =PVRT

=11.00 atm211.00 L2

10.08206 L@atm>mol@K21273 K2 = 4.46 * 10-2 mol

The gas constant, R, can also be expressed as 8.314 J>mol@K (Table 10.2), and this is the value we must use in Equation 19.3 because we want our answer to be expressed in terms of J rather than in L-atm. Thus, for the expansion of the gas from 1.00 L to 2.00 L, we have

∆Ssys = 14.46 * 10-2 mol2a8.314 J

mol@Kb a ln

2.00 L1.00 L

b = 0.26 J>K

In Section 19.3 we will see that this increase in entropy is a measure of the increased randomness of the molecules due to the expansion.

Related Exercises: 19.29, 19.30, 19.106

secTiOn 19.3 The Molecular interpretation of entropy and the Third Law of Thermodynamics 821

The surroundings immediately in contact with the ice are your hand, which we assume is at body temperature, 37 °C = 310 K. The quantity of heat lost by your hand is -6.01 * 103 J>mol, which is equal in magnitude to the quantity of heat gained by the ice but has the opposite sign. Hence, the entropy change of the surroundings is

∆Ssurr =qrev

T=

11 mol21-6.01 * 103 J>mol2310 K

= -19.4 J>K

We can consider that everything in the universe is either the system of interest, or its surroundings. Therefore ∆Suniv = ∆Ssys + ∆Ssurr. Thus, the overall entropy change of the universe is positive in our example:

∆Suniv = ∆Ssys + ∆Ssurr = 122.0 J>K2 + 1-19.4 J>K2 = 2.6 J>K

If the temperature of the surroundings were not 310 K but rather some temperature infinitesimally above 273 K, the melting would be reversible instead of irreversible. In that case the entropy change of the surroundings would equal -22.0 J>K and ∆Suniv would be zero.

In general, any irreversible process results in an increase in the entropy of the universe, whereas any reversible process results in no change in the entropy of the universe, a statement known as the second law of thermodynamics. The second law of thermodynamics can be expressed in terms of either of the following two equations:

Reversible Process: ∆Suniv = ∆Ssys + ∆Ssurr = 0

Irreversible Process: ∆Suniv = ∆Ssys + ∆Ssurr 7 0 [19.4]

Because spontaneous processes are irreversible, we can say that the entropy of the universe increases in any spontaneous process. This profound generalization is yet another way of expressing the second law of thermodynamics.

Give It Some ThoughtThe rusting of iron is spontaneous and is accompanied by a decrease in the entropy of the system (the iron and oxygen). What can we conclude about the entropy change of the surroundings?

The second law of thermodynamics tells us the essential character of any sponta-neous change—it is always accompanied by an increase in the entropy of the universe. We can use this criterion to predict whether a given process is spontaneous or not. Before seeing how this is done, however, we will find it useful to explore entropy from a molecular perspective.

A word on notation before we proceed: Throughout most of the remainder of this chapter, we will focus on systems rather than surroundings. To simplify the notation, we will usually refer to the entropy change of the system as ∆S rather than explicitly indicating ∆Ssys.

19.3 | The molecular interpretation of entropy and the Third law of Thermodynamics

As chemists, we are interested in molecules. What does entropy have to do with them and with their transformations? What molecular property does entropy reflect? Ludwig Boltzmann (1844–1906) gave another conceptual meaning to the notion of entropy, and to understand his contribution, we need to examine the ways in which we can interpret entropy at the molecular level.

Expansion of a Gas at the Molecular LevelIn discussing Figure 19.2, we talked about the expansion of a gas into a vacuum as a spontaneous process. We now understand that it is an irreversible process and that the entropy of the universe increases during the expansion. How can we explain


the spontaneity of this process at the molecular level? We can get a sense of what makes this expansion spontaneous by envisioning the gas as a collection of particles in constant motion, as we did in discussing the kinetic-molecular theory of gases.

(Section 10.7) When the stopcock in Figure 19.2 is opened, we can view the expansion of the gas as the ultimate result of the gas molecules moving randomly throughout the larger volume.

Let’s look at this idea more closely by tracking two of the gas molecules as they move around. Before the stopcock is opened, both molecules are confined to the left flask, as shown in ▲ Figure 19.6(a). After the stopcock is opened, the molecules travel randomly throughout the entire apparatus. As Figure 19.6(b) shows, there are four possible arrangements for the two molecules once both flasks are available to them. Because the molecular motion is random, all four arrangements are equally likely. Note that now only one arrangement corresponds to the situation before the stopcock was opened: both molecules in the left flask.

Figure 19.6(b) shows that with both flasks available to the molecules, the prob-ability of the red molecule being in the left flask is two in four (top right and bottom left arrangements), and the probability of the blue molecule being in the left flask is the same (top left and bottom left arrangements). Because the probability is 2>4 = 1>2 that each molecule is in the left flask, the probability that both are there is 11>222 = 1>4. If we apply the same analysis to three gas molecules, we find that the probability that all three are in the left flask at the same time is 11>223 = 1>8.

Now let’s consider a mole of gas. The probability that all the molecules are in the left flask at the same time is 11>22N, where N = 6.02 * 1023. This is a vanishingly small number! Thus, there is essentially zero likelihood that all the gas molecules will be in the left flask at the same time. This analysis of the microscopic behavior of the gas molecules leads to the expected macroscopic behavior: The gas spontaneously expands to fill both the left and right flasks, and it does not spontaneously all go back in the left flask.

This molecular view of gas expansion shows the tendency of the molecules to “spread out” among the different arrangements they can take. Before the stopcock is opened, there is only one possible arrangement: all molecules in the left flask. When the stopcock is opened, the arrangement in which all the molecules are in the left flask is but one of an extremely large number of possible arrangements. The most probable arrangements by far are those in which there are essentially equal numbers of molecules in the two flasks. When the gas spreads throughout the apparatus, any given molecule can be in either flask rather than confined to the left flask. We say that with the stopcock opened, the arrangement of gas mol-ecules is more random or disordered than when the molecules are all confined in the left flask.

We will see this notion of increasing randomness helps us understand entropy at the molecular level.

▲ Figure 19.6 Possible arrangements of two gas molecules in two flasks. (a) Before the stopcock is opened, both molecules are in the left flask. (b) After the stopcock is opened, there are four possible arrangements of the two molecules.

(a)The two molecules are colored

red and blue to keep track of them.

(b)Four possible arrangements (microstates)

once the stopcock is opened.


Boltzmann’s Equation and MicrostatesThe science of thermodynamics developed as a means of describing the properties of matter in our macroscopic world without regard to microscopic structure. In fact, ther-modynamics was a well-developed field before the modern view of atomic and molecu-lar structure was even known. The thermodynamic properties of water, for example, addressed the behavior of bulk water (or ice or water vapor) as a substance without considering any specific properties of individual H2O molecules.

To connect the microscopic and macroscopic descriptions of matter, scientists have developed the field of statistical thermodynamics, which uses the tools of statistics and probability to link the microscopic and macroscopic worlds. Here we show how entropy, which is a property of bulk matter, can be connected to the behavior of atoms and molecules. Because the mathematics of statistical thermodynamics is complex, our discussion will be largely conceptual.

In our discussion of two gas molecules in the two-flask system in Figure 19.6, we saw that the number of possible arrangements helped explain why the gas expands.

Suppose we now consider one mole of an ideal gas in a particular thermodynamic state, which we can define by specifying the temperature, T, and volume, V, of the gas. What is happening to this gas at the microscopic level, and how does what is going on at the microscopic level relate to the entropy of the gas?

Imagine taking a snapshot of the positions and speeds of all the molecules at a given instant. The speed of each molecule tells us its kinetic energy. That particular set of 6 * 1023 positions and kinetic energies of the individual gas molecules is what we call a microstate of the system. A microstate is a single possible arrangement of the positions and kinetic energies of the molecules when the molecules are in a specific thermodynamic state. We could envision continuing to take snapshots of our system to see other possible microstates.

As you no doubt see, there would be such a staggeringly large number of micro-states that taking individual snapshots of all of them is not feasible. Because we are examining such a large number of particles, however, we can use the tools of statistics and probability to determine the total number of microstates for the thermodynamic state. (That is where the statistical part of the name statistical thermodynamics comes in.) Each thermodynamic state has a characteristic number of microstates associated with it, and we will use the symbol W for that number.

Students sometimes have difficulty distinguishing between the state of a system and the microstates associated with the state. The difference is that state is used to describe the macroscopic view of our system as characterized, for example, by the pressure or temper-ature of a sample of gas. A microstate is a particular microscopic arrange-ment of the atoms or molecules of the system that corresponds to the given state of the system. Each of the snapshots we described is a microstate—the positions and kinetic energies of individual gas molecules will change from snapshot to snapshot, but each one is a possible arrangement of the collec-tion of molecules corresponding to a single state. For macroscopically sized systems, such as a mole of gas, there is a very large number of microstates for each state—that is, W is generally an extremely large number.

The connection between the number of microstates of a system, W, and the entropy of the system, S, is expressed in a beautifully sim-ple equation developed by Boltzmann and engraved on his tombstone (▶ Figure 19.7):

S = k ln W [19.5]In this equation, k is the Boltzmann constant, 1.38 * 10-23 J>K. Thus, entropy is a measure of how many microstates are associated with a particular macroscopic state.

Give It Some ThoughtWhat is the entropy of a system that has only a single microstate?

▲ Figure 19.7 Ludwig Boltzmann’s gravestone. Boltzmann’s gravestone in Vienna is inscribed with his famous relationship between the entropy of a state, S, and the number of available microstates, W. (In Boltzmann’s time, “log” was used to represent the natural logarithm.)


From Equation 19.5, we see that the entropy change accompanying any process is

∆S = k ln Wfinal - k ln Winitial = k ln Wfinal

Winitial [19.6]

Any change in the system that leads to an increase in the number of microstates 1Wfinal 7 Winitial2 leads to a positive value of ∆S: Entropy increases with the number of microstates of the system.

Let’s consider two modifications to our ideal-gas sample and see how the entropy changes in each case. First, suppose we increase the volume of the system, which is analogous to allowing the gas to expand isothermally. A greater volume means a greater number of positions available to the gas atoms and therefore a greater number of microstates. The entropy therefore increases as the volume increases, as we saw in the “A Closer Look” box in Section 19.2.

Second, suppose we keep the volume fixed but increase the temperature. How does this change affect the entropy of the system? Recall the distribution of molecular speeds presented in Figure 10.13(a). An increase in temperature increases the most probable speed of the molecules and also broadens the distribution of speeds. Hence, the mole-cules have a greater number of possible kinetic energies, and the number of microstates increases. Thus, the entropy of the system increases with increasing temperature.

Molecular Motions and EnergyWhen a substance is heated, the motion of its molecules increases. In Section 10.7, we found that the average kinetic energy of the molecules of an ideal gas is directly propor-tional to the absolute temperature of the gas. That means the higher the temperature, the faster the molecules move and the more kinetic energy they possess. Moreover, hot-ter systems have a broader distribution of molecular speeds, as Figure 10.13(a) shows.

The particles of an ideal gas are idealized points with no volume and no bonds, however, points that we visualize as flitting around through space. Any real molecule can undergo three kinds of more complex motion. The entire molecule can move in one direction, which is the simple motion we visualize for an ideal particle and see in a macroscopic object, such as a thrown baseball. We call such movement translational motion. The molecules in a gas have more freedom of translational motion than those in a liquid, which have more freedom of translational motion than the molecules of a solid.

A real molecule can also undergo vibrational motion, in which the atoms in the molecule move periodically toward and away from one another, and rotational motion, in which the molecule spins about an axis. ▼ Figure 19.8 shows the vibrational motions and one of the rotational motions possible for the water molecule. These dif-ferent forms of motion are ways in which a molecule can store energy.

Give It Some ThoughtCan an argon atom undergo vibrational motion?

The vibrational and rotational motions possible in real molecules lead to arrange-ments that a single atom cannot have. A collection of real molecules therefore has a

Vibrations Rotation

▲ Figure 19.8 Vibrational and rotational motions in a water molecule.

Go FiGureDescribe another possible rotational motion for this molecule.


greater number of possible microstates than does the same number of ideal-gas particles. In general, the number of microstates possible for a system increases with an increase in volume, an increase in temperature, or an increase in the number of molecules because any of these changes increases the possible positions and kinetic energies of the molecules mak-ing up the system. We will also see that the number of microstates increases as the com-plexity of the molecule increases because there are more vibrational motions available.

Chemists have several ways of describing an increase in the number of microstates possible for a system and therefore an increase in the entropy for the system. Each way seeks to capture a sense of the increased freedom of motion that causes molecules to spread out when not restrained by physical barriers or chemical bonds.

The most common way of describing an increase in entropy is the increase in the ran-domness, or disorder, of the system. Another way likens an entropy increase to an increased dispersion (spreading out) of energy because there is an increase in the number of ways the positions and energies of the molecules can be distributed throughout the system. Each description (randomness or energy dispersal) is conceptually helpful if applied correctly.

Making Qualitative Predictions about ∆SIt is usually not difficult to estimate qualitatively how the entropy of a system changes during a simple process. As noted earlier, an increase in either the temperature or the volume of a system leads to an increase in the number of microstates, and hence an in-crease in the entropy. One more factor that correlates with number of microstates is the number of independently moving particles.

We can usually make qualitative predictions about entropy changes by focusing on these factors. For example, when water vaporizes, the molecules spread out into a larger vol-ume. Because they occupy a larger volume, there is an increase in their freedom of motion, giving rise to a greater number of possible microstates, and hence an increase in entropy.

Now consider the phases of water. In ice, hydrogen bonding leads to the rigid struc-ture shown in ▼ Figure 19.9. Each molecule in the ice is free to vibrate, but its translational and rotational motions are much more restricted than in liquid water. Although there are hydrogen bonds in liquid water, the molecules can more readily move about relative to

Rigid, crystalline structure

Motion restricted to vibration only

Smallest number of microstates

Ice

Increased freedom with respectto translation

Free to vibrate and rotate

Larger number of microstates

Liquid water

Increasing entropy

Molecules spread out, essentiallyindependent of one another

Complete freedom for translation,vibration, and rotation

Largest number of microstates

Water vapor

▲ Figure 19.9 Entropy and the phases of water. The larger the number of possible microstates, the higher the entropy of the system.

Go FiGureIn which phase are water molecules least able to have rotational motion?


one another (translation) and tumble around (rotation). During melting, therefore, the number of possible microstates increases and so does the entropy. In water vapor, the molecules are essentially independent of one another and have their full range of transla-tional, vibrational, and rotational motions. Thus, water vapor has an even greater number of possible microstates and therefore a higher entropy than liquid water or ice.

When an ionic solid dissolves in water, a mixture of water and ions replaces the pure solid and pure water, as shown for KCl in ◀ Figure 19.10. The ions in the liquid move in a volume that is larger than the volume in which they were able to move in the crystal lattice and so undergo more motion. This increased motion might lead us to conclude that the entropy of the system has increased. We have to be careful, however, because some of the water molecules have lost some freedom of motion because they are now held around the ions as water of hydration. (Section 13.1) These water molecules are in a more ordered state than before because they are now confined to the immediate environment of the ions. Therefore, the dissolving of a salt involves both a disordering process (the ions become less confined) and an ordering process (some water molecules become more confined). The disordering processes are usually dominant, and so the overall effect is an increase in the randomness of the system when most salts dissolve in water.

Now, imagine arranging biomolecules into a highly organized biochemical system, such as the nucleosome in the chapter-opening figure. We might expect that the creation of this well-ordered structure would lead to a decrease in the entropy of the system. But this is frequently not the case. Waters of hydration and counterions can be expelled from the interface as two large biomolecules interact, and so the entropy of the system can actually increase—if you consider the water and counterions to be part of the system.

The same ideas apply to chemical reactions. Consider the reaction between nitric oxide gas and oxygen gas to form nitrogen dioxide gas:

2 NO1g2 + O21g2 ¡ 2 NO21g2 [19.7]

which results in a decrease in the number of molecules—three molecules of gaseous reactants form two molecules of gaseous products (◀ Figure 19.11). The formation of new N ¬ O bonds reduces the motions of the atoms in the system. The formation of new bonds decreases the number of degrees of freedom, or forms of motion, available to the atoms. That is, the atoms are less free to move in random fashion because of the formation of new bonds. The decrease in the number of molecules and the resultant decrease in motion result in fewer possible microstates and therefore a decrease in the entropy of the system.

In summary, we generally expect the entropy of a system to increase for processes in which

1. Gases form from either solids or liquids. 2. Liquids or solutions form from solids. 3. The number of gas molecules increases during a chemical reaction.

+ −− + −

−−

+

+

+

− + − + −

− + − + − + − + −

− + − + − + −

▲ Figure 19.10 Entropy changes when an ionic solid dissolves in water. The ions become more spread out and disordered, but the water molecules that hydrate the ions become less disordered.

▲ Figure 19.11 Entropy decreases when NO 1g 2 is oxidized by O2 1g 2 to NO2 1g 2 . A decrease in the number of gaseous molecules leads to a decrease in the entropy of the system.

Go FiGureWhat major factor leads to a decrease in entropy as this reaction takes place?

2 NO2(g)2 NO(g) + O2(g)

samPle exerCise 19.3 Predicting the sign of ∆S

Predict whether ∆S is positive or negative for each process, assuming each occurs at constant temperature:(a) H2O1l2 ¡ H2O1g2(b) Ag+1aq2 + Cl-1aq2 ¡ AgCl1s2(c) 4 Fe1s2 + 3 O21g2 ¡ 2 Fe2O31s2(d) N21g2 + O21g2 ¡ 2 NO1g2

soluTionAnalyze We are given four reactions and asked to predict the sign of ∆S for each.Plan We expect ∆S to be positive if there is an increase in tem-perature, increase in volume, or increase in number of gas par-ticles. The question states that the temperature is constant, and so we need to concern ourselves only with volume and number of particles.

Solve

(a) Evaporation involves a large increase in volume as liquid changes to gas. One mole of water (18 g) occupies about 18 mL as a liq-uid and if it could exist as a gas at STP it would occupy 22.4 L. Because the molecules are distributed throughout a much larger volume in the gaseous state, an increase in motional freedom accompanies vaporization and ∆S is positive.


(b) In this process, ions, which are free to move throughout the vol-ume of the solution, form a solid, in which they are confined to a smaller volume and restricted to more highly constrained posi-tions. Thus, ∆S is negative.

(c) The particles of a solid are confined to specific locations and have fewer ways to move (fewer microstates) than do the molecules of a gas. Because O2 gas is converted into part of the solid product Fe2O3, ∆S is negative.

(d) The number of moles of reactant gases is the same as the num-ber of moles of product gases, and so the entropy change is expected to be small. The sign of ∆S is impossible to predict based on our discussions thus far, but we can predict that ∆S will be close to zero.

Practice exercise 1Indicate whether each process produces an increase or decrease in the entropy of the system:(a) CO21s2 ¡ CO21g2(b) CaO1s2 + CO21g2 ¡ CaCO31s2(c) HCl1g2 + NH31g2 ¡ NH4Cl1s2(d) 2 SO21g2 + O21g2 ¡ 2 SO31g2Practice exercise 2Since the entropy of the universe increases for spontaneous pro-cesses, does it mean that the entropy of the universe decreases for nonspontaneous processes?

soluTionAnalyze We need to select the system in each pair that has the greater entropy.

Plan We examine the state of each system and the complexity of the molecules it contains.

Solve (a) HCl(g) has the higher entropy because the particles in gases are more disordered and have more freedom of motion than the particles in solids. (b) When these two systems are at the same pres-sure, the sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol sample has twice the number of microstates and twice the entropy. (c) The HCl system has the higher entropy because the number of ways in which an HCl molecule can store energy is greater than the number of ways

samPle exerCise 19.4 Predicting relative entropies

In each pair, choose the system that has greater entropy and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K.

in which an Ar atom can store energy. (Molecules can rotate and vibrate; atoms cannot.)

Practice exercise 1Which system has the greatest entropy? (a) 1 mol of H21g2 at STP, (b) 1 mol of H21g2 at 100 °C and 0.5 atm, (c) 1 mol of H2O1s2 at 0 °C, (d) 1 mol of H2O1l2 at 25 °C.

Practice exercise 2Choose the system with the greater entropy in each case:(a) 1 mol of H21g2 at STP or 1 mol of SO21g2 at STP, (b) 1 mol of N2O41g2 at STP or 2 mol of NO21g2 at STP.

The Third Law of ThermodynamicsIf we decrease the thermal energy of a system by lowering the temperature, the energy stored in translational, vibrational, and rotational motion decreases. As less energy is stored, the entropy of the system decreases because it has fewer and fewer microstates available. If we keep lowering the temperature, do we reach a state in which these mo-tions are essentially shut down, a point described by a single microstate? This question is addressed by the third law of thermodynamics, which states that the entropy of a pure, perfect crystalline substance at absolute zero is zero: S10 K2 = 0.

Consider a pure, perfect crystalline solid. At absolute zero, the individual atoms or molecules in the lattice would be perfectly ordered in position. Because none of them would have thermal motion, there is only one possible microstate. As a result, Equation 19.5 becomes S = k ln W = k ln 1 = 0. As the temperature is increased from abso-lute zero, the atoms or molecules in the crystal gain energy in the form of vibrational motion about their lattice positions. This means that the degrees of freedom and the entropy both increase. What happens to the entropy, however, as we continue to heat the crystal? We consider this important question in the next section.

Give It Some ThoughtIf you are told that the entropy of a system is zero, what do you know about the system?


19.4 | entropy Changes in Chemical reactions

In Section 5.5 we discussed how calorimetry can be used to measure ∆H for chemical reactions. No comparable method exists for measuring ∆S for a reaction. However, be-cause the third law establishes a zero point for entropy, we can use experimental mea-surements to determine the absolute value of the entropy, S. To see schematically how this is done, let’s review in greater detail the variation in the entropy of a substance with temperature.

We know that the entropy of a pure, perfect crystalline solid at 0 K is zero and that the entropy increases as the temperature of the crystal is increased. ▼ Figure 19.12 shows that the entropy of the solid increases steadily with increasing temperature up to the melting point of the solid. When the solid melts, the atoms or molecules are free

Chemistry and life

Entropy and Human Society

Any living organism is a complex, highly organized, well-ordered system, even at the molecular level like the nucleosome we saw at the beginning of this chapter. Our entropy content is much lower than it would be if we were completely decomposed into carbon dioxide, water, and several other simple chemicals. Does this mean that life is a violation of the second law? No, because the thousands of chemical reactions necessary to produce and maintain life have caused a large increase in the entropy of the rest of the universe. Thus, as the second law requires, the overall entropy change during the lifetime of a hu-man, or any other living system, is positive.

The second law of thermodynamics applies also to the way we humans order our surroundings. In addition to being complex living systems ourselves, we humans are masters of producing order in the world around us. We manipulate and order matter at the nanoscale level in order to produce the technological breakthroughs that have

become so commonplace in the twenty-first century. We use tremen-dous quantities of raw materials to produce highly ordered materials. In so doing, we expend a great deal of energy to, in essence, “fight” the second law of thermodynamics.

For every bit of order we produce, however, we produce an even greater amount of disorder. Petroleum, coal, and natural gas are burned to provide the energy necessary for us to achieve highly ordered structures, but their combustion increases the entropy of the universe by releasing CO21g2, H2O1g2, and heat. Oxide and sulfide ores release CO21g2 and SO21g2 that spread throughout our atmo-sphere. Thus, even as we strive to create more impressive discoveries and greater order in our society, we drive the entropy of the universe higher, as the second law says we must.

We humans are, in effect, using up our storehouse of energy-rich materials to create order and advance technology. As noted in Chapter 5, we must learn to harness new energy sources, such as solar energy, before we exhaust the supplies of readily available energy of other kinds.

00

Temperature (K)

Ent

ropy

, S

Melting

LiquidSolid Gas

Boiling

▲ Figure 19.12 Entropy increases with increasing temperature.

Go FiGureWhy does the plot show vertical jumps at the melting and boiling points?

secTiOn 19.4 entropy changes in chemical reactions 829

to move about the entire volume of the sample. The added degrees of freedom increase the randomness of the substance, thereby increasing its entropy. We therefore see a sharp increase in the entropy at the melting point. After all the solid has melted, the temperature again increases and with it, the entropy.

At the boiling point of the liquid, another abrupt increase in entropy occurs. We can understand this increase as resulting from the increased volume available to the atoms or molecules as they enter the gaseous state. When the gas is heated further, the entropy increases steadily as more energy is stored in the translational motion of the gas atoms or molecules.

Another change occurring at higher temperatures is the skewing of molecular speeds toward higher values (Figure 10.13(a)). The expansion of the range of speeds leads to increased kinetic energy and increased disorder and, hence, increased entropy. The conclusions we reach in examining Figure 10.13 are consistent with what we noted earlier: Entropy generally increases with increasing temperature because the increased motional energy leads to a greater number of possible microstates.

Entropy versus temperature graphs such as Figure 19.12 can be obtained by care-fully measuring how the heat capacity of a substance (Section 5.5) varies with temperature, and we can use the data to obtain the absolute entropies at different tem-peratures. (The theory and methods used for these measurements and calculations are beyond the scope of this text.) Entropies are usually tabulated as molar quantities, in units of joules per mole-kelvin 1J>mol@K2.

Molar entropies for substances in their standard states are known as standard molar entropies and denoted S°. The standard state for any substance is defined as the pure substance at 1 atm pressure.* ▶ Table 19.1 lists the values of S° for a number of substances at 298 K; Appendix C gives a more extensive list.

We can make several observations about the S° values in Table 19.1:

1. Unlike enthalpies of formation, standard molar entropies of elements at the ref-erence temperature of 298 K are not zero.

2. The standard molar entropies of gases are greater than those of liquids and solids, consistent with our interpretation of experimental observations, as represented in Figure 19.12.

3. Standard molar entropies generally increase with increasing molar mass. 4. Standard molar entropies generally increase with an increasing number of atoms

in the formula of a substance.

Point 4 is related to the molecular motion discussed in Section 19.3. In general, the number of degrees of freedom for a molecule increases with increasing number of atoms, and thus the number of possible microstates also increases. Figure 19.13 com-pares the standard molar entropies of three hydrocarbons in the gas phase. Notice how the entropy increases as the number of atoms in the molecule increases.

The entropy change in a chemical reaction equals the sum of the entropies of the products minus the sum of the entropies of the reactants:

∆S° = anS°1products2 - amS°1reactants2 [19.8]

As in Equation 5.31, the coefficients n and m are the coefficients in the balanced chemi-cal equation for the reaction.

*The standard pressure used in thermodynamics is no longer 1 atm but rather is based on the SI unit for pres-sure, the pascal (Pa). The standard pressure is 105 Pa, a quantity known as a bar: 1 bar = 105 Pa = 0.987 atm.Because 1 bar differs from 1 atm by only 1.3%, we will continue to refer to the standard pressure as 1 atm.

Table 19.1 standard molar entropies of selected substances at 298 K

substance S° 1J ,mol@K 2H21g2 130.6

N21g2 191.5

O21g2 205.0

H2O1g2 188.8

NH31g2 192.5

CH3OH1g2 237.6

C6H61g2 269.2

H2O1l2 69.9

CH3OH1l2 126.8

C6H61l2 172.8

Li1s2 29.1

Na1s2 51.4

K1s2 64.7

Fe(s) 27.23

FeCl31s2 142.3

NaCl(s) 72.3

samPle exerCise 19.5 Calculating ∆S° from Tabulated entropies

Calculate the change in the standard entropy of the system, ∆S°, for the synthesis of ammonia from N21g2 and H21g2 at 298 K:

N21g2 + 3 H21g2 ¡ 2 NH31g2


Entropy Changes in the SurroundingsWe can use tabulated absolute entropy values to calculate the standard entropy change in a system, such as a chemical reaction, as just described. But what about the entropy change in the surroundings? We encountered this situation in Section 19.2, but it is good to revisit it now that we are examining chemical reactions.

We should recognize that the surroundings for any system serve essentially as a large, constant-temperature heat source (or heat sink if the heat flows from the sys-tem to the surroundings). The change in entropy of the surroundings depends on how much heat is absorbed or given off by the system.

For an isothermal process, the entropy change of the surroundings is given by

∆Ssurr =-qsys

T

▲ Figure 19.13 Entropy increases with increasing molecular complexity.

Go FiGureWhat might you expect for the value of S° for butane, C4H10?

Methane, CH4S° = 186.3 J/mol-K

Ethane, C2H6S° = 229.6 J/mol-K

Propane, C3H8S° = 270.3 J/mol-K

∆S° = 12 mol21192.5 J>mol@K2 - 311 mol21191.5 J>mol@K2 + 13 mol21130.6 J>mol@K24 = -198.3 J>K

Check: The value for ∆S° is negative, in agreement with our qualitative prediction based on the decrease in the number of molecules of gas during the reaction.

Practice exercise 1Using the standard molar entropies in Appendix C, calculate the standard entropy change, ∆S°, for the “water-splitting” reaction at 298 K:

2 H2O1l2 ¡ 2 H21g2 + O21g2(a) 326.3 J>K (b) 265.7 J>K (c) 163.2 J>K (d) 88.5 J>K (e) -326.3 J>K.

Practice exercise 2Using the standard molar entropies in Appendix C, calculate the standard entropy change, ∆S°, for the following reaction at 298 K:

Al2O31s2 + 3 H21g2 ¡ 2 Al1s2 + 3 H2O1g2

soluTionAnalyze We are asked to calculate the standard entropy change for the synthesis of NH31g2 from its constituent elements.Plan We can make this calculation using Equation 19.8 and the standard molar entropy values in Table 19.1 and Appendix C.Solve Using Equation 19.8, we have ∆S° = 2S°1NH32 - 3S°1N22 + 3S°1H224Substituting the appropriate S° values from Table 19.1 yields

secTiOn 19.5 Gibbs Free energy 831

Because in a constant-pressure process, qsys is simply the enthalpy change for the reac-tion, ∆H, we can write

∆Ssurr =- ∆Hsys

T 3at constant P4 [19.9]

For the ammonia synthesis reaction in Sample Exercise 19.5, qsys is the enthalpy change for the reaction under standard conditions, ∆H°, so the changes in entropy will be standard entropy changes, ∆S°. Therefore, using the procedures described in Sec-tion 5.7, we have

∆H°rxn = 2 ∆Hf°3NH31g24 - 3 ∆Hf°3H21g24 - ∆Hf°3N21g24 = 21-46.19 kJ2 - 310 kJ2 - 10 kJ2 = -92.38 kJ

The negative value tells us that at 298 K the formation of ammonia from H21g2 and N21g2 is exothermic. The surroundings absorb the heat given off by the system, which means an increase in the entropy of the surroundings:

∆S°surr =92.38 kJ298 K

= 0.310 kJ>K = 310 J>K

Notice that the magnitude of the entropy gained by the surroundings is greater than that lost by the system, calculated as -198.3 J>K in Sample Exercise 19.5.

The overall entropy change for the reaction is

∆S°univ = ∆S°sys + ∆S°surr = -198.3 J>K + 310 J>K = 112 J>K

Because ∆S°univ is positive for any spontaneous reaction, this calculation indicates that when NH31g2, H21g2, and N21g2 are together at 298 K in their standard states (each at 1 atm pressure), the reaction moves spontaneously toward formation of NH31g2.

Keep in mind that while the thermodynamic calculations indicate that forma-tion of ammonia is spontaneous, they do not tell us anything about the rate at which ammonia is formed. Establishing equilibrium in this system within a reasonable period requires a catalyst, as discussed in Section 15.7.

Give It Some ThoughtIf a process is exothermic, does the entropy of the surroundings (a) always increase, (b) always decrease, or (c) sometimes increase and sometimes decrease, depending on the process?

19.5 | Gibbs Free energyWe have seen examples of endothermic processes that are spontaneous, such as the dissolution of ammonium nitrate in water. (Section 13.1) We learned in our discussion of the solution process that a spontaneous process that is endother-mic must be accompanied by an increase in the entropy of the system. However, we have also encountered processes that are spontaneous and yet proceed with a decrease in the entropy of the system, such as the highly exothermic formation of sodium chloride from its constituent elements. (Section 8.2) Spontaneous pro-cesses that result in a decrease in the system’s entropy are always exothermic. Thus, the spontaneity of a reaction seems to involve two thermodynamic concepts, en-thalpy and entropy.

How can we use ∆H and ∆S to predict whether a given reaction occurring at con-stant temperature and pressure will be spontaneous? The means for doing so was first developed by the American mathematician J. Willard Gibbs (1839–1903). Gibbs pro-posed a new state function, now called the Gibbs free energy (or just free energy), G, and defined as

G = H - TS [19.10]


where T is the absolute temperature. For an isothermal process, the change in the free energy of the system, ∆G, is

∆G = ∆H - T∆S [19.11]

Under standard conditions, this equation becomes

∆G° = ∆H° - T∆S° [19.12]

To see how the state function G relates to reaction spontaneity, recall that for a reaction occurring at constant temperature and pressure

∆Suniv = ∆Ssys + ∆Ssurr = ∆Ssys + a- ∆Hsys

Tb

where Equation 19.9 substitutes for ∆Ssurr. Multiplying both sides by -T gives

-T∆Suniv = ∆Hsys - T∆Ssys [19.13]

Comparing Equations 19.11 and 19.13, we see that in a process occurring at con-stant temperature and pressure, the free-energy change, ∆G, is equal to -T∆Suniv. We know that for spontaneous processes, ∆Suniv is always positive and, therefore, -T∆Suniv is always negative. Thus, the sign of ∆G provides us with extremely valuable informa-tion about the spontaneity of processes that occur at constant temperature and pres-sure. If both T and P are constant, the relationship between the sign of ∆G and the spontaneity of a reaction is:

1. If ∆G 6 0, the reaction is spontaneous in the forward direction. 2. If ∆G = 0, the reaction is at equilibrium. 3. If ∆G 7 0, the reaction in the forward direction is nonspontaneous (work must be

done to make it occur) but the reverse reaction is spontaneous.

It is more convenient to use ∆G as a criterion for spontaneity than to use ∆Suniv because ∆G relates to the system alone and avoids the complication of having to exam-ine the surroundings.

An analogy is often drawn between the free-energy change during a spontaneous reac-tion and the potential-energy change when a boulder rolls down a hill (◀ Figure 19.14). Potential energy in a gravitational field “drives” the boulder until it reaches a state of mini-mum potential energy in the valley. Similarly, the free energy of a chemical system decreases until it reaches a minimum value. When this minimum is reached, a state of equilibrium exists. In any spontaneous process carried out at constant temperature and pressure, the free energy always decreases.

To illustrate these ideas, let’s return to the Haber process for the synthesis of ammonia from nitrogen and hydrogen, which we discussed extensively in Chapter 15:

N21g2 + 3 H21g2 ∆ 2 NH31g2Imagine that we have a reaction vessel that allows us to maintain a constant tem-perature and pressure and that we have a catalyst that allows the reaction to proceed at a reasonable rate. What happens when we load the vessel with a certain number of moles of N2 and three times that number of moles of H2? As we saw in Figure 15.3 (p. 632), the N2 and H2 react spontaneously to form NH3 until equilibrium is achieved. Similarly, Figure 15.3 shows that if we load the vessel with pure NH3, it decomposes spontaneously to N2 and H2 until equilibrium is reached. In each case the free energy of the system gets progressively lower and lower as the reaction moves toward equilibrium, which represents a minimum in the free energy. We illustrate these cases in ▶ Figure 19.15.

Give It Some ThoughtWhat are the criteria for spontaneity(a) in terms of entropy and(b) in terms of free energy?

Course of reaction

Equilibriummixture

Reactants

Products

Free

ene

rgy

Position

Equilibriumposition in

valley

Pote

ntia

l ene

rgy

▲ Figure 19.14 Potential energy and free energy. An analogy is shown between the gravitational potential-energy change of a boulder rolling down a hill and the free-energy change in a spontaneous reaction. Free energy always decreases in a spontaneous process when pressure and temperature are held constant.

Go FiGureAre the processes that move a system toward equilibrium spontaneous or nonspontaneous?


This is a good time to remind ourselves of the significance of the reaction quotient, Q, for a system that is not at equilibrium. (Section 15.6) Recall that when Q 6 K, there is an excess of reactants relative to products and the reaction proceeds spontaneously in the forward direction to reach equilibrium, as noted in Figure 19.15. When Q 7 K, the reaction proceeds spontaneously in the reverse direction. At equilibrium Q = K.

▲ Figure 19.15 Free energy and approaching equilibrium. In the reaction N21g2 + 3 H21g2 ∆ 2 NH31g2, if the reaction mixture has too much N2 and H2 relative to NH3 (left), Q 6 K and NH3 forms spontaneously. If there is more NH3 in the mixture relative to the reactants N2 and H2 (right), Q 7 K and the NH3 decomposes spontaneously into N2 and H2.

Go FiGureWhy are the spontaneous processes shown sometimes said to be “downhill” in free energy?

Equilibriummixture

(Q = K, ∆G = 0)

PureN2 + H2

PureNH3

Q > KQ < K

Free

ene

rgy Spontaneous

N2(g) + 3 H2(g) 2 NH3(g)

Spontaneous

soluTionAnalyze We are asked to calculate ∆G° for the indicated reaction (given ∆H°, ∆S°, and T) and to predict whether the reaction is spon-taneous under standard conditions at 298 K.Plan To calculate ∆G°, we use Equation 19.12, ∆G° = ∆H° - T∆S°. To determine whether the reaction is spontaneous under standard conditions, we look at the sign of ∆G°.

samPle exerCise 19.6 Calculating Free-energy Change from ∆H°, T, and ∆S°Calculate the standard free-energy change for the formation of NO(g) from N21g2 and O21g2 at 298 K:

N21g2 + O21g2 ¡ 2 NO1g2given that ∆H° = 180.7 kJ and ∆S° = 24.7 J>K. Is the reaction spontaneous under these conditions?

Solve

∆G° = ∆H° - T∆S° = 180.7 kJ - 1298 K2124.7 J>K2a 1 kJ

103 Jb

= 180.7 kJ - 7.4 kJ = 173.3 kJ


Standard Free Energy of FormationRecall that we defined standard enthalpies of formation, ∆H °f , as the enthalpy change when a substance is formed from its elements under defined standard conditions.

(Section 5.7) We can define standard free energies of formation, ∆Gf°, in a simi-lar way: ∆Gf° for a substance is the free-energy change for its formation from its ele-ments under standard conditions. As is summarized in ◀ Table 19.2, standard state means 1 atm pressure for gases, the pure solid for solids, and the pure liquid for liquids. For substances in solution, the standard state is normally a concentration of 1 M. (In very accurate work it may be necessary to make certain corrections, but we need not worry about these.)

The temperature usually chosen for purposes of tabulating data is 25 °C, but we will calculate ∆G at other temperatures as well. Just as for the standard heats of forma-tion, the free energies of elements in their standard states are set to zero. This arbitrary choice of a reference point has no effect on the quantity in which we are interested, which is the difference in free energy between reactants and products.

A listing of standard free energies of formation is given in Appendix C.

Give It Some ThoughtWhat does the superscript ° indicate when associated with a thermodynamic quantity, as in ∆H°, ∆S°, or ∆G°?

Standard free energies of formation are useful in calculating the standard free-energy change for chemical processes. The procedure is analogous to the calculation of ∆H° (Equation 5.31) and ∆S° (Equation 19.8):

∆G° = an∆Gf°1products2 - am∆Gf°1reactants2 [19.14]

Because ∆G° is positive, the reaction is not spontaneous under stan-dard conditions at 298 K.Comment Notice that we had to convert the units of the T∆S° term to kJ so that they could be added to the ∆H° term, whose units are kJ.

Practice exercise 1Which of these statements is true? (a) All spontaneous reactions have a negative enthalpy change, (b) All spontaneous reactions have a positive entropy change, (c) All spontaneous reactions have a

positive free-energy change, (d) All spontaneous reactions have a negative free-energy change, (e) All spontaneous reactions have a negative entropy change.

Practice exercise 2Calculate ∆G° for a reaction for which ∆H° = 24.6 kJ and ∆S° = 132 J>K at 298 K. Is the reaction spontaneous under these conditions?

Table 19.2 Conventions used in establishing standard Free energies

state of matter standard state

Solid Pure solidLiquid Pure liquidGas 1 atm pressureSolution 1 M concentrationElement ∆Gf° = 0 for

element in standard state

soluTionAnalyze We are asked to calculate the free-energy change for a reac-tion and then to determine the free-energy change for the reverse reaction.

Plan We look up the free-energy values for the products and reactants and use Equation 19.14: We multiply the molar quantities by the coefficients in the balanced equation and subtract the total for the reactants from that for the products.

Solve

(a) Cl21g2 is in its standard state, so ∆Gf° is zero for this reactant. P41g2, however, is not in its standard state, so ∆Gf° is not zero for

samPle exerCise 19.7 Calculating standard Free-energy Change from Free energies of Formation

(a) Use data from Appendix C to calculate the standard free-energy change for the reaction P41g2 + 6 Cl21g2 ¡ 4 PCl31g2 at 298 K. (b) What is ∆G° for the reverse of this reaction?

this reactant. From the balanced equation and values from Appendix C, we have

∆G°rxn = 4 ∆Gf°3PCl31g24 - ∆Gf°3P41g24 - 6 ∆Gf°3Cl21g24 = 14 mol21-269.6 kJ>mol2 - 11 mol2124.4 kJ>mol2 - 0 = -1102.8 kJ

That ∆G° is negative tells us that a mixture of P41g2, Cl21g2, and PCl31g2 at 25 °C, each present at a partial pressure of 1 atm, would react spontaneously in the forward direction to form more PCl3. Remember, however, that the value of ∆G° tells us nothing about the rate at which the reaction occurs.


(b) When we reverse the reaction, we reverse the roles of the reac-tants and products. Thus, reversing the reaction changes the sign of ∆G in Equation 19.14, just as reversing the reaction changes the sign of ∆H. (Section 5.4) Hence, using the result from part (a), we have

4 PCl31g2 ¡ P41g2 + 6 Cl21g2 ∆G° = +1102.8 kJ

Practice exercise 1The following chemical equations describe the same chemical re-action. How do the free energies of these two chemical equations compare?

(1) 2 H2O1l2 ¡ 2 H21g2 + O21g2(2) H2O1l2 ¡ H21g2 + 1>2 O21g2(a) ∆G°1 = ∆G°2, (b) ∆G°1 = 2 ∆G°2, (c) 2∆G°1 = ∆G°2,(d) None of the above.

Practice exercise 2Use data from Appendix C to calculate ∆G° at 298 K for the combus-tion of methane: CH41g2 + 2 O21g2 ¡ CO21g2 + 2 H2O1g2.

In Section 5.7 we used Hess’s law to calculate ∆H° for the combustion of propane gas at 298 K:

C3H81g2 + 5 O21g2 ¡ 3 CO21g2 + 4 H2O1l2 ∆H° = -2220 kJ

(a) Without using data from Appendix C, predict whether ∆G° for this reaction is more negative or less negative than ∆H°. (b) Use data from Appendix C to calculate ∆G° for the reaction at 298 K. Is your prediction from part (a) correct?

soluTionAnalyze In part (a) we must predict the value for ∆G° relative to that for ∆H° on the basis of the balanced equation for the reaction. In part (b) we must calculate the value for ∆G° and compare this value with our qualitative prediction.Plan The free-energy change incorporates both the change in enthalpy and the change in entropy for the reaction (Equation 19.11), so under standard conditions

∆G° = ∆H° - T∆S°

To determine whether ∆G° is more negative or less negative than ∆H°, we need to determine the sign of the term T∆S°. Because T is the absolute temperature, 298 K, it is always a positive num-ber. We can predict the sign of ∆S° by looking at the reaction.Solve

(a) The reactants are six molecules of gas, and the products are three molecules of gas and four molecules of liquid. Thus, the number of molecules of gas has decreased significantly during the reaction. By using the general rules discussed in Section 19.3, we expect a decrease in the number of gas molecules to lead to a decrease in the entropy of the system—the products have fewer possible microstates than the reactants. We therefore expect ∆S° and T∆S° to be negative. Because we are subtracting T∆S°, which is a negative number, we predict that ∆G° is less negative than ∆H°.

(b) Using Equation 19.14 and values from Appendix C, we have

∆G° = 3 ∆Gf°3CO21g24 + 4 ∆Gf°3H2O1l24 - ∆Gf°3C3H81g24 - 5 ∆Gf°3O21g24 = 3 mol1-394.4 kJ>mol2 + 4 mol1-237.13 kJ>mol2 -

1 mol1-23.47 kJ>mol2 - 5 mol10 kJ>mol2 = -2108 kJ

Notice that we have been careful to use the value of ∆Gf° for H2O1l2. As in calculating ∆H val-ues, the phases of the reactants and products are important. As we predicted, ∆G° is less negative than ∆H° because of the decrease in entropy during the reaction.

Practice exercise 1If a reaction is exothermic and its entropy change is positive, which statement is true? (a) The reaction is spontaneous at all temperatures, (b) The reaction is nonspontaneous at all temperatures, (c) The reaction is spontaneous only at higher temperatures, (d) The reaction is spontaneous only at lower temperatures.

Practice exercise 2For the combustion of propane at 298 K, C3H81g2 + 5 O21g2 ¡ 3 CO21g2 + 4 H2O1g2, do you expect ∆G° to be more negative or less negative than ∆H°?

samPle exerCise 19.8 Predicting and Calculating ∆G°


Equations 19.17 allow us to use the sign of ∆G to conclude whether a reaction is spontaneous, nonspontaneous, or at equilibrium. When ∆G 6 0, a process is irreversible and, therefore, spontaneous. When ∆G = 0, the process is reversible and, therefore, at equilibrium. If a process has ∆G 7 0, then the reverse process will have ∆G 6 0; thus, the process as written is nonspontaneous but its reverse reaction will be irreversible and spontaneous.

The magnitude of ∆G is also significant. A reaction for which ∆G is large and negative, such as the burning of gasoline, is much more capable of doing work on the surroundings than is a reaction for which ∆G is small and negative, such as ice melting at room tempera-ture. In fact, thermodynamics tells us that the change in free energy for a process, ∆G, equals the maximum useful work that can be done by the system on its surroundings in a spontaneous process occurring at constant temperature and pressure:

∆G = -wmax [19.18]

(Remember our sign convention from Table 5.1: Work done by a system is negative.) In other words, ∆G gives the theoretical limit to how much work can be done by a process.

The relationship in Equation 19.18 explains why ∆G is called free energy—it is the portion of the energy change of a spontaneous reac-tion that is free to do useful work. The remainder of the energy enters the environment as heat. For example, the theoretical maximum work obtained for the combustion of gasoline is given by the value of ∆G for the combustion reaction. On average, standard internal combus-tion engines are inefficient in utilizing this potential work—more than 60% of the potential work is lost (primarily as heat) in converting the chemical energy of the gasoline to mechanical energy to move the ve-hicle. When other losses are considered—idling time, braking, aero-dynamic drag, and so forth—only about 15% of the potential work from the gasoline is used to move the car. Advances in automobile design—such as hybrid technology, efficient diesel engines, and new lightweight materials—have the potential to increase the percentage of useful work obtained from the gasoline.

For nonspontaneous processes 1∆G 7 02, the free-energy change is a measure of the minimum amount of work that must be done to cause the process to occur. In actuality, we always need to do more than this theoretical minimum amount because of the ineffi-ciencies in the way the changes occur.

a Closer look

What’s “Free” about Free Energy?

The Gibbs free energy is a remarkable thermodynamic quantity. Be-cause so many chemical reactions are carried out under conditions of near-constant pressure and temperature, chemists, biochemists, and engineers consider the sign and magnitude of ∆G as exceptionally useful tools in the design of chemical and biochemical reactions. We will see examples of the usefulness of ∆G throughout the remainder of this chapter and this text.

When first learning about ∆G, two common questions often arise: Why is the sign of ∆G an indicator of the spontaneity of reac-tions? And what is “free” about free energy?

In Section 19.2 we saw that the second law of thermodynam-ics governs the spontaneity of processes. In order to apply the sec-ond law (Equation 19.4), however, we must determine ∆Suniv, which is often difficult to evaluate. When T and P are constant, however, we can relate ∆Suniv to the changes in entropy and enthalpy of just the system by substituting the Equation 19.9 expression for ∆Ssurr in Equation 19.4:

∆Suniv = ∆Ssys + ∆Ssurr = ∆Ssys + a- ∆Hsys

Tb 1constant T, P2

[19.15]

Thus, at constant temperature and pressure, the second law becomes

Reversible process: ∆Suniv = ∆Ssys -∆Hsys

T= 0

Irreversible process: ∆Suniv = ∆Ssys -∆Hsys

T7 0 [19.16]1constant T, P2

Now we can see the relationship between ∆Gsys (which we call simply ∆G) and the second law. From Equation 19.11 we know that ∆G = ∆Hsys - T∆Ssys. If we multiply Equations 19.16 by -T and re-arrange, we reach the following conclusion:

Reversible process: ∆G = ∆Hsys - T ∆Ssys = 0Irreversible process: ∆G = ∆Hsys - T ∆Ssys 6 0 [19.17]

1constant T, P2

19.6 | Free energy and TemperatureTabulations of ∆Gf°, such as those in Appendix C, make it possible to calculate ∆G° for reactions at the standard temperature of 25 °C, but we are often interested in examin-ing reactions at other temperatures. To see how ∆G is affected by temperature, let’s look again at Equation 19.11:

∆G = ∆H - T∆S = ∆H + 1-T∆S2Enthalpy Entropy term term

Notice that we have written the expression for ∆G as a sum of two contributions, an enthalpy term, ∆H, and an entropy term, -T∆S. Because the value of -T∆S depends directly on the absolute temperature T, ∆G varies with temperature. We know that the enthalpy term, ∆H, can be either positive or negative and that T is positive at all tem-peratures other than absolute zero. The entropy term, -T∆S, can also be positive or

secTiOn 19.6 Free energy and Temperature 837

negative. When ∆S is positive, which means the final state has greater randomness (a greater number of microstates) than the initial state, the term -T∆S is negative. When ∆S is negative, -T∆S is positive.

The sign of ∆G, which tells us whether a process is spontaneous, depends on the signs and magnitudes of ∆H and -T∆S. The various combinations of ∆H and -T∆S signs are given in ▲ Table 19.3.

Note in Table 19.3 that when ∆H and -T∆S have opposite signs, the sign of ∆G depends on the magnitudes of these two terms. In these instances temperature is an important consideration. Generally, ∆H and ∆S change very little with temperature. However, the value of T directly affects the magnitude of -T∆S. As the temperature increases, the magnitude of -T∆S increases, and this term becomes relatively more important in determining the sign and magnitude of ∆G.

As an example, let’s consider once more the melting of ice to liquid water at 1 atm:

H2O1s2 ¡ H2O1l2 ∆H 7 0, ∆S 7 0

This process is endothermic, which means that ∆H is positive. Because the entropy increases during the process, ∆S is positive, which makes -T∆S negative. At temperatures below 0 °C 1273 K2, the magnitude of ∆H is greater than that of -T∆S. Hence, the posi-tive enthalpy term dominates, and ∆G is positive. This positive value of ∆G means that ice melting is not spontaneous at T 6 0 °C, just as our everyday experience tells us; rather, the reverse process, the freezing of liquid water into ice, is spontaneous at these temperatures.

What happens at temperatures greater than 0 °C? As T increases, so does the mag-nitude of -T∆S. When T 7 0 °C, the magnitude of -T∆S is greater than the mag-nitude of ∆H, which means that the -T∆S term dominates and ∆G is negative. The negative value of ∆G tells us that ice melting is spontaneous at T 7 0 °C.

At the normal melting point of water, T = 0 °C, the two phases are in equilibrium. Recall that ∆G = 0 at equilibrium; at T = 0 °C, ∆H and -T∆S are equal in magni-tude and opposite in sign, so they cancel and give ∆G = 0.

Give It Some ThoughtThe normal boiling point of benzene is 80 °C. At 100 °C and 1 atm, which term is greater in magnitude for the vaporization of benzene, ∆H or T∆S?

Our discussion of the temperature dependence of ∆G is also relevant to standard free-energy changes. We can calculate the values of ∆H° and ∆S° at 298 K from the data in Appendix C. If we assume that these values do not change with temperature, we can then use Equation 19.12 to estimate ∆G at temperatures other than 298 K.

Table 19.3 how signs of ∆H and ∆S affect reaction spontaneity

∆H ∆S -T∆S ∆G = ∆H - T∆S reaction Characteristics example

- + - - Spontaneous at all temperatures 2 O31g2 ¡ 3 O21g2+ - + + Nonspontaneous at all temperatures 3 O21g2 ¡ 2 O31g2- - + + or - Spontaneous at low T; nonspontaneous

at high TH2O1l2 ¡ H2O1s2

+ + - + or - Spontaneous at high T; nonspontaneous at low T

H2O1s2 ¡ H2O1l2

samPle exerCise 19.9 determining the effect of Temperature on spontaneity

The Haber process for the production of ammonia involves the equilibrium

N21g2 + 3 H21g2 ∆ 2 NH31g2Assume that ∆H° and ∆S° for this reaction do not change with temperature. (a) Predict the direc-tion in which ∆G for the reaction changes with increasing temperature. (b) Calculate ∆G at 25 °C and at 500 °C.


19.7 | Free energy and the equilibrium Constant

In Section 19.5 we saw a special relationship between ∆G and equilibrium: For a system at equilibrium, ∆G = 0. We have also seen how to use tabulated thermodynamic data to calculate values of the standard free-energy change, ∆G°. In this final section, we learn two more ways in which we can use free energy to analyze chemical reactions: us-ing ∆G° to calculate ∆G under nonstandard conditions and relating the values of ∆G° and K for a reaction.

Free Energy under Nonstandard ConditionsThe set of standard conditions for which ∆G° values pertain is given in Table 19.2. Most chemical reactions occur under nonstandard conditions. For any chemical pro-cess, the relationship between the free-energy change under standard conditions, ∆G°, and the free-energy change under any other conditions, ∆G, is given by

∆G = ∆G° + RT ln Q [19.19]

In this equation R is the ideal-gas constant, 8.314 J>mol@K; T is the absolute temperature; and Q is the reaction quotient for the reaction mixture of interest.

(Section 15.6) Recall that the reaction quotient Q is calculated like an equilibrium constant, except that you use the concentrations at any point of interest in the reaction; if Q = K, then the reaction is at equilibrium. Under standard conditions, the concen-trations of all the reactants and products are equal to 1 M. Thus, under standard con-ditions Q = 1, ln Q = 0, and Equation 19.19 reduces to ∆G = ∆G° under standard conditions, as it should.

soluTionAnalyze In part (a) we are asked to predict the direction in which ∆G changes as temperature increases. In part (b) we need to determine ∆G for the reaction at two temperatures.Plan We can answer part (a) by determining the sign of ∆S for the re-action and then using that information to analyze Equation 19.12. In part (b) we first calculate ∆H° and ∆S° for the reaction using data in Appendix C and then use Equation 19.12 to calculate ∆G.Solve

(a) The temperature dependence of ∆G comes from the entropy term in Equation 19.12, ∆G = ∆H - T∆S. We expect ∆S for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ∆S is negative, -T∆S is positive and increases with increasing temperature. As a result, ∆G becomes less negative (or more positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing temperature.

(b) We calculated ∆H° for this reaction in Sample Exercise 15.14 and ∆S° in Sample Exercise 19.5: ∆H° = -92.38 kJ and ∆S° = -198.3 J>K. If we assume that these values do not change with temperature, we can calculate ∆G at any temperature by us-ing Equation 19.12. At T = 25 °C = 298 K, we have

∆G° = -92.38 kJ - 1298 K21-198.3 J>K2a 1 kJ1000 J

b = -92.38 kJ + 59.1 kJ = -33.3 kJ

At T = 500 °C = 773 K, we have

∆G = -92.38 kJ - 1773 K21-198.3 J>K2a 1 kJ1000 J

b = -92.38 kJ + 153 kJ = 61 kJ

Notice that we had to convert the units of -T∆S° to kJ in both cal-culations so that this term can be added to the ∆H° term, which has units of kJ.Comment Increasing the temperature from 298 K to 773 K changes ∆G from -33.3 kJ to +61 kJ. Of course, the result at 773 K assumes that ∆H° and ∆S° do not change with temperature. Although these values do change slightly with temperature, the result at 773 K should be a reasonable approximation.The positive increase in ∆G with increasing T agrees with our predic-tion in part (a). Our result indicates that in a mixture of N21g2, H21g2, and NH31g2, each present at a partial pressure of 1 atm, the N21g2 and H21g2 react spontaneously at 298 K to form more NH31g2. At 773 K, the positive value of ∆G tells us that the reverse reaction is spontaneous. Thus, when the mixture of these gases, each at a partial pressure of 1 atm, is heated to 773 K, some of the NH31g2 spontaneously decomposes into N21g2 and H21g2.

Practice exercise 1What is the temperature above which the Haber ammonia process becomes nonspontaneous?(a) 25 °C, (b) 47 °C, (c) 61 °C, (d) 193 °C, (e) 500 °C.

Practice exercise 2(a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ∆H° and ∆S° at 298 K for the reaction 2 SO21g2 + O21g2 ¡ 2 SO31g2. (b) Use your values from part (a) to estimate ∆G at 400 K.

secTiOn 19.7 Free energy and the equilibrium constant 839

samPle exerCise 19.10 relating ∆G to a Phase Change at equilibrium

(a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl41l2. (b) What is the value of ∆G° for the equilibrium in part (a)? (c) Use data from Appendix C and Equation 19.12 to estimate the normal boiling point of CCl4.

soluTionAnalyze (a) We must write a chemical equation that describes the physical equilibrium between liquid and gaseous CCl4 at the normal boiling point. (b) We must determine the value of ∆G° for CCl4, in equilibrium with its vapor at the normal boiling point. (c) We must estimate the normal boil-ing point of CCl4, based on available thermodynamic data.

Plan (a) The chemical equation is the change of state from liquid to gas. For (b), we need to analyze Equation 19.19 at equilibrium 1∆G = 02, and for (c) we can use Equation 19.12 to calculate T when ∆G = 0.

Solve

(a) The normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm: CCl41l2 ∆ CCl41g2 P = 1 atm

(b) At equilibrium, ∆G = 0. In any normal boiling-point equilibrium, both liquid and vapor are in their standard state of pure liquid and vapor at 1 atm (Table 19.2). Consequently, Q = 1, ln Q = 0, and ∆G = ∆G ° for this process. We conclude that ∆G° = 0 for the equilibrium representing the normal boiling point of any liquid. (We would also find that ∆G° = 0 for the equilibria relevant to normal melting points and normal sublimation points.)

(c) Combining Equation 19.12 with the result from part (b), we see that the equality at the normal boiling point, Tb, of CCl41l2 (or any other pure liquid) is ∆G° = ∆H° - Tb ∆S° = 0

Solving the equation for Tb, we obtain Tb = ∆H°>∆S° Strictly speaking, we need the values of ∆H° and ∆S° for the

CCl41l2>CCl41g2 equilibrium at the normal boiling point to do this calculation. However, we can estimate the boiling point by using the values of ∆H° and ∆S° for the phases of CCl4 at 298 K, which we obtain from Appendix C and Equations 5.31 and 19.8:

∆H° = 11 mol21-106.7 kJ>mol2 - 11 mol21-139.3 kJ>mol2 = +32.6 kJ ∆S° = 11 mol21309.4 J>mol@K2 - 11 mol21214.4 J>mol@K2 = +95.0 J>K

As expected, the process is endothermic 1∆H 7 02 and produces a gas, thus increasing the entropy 1∆S 7 02. We now use these values to estimate Tb for CCl41l2: Tb =

∆H°∆S°

= a 32.6 kJ95.0 J>K

b a 1000 J1 kJ

b = 343 K = 70 °C

Note that we have used the conversion factor between joules and kilojoules to make the units of ∆H° and ∆S° match.Check The experimental normal boiling point of CCl41l2 is 76.5 °C. The small deviation of our estimate from the experimental value is due to the assumption that ∆H° and ∆S° do not change with temperature.

Practice exercise 1If the normal boiling point of a liquid is 67 °C, and the standard molar entropy change for the boiling process is +100 J>K, estimate the standard molar enthalpy change for the boiling process.(a) +6700 J, (b) -6700 J, (c) +34,000 J, (d) -34,000 J.

Practice exercise 2Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br21l2. (The experimental value is given in Figure 11.5.)

When the concentrations of reactants and products are nonstandard, we must cal-culate Q in order to determine ∆G. We illustrate how this is done in Sample Exer-cise 19.11. At this stage in our discussion, therefore, it becomes important to note the units used to calculate Q when using Equation 19.19. The convention used for standard states is used when applying this equation: In determining the value of Q, the concen-trations of gases are always expressed as partial pressures in atmospheres and solutes are expressed as their concentrations in molarities.


Relationship between ∆G° and KWe can now use Equation 19.19 to derive the relationship between ∆G° and the equi-librium constant, K. At equilibrium, ∆G = 0 and Q = K. Thus, at equilibrium, Equa-tion 19.19 transforms as follows:

∆G = ∆G° + RT ln Q 0 = ∆G° + RT ln K

∆G° = -RT ln K [19.20]

Equation 19.20 is a very important one, with broad significance in chemistry. By relating K to ∆G°, we can also relate K to entropy and enthalpy changes for a reaction.

Calculate ∆G at 298 K for a mixture of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process:

N21g2 + 3 H21g2 ∆ 2 NH31g2

soluTionAnalyze We are asked to calculate ∆G under nonstandard conditions.Plan We can use Equation 19.19 to calculate ∆G. Doing so requires that we calculate the value of the reaction quotient Q for the specified partial pressures, for which we use the partial-pressures form of Equation 15.23: Q = 3D4d3E4e>3A]a3B4b. We then use a table of standard free energies of formation to evaluate ∆G°.Solve The partial-pressures form of Equation 15.23 gives

Q =PNH3

2

PN2 PH2

3 =10.5022

11.0213.023 = 9.3 * 10-3

In Sample Exercise 19.9 we calculated ∆G° = -33.3 kJ for this reaction. We will have to change the units of this quantity in applying Equation 19.19, however. For the units in Equation 19.19 to work out, we will use kJ>mol as our units for ∆G°, where “per mole” means “per mole of the reac-tion as written.” Thus, ∆G° = -33.3 kJ>mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3.We now use Equation 19.19 to calculate ∆G for these nonstandard conditions:

∆G = ∆G° + RT ln Q = 1-33.3 kJ>mol2 + 18.314 J>mol@K21298 K211 kJ>1000 J2 ln19.3 * 10-32 = 1-33.3 kJ>mol2 + 1-11.6 kJ>mol2 = -44.9 kJ>mol

Comment We see that ∆G becomes more negative as the pressures of N2, H2, and NH3 are changed from 1.0 atm (standard conditions, ∆G°) to 1.0 atm, 3.0 atm, and 0.50 atm, respectively. The larger negative value for ∆G indicates a larger “driving force” to produce NH3.We would make the same prediction based on Le Châtelier’s principle. (Section 15.7) Rela-tive to standard conditions, we have increased the pressure of a reactant 1H22 and decreased the pressure of the product 1NH32. Le Châtelier’s principle predicts that both changes shift the reac-tion to the product side, thereby forming more NH3.

Practice exercise 1Which of the following statements is true? (a) The larger the Q, the larger the ∆G°. (b) If Q = 0, the system is at equilibrium. (c) If a reaction is spontaneous under standard condi-tions, it is spontaneous under all conditions. (d) The free-energy change for a reaction is independent of temperature. (e) If Q 7 1, ∆G 7 ∆G°.

Practice exercise 2Calculate ∆G at 298 K for the Haber reaction if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3.

samPle exerCise 19.11 Calculating the Free-energy Change under

nonstandard Conditions


We can also solve Equation 19.20 for K, to yield an expression that allows us to calculate K if we know the value of ∆G°:

ln K =∆G°-RT

K = e-∆G°>RT [19.21]

As usual, we must be careful in our choice of units. In Equations 19.20 and 19.21 we again express ∆G° in kJ>mol. In the equilibrium-constant expression, we use atmospheres for gas pressures, molarities for solutions; and solids, liquids, and solvents do not appear in the expression. (Section 15.4) Thus, the equi-librium constant is Kp for gas-phase reactions and Kc for reactions in solution.

(Section 15.2)From Equation 19.20 we see that if ∆G° is negative, ln K must be positive, which

means K 7 1. Therefore, the more negative ∆G° is, the larger K is. Conversely, if ∆G° is positive, ln K is negative, which means K 6 1. Finally, if ∆G° is zero, K = 1.

Give It Some ThoughtCan K = 0?

The standard free-energy change for the Haber process at 25 °C was obtained in Sample Exercise 19.9 for the Haber reaction:

N21g2 + 3 H21g2 ∆ 2 NH31g2 ∆G° = -33.3 kJ>mol = -33,300 J>mol

Use this value of ∆G° to calculate the equilibrium constant for the process at 25 °C.

soluTionAnalyze We are asked to calculate K for a reaction, given ∆G°.Plan We can use Equation 19.21 to calculate K.Solve Remembering to use the absolute temperature for T in Equation 19.21 and the form of R that matches our units, we have

K = e-∆G°>RT = e-1-33,300 J>mol2>18.314 J>mol@K21298 K2 = e13.4 = 7 * 105

Comment This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25 °C. The equilibrium constants for the Haber reaction at temperatures in the range 300 °C to 600 °C, given in Table 15.2, are much smaller than the value at 25 °C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber pro-cess is carried out at high temperatures because the reaction is extremely slow at room temperature.Remember Thermodynamics can tell us the direction and extent of a reaction but tells us noth-ing about the rate at which it will occur. If a catalyst were found that would permit the reaction to proceed at a rapid rate at room temperature, high pressures would not be needed to force the equilibrium toward NH3.

Practice exercise 1The Ksp for a very insoluble salt is 4.2 * 10-47 at 298 K. What is ∆G° for the dissolution of the salt in water?(a) -265 kJ>mol, (b) -115 kJ>mol, (c) -2.61 kJ>mol, (d) +115 kJ>mol,(e) +265 kJ>mol.

Practice exercise 2Use data from Appendix C to calculate ∆G° and K at 298 K for the reaction H21g2 + Br21l2 ∆ 2 HBr1g2.

samPle exerCise 19.12 Calculating an equilibrium Constant from ∆G°


Chemistry and life

Driving Nonspontaneous Reactions: Coupling Reactions

Many desirable chemical reactions, including a large number that are central to living systems, are nonspontaneous as written. For example, consider the extraction of copper metal from the mineral chalcocite, which contains Cu2S. The decomposition of Cu2S to its elements is nonspontaneous:

Cu2S1s2 ¡ 2 Cu1s2 + S1s2 ∆G° = +86.2 kJ

Because ∆G° is very positive, we cannot obtain Cu(s) directly via this reaction. Instead, we must find some way to “do work” on the reac-tion to force it to occur as we wish. We can do this by coupling the reaction to another one so that the overall reaction is spontaneous. For example, we can envision the S(s) reacting with O21g2 to form SO21g2:

S1s2 + O21g2 ¡ SO21g2 ∆G° = -300.4 kJ

By coupling (adding together) these reactions, we can extract much of the copper metal via a spontaneous reaction:

Cu2S1s2 + O21g2 ¡ 2 Cu1s2 + SO21g2 ∆G° = 1+86.2 kJ2 + 1-300.4 kJ2 = -214.2 kJ

In essence, we have used the spontaneous reaction of S(s) with O21g2 to provide the free energy needed to extract the copper metal from the mineral.

Biological systems employ the same principle of using spon-taneous reactions to drive nonspontaneous ones. Many of the biochemical reactions that are essential for the formation and

maintenance of highly ordered biological structures are not spon-taneous. These necessary reactions are made to occur by coupling them with spontaneous reactions that release energy. The meta-bolism of food is the usual source of the free energy needed to do the work of maintaining biological systems. For example, complete oxidation of the sugar glucose, C6H12O6, to CO2 and H2O yields sub-stantial free energy:

C6H12O61s2 + 6 O21g2 ¡ 6 CO21g2 + 6 H2O1l2 ∆G° = -2880 kJ

This energy can be used to drive nonspontaneous reactions in the body. However, a means is necessary to transport the energy released by glucose metabolism to the reactions that require energy. One way, shown in ▼ Figure 19.16, involves the interconversion of adenos-ine triphosphate (ATP) and adenosine diphosphate (ADP), molecules that are related to the building blocks of nucleic acids. The conversion of ATP to ADP releases free energy 1∆G° = -30.5 kJ2 that can be used to drive other reactions.

In the human body the metabolism of glucose occurs via a complex series of reactions, most of which release free energy. The free energy released during these steps is used in part to reconvert lower-energy ADP back to higher-energy ATP. Thus, the ATP–ADP interconversions are used to store energy during metabolism and to release it as needed to drive nonspontaneous reactions in the body. If you take a course in general biology or biochemistry, you will have the opportunity to learn more about the remarkable se-quence of reactions used to transport free energy throughout the human body.

Related Exercises: 19.102, 19.103

▲ Figure 19.16 Schematic representation of free-energy changes during cell metabolism. The oxidation of glucose to CO2 and H2O produces free energy that is then used to convert ADP into the more energetic ATP. The ATP is then used, as needed, as an energy source to drive nonspontaneous reactions, such as the conversion of simple molecules into more complex cell constituents.

Free energyreleasedby oxidationof glucoseconvertsADP to ATP

Free energyreleased byATP convertssimple moleculesto morecomplex ones

Glucose(C6H12O6)

Cell constituents

CO2 + H2O Simplermolecules

Glucoseoxidation

Cellulardevelopment

Highfree energy

Cell

ATP

ADPLow

free energy


Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions:

NaCl1s2 ∆ Na+1aq2 + Cl-1aq2AgCl1s2 ∆ Ag+1aq2 + Cl-1aq2

(a) Calculate the value of ∆G° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the en-tropy term of the standard free-energy change? (c) Use the values of ∆G° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will ∆G° for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts?

soluTion(a) We will use Equation 19.14 along with ∆Gf° values from Appendix C to calculate the ∆G°soln

values for each equilibrium. (As we did in Section 13.1, we use the subscript “soln” to indi-cate that these are thermodynamic quantities for the formation of a solution.) We find

∆G°soln1NaCl2 = 1-261.9 kJ>mol2 + 1-131.2 kJ>mol2 - 1-384.0 kJ>mol2 = -9.1 kJ>mol

∆G°soln1AgCl2 = 1+77.11 kJ>mol2 + 1-131.2 kJ>mol2 - 1-109.70 kJ>mol2 = +55.6 kJ>mol

(b) We can write ∆G°soln as the sum of an enthalpy term, ∆H°soln, and an entropy term, -T∆S°soln: ∆G°soln = ∆H°soln + 1-T∆S°soln2. We can calculate the values of ∆H°soln and ∆S°soln by using Equations 5.31 and 19.8. We can then calculate -T∆S°soln at T = 298 K. All these calculations are now familiar to us. The results are summarized in the following table:

salt ∆H°soln ∆S°soln T∆S°soln

NaCl +3.6 kJ>mol +43.2 kJ>mol@K -12.9 kJ>molAgCl +65.7 kJ>mol +34.3 kJ>mol@K -10.2 kJ>mol

The entropy terms for the solution of the two salts are very similar. That seems sensible because each solution process should lead to a similar increase in randomness as the salt dis-solves, forming hydrated ions. (Section 13.1) In contrast, we see a very large difference in the enthalpy term for the solution of the two salts. The difference in the values of ∆G°soln is dominated by the difference in the values of ∆H°soln.

(c) The solubility product, Ksp, is the equilibrium constant for the solution process. (Section 17.4) As such, we can relate Ksp directly to ∆G°soln by using Equation 19.21:

Ksp = e-∆G°so ln >RT

We can calculate the Ksp values in the same way we applied Equation 19.21 in Sample Exercise 19.12. We use the ∆G°soln values we obtained in part (a), remembering to convert them from kJ>mol to J>mol:

NaCl: Ksp = 3Na+1aq243Cl-1aq24 = e-1-91002>318.3142129824 = e+3.7 = 40 AgCl: Ksp = 3Ag+1aq243Cl-1aq24 = e-1+55,6002>318.3142129824 = e-22.4 = 1.9 * 10-10

The value calculated for the Ksp of AgCl is very close to that listed in Appendix D.(d) A soluble salt is one that dissolves appreciably in water. (Section 4.2) The Ksp value for

NaCl is greater than 1, indicating that NaCl dissolves to a great extent. The Ksp value for AgCl is very small, indicating that very little dissolves in water. Silver chloride should indeed be considered an insoluble salt.

(e) As we expect, the solution process has a positive value of ∆S for both salts (see the table in part b). As such, the entropy term of the free-energy change, -T∆S°soln, is negative. If we assume that ∆H°soln and ∆S°soln do not change much with temperature, then an increase in T will serve to make ∆G°soln more negative. Thus, the driving force for dissolution of the salts will increase with increasing T, and we therefore expect the solubility of the salts to increase with increasing T. In Figure 13.18 we see that the solubility of NaCl (and that of nearly any other salt) increases with increasing temperature. (Section 13.3)

samPle inTeGraTive exerCise Putting Concepts Together


Chapter summary and Key TermsSPONTANEOUS PROCESSES (SECTION 19.1) Most reactions and chemical processes have an inherent directionality: They are spontane-ous in one direction and nonspontaneous in the reverse direction. The spontaneity of a process is related to the thermodynamic path the sys-tem takes from the initial state to the final state. In a reversible process, both the system and its surroundings can be restored to their original state by exactly reversing the change. In an irreversible process the sys-tem cannot return to its original state without a permanent change in the surroundings. Any spontaneous process is irreversible. A process that occurs at a constant temperature is said to be isothermal.ENTROPy AND THE SECOND LAW OF THERMODyNAMICS (SECTION 19.2) The spontaneous nature of processes is related to a thermodynamic state function called entropy, denoted S. For a process that occurs at constant temperature, the entropy change of the system is given by the heat absorbed by the system along a reversible path, divided by the temperature: ∆S = qrev>T. For any process, the entropy change of the universe equals the entropy change of the system plus the entropy change of the surroundings: ∆Suniv = ∆Ssys + ∆Ssurr. The way entropy controls the spontaneity of processes is given by the second law of thermodynamics, which states that in a reversible process ∆Suniv = 0; in an irreversible (spontaneous) process ∆Suniv 7 0. Entropy values are usually expressed in units of joules per kelvin, J>K.THE MOLECULAR INTERPRETATION OF ENTROPy AND THE THIRD LAW OF THERMODyNAMICS (SECTION 19.3) A particular combina-tion of motions and locations of the atoms and molecules of a system at a particular instant is called a microstate. The entropy of a system is a mea-sure of its randomness or disorder. The entropy is related to the number of microstates, W, corresponding to the state of the system: S = k ln W. Molecules can undergo three kinds of motion: In translational motion the entire molecule moves in space. Molecules can also undergo vibrational motion, in which the atoms of the molecule move toward and away from one another in periodic fashion, and rotational motion, in which the entire molecule spins like a top. The number of available microstates, and therefore the entropy, increases with an increase in volume, temperature, or motion of molecules because any of these changes increases the pos-sible motions and locations of the molecules. As a result, entropy gen-erally increases when liquids or solutions are formed from solids, gases are formed from either solids or liquids, or the number of molecules of gas increases during a chemical reaction. The third law of thermodynamics states that the entropy of a pure crystalline solid at 0 K is zero.

ENTROPy CHANGES IN CHEMICAL REACTIONS (SECTION 19.4) The third law allows us to assign entropy values for substances at different temperatures. Under standard conditions the entropy of a mole of a substance is called its standard molar entropy, denoted S°. From tabulated values of S°, we can calculate the entropy change for any process under standard conditions. For an isothermal process, the entropy change in the surroundings is equal to - ∆H>T.

GIBBS FREE ENERGy (SECTION 19.5) The Gibbs free energy (or just free energy), G, is a thermodynamic state function that combines the two state functions enthalpy and entropy: G = H - TS. For processes that occur at constant temperature, ∆G = ∆H - T∆S. For a process occurring at constant temperature and pressure, the sign of ∆G relates to the spontaneity of the process. When ∆G is negative, the process is spontaneous. When ∆G is positive, the process is nonspontaneous but the reverse process is spontane-ous. At equilibrium the process is reversible and ∆G is zero. The free energy is also a measure of the maximum useful work that can be performed by a system in a spontaneous process. The standard free-energy change, ∆G°, for any process can be calculated from tabulations of standard free energies of formation, ∆G°f , which are defined in a fashion analogous to standard enthalpies of formation, ∆G°f. The value of ∆G°f for a pure element in its standard state is defined to be zero.

FREE ENERGy, TEMPERATURE, AND THE EqUILIBRIUM CON-STANT (SECTIONS 19.6 AND 19.7) The values of ∆H and ∆S for a chemical process generally do not vary much with tempera-ture. Therefore, the dependence of ∆G with temperature is gov-erned mainly by the value of T in the expression ∆G = ∆H - T∆S. The entropy term -T∆S has the greater effect on the temperature dependence of ∆G and, hence, on the spontaneity of the process. For example, a process for which ∆H 7 0 and ∆S 7 0, such as the melting of ice, can be nonspontaneous 1∆G 7 02 at low tem-peratures and spontaneous 1∆G 7 02 at higher temperatures. Under nonstandard conditions ∆G is related to ∆G° and the value of the reaction quotient, Q: ∆G = ∆G° + RT ln Q. At equilibrium 1∆G = 0, Q = K2, ∆G° = -RT ln K. Thus, the standard free-energy change is directly related to the equilibrium constant for the reaction. This relationship expresses the temperature dependence of equilibrium constants.

learning outcomes after studying this chapter, you should be able to:

• Explain the meaning of spontaneous process, reversible process, irreversible process, and isothermal process. (Section 19.1)

• Define entropy and state the second law of thermodynamics. (Section 19.2)

• Explain how the entropy of a system is related to the number of possible microstates. (Section 19.3)

• Describe the kinds of molecular motion that a molecule can pos-sess. (Section 19.3)

• Predict the sign of ∆S for physical and chemical processes. (Section 19.3)

• State the third law of thermodynamics. (Section 19.3)

• Calculate standard entropy changes for a system from standard molar entropies. (Section 19.4)

• Calculate entropy changes in the surroundings for isothermal processes. (Section 19.4)

• Calculate the Gibbs free energy from the enthalpy change and entropy change at a given temperature. (Section 19.5)

• Use free-energy changes to predict whether reactions are spontaneous. (Section 19.5)

• Calculate standard free-energy changes using standard free energies of formation. (Section 19.5)

• Predict the effect of temperature on spontaneity given ∆H and∆S. (Section 19.6)

• Calculate ∆G under nonstandard conditions. (Section 19.7)

• Relate ∆G ° and equilibrium constant. (Section 19.7)

exercises 845

Vaporized C2H4F2

Lique�ed C2H4F2

19.3 (a) What are the signs of ∆S and ∆H for the process depicted here? (b) If energy can flow in and out of the system to main-tain a constant temperature during the process, what can you say about the entropy change of the surroundings as a result of this process? [Sections 19.2 and 19.5]

visualizing Concepts

19.1 Two different gases occupy the two bulbs shown here. Con-sider the process that occurs when the stopcock is opened, assuming the gases behave ideally. (a) Draw the final (equi-librium) state. (b) Predict the signs of ∆H and ∆S for the process. (c) Is the process that occurs when the stopcock is opened a reversible one? (d) How does the process affect the entropy of the surroundings? [Sections 19.1 and 19.2]

19.2 As shown here, one type of computer keyboard cleaner con-tains liquefied 1,1-difluoroethane 1C2H4F22, which is a gas at atmospheric pressure. When the nozzle is squeezed, the 1,1-difluoroethane vaporizes out of the nozzle at high pres-sure, blowing dust out of objects. (a) Based on your experi-ence, is the vaporization a spontaneous process at room temperature? (b) Defining the 1,1-difluoroethane as the sys-tem, do you expect qsys for the process to be positive or nega-tive? (c) Predict whether ∆S is positive or negative for this process. (d) Given your answers to (a), (b), and (c), do you think the operation of this product depends more on enthalpy or entropy? [Sections 19.1 and 19.2]

Key equations

• ∆S =qrev

T 1constant T2 [19.2] Relating entropy change to the heat absorbed or released in a

reversible process

• Reversible process: ∆Suniv = ∆Ssys + ∆Ssurr = 0 ¶

Irreversible process: ∆Suniv = ∆Ssys + ∆Ssurr 7 0 [19.4] The second law of thermodynamics

• S = k ln W [19.5] Relating entropy to the number of microstates

• ∆S° = anS°1products2 - amS°1reactants2 [19.8] Calculating the standard entropy change from standard molar entropies

• ∆Ssurr =- ∆Hsys

T [19.9] The entropy change of the surroundings for a process at constant

temperature and pressure

• ∆G = ∆H - T∆S [19.11] Calculating the Gibbs free-energy change from enthalpy and en-tropy changes at constant temperature

• ∆G° = an∆G°f 1products2 - am∆G°f 1reactants2 [19.14] Calculating the standard free-energy change from standard free energies of formation

• Reversible process: ∆G = ∆Hsys - T∆Ssys = 0 ¶

Irreversible process: ∆G = ∆Hsys - T∆Ssys 6 0 [19.17] Relating the free-energy change to the reversibility of a process at constant temperature and pressure

• ∆G = -wmax [19.18] Relating the free-energy change to the maximum work a system can perform

• ∆G = ∆G° + RT ln Q [19.19] Calculating free-energy change under nonstandard conditions

• ∆G° = -RT ln K [19.20] Relating the standard free-energy change and the equilibrium constant

exercises


19.7 The accompanying diagram shows how ∆H (red line) and T∆S (blue line) change with temperature for a hypothetical reaction. (a) What is the significance of the point at 300 K, where ∆H and T∆S are equal? (b) In what temperature range is this reaction spontaneous? [Section 19.6]

Temperature

T∆S

∆H

300 K

19.8 The accompanying diagram shows how ∆G for a hypotheti-cal reaction changes as temperature changes. (a) At what temperature is the system at equilibrium? (b) In what tem-perature range is the reaction spontaneous? (c) Is ∆H posi-tive or negative? (d) Is ∆S positive or negative? [Sections 19.5 and 19.6]

Temperature250 K

∆G 0

19.9 Consider a reaction A21g2 + B21g2 ∆ 2 AB1g2, with atoms of A shown in red in the diagram and atoms of B shown in blue. (a) If Kc = 1, which box represents the system at equilibrium? (b) If Kc = 1, which box represents the system at Q 6 Kc? (c) Rank the boxes in order of increasing magni-tude of ∆G for the reaction. [Sections 19.5 and 19.7]

(2) (3)(1)

19.4 Predict the signs of ∆H and ∆S for this reaction. Explain your choice. [Section 19.3]

19.5 The accompanying diagram shows how entropy varies with temperature for a substance that is a gas at the highest temperature shown. (a) What processes correspond to the entropy increases along the vertical lines labeled 1 and 2? (b) Why is the entropy change for 2 larger than that for 1? (c) If this substance is a perfect crystal at T = 0 K, what is the value of S at this temperature? [Section 19.3]

Temperature (K)

Ent

ropy

, S

1

2

19.6 Isomers are molecules that have the same chemical formula but different arrangements of atoms, as shown here for two isomers of pentane, C5H12. (a) Do you expect a sig-nificant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. [Section 19.4]

CH3 CH2 CH2CH2 CH3 C

CH3

CH3CH3

CH3

n-Pentane Neopentane

exercises 847

19.15 Consider the vaporization of liquid water to steam at a pressure of 1 atm. (a) Is this process endothermic or exo-thermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontane-ous process? (d) At what temperature are the two phases in equilibrium?

19.16 The normal freezing point of n-octane 1C8H182 is -57 °C. (a) Is the freezing of n-octane an endothermic or exother-mic process? (b) In what temperature range is the freezing of n-octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any tem-perature at which liquid n-octane and solid n-octane are in equilibrium? Explain.

19.17 Indicate whether each statement is true or false. (a) If a system undergoes a reversible process, the entropy of the universe increases. (b) If a system undergoes a reversible process, the change in entropy of the system is exactly matched by an equal and opposite change in the entropy of the surroundings. (c) If a system undergoes a reversible process, the entropy change of the system must be zero. (d) Most spontaneous processes in nature are reversible.

19.18 Indicate whether each statement is true or false. (a) All spon-taneous processes are irreversible. (b) The entropy of the uni-verse increases for spontaneous processes. (c) The change in entropy of the surroundings is equal in magnitude and oppo-site in sign for the change in entropy of the system, for an ir-reversible process. (d) The maximum amount of work can be gotten out of a system that undergoes an irreversible process, as compared to a reversible process.

19.19 Consider a process in which an ideal gas changes from state 1 to state 2 in such a way that its temperature changes from 300 K to 200 K. (a) Does the temperature change depend on whether the process is reversible or irreversible? (b) Is this process isothermal? (c) Does the change in the internal en-ergy, ∆E, depend on the particular pathway taken to carry out this change of state?

19.20 A system goes from state 1 to state 2 and back to state 1. (a) Is ∆E the same in magnitude for both the forward and reverse processes? (b) Without further information, can you conclude that the amount of heat transferred to the system as it goes from state 1 to state 2 is the same or different as compared to that upon going from state 2 back to state 1? (c) Suppose the changes in state are reversible processes. Is the work done by the system upon going from state 1 to state 2 the same or different as compared to that upon going from state 2 back to state 1?

19.21 Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is ∆E zero for the process?

19.22 Consider what happens when a sample of the explosive TNT (Section 8.8: “Chemistry Put to Work: Explosives and Alfred Nobel”) is detonated under atmospheric pressure. (a) Is the detonation a spontaneous process? (b) What is the sign of q for this process? (c) Is it possible to tell whether w is positive, negative, or zero for the process? Explain. (d) Can you deter-mine the sign of ∆E for the process? Explain.

19.10 The accompanying diagram shows how the free energy, G, changes during a hypothetical reaction A1g2 + B1g2 ¡ C1g2. On the left are pure reactants A and B, each at 1 atm, and on the right is the pure product, C, also at 1 atm. Indicate whether each of the following statements is true or false. (a) The minimum of the graph corresponds to the equilibrium mixture of reactants and products for this reaction. (b) At equilibrium, all of A and B have reacted to give pure C. (c) The entropy change for this re-action is positive. (d) The “x” on the graph corresponds to ∆G for the reaction. (e) ∆G for the reaction corresponds to the differ-ence between the top left of the curve and the bottom of the curve. [Section 19.7]

G x

Progress of reaction

spontaneous Processes (section 19.1)

19.11 Which of the following processes are spontaneous and which are nonspontaneous: (a) the ripening of a banana, (b) dissolution of sugar in a cup of hot coffee, (c) the reac-tion of nitrogen atoms to form N2 molecules at 25 °C and 1 atm, (d) lightning, (e) formation of CH4 and O2 mole-cules from CO2 and H2O at room temperature and 1 atm of pressure?

19.12 Which of the following processes are spontaneous: (a) the melting of ice cubes at -10 °C and 1 atm pressure; (b) sepa-rating a mixture of N2 and O2 into two separate samples, one that is pure N2 and one that is pure O2; (c) alignment of iron filings in a magnetic field; (d) the reaction of hydrogen gas with oxygen gas to form water vapor at room temperature; (e) the dissolution of HCl(g) in water to form concentrated hydrochloric acid?

19.13 (a) Can endothermic chemical reactions be spontaneous? (b) Can a process that is spontaneous at one temperature be non-spontaneous at a different temperature?

19.14 The crystalline hydrate Cd1NO322 # 4 H2O1s2 loses wa-ter when placed in a large, closed, dry vessel at room temperature:

Cd1NO322 # 4 H2O1s2 ¡ Cd1NO3221s2 + 4 H2O1g2 This process is spontaneous and ∆H° is positive at room

temperature. (a) What is the sign of ∆S° at room tem-perature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does ∆S° change for this reaction at room temperature?


process? (b) Explain why no work is done by the system dur-ing this process. (c) What is the “driving force” for the expan-sion of the gas: enthalpy or entropy?

19.32 (a) What is the difference between a state and a microstate of a system? (b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state? (c) In a particular spontaneous process, the number of microstates available to the system decreases. What can you conclude about the sign of ∆Ssurr?

19.33 Would each of the following changes increase, decrease, or have no effect on the number of microstates available to a sys-tem: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas?

19.34 (a) Using the heat of vaporization in Appendix B, calculate the entropy change for the vaporization of water at 25 °C and at 100 °C. (b) From your knowledge of microstates and the structure of liquid water, explain the difference in these two values.

19.35 (a) What do you expect for the sign of ∆S in a chemical reac-tion in which two moles of gaseous reactants are converted to three moles of gaseous products? (b) For which of the processes in Exercise 19.11 does the entropy of the system increase?

19.36 (a) In a chemical reaction two gases combine to form a solid. What do you expect for the sign of ∆S? (b) How does the entropy of the system change in the processes described in Exercise 19.12?

19.37 Does the entropy of the system increase, decrease, or stay the same when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes?

19.38 Does the entropy of the system increase, decrease, or stay the same when (a) the temperature of the system increases, (b) the volume of a gas increases, (c) equal volumes of ethanol and water are mixed to form a solution?

19.39 Indicate whether each statement is true or false. (a) The third law of thermodynamics says that the entropy of a perfect, pure crystal at absolute zero increases with the mass of the crystal. (b) “Translational motion” of molecules refers to their change in spatial location as a function of time. (c) “Rotational” and “vibrational” motions contribute to the entropy in atomic gases like He and Xe. (d) The larger the number of atoms in a molecule, the more degrees of freedom of rotational and vi-brational motion it likely has.

19.40 Indicate whether each statement is true or false. (a) Unlike enthalpy, where we can only ever know changes in H, we can know absolute values of S. (b) If you heat a gas such as CO2, you will increase its degrees of translational, rotational and vi-brational motions. (c) CO21g2 and Ar(g) have nearly the same molar mass. At a given temperature, they will have the same number of microstates.

19.41 For each of the following pairs, choose the substance with the higher entropy per mole at a given temperature: (a) Ar(l) or Ar(g), (b) He(g) at 3 atm pressure or He(g) at 1.5 atm pres-sure, (c) 1 mol of Ne(g) in 15.0 L or 1 mol of Ne(g) in 1.50 L, (d) CO21g2 or CO21s2.

19.42 For each of the following pairs, indicate which substance possesses the larger standard entropy: (a) 1 mol of P41g2 at 300 °C, 0.01 atm, or 1 mol of As41g2 at 300 °C, 0.01 atm; (b) 1 mol of H2O1g2 at 100 °C, 1 atm, or 1 mol of H2O1l2 at

entropy and the second law of Thermodynamics (section 19.2)

19.23 Indicate whether each statement is true or false. (a) ∆S for an isothermal process depends on both the temperature and the amount of heat reversibly transferred. (b) ∆S is a state function. (c) The second law of thermodynamics says that the entropy of the system increases for all spontaneous processes.

19.24 Suppose we vaporize a mole of liquid water at 25 °C and an-other mole of water at 100 °C. (a) Assuming that the enthalpy of vaporization of water does not change much between 25 °C and 100 °C, which process involves the larger change in en-tropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

19.25 The normal boiling point of Br21l2 is 58.8 °C, and its molar enthalpy of vaporization is ∆Hvap = 29.6 kJ>mol. (a) When Br21l2 boils at its normal boiling point, does its entropy in-crease or decrease? (b) Calculate the value of ∆S when 1.00 mol of Br21l2 is vaporized at 58.8 °C.

19.26 The element gallium (Ga) freezes at 29.8 °C, and its molar enthalpy of fusion is ∆Hfus = 5.59 kJ>mol. (a) When mol-ten gallium solidifies to Ga(s) at its normal melting point, is ∆S positive or negative? (b) Calculate the value of ∆S when 60.0 g of Ga(l) solidifies at 29.8 °C.

19.27 Indicate whether each statement is true or false. (a) The second law of thermodynamics says that entropy is con-served. (b) If the entropy of the system increases during a reversible process, the entropy change of the surroundings must decrease by the same amount. (c) In a certain sponta-neous process the system undergoes an entropy change of 4.2 J>K; therefore, the entropy change of the surroundings must be -4.2 J>K.

19.28 (a) Does the entropy of the surroundings increase for spon-taneous processes? (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of ∆Ssurr? (c) During a certain reversible process, the surroundings undergo an entropy change, ∆Ssurr = -78 J>K. What is the entropy change of the system for this process?

19.29 (a) What sign for ∆S do you expect when the volume of 0.200 mol of an ideal gas at 27 °C is increased isothermally from an initial volume of 10.0 L? (b) If the final volume is 18.5 L, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

19.30 (a) What sign for ∆S do you expect when the pressure on 0.600 mol of an ideal gas at 350 K is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calcu-late the entropy change? Explain.

The molecular interpretation of entropy and the Third law of Thermodynamics (section 19.3)

19.31 For the isothermal expansion of a gas into a vacuum, ∆E = 0, q = 0, and w = 0. (a) Is this a spontaneous

exercises 849

[19.52] Three of the forms of elemental carbon are graphite, dia-mond, and buckminsterfullerene. The entropies at 298 K for graphite and diamond are listed in Appendix C. (a) Account for the difference in the S° values of graphite and diamond in light of their structures (Figure 12.29, p. 503). (b) What would you expect for the S° value of buckminsterfullerene (Figure 12.47, p. 516) relative to the values for graphite and diamond? Explain.

19.53 Using S° values from Appendix C, calculate ∆S° values for the following reactions. In each case account for the sign of ∆S°.(a) C2H41g2 + H21g2 ¡ C2H61g2(b) N2O41g2 ¡ 2 NO21g2(c) Be1OH221s2 ¡ BeO1s2 + H2O1g2(d) 2 CH3OH1g2 + 3 O21g2 ¡ 2 CO21g2 + 4 H2O1g2

19.54 Calculate ∆S° values for the following reactions by using tab-ulated S° values from Appendix C. In each case explain the sign of ∆S°.(a) HNO31g2 + NH31g2 ¡ NH4NO31s2(b) 2 Fe2O31s2 ¡ 4 Fe1s2 + 3 O21g2(c) CaCO31s, calcite2 + 2HCl1g2 ¡

CaCl21s2 + CO21g2 + H2O1l2(d) 3 C2H61g2 ¡ C6H61l2 + 6 H21g2

Gibbs Free energy (sections 19.5 and 19.6)

19.55 (a) For a process that occurs at constant temperature, does the change in Gibbs free energy depend on changes in the en-thalpy and entropy of the system? (b) For a certain process that occurs at constant T and P, the value of ∆G is positive. Is the process spontaneous? (c) If ∆G for a process is large, is the rate at which it occurs fast?

19.56 (a) Is the standard free-energy change, ∆G°, always larger than ∆G? (b) For any process that occurs at constant tem-perature and pressure, what is the significance of ∆G = 0? (c) For a certain process, ∆G is large and negative. Does this mean that the process necessarily has a low activation barrier?

19.57 For a certain chemical reaction, ∆H° = -35.4 kJ and ∆S° = -85.5 J>K. (a) Is the reaction exothermic or endo-thermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate ∆G° for the reaction at 298 K. (d) Is the reaction spontaneous at 298 K under standard conditions?

19.58 A certain reaction has ∆H° = +23.7 kJ and ∆S° = +52.4 J>K. (a) Is the reaction exothermic or endothermic? (b) Does the re-action lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate ∆G° for the reaction at 298 K. (d) Is the reaction spontaneous at 298 K under stan-dard conditions?

19.59 Using data in Appendix C, calculate ∆H°, ∆S°, and ∆G° at 298 K for each of the following reactions.(a) H21g2 + F21g2 ¡ 2 HF1g2(b) C1s, graphite2 + 2 Cl21g2 ¡ CCl41g2(c) 2 PCl31g2 + O21g2 ¡ 2 POCl31g2(d) 2 CH3OH1g2 + H21g2 ¡ C2H61g2 + 2 H2O1g2

100 °C, 1 atm; (c) 0.5 mol of N21g2 at 298 K, 20-L volume, or 0.5 mol CH41g2 at 298 K, 20-L volume; (d) 100 g Na2SO41s2 at 30 °C or 100 g Na2SO41aq2 at 30 °C.

19.43 Predict the sign of the entropy change of the system for each of the following reactions:(a) N21g2 + 3 H21g2 ¡ 2 NH31g2(b) CaCO31s2 ¡ CaO1s2 + CO21g2(c) 3 C2H21g2 ¡ C6H61g2(d) Al2O31s2 + 3 H21g2 ¡ 2 Al1s2 + 3 H2O1g2

19.44 Predict the sign of ∆Ssys for each of the following processes: (a) Molten gold solidifies. (b) Gaseous Cl2 dissociates in the stratosphere to form gaseous Cl atoms. (c) Gaseous CO reacts with gaseous H2 to form liquid methanol, CH3OH. (d) Cal-cium phosphate precipitates upon mixing Ca1NO3221aq2 and 1NH423PO41aq2.

entropy Changes in Chemical reactions (section 19.4)

19.45 (a) Using Figure 19.12 as a model, sketch how the entropy of water changes as it is heated from -50 °C to 110 °C at sea level. Show the temperatures at which there are vertical in-creases in entropy. (b) Which process has the larger entropy change: melting ice or boiling water? Explain.

19.46 Propanol 1C3H7OH2 melts at -126.5 °C and boils at 97.4 °C. Draw a qualitative sketch of how the entropy changes as pro-panol vapor at 150 °C and 1 atm is cooled to solid propanol at -150 °C and 1 atm.

19.47 In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) C2H21g2 or C2H61g2, (b) CO21g2 or CO(g)?

19.48 Cyclopropane and propylene are isomers that both have the formula C3H6. Based on the molecular structures shown, which of these isomers would you expect to have the higher standard molar entropy at 25 °C?

CC

H

H

H H

H

C

H

Propylene

C

H

H

C

H

H

H

C

H

Cyclopropane

19.49 Use Appendix C to compare the standard entropies at 25 °C for the following pairs of substances: (a) Sc(s) and Sc(g), (b) NH31g2 and NH31aq2, (c) 1 mol P41g2 and 2 mol P21g2, (d) C(graphite) and C(diamond).

19.50 Using Appendix C, compare the standard entropies at 25 °C for the following pairs of substances: (a) CuO(s) and Cu2O1s2, (b) 1 mol N2O41g2 and 2 mol NO21g2, (c) SiO21s2 and CO21g2, (d) CO(g) and CO21g2.

[19.51] The standard entropies at 298 K for certain group 4A ele-ments are: C(s, diamond) = 2.43 J>mol@K, Si1s2 = 18.81 J>mol@K, Ge1s2 = 31.09 J>mol@K, and Sn1s2 = 51.818 J>mol@K. All but Sn have the same (diamond) structure. How do you account for the trend in the S° values?


19.68 A certain constant-pressure reaction is barely nonspontane-ous at 45 °C. The entropy change for the reaction is 72 J>K. Estimate ∆H.

19.69 For a particular reaction, ∆H = -32 kJ and ∆S = -98 J>K. Assume that ∆H and ∆S do not vary with temperature. (a) At what temperature will the reaction have ∆G = 0? (b) If T is increased from that in part (a), will the reaction be spontane-ous or nonspontaneous?

19.70 Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of ace-tic acid proceeds according to:

CH3COOH1l2 ¡ CH3OH1g2 + CO1g2 By using data from Appendix C, calculate the minimum

temperature at which this process will be spontaneous under standard conditions. Assume that ∆H° and ∆S° do not vary with temperature.

19.71 Consider the following reaction between oxides of nitrogen:

NO21g2 + N2O1g2 ¡ 3 NO1g2 (a) Use data in Appendix C to predict how ∆G for the reac-

tion varies with increasing temperature. (b) Calculate ∆G at 800 K, assuming that ∆H° and ∆S° do not change with tem-perature. Under standard conditions is the reaction sponta-neous at 800 K? (c) Calculate ∆G at 1000 K. Is the reaction spontaneous under standard conditions at this temperature?

19.72 Methanol 1CH3OH2 can be made by the controlled oxidation of methane:

CH41g2 + 12O21g2 ¡ CH3OH1g2

(a) Use data in Appendix C to calculate ∆H° and ∆S° for this reaction. (b) Will ∆G for the reaction increase, decrease, or stay unchanged with increasing temperature? (c) Calculate ∆G° at 298 K. Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds in-volved are likely to be stable?

19.73 (a) Use data in Appendix C to estimate the boiling point of benzene, C6H61l2. (b) Use a reference source, such as the CRC Handbook of Chemistry and Physics, to find the experi-mental boiling point of benzene. How do you explain any deviation between your answer in part (a) and the experi-mental value?

19.74 (a) Using data in Appendix C, estimate the temperature at which the free-energy change for the transformation from I21s2 to I21g2 is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experi-mental melting and boiling points of I2. (c) Which of the val-ues in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

19.75 Acetylene gas, C2H21g2, is used in welding. (a) Write a bal-anced equation for the combustion of acetylene gas to CO21g2 and H2O1l2. (b) How much heat is produced in burning 1 mol of C2H2 under standard conditions if both reactants and prod-ucts are brought to 298 K? (c) What is the maximum amount of useful work that can be accomplished under standard con-ditions by this reaction?

19.60 Use data in Appendix C to calculate ∆H°, ∆S°, and ∆G° at 25 °C for each of the following reactions.(a) 4 Cr1s2 + 3 O21g2 ¡ 2 Cr2O31s2(b) BaCO31s2 ¡ BaO1s2 + CO21g2(c) 2 P1s2 + 10 HF1g2 ¡ 2 PF51g2 + 5 H21g2(d) K1s2 + O21g2 ¡ KO21s2

19.61 Using data from Appendix C, calculate ∆G° for the following reactions. Indicate whether each reaction is spontaneous at 298 K under standard conditions.(a) 2 SO21g2 + O21g2 ¡ 2 SO31g2(b) NO21g2 + N2O1g2 ¡ 3 NO1g2(c) 6 Cl21g2 + 2 Fe2O31s2 ¡ 4 FeCl31s2 + 3 O21g2(d) SO21g2 + 2 H21g2 ¡ S1s2 + 2 H2O1g2

19.62 Using data from Appendix C, calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous at 298 K under standard conditions.(a) 2 Ag1s2 + Cl21g2 ¡ 2 AgCl1s2(b) P4O101s2 + 16 H21g2 ¡ 4 PH31g2 + 10 H2O1g2(c) CH41g2 + 4 F21g2 ¡ CF41g2 + 4 HF1g2(d) 2 H2O21l2 ¡ 2 H2O1l2 + O21g2

19.63 Octane 1C8H182 is a liquid hydrocarbon at room temperature that is the primary constituent of gasoline. (a) Write a balanced equation for the combustion of C8H181l2 to form CO21g2 and H2O1l2. (b) Without using thermochemical data, predict whether ∆G° for this reaction is more negative or less negative than ∆H°.

19.64 Sulfur dioxide reacts with strontium oxide as follows:

SO21g2 + SrO1g2 ¡ SrSO31s2 (a) Without using thermochemical data, predict whether

∆G° for this reaction is more negative or less negative than ∆H°. (b) If you had only standard enthalpy data for this reac-tion, estimate of the value of ∆G° at 298 K, using data from Appendix C on other substances.

19.65 Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontanous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low T but not spontaneous at high T; (iv) spontaneous at high T but not spontaneous at low T.(a) N21g2 + 3 F21g2 ¡ 2 NF31g2

∆H° = -249 kJ; ∆S° = -278 J>K(b) N21g2 + 3 Cl21g2 ¡ 2 NCl31g2

∆H° = 460 kJ; ∆S° = -275 J>K(c) N2F41g2 ¡ 2 NF21g2

∆H° = 85 kJ; ∆S° = 198 J>K 19.66 From the values given for ∆H° and ∆S°, calculate ∆G° for

each of the following reactions at 298 K. If the reaction is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous?(a) 2 PbS1s2 + 3 O21g2 ¡ 2 PbO1s2 + 2 SO21g2

∆H° = -844 kJ; ∆S° = -165 J>K(b) 2 POCl31g2 ¡ 2 PCl31g2 + O21g2

∆H° = 572 kJ; ∆S° = 179 J>K 19.67 A particular constant-pressure reaction is barely spontaneous

at 390 K. The enthalpy change for the reaction is +23.7 kJ. Estimate∆S for the reaction.

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additional exercises 851

(a) H21g2 + I21g2 ∆ 2 HI1g2(b) C2H5OH1g2 ∆ C2H41g2 + H2O1g2(c) 3 C2H21g2 ∆ C6H61g2

19.82 Using data from Appendix C, write the equilibrium-constant expression and calculate the value of the equilibrium con-stant and the free-energy change for these reactions at 298 K:(a) NaHCO31s2 ∆ NaOH1s2 + CO21g2(b) 2 HBr1g2 + Cl21g2 ∆ 2 HCl1g2 + Br21g2(c) 2 SO21g2 + O21g2 ∆ 2 SO31g2

19.83 Consider the decomposition of barium carbonate:

BaCO31s2 ∆ BaO1s2 + CO21g2 Using data from Appendix C, calculate the equilibrium pres-

sure of CO2 at (a) 298 K and (b) 1100 K. 19.84 Consider the reaction

PbCO31s2 ∆ PbO1s2 + CO21g2 Using data in Appendix C, calculate the equilibrium pressure

of CO2 in the system at (a) 400 °C and (b) 180 °C. 19.85 The value of Ka for nitrous acid 1HNO22 at 25 °C is given in Ap-

pendix D. (a) Write the chemical equation for the equilibrium that corresponds to Ka. (b) By using the value of Ka, calculate ∆G° for the dissociation of nitrous acid in aqueous solution. (c) What is the value of ∆G at equilibrium? (d) What is the value of ∆G when 3H+4 = 5.0 * 10-2 M, 3NO2 -4 = 6.0 * 10-4 M, and 3HNO24 = 0.20 M?

19.86 The Kb for methylamine 1CH3NH22 at 25 °C is given in Ap-pendix D. (a) Write the chemical equation for the equilib-rium that corresponds to Kb. (b) By using the value of Kb, calculate ∆G° for the equilibrium in part (a). (c) What is the value of ∆G at equilibrium? (d) What is the value of ∆G when 3H+4 = 6.7 * 10-9 M, 3CH3NH3 +4 = 2.4 * 10-3 M, and 3CH3NH24 = 0.098 M?

19.76 The fuel in high-efficiency natural gas vehicles consists pri-marily of methane 1CH42. (a) How much heat is produced in burning 1 mol of CH41g2 under standard conditions if reactants and products are brought to 298 K and H2O1l2 is formed? (b) What is the maximum amount of useful work that can be accomplished under standard conditions by this system?

Free energy and equilibrium (section 19.7)

19.77 Indicate whether ∆G increases, decreases, or stays the same for each of the following reactions as the partial pressure of O2 is increased:(a) 2 CO1g2 + O21g2 ¡ 2 CO21g2(b) 2 H2O21l2 ¡ 2 H2O1l2 + O21g2(c) 2 KClO31s2 ¡ 2 KCl1s2 + 3 O21g2

19.78 Indicate whether ∆G increases, decreases, or does not change when the partial pressure of H2 is increased in each of the fol-lowing reactions:(a) N21g2 + 3 H21g2 ¡ 2 NH31g2(b) 2 HBr1g2 ¡ H21g2 + Br21g2(c) 2 H21g2 + C2H21g2 ¡ C2H61g2

19.79 Consider the reaction 2 NO21g2 ¡ N2O41g2. (a) Using data from Appendix C, calculate ∆G° at 298 K. (b) Calcu-late ∆G at 298 K if the partial pressures of NO2 and N2O4 are 0.40 atm and 1.60 atm, respectively.

19.80 Consider the reaction 3 CH41g2 ¡ C3H81g2 + 2 H21g2. (a) Using data from Appendix C, calculate ∆G° at 298 K. (b) Calculate ∆G at 298 K if the reaction mixture consists of 40.0 atm of CH4, 0.0100 atm of C3H81g2, and 0.0180 atm of H2.

19.81 Use data from Appendix C to calculate the equilibrium con-stant, K, and ∆G° at 298 K for each of the following reactions:

Co(s) is lowered from 60 °C to 25 °C. (c) Ethyl alcohol evapo-rates from a beaker. (d) A diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form CO21g2 and H2O1g2.

19.90 The reaction 2 Mg1s2 + O21g2 ¡ 2 MgO1s2 is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for ∆S°. Did your classmate make a mistake in the calculation? Explain.

[19.91] Suppose four gas molecules are placed in the left flask in Figure 19.6(a). Initially, the right flask is evacuated and the stopcock is closed. (a) After the stopcock is opened, how many different arrangements of the molecules are possible? (b) How many of the arrangements from part (a) have all the molecules in the left flask? (c) How does the answer to part (b) explain the spontaneous expansion of the gas?

[19.92] Consider a system that consists of two standard playing dice, with the state of the system defined by the sum of the values shown on the top faces. (a) The two arrangements of top

19.87 (a) Which of the thermodynamic quantities T, E, q, w, and S are state functions? (b) Which depend on the path taken from one state to another? (c) How many reversible paths are there between two states of a system? (d) For a reversible isother-mal process, write an expression for ∆E in terms of q and w and an expression for ∆S in terms of q and T.

19.88 Indicate whether each of the following statements is true or false. If it is false, correct it. (a) The feasibility of manufac-turing NH3 from N2 and H2 depends entirely on the value of ∆H for the process N21g2 + 3 H21g2 ¡ 2 NH31g2. (b) The reaction of Na(s) with Cl21g2 to form NaCl(s) is a spontaneous process. (c) A spontaneous process can in prin-ciple be conducted reversibly. (d) Spontaneous processes in general require that work be done to force them to proceed. (e) Spontaneous processes are those that are exothermic and that lead to a higher degree of order in the system.

19.89 For each of the following processes, indicate whether the signs of ∆S and ∆H are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of

additional exercises


up the normal boiling point of Br2 in a chemistry hand-book or at the WebElements Web site (www.webelements.com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

[19.96] For the majority of the compounds listed in Appendix C, the value of ∆G°f is more positive (or less negative) than the value of ∆H°f. (a) Explain this observation, using NH31g2, CCl41l2, and KNO31s2 as examples. (b) An exception to this observa-tion is CO(g). Explain the trend in the ∆H°f and ∆G°f values for this molecule.

19.97 Consider the following three reactions: (i) Ti1s2 + 2 Cl21g2 ¡ TiCl41g2 (ii) C2H61g2 + 7 Cl21g2 ¡ 2 CCl41g2 + 6 HCl1g2 (iii) BaO1s2 + CO21g2 ¡ BaCO31s2 (a) For each of the reactions, use data in Appendix C to cal-

culate ∆H°, ∆G°, K, and ∆S° at 25 °C. (b) Which of these reactions are spontaneous under standard conditions at 25 °C? (c) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

19.98 Using the data in Appendix C and given the pressures listed, calculate Kp and ∆G for each of the following reactions:(a) N21g2 + 3 H21g2 ¡ 2 NH31g2

PN2= 2.6 atm, PH2

= 5.9 atm, PNH3= 1.2 atm

(b) 2 N2H41g2 + 2 NO21g2 ¡ 3 N21g2 + 4 H2O1g2 PN2H4

= PNO2= 5.0 * 10-2 atm,

PN2= 0.5 atm, PH2O = 0.3 atm

(c) N2H41g2 ¡ N21g2 + 2 H21g2PN2H4

= 0.5 atm, PN2= 1.5 atm, PH2

= 2.5 atm 19.99 (a) For each of the following reactions, predict the sign of

∆H° and ∆S° without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have K 7 1. (c) In each case indicate whether K should increase or decrease with increasing temperature.

(i) 2 Mg1s2 + O21g2 ∆ 2 MgO1s2 (ii) 2 KI1s2 ∆ 2 K1g2 + I21g2 (iii) Na21g2 ∆ 2 Na1g2 (iv) 2 V2O51s2 ∆ 4 V1s2 + 5 O21g2 19.100 Acetic acid can be manufactured by combining methanol with

carbon monoxide, an example of a carbonylation reaction:

CH3OH1l2 + CO1g2 ¡ CH3COOH1l2

(a) Calculate the equilibrium constant for the reaction at 25 °C. (b) Industrially, this reaction is run at temperatures above 25 °C. Will an increase in temperature produce an in-crease or decrease in the mole fraction of acetic acid at equi-librium? Why are elevated temperatures used? (c) At what temperature will this reaction have an equilibrium constant equal to 1? (You may assume that ∆H° and ∆S° are tempera-ture independent, and you may ignore any phase changes that might occur.)

19.101 The oxidation of glucose 1C6H12O62 in body tissue produces CO2 and H2O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol 1C2H5OH2

faces shown here can be viewed as two possible microstates of the system. Explain. (b) To which state does each microstate correspond? (c) How many possible states are there for the system? (d) Which state or states have the highest entropy? Explain. (e) Which state or states have the lowest entropy? Explain. (f) Calculate the absolute entropy of the two-dice system.

19.93 Ammonium nitrate dissolves spontaneously and endother-mally in water at room temperature. What can you deduce about the sign of ∆S for this solution process?

[19.94] A standard air conditioner involves a refrigerant that is typically now a fluorinated hydrocarbon, such as CH2F2. An air-conditioner refrigerant has the property that it read-ily vaporizes at atmospheric pressure and is easily com-pressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram).

Expansion (low pressure)

Liquid Vapor

Liquid Vapor

Compression (high pressure)

Expansion chamber

Compression chamber

During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of q for the expansion? (b) What is the sign of q for the compres-sion? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refriger-ant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a revers-ible process? (e) Suppose that a house and its exterior are both initially at 31 °C. Some time after the air conditioner is turned on, the house is cooled to 24 °C. Is this process spontaneous or nonspontaneous?

[19.95] Trouton’s rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about 88 J>mol@K. (a) Estimate the normal boiling point of bromine, Br2, by determining ∆H°vap for Br2 using data from Appendix C. Assume that ∆H°vap remains constant with temperature and that Trouton’s rule holds. (b) Look

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integrative exercises 853

is much greater (0.15 M). The plasma and intracellular fluid are separated by the cell membrane, which we assume is per-meable only to K+. (a) What is ∆G for the transfer of 1 mol of K+ from blood plasma to the cellular fluid at body tempera-ture 37 °C? (b) What is the minimum amount of work that must be used to transfer this K+?

[19.105] One way to derive Equation 19.3 depends on the observation that at constant T the number of ways, W, of arranging m ideal-gas particles in a volume V is proportional to the vol-ume raised to the m power:

W ∝Vm

Use this relationship and Boltzmann’s relationship between entropy and number of arrangements (Equation 19.5) to de-rive the equation for the entropy change for the isothermal expansion or compression of n moles of an ideal gas.

[19.106] About 86% of the world’s electrical energy is produced by us-ing steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat avail-able to do the work (for example, from hot steam produced by combustion of a fuel such as coal or methane). This effi-ciency is given by the ratio 1Thigh - Tlow2>Thigh, where Thigh is the temperature of the heat going into the engine and Tlow is that of the heat leaving the engine. (a) What is the maxi-mum possible efficiency of a heat engine operating between an input temperature of 700 K and an exit temperature of 288 K? (b) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near 100% effi-ciency? (d) It is often said that if the energy of combustion of a fuel such as methane were captured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater frac-tion of the energy could be put to useful work. Make a quali-tative drawing like that in Figure 5.10 (p. 175) that illustrates the fact that in principle the fuel cell route will produce more useful work than the heat engine route from combustion of methane.

and CO2. (a) Using data given in Appendix C, compare the equilibrium constants for the following reactions:

C6H12O61s2 + 6 O21g2 ∆ 6 CO21g2 + 6 H2O1l2C6H12O61s2 ∆ 2 C2H5OH1l2 + 2 CO21g2

(b) Compare the maximum work that can be obtained from these processes under standard conditions.

[19.102] The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane 1C2H62, is a very important industrial chemical pro-cess. In principle, methane can be converted into ethane and hydrogen:

2 CH41g2 ¡ C2H61g2 + H21g2 In practice, this reaction is carried out in the presence of

oxygen:

2 CH41g2 + 12 O21g2 ¡ C2H61g2 + H2O1g2

(a) Using the data in Appendix C, calculate K for these reac-tions at 25 °C and 500 °C. (b) Is the difference in ∆G° for the two reactions due primarily to the enthalpy term 1∆H2 or the entropy term 1-T∆S2? (c) Explain how the preceding reac-tions are an example of driving a nonspontaneous reaction, as discussed in the “Chemistry and Life” box in Section 19.7. (d) The reaction of CH4 and O2 to form C2H6 and H2O must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

[19.103] Cells use the hydrolysis of adenosine triphosphate (ATP) as a source of energy (Figure 19.16). The conversion of ATP to ADP has a standard free-energy change of -30.5 kJ>mol. If all the free energy from the metabolism of glucose,

C6H12O61s2 + 6 O21g2 ¡ 6 CO21g2 + 6 H2O1l2 goes into the conversion of ADP to ATP, how many moles of

ATP can be produced for each mole of glucose? [19.104] The potassium-ion concentration in blood plasma is about

5.0 * 10-3 M, whereas the concentration in muscle-cell fluid

(a) Calculate ∆Svap for each of the liquids. Do all the liquids obey Trouton’s rule? (b) With reference to intermolecu-lar forces (Section 11.2), can you explain any exceptions to the rule? (c) Would you expect water to obey Trouton’s rule? By using data in Appendix B, check the accuracy of your conclusion. (d) Chlorobenzene 1C6H5Cl2 boils at 131.8 °C. Use Trouton’s rule to estimate ∆Hvap for this substance.

19.108 In chemical kinetics the entropy of activation is the entropy change for the process in which the reactants reach the ac-tivated complex. The entropy of activation for bimolecular processes is usually negative. Explain this observation with reference to Figure 14.15.

19.107 Most liquids follow Trouton’s rule (see Exercise 19.95), which states that the molar entropy of vaporization is approximately 88 { 5 J>mol@K. The normal boiling points and enthalpies of vaporization of several organic liquids are as follows:

substancenormal Boiling Point 1°C2 ∆Hvap 1kJ>mol2

Acetone, 1CH322CO 56.1 29.1

Dimethyl ether, 1CH322O -24.8 21.5

Ethanol, C2H5OH 78.4 38.6Octane, C8H18 125.6 34.4Pyridine, C5H5N 115.3 35.1

integrative exercises


Write the formation reaction for AgNO31s2. Based on this reaction, do you expect the entropy of the system to in-crease or decrease upon the formation of AgNO31s2? (b) Use ∆H°f and ∆G°f of AgNO31s2 to determine the entropy change upon formation of the substance. Is your answer consistent with your reasoning in part (a)? (c) Is dissolv-ing AgNO3 in water an exothermic or endothermic pro-cess? What about dissolving MgSO4 in water? (d) For both AgNO3 and MgSO4, use the data to calculate the entropy change when the solid is dissolved in water. (e) Discuss the results from part (d) with reference to material pre-sented in this chapter and in the “A Closer Look” box on page 820.

[19.114] Consider the following equilibrium:

N2O41g2 ∆ 2 NO21g2 Thermodynamic data on these gases are given in Appendix C.

You may assume that ∆H° and ∆S° do not vary with temper-ature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what tem-perature will an equilibrium mixture of 1 atm total pressure contain twice as much NO2 as N2O4? (c) At what temperature will an equilibrium mixture of 10 atm total pressure con-tain twice as much NO2 as N2O4? (d) Rationalize the results from parts (b) and (c) by using Le Châtelier’s principle. [Sec-tion 15.7]

[19.115] The reaction

SO21g2 + 2 H2S1g2 ∆ 3 S1s2 + 2 H2O1g2 is the basis of a suggested method for removal of SO2 from

power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at 298 K? (b) In principle, is this re-action a feasible method of removing SO2? (c) If PSO2

= PH2S and the vapor pressure of water is 25 torr, calculate the equi-librium SO2 pressure in the system at 298 K. (d) Would you expect the process to be more or less effective at higher temperatures?

19.116 When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here:

Suppose you stretch a rubber band. (a) Do you expect the en-tropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be ab-sorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on!). What do you observe? Are your observa-tions consistent with your answer to part (b)?

19.109 At what temperatures is the following reaction, the reduction of magnetite by graphite to elemental iron, spontaneous?

Fe3O41s2 + 2 C 1s, graphite2 ¡ 2 CO21g2 + 3 Fe 1s2

19.110 The following processes were all discussed in Chapter 18, “Chemistry of the Environment.” Estimate whether the en-tropy of the system increases or decreases during each pro-cess: (a) photodissociation of O21g2, (b) formation of ozone from oxygen molecules and oxygen atoms, (c) diffusion of CFCs into the stratosphere, (d) desalination of water by re-verse osmosis.

[19.111] An ice cube with a mass of 20 g at -20 °C (typical freezer temperature) is dropped into a cup that holds 500 mL of hot water, initially at 83 °C. What is the final temperature in the cup? The density of liquid water is 1.00 g>mL; the specific heat capacity of ice is 2.03 J>g@C; the specific heat capacity of liquid water is 4.184 J>g@C; the enthalpy of fusion of water is 6.01 kJ>mol.

19.112 Carbon disulfide 1CS22 is a toxic, highly flammable sub-stance. The following thermodynamic data are available for CS21l2 and CS21g2 at 298 K:

∆Hf°1kJ>mol2 ∆Gf°1kJ>mol2CS21l2 89.7 65.3

CS21g2 117.4 67.2

(a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the C ¬ S bonds? (b) Use the VSEPR method to predict the structure of the CS2 mol-ecule. (c) Liquid CS2 burns in O2 with a blue flame, form-ing CO21g2 and SO21g2. Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix C, calculate ∆H° and ∆G° for the reaction in part (c). Is the reaction exothermic? Is it spontaneous at 298 K? (e) Use the data in the table to calculate ∆S° at 298 K for the vaporization of CS21l2. Is the sign of ∆S° as you would ex-pect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of CS21l2. Do you predict that the substance will be a liquid or a gas at 298 K and 1 atm?

[19.113] The following data compare the standard enthalpies and free energies of formation of some crystalline ionic substances and aqueous solutions of the substances:

substance ∆Hf°1kJ>mol2 ∆Gf°1kJ>mol2AgNO31s2 -124.4 -33.4

AgNO31aq2 -101.7 -34.2

MgSO41s2 -1283.7 -1169.6

MgSO41aq2 -1374.8 -1198.4

Design an experiment 855

design an experiment You are measuring the equilibrium constant for a drug candidate binding to its DNA target over a series of different temperatures. You chose your drug candidate based on computer-aided molecular modeling, which indicates that the drug molecule likely would make many hydrogen bonds and favorable dipole–dipole interactions with the DNA site. You perform a set of experiments in buffer solution for the drug–DNA complex and gen-erate a table of K’s at different T’s. (a) Derive an equation that relates equilibrium con-stant to standard enthalpy and entropy changes. (Hint: equilibrium constant, enthalpy and entropy are all related to free energy). (b) Show how you can graph your K and T data to calculate the standard entropy and enthalpy changes for the drug candidate + DNA bind-ing interaction. (c) You are surprised to learn that the enthalpy change for the binding reaction is close to zero, and the entropy change is large and positive. Suggest an explana-tion and design an experiment to test it. (Hint: think about water and ions). (d) You try another drug candidate with the DNA target and find that this drug candidate has a large negative enthalpy change upon DNA binding, and the entropy change is small and posi-tive. Suggest an explanation, at the molecular level, and design an experiment to test your hypothesis.

20 ElectrochemistryThe electricity that powers much of modern society has many favorable characteristics, but it has a serious shortcoming: it cannot easily be stored. The electricity that flows into power company lines is consumed as it is generated, but for many other applications, stored electrical energy is needed.

20.4 Cell Potentials under standard Conditions We see that an important characteristic of a voltaic cell is its cell potential, which is the difference in the electrical potentials at the two electrodes and is measured in units of volts. Half-cell potentials are tabulated for reduction half-reactions under standard conditions (standard reduction potentials).

20.5 Free energy and redox reaCtions We relate the Gibbs free energy, ∆G°, to cell potential.

20.6 Cell Potentials under nonstandard Conditions We calculate cell potentials under nonstandard conditions using standard cell potentials and the Nernst equation.

20.1 oxidation states and oxidation–reduCtion reaCtions We review oxidation states and oxidation–reduction (redox) reactions.

20.2 BalanCing redox equations We learn how to balance redox equations using the method of half-reactions.

20.3 VoltaiC Cells We consider voltaic cells, which produce electricity from spontaneous redox reactions. Solid electrodes serve as the surfaces at which oxidation and reduction take place. The electrode where oxidation occurs is the anode, and the electrode where reduction occurs is the cathode.

WhaT’s ahEad

▶ An AdvAnced Li-ion bAttery pack built for use in Mercedes S-class hybrid automobiles.

In such cases electrical energy is converted into chemical energy, which can be stored and is portable, and then converted back to electricity when needed. Batteries are the most familiar devices for converting between electrical and chemical energies. Objects like laptop computers, cell phones, pacemakers, portable music players, cordless power tools, wristwatches, and countless other devices rely on batteries to provide the electric-ity needed for operation. A considerable amount of effort is currently being focused on research and development of new batteries, such as the one shown at right, particularly for powering electric vehicles. For that application new batteries are needed that are lighter, can charge faster, deliver more power, and have longer lifetimes. Cost and toxic-ity of the materials used in the battery are also important. At the heart of such develop-ments are the oxidation–reduction reactions that power batteries.

As discussed in Chapter 4, oxidation is the loss of electrons in a chemical reac-tion, and reduction is the gain of electrons. (Section 4.4) Thus, oxidation–reduction (redox) reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced. Redox reactions are involved not only in the operation of batteries but also in a wide variety of important natural processes, including the rusting of iron, the browning of foods, and the respiration of animals. Electrochemistry is the study of the relationships between electricity and chemical reactions. It includes the study of both spontaneous and nonspontaneous processes.

20.9 eleCtrolysis Finally, we focus on nonspontaneous redox reactions, examining electrolytic cells, which use electricity to perform chemical reactions.

20.7 Batteries and Fuel Cells We describe batteries and fuel cells, which are commercially important energy sources that use electrochemical reactions.

20.8 Corrosion We discuss corrosion, a spontaneous electrochemical process involving metals.

858 cHapTer 20 electrochemistry

20.1 | Oxidation states and Oxidation–Reduction Reactions

We determine whether a given chemical reaction is an oxidation–reduction reaction by keeping track of the oxidation numbers (oxidation states) of the elements involved in the reaction. (Section 4.4) This procedure identifies whether the oxidation number changes for any elements involved in the reaction. For example, consider the reaction that occurs spontaneously when zinc metal is added to a strong acid (▼ Figure 20.1):

Zn1s2 + 2H+1aq2 ¡ Zn2+1aq2 + H21g2 [20.1]

The chemical equation for this reaction can be written as

Zn(s) Zn2+(aq) H2(g)

00 +2+1

+2 H+(aq)+

H+ reduced

Zn oxidized

[20.2]

The oxidation numbers below the equation show that the oxidation number of Zn changes from 0 to +2 while that of H changes from +1 to 0. Thus, this is an oxidation–reduction reaction. Electrons are transferred from zinc atoms to hydrogen ions; Zn is oxidized and H+ is reduced.

Zn(s) 2 HCl(aq)

+

ZnCl2(aq)

2+

−−

−

+

H2(g)

+

+−

▲ Figure 20.1 oxidation of zinc by hydrochloric acid.

GO FiGuREExplain (a) the vigorous bubbling in the beaker on the right and (b) the formation of steam above that beaker.

SecTioN 20.1 oxidation States and oxidation–reduction reactions 859

In a reaction such as Equation 20.2, a clear transfer of electrons occurs. In some reactions, however, the oxidation numbers change, but we cannot say that any sub-stance literally gains or loses electrons. For example, in the combustion of hydrogen gas,

2 H2(g) O2(g)

0

2 H2O(g)

0 +1 −2

+

[20.3]

hydrogen is oxidized from the 0 to the +1 oxidation state and oxygen is reduced from the 0 to the -2 oxidation state, indicating that Equation 20.3 is an oxidation–reduction reaction. Water is not an ionic substance, therefore, a complete transfer of electrons from hydrogen to oxygen does not occur as water is formed. So, while keeping track of oxidation states offers a convenient form of “bookkeeping,” you should not generally equate the oxi-dation state of an atom with its actual charge in a chemical compound. (Section 8.5, “Oxidation Numbers, Formal Charges, and Actual Partial Charges”)

Give it Some thoughtWhat are the oxidation numbers of the elements in the nitrite ion, NO2

- ?

In any redox reaction, both oxidation and reduction must occur. If one substance is oxidized, another must be reduced. The substance that makes it possible for another substance to be oxidized is called either the oxidizing agent or the oxidant. The oxidiz-ing agent acquires electrons from the other substance and so is itself reduced. A reducing agent, or reductant, is a substance that gives up electrons, thereby causing another substance to be reduced. The reducing agent is therefore oxidized in the process. In Equation 20.2, H+1aq2, the species that is reduced, is the oxidizing agent and Zn(s), the species that is oxidized, is the reducing agent.

The nickel–cadmium (nicad) battery uses the following redox reaction to generate electricity:

Cd1s2 + NiO21s2 + 2H2O1l2 ¡ Cd1OH221s2 + Ni1OH221s2Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent.

sOluTiOnAnalyze We are given a redox equation and asked to identify the substance oxidized and the sub-stance reduced and to label the oxidizing agent and the reducing agent.Plan First, we use the rules outlined earlier (Section 4.4) to assign oxidation states, or num-bers, to all the atoms and determine which elements change oxidation state. Second, we apply the definitions of oxidation and reduction.

Solve Cd(s) NiO2(s) 2H2O(l) Ni(OH)2(s)

0

Cd(OH)2(s)

2 2 1 2 2 14 2 1 2

+ + +

+ + + + + +− − − −

The oxidation state of Cd increases from 0 to +2, and that of Ni decreases from +4 to +2. Thus, the Cd atom is oxidized (loses electrons) and is the reducing agent. The oxidation state of Ni decreases as NiO2 is converted into Ni1OH22. Thus, NiO2 is reduced (gains electrons) and is the oxidizing agent.Comment A common mnemonic for remembering oxidation and reduction is “LEO the lion says GER”: losing electrons is oxidation; gaining electrons is reduction.

Practice Exercise 1What is the reducing agent in the following reaction?

2 Br-1aq2 + H2O21aq2 + 2 H+1aq2 ¡ Br21aq2 + 2 H2O1l2(a) Br-1aq2 (b) H2O21aq2 (c) H+1aq2 (d) Br21aq2 (e) Na+1aq2Practice Exercise 2Identify the oxidizing and reducing agents in the reaction

2 H2O1l2 + Al1s2 + MnO4-1aq2 ¡ Al1OH24

-1aq2 + MnO21s2

samPlE ExERcisE 20.1 identifying Oxidizing and Reducing agents


20.2 | Balancing Redox EquationsWhenever we balance a chemical equation, we must obey the law of conservation of mass: The amount of each element must be the same on both sides of the equation. (Atoms are neither created nor destroyed in any chemical reaction.) As we balance oxidation–reduction reactions, there is an additional requirement: The gains and losses of electrons must be balanced. If a substance loses a certain number of electrons during a reaction, another substance must gain that same number of electrons. (Electrons are neither created nor destroyed in any chemical reaction.)

In many simple chemical equations, such as Equation 20.2, balancing the elec-trons is handled “automatically”—that is, we balance the equation without explicitly accounting for the transfer of electrons. Many redox equations are more complex than Equation 20.2, however, and cannot be balanced easily without taking into account the number of electrons lost and gained. In this section, we examine the method of half-reactions, a systematic procedure for balancing redox equations.

Half-ReactionsAlthough oxidation and reduction must take place simultaneously, it is often convenient to consider them as separate processes. For example, the oxidation of Sn2+ by Fe3+,

Sn2+1aq2 + 2 Fe3+1aq2 ¡ Sn4+1aq2 + 2 Fe2+1aq2can be considered as consisting of two processes: oxidation of Sn2+ and reduction of Fe3+:

Oxidation: Sn2+1aq2 ¡ Sn4+1aq2 + 2e- [20.4]

Reduction: 2 Fe3+1aq2 + 2e- ¡ 2 Fe2+1aq2 [20.5]

Notice that electrons are shown as products in the oxidation process and as reactants in the reduction process.

Equations that show either oxidation or reduction alone, such as Equations 20.4 and 20.5, are called half-reactions. In the overall redox reaction, the number of elec-trons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction. When this condition is met and each half-reaction is bal-anced, the electrons on the two sides cancel when the two half-reactions are added to give the balanced oxidation–reduction equation.

Balancing Equations by the Method of Half-ReactionsIn the half-reaction method, we usually begin with a “skeleton” ionic equation show-ing only the substances undergoing oxidation and reduction. In such cases, we assign oxidation numbers only when we are unsure whether the reaction involves oxidation–reduction. We will find that H+ (for acidic solutions), OH- (for basic solutions), and H2O are often involved as reactants or products in redox reactions. Unless H+, OH-, or H2O is being oxidized or reduced, these species do not appear in the skeleton equation. Their presence, however, can be deduced as we balance the equation.

For balancing a redox reaction that occurs in acidic aqueous solution, the proce-dure is as follows:

1. Divide the equation into one oxidation half-reaction and one reduction half- reaction.

2. Balance each half-reaction. (a) First, balance elements other than H and O. (b) Next, balance O atoms by adding H2O as needed. (c) Then balance H atoms by adding H+ as needed. (d) Finally, balance charge by adding e- as needed. This specific sequence 1a2-1d2 is important, and it is summarized in the diagram

in the margin. At this point, you can check whether the number of electrons in each half-reaction corresponds to the changes in oxidation state.

(a)

(b)

(c)

(d)

“Other” atoms Balance atomsother than H, O

Balance Oby adding H2O

Balance Hby adding H+

Balance electrons

O

H

e–

SecTioN 20.2 Balancing redox equations 861

3. Multiply half-reactions by integers as needed to make the number of electrons lost in the oxidation half-reaction equal the number of electrons gained in the reduc-tion half-reaction.

4. Add half-reactions and, if possible, simplify by canceling species appearing on both sides of the combined equation.

5. Check to make sure that atoms and charges are balanced.

As an example, let’s consider the reaction between permanganate ion 1MnO4-2

and oxalate ion 1C2O42-2 in acidic aqueous solution (▼ Figure 20.2). When MnO4

- is added to an acidified solution of C2O4

2-, the deep purple color of the MnO4- ion fades,

bubbles of CO2 form, and the solution takes on the pale pink color of Mn2+. We can write the skeleton equation as

MnO4-1aq2 + C2O4

2-1aq2 ¡ Mn2+1aq2 + CO21aq2 [20.6]

Experiments show that H+ is consumed and H2O is produced in the reaction. We will see that their involvement in the reaction is deduced in the course of balancing the equation.

To complete and balance Equation 20.6, we first write the two half-reactions (step 1). One half-reaction must have Mn on both sides of the arrow, and the other must have C on both sides of the arrow:

MnO4-1aq2 ¡ Mn2+1aq2

C2O42-1aq2 ¡ CO21g2

We next complete and balance each half-reaction. First, we balance all the atoms except H and O (step 2a). In the permanganate half-reaction, we have one manga-nese atom on each side of the equation and so need to do nothing. In the oxalate half- reaction, we add a coefficient 2 on the right to balance the two carbons on the left:

MnO4-1aq2 ¡ Mn2+1aq2

C2O42-1aq2 ¡ 2 CO21g2

Next we balance O (step 2b). The permanganate half-reaction has four oxygens on the left and none on the right; to balance these four oxygen atoms, we add four H2O mol-ecules on the right:

MnO4-1aq2 ¡ Mn2+1aq2 + 4 H2O1l2

▲ Figure 20.2 titration of an acidic solution of na2c2o4 with KMno4(aq).

GO FiGuREWhich species is reduced in this reaction? Which species is the reducing agent?

(a) (b)

At end point, purple color of MnO4

− remains because all C2O4

2− consumed

The purple color of MnO4−

disappears immediately as reaction with C2O4

2− occurs

MnO4−(aq)

C2O42−(aq)


The eight hydrogen atoms now in the products must be balanced by adding 8 H+ to the reactants (step 2c):

8 H+1aq2 + MnO4-1aq2 ¡ Mn2+1aq2 + 4 H2O1l2

Now there are equal numbers of each type of atom on the two sides of the equa-tion, but the charge still needs to be balanced. The charge of the reactants is 811+2 +111-2 = 7+ , and that of the products is 112+2 +4102 = 2+ . To balance the charge, we add five electrons to the reactant side (step 2d):

5 e- + 8 H+1aq2 + MnO4-1aq2 ¡ Mn2+1aq2 + 4 H2O1l2

We can use oxidation states to check our result. In this half-reaction Mn goes from the +7 oxidation state in MnO4

- to the +2 oxidation state of Mn2+. Therefore, each Mn atom gains five electrons, in agreement with our balanced half-reaction.

In the oxalate half-reaction, we have C and O balanced (step 2a). We balance the charge (step 2d) by adding two electrons to the products:

C2O42-1aq2 ¡ 2 CO21g2 + 2 e-

We can check this result using oxidation states. Carbon goes from the +3 oxidation state in C2O4

2- to the +4 oxidation state in CO2. Thus, each C atom loses one electron; therefore, the two C atoms in C2O4

2- lose two electrons, in agreement with our bal-anced half-reaction.

Now we multiply each half-reaction by an appropriate integer so that the number of electrons gained in one half-reaction equals the number of electrons lost in the other (step 3). We multiply the MnO4

- half-reaction by 2 and the C2O42- half-reaction by 5:

10 e- + 16 H+1aq2 + 2 MnO4-1aq2 ¡ 2 Mn2+1aq2 + 8 H2O1l2

5 C2O42-1aq2 ¡ 10 CO21g2 + 10 e-

16 H+1aq2 + 2 MnO4-1aq2 + 5 C2O4

2-1aq2 ¡2 Mn2+1aq2 + 8 H2O1l2 + 10 CO21g2

The balanced equation is the sum of the balanced half-reactions (step 4). Note that the electrons on the reactant and product sides of the equation cancel each other.

We check the balanced equation by counting atoms and charges (step 5). There are 16 H, 2 Mn, 28 O, 10 C, and a net charge of 4+ on each side of the equation, confirming that the equation is correctly balanced.

Give it Some thoughtDo free electrons appear anywhere in the balanced equation for a redox reaction?

Complete and balance this equation by the method of half-reactions:

Cr2O72-1aq2 + Cl-1aq2 ¡ Cr3+1aq2 + Cl21g2 1acidic solution2

sOluTiOnAnalyze We are given an incomplete, unbalanced (skeleton) equation for a redox reaction occur-ring in acidic solution and asked to complete and balance it.Plan We use the half-reaction procedure we just learned.Solve

Step 1: We divide the equation into two half-reactions:

Cr2O72-1aq2 ¡ Cr3+1aq2

Cl-1aq2 ¡ Cl21g2

samPlE ExERcisE 20.2 Balancing Redox Equations in acidic solution

SecTioN 20.2 Balancing redox equations 863

Balancing Equations for Reactions Occurring in Basic SolutionIf a redox reaction occurs in basic solution, the equation must be balanced by using OH- and H2O rather than H+ and H2O. Because the water molecule and the hydroxide ion both contain hydrogen, this approach can take more moving back and forth from one side of the equation to the other to arrive at the appropriate half-reaction. An alternate approach is to first balance the half-reactions as if they occurred in acidic solution, count the number of H+ in each half-reaction, and then add the same number of OH- to each side of the half-reaction. This way, the reaction is mass-balanced because you are adding the same thing to both sides. In essence, what you are doing is “neutralizing” the protons to form water 1H+ + OH- ¡ H2O2 on the side containing H+, and the other side ends up with the OH-. The resulting water molecules can be canceled as needed.

Step 2: We balance each half-reaction. In the first half-reaction the presence of one Cr2O72-

among the reactants requires two Cr3+ among the products. The seven oxygen atoms in Cr2O7

2- are balanced by adding seven H2O to the products. The 14 hydrogen atoms in 7 H2O are then balanced by adding 14 H+ to the reactants:

14 H+1aq2 + Cr2O72-1aq2 ¡ 2 Cr3+1aq2 + 7 H2O1l2

We then balance the charge by adding electrons to the left side of the equation so that the total charge is the same on the two sides:

6 e- + 14 H+1aq2 + Cr2O72-1aq2 ¡ 2 Cr3+1aq2 + 7 H2O1l2

We can check this result by looking at the oxidation state changes. Each chromium atom goes from +6 to +3, gaining three electrons; therefore, the two Cr atoms in Cr2O7

2- gain six elec-trons, in agreement with our half-reaction.In the second half-reaction, two Cl- are required to balance one Cl2:

2 Cl-1aq2 ¡ Cl21g2We add two electrons to the right side to attain charge balance:

2 Cl-1aq2 ¡ Cl21g2 + 2 e-

This result agrees with the oxidation state changes. Each chlorine atom goes from -1 to 0, los-ing one electron; therefore, the two chlorine atoms lose two electrons.Step 3: We equalize the number of electrons transferred in the two half-reactions. To do so, we multiply the Cl half-reaction by 3 so that the number of electrons gained in the Cr half-reaction (6) equals the number lost in the Cl half-reaction, allowing the electrons to cancel when the half-reactions are added:

6 Cl-1aq2 ¡ 3 Cl21g2 + 6 e-

Step 4: The equations are added to give the balanced equation:

14 H+1aq2 + Cr2O72-1aq2 + 6 Cl-1aq2 ¡ 2 Cr3+1aq2 + 7 H2O1l2 + 3 Cl21g2

Step 5: There are equal numbers of atoms of each kind on the two sides of the equation (14 H, 2 Cr, 7 O, 6 Cl). In addition, the charge is the same on the two sides 16+2. Thus, the equation is balanced.

Practice Exercise 1If you complete and balance the following equation in acidic solution

Mn2+1aq2 + NaBiO31s2 ¡ Bi3+1aq2 + MnO4-1aq2 + Na+1aq2

how many water molecules are there in the balanced equation (for the reaction balanced with the smallest whole-number coefficients)? (a) Four on the reactant side, (b) Three on the product side, (c) One on the reactant side, (d) Seven on the product side, (e) Two on the product side.

Practice Exercise 2Complete and balance the following equation in acidic solution using the method of half-reactions.

Cu1s2 + NO3-1aq2 ¡ Cu2+1aq2 + NO21g2


Step 2: We balance each half- reaction as if it took place in acidic solution:

sOluTiOnAnalyze We are given an incomplete equation for a basic redox reaction and asked to balance it.Plan We go through the first steps of our procedure as if the reaction were occurring in acidic solution. We then add the appropriate number of OH- to each side of the equation, combining H+ and OH- to form H2O. We complete the process by simplifying the equation.

samPlE ExERcisE 20.3 Balancing Redox Equations in Basic solution

Complete and balance this equation for a redox reaction that takes place in basic solution:

CN-1aq2 + MnO4-1aq2 ¡ CNO-1aq2 + MnO21s2 1basic solution2

CN-1aq2 ¡ CNO-1aq2MnO4

-1aq2 ¡ MnO21s2

Solve

Step 1: We write the incomplete, unbalanced half-reactions:

CN-1aq2 + H2O1l2 ¡ CNO-1aq2 + 2 H+1aq2 + 2 e-

3 e- + 4 H+1aq2 + MnO4-1aq2 ¡ MnO21s2 + 2 H2O1l2

Now we must take into account that the reaction occurs in basic solution, adding OH- to both sides of both half-reactions to neutralize H+:

CN-1aq2 + H2O1l2 + 2 OH-1aq2 ¡ CNO-1aq2 + 2 H+1aq2 + 2 e- + 2 OH-1aq23 e- + 4 H+1aq2 + MnO4

-1aq2 + 4 OH-1aq2 ¡ MnO21s2 + 2 H2O1l2 + 4 OH-1aq2

We “neutralize” H+ and OH- by forming H2O when they are on the same side of either half-reaction:

CN-1aq2 + H2O1l2 + 2 OH-1aq2 ¡ CNO-1aq2 + 2 H2O1l2 + 2 e-

3 e- + 4 H2O1l2 + MnO4-1aq2 ¡ MnO21s2 + 2 H2O1l2 + 4 OH-1aq2

Next, we cancel water molecules that appear as both reactants and products:

CN-1aq2 + 2 OH-1aq2 ¡ CNO-1aq2 + H2O1l2 + 2 e-

3 e- + 2 H2O1l2 + MnO4-1aq2 ¡ MnO21s2 + 4 OH-1aq2

Both half-reactions are now balanced. You can check the atoms and the overall charge.

Step 3: We multiply the cyanide half-reaction by 3, which gives 6 electrons on the product side, and multiply the permanganate half-reaction by 2, which gives 6 elec-trons on the reactant side:

3 CN-1aq2 + 6 OH-1aq2 ¡ 3 CNO-1aq2 + 3 H2O1l2 + 6 e-

6 e- + 4 H2O1l2 + 2 MnO4-1aq2 ¡ 2 MnO21s2 + 8 OH-1aq2

Step 4: We add the two half-reactions together and simplify by canceling species that appear as both reac-tants and products: 3 CN-1aq2 + H2O1l2 + 2 MnO4

-1aq2 ¡ 3 CNO-1aq2 + 2 MnO21s2 + 2 OH-1aq2Step 5: Check that the atoms and charges are balanced.

There are 3 C, 3 N, 2 H, 9 O, 2 Mn, and a charge of 5− on both sides of the equation.Comment It is important to remember that this procedure does not imply that H+ ions are involved in the chemical reaction. Recall that in aqueous solutions at 20 °C, Kw = 3H+43OH-4 = 1.0 * 10-14. Thus, 3H+4 is very small in this basic solution. (Section 16.3)

Practice Exercise 1If you complete and balance the following oxidation–reduction reaction in basic solution

NO2-1aq2 + Al1s2 ¡ NH31aq2 + Al1OH24

-1aq2how many hydroxide ions are there in the balanced equation (for the reaction balanced with the smallest whole-number coefficients)? (a) One on the reactant side, (b) One on the product side, (c) Four on the reactant side, (d) Seven on the product side, (e) None.

Practice Exercise 2Complete and balance the following oxidation–reduction reaction in basic solution:

Cr1OH231s2 + ClO-1aq2 ¡ CrO42-1aq2 + Cl21g2

SecTioN 20.3 Voltaic cells 865

20.3 | Voltaic cellsThe energy released in a spontaneous redox reaction can be used to perform electrical work. This task is accomplished through a voltaic (or galvanic) cell, a device in which the transfer of electrons takes place through an external pathway rather than directly between reactants present in the same reaction vessel.

One such spontaneous reaction occurs when a strip of zinc is placed in contact with a solution containing Cu2+. As the reaction proceeds, the blue color of Cu2+1aq2 ions fades and copper metal deposits on the zinc. At the same time, the zinc begins to dissolve. These transformations, shown in ▲ Figure 20.3, are summarized by the equation

Zn1s2 + Cu2+1aq2 ¡ Zn2+1aq2 + Cu1s2 [20.7]

▶ Figure 20.4 shows a voltaic cell that uses the redox reaction given in Equation 20.7. Although the setup in Figure 20.4 is more complex than that in Figure 20.3, the reac-tion is the same in both cases. The significant difference is that in the voltaic cell the Zn metal and Cu2+1aq2 are not in direct contact with each other. Instead, Zn metal is in contact with Zn2+1aq2 in one compartment, and Cu metal is in contact with Cu2+1aq2 in the other compartment. Consequently, Cu2+1aq2 reduction can occur only by the flow of electrons through an external circuit, namely, a wire connect-ing the Zn and Cu strips. Electrons flowing through a wire and ions moving in solu-tion both constitute an electrical current. This flow of electrical charge can be used to accomplish electrical work.

▲ Figure 20.3 A spontaneous oxidation–reduction reaction involving zinc and copper.

GO FiGuREWhy does the intensity of the blue solution color lessen as the reaction proceeds?

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Atoms inZn strip

Cu2+ ionsin solution

2 e−

Zn2+ ion

Cu atomElectrons movefrom Zn to Cu2+

Zn oxidized

Cu2+ reduced

Cu electrode in1 M CuSO4 solution

Solutions in contact with eachother through porous glass disc

Zn electrode in1 M ZnSO4 solution

▲ Figure 20.4 A cu–Zn voltaic cell based on the reaction in equation 20.7.

GO FiGuREWhich metal, Cu or Zn, is oxidized in this voltaic cell?


The two solid metals connected by the external circuit are called electrodes. By defi-nition, the electrode at which oxidation occurs is the anode and the electrode at which reduction occurs is the cathode.* The electrodes can be made of materials that participate in the reaction, as in the present example. Over the course of the reaction, the Zn elec-trode gradually disappears and the copper electrode gains mass. More typically, the elec-trodes are made of a conducting material, such as platinum or graphite, that does not gain or lose mass during the reaction but serves as a surface at which electrons are transferred.

Each compartment of a voltaic cell is called a half-cell. One half-cell is the site of the oxidation half-reaction, and the other is the site of the reduction half-reaction. In our present example, Zn is oxidized and Cu2+ is reduced:

Anode 1oxidation half@reaction2 Zn1s2 ¡ Zn2+1aq2 + 2 e-

Cathode 1reduction half@reaction2 Cu2+1aq2 + 2 e- ¡ Cu1s2Electrons become available as zinc metal is oxidized at the anode. They flow through

the external circuit to the cathode, where they are consumed as Cu2+1aq2 is reduced. Because Zn(s) is oxidized in the cell, the zinc electrode loses mass, and the concentration of the Zn2+ solution increases as the cell operates. At the same time, the Cu electrode gains mass, and the Cu2+ solution becomes less concentrated as Cu2+ is reduced to Cu(s).

For a voltaic cell to work, the solutions in the two half-cells must remain electrically neu-tral. As Zn is oxidized in the anode half-cell, Zn2+ ions enter the solution, upsetting the initial Zn2+>SO4

2- charge balance. To keep the solution electrically neutral, there must be some means for Zn2+ cations to migrate out of the anode half-cell and for anions to migrate in. Similarly, the reduction of Cu2+ at the cathode removes these cations from the solution, leav-ing an excess of SO4

2- anions in that half-cell. To maintain electrical neutrality, some of these anions must migrate out of the cathode half-cell, and positive ions must migrate in. In fact, no measurable electron flow occurs between electrodes unless a means is provided for ions to migrate through the solution from one half-cell to the other, thereby completing the circuit.

In Figure 20.4, a porous glass disc separating the two half-cells allows ions to migrate and maintain the electrical neutrality of the solutions. In ▼ Figure 20.5, a salt

*To help remember these definitions, note that anode and oxidation both begin with a vowel, and cathode and reduction both begin with a consonant.

Salt bridge(allows ionmigration)

Cathode (+)Anode (−)

Zn Cu

NO3− Na+

Zn Cu

Zn2+

Zn2+ Cu2+

Cu2+

2e−

e− e−

2e−

Zn(s) Zn2+(aq) + 2 e− Cu2+(aq) + 2 e− Cu(s)

Voltmeter1.10 V

Cations migratetoward cathode

Anions migratetoward anode

▲ Figure 20.5 A voltaic cell that uses a salt bridge to complete the electrical circuit.

GO FiGuREHow is electrical balance maintained in the left beaker as Zn2+ are formed at the anode?

SecTioN 20.3 Voltaic cells 867

bridge serves this purpose. The salt bridge consists of a U-shaped tube containing an electrolyte solution, such as NaNO31aq2, whose ions will not react with other ions in the voltaic cell or with the electrodes. The electrolyte is often incorporated into a paste or gel so that the electrolyte solution does not pour out when the U-tube is inverted. As oxidation and reduction proceed at the electrodes, ions from the salt bridge migrate into the two half-cells—cations migrating to the cathode half-cell and anions migrating to the anode half-cell—to neutralize charge in the half-cell solutions. Whichever device is used to allow ions to migrate between half-cells, anions always migrate toward the anode and cations toward the cathode.

▶ Figure 20.6 summarizes the various relationships in a voltaic cell. Notice in par-ticular that electrons flow from the anode through the external circuit to the cathode. Because of this directional flow, the anode in a voltaic cell is labeled with a negative sign and the cathode is labeled with a positive sign. We can envision the electrons as being attracted to the positive cathode from the negative anode through the external circuit. ▲ Figure 20.6 Summary of reactions

occurring in a voltaic cell. The half-cells can be separated by either a porous glass disc (as in Figure 20.4) or by a salt bridge (as in Figure 20.5).

Electron flow

Porous barrieror salt bridge Cathode

(+)Anode(−)

Anions

Cations

e− e−

Anode half-cell,oxidation occurs

Cathode half-cell,reduction occurs

Voltmeter

The oxidation–reduction reaction

Cr2O72-1aq2 + 14 H+1aq2 + 6 I-1aq2 ¡ 2 Cr3+1aq2 + 3 I21s2 + 7 H2O1l2

is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solu-tion of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the di-rection of ion migration, and the signs of the electrodes.

sOluTiOnAnalyze We are given the equation for a spontaneous reaction that takes place in a voltaic cell and a description of how the cell is constructed. We are asked to write the half-reactions occur-ring at the anode and at the cathode, as well as the directions of electron and ion movements and the signs assigned to the electrodes.Plan Our first step is to divide the chemical equation into half-reactions so that we can identify the oxidation and the reduction processes. We then use the definitions of anode and cathode and the other terminologies summarized in Figure 20.6.Solve In one half-reaction, Cr2O7

2- 1aq2 is converted into Cr3+1aq2. Starting with these ions and then completing and balancing the half-reaction, we have

Cr2O72-1aq2 + 14 H+1aq2 + 6 e- ¡ 2 Cr3+1aq2 + 7 H2O1l2

In the other half-reaction, I-1aq2 is converted to I21s2:6 I-1aq2 ¡ 3 I21s2+6 e-

Now we can use the summary in Figure 20.6 to help us describe the voltaic cell. The first half-reaction is the reduction process (electrons on the reactant side of the equation). By definition, the reduction process occurs at the cathode. The second half-reaction is the oxidation process (electrons on the product side of the equation), which occurs at the anode.The I- ions are the source of electrons, and the Cr2O7

2- ions accept the electrons. Hence, the electrons flow through the external circuit from the electrode immersed in the KI solution (the anode) to the electrode immersed in the K2Cr2O7–H2SO4 solution (the cathode). The electrodes themselves do not react in any way; they merely provide a means of transferring electrons from or to the solutions. The cations move through the solutions toward the cathode, and the anions move toward the anode. The anode (from which the electrons move) is the negative electrode, and the cathode (toward which the electrons move) is the positive electrode.

Practice Exercise 1The following two half-reactions occur in a voltaic cell:

Ni1s2 ¡ Ni2+1aq2 + 2 e- 1electrode = Ni2Cu2+1aq2 + 2 e- ¡ Cu1s2 1electrode = Cu2

Which one of the following descriptions most accurately describes what is occurring in the half-cell containing the Cu electrode and Cu2+1aq2 solution?(a) The electrode is losing mass and cations from the salt bridge are flowing into the half-cell.(b) The electrode is gaining mass and cations from the salt bridge are flowing into the half-cell.

samPlE ExERcisE 20.4 describing a Voltaic cell


20.4 | cell Potentials under standard conditions

Why do electrons transfer spontaneously from a Zn atom to a Cu2+ ion, either directly as in Figure 20.3 or through an external circuit as in Figure 20.4? In a simple sense, we can compare the electron flow to the flow of water in a waterfall (▼ Figure 20.7). Water flows spontaneously over a waterfall because of a difference in potential energy between the top of the falls and the bottom. (Section 5.1) In a similar fashion, electrons flow spontaneously through an external circuit from the anode of a voltaic cell to the cathode because of a difference in potential energy. The potential energy of electrons is higher in the anode than in the cathode. Thus, electrons flow spontaneously toward the electrode with the more positive electrical potential.

The difference in potential energy per electrical charge (the potential difference) between two electrodes is measured in volts. One volt (V) is the potential difference required to impart 1 joule (J) of energy to a charge of 1 coulomb (C):

1 V = 1 JC

Recall that one electron has a charge of 1.60 * 10-19 C. (Section 2.2)

(c) The electrode is losing mass and anions from the salt bridge are flowing into the half-cell.(d) The electrode is gaining mass and anions from the salt bridge are flowing into the half-cell.

Practice Exercise 2The two half-reactions in a voltaic cell are

Zn1s2 ¡ Zn2+1aq2 + 2 e- 1electrode = Zn2ClO3

-1aq2 + 6 H+1aq2 + 6 e- ¡ Cl-1aq2 + 3 H2O1l2 1electrode = Pt2(a) Indicate which reaction occurs at the anode and which at the cathode. (b) Does the zinc electrode gain, lose, or retain the same mass as the reaction proceeds? (c) Does the platinum electrode gain, lose, or retain the same mass as the reaction proceeds? (d) Which electrode is positive?

▲ Figure 20.7 Water analogy for electron flow.

Cathode

Anode

+

−

Flow

of e

lect

rons

Low potential energy

High potential energy

SecTioN 20.4 cell potentials Under Standard conditions 869

The potential difference between the two electrodes of a voltaic cell is called the cell potential, denoted Ecell. Because the potential difference provides the driving force that pushes electrons through the external circuit, we also call it the electromotive (“causing electron motion”) force, or emf. Because Ecell is measured in volts, it is also commonly called the voltage of the cell.

The cell potential of any voltaic cell is positive. The magnitude of the cell poten-tial depends on the reactions that occur at the cathode and anode, the concentra-tions of reactants and products, and the temperature, which we will assume to be 25 °C unless otherwise noted. In this section, we focus on cells that are operated at 25 °C under standard conditions. Recall from Table 19.2 that standard conditions include 1 M concentration for reactants and products in solution and 1 atm pressure for gaseous reactants and products. The cell potential under standard conditions is called either the standard cell potential or standard emf and is denoted E°cell. For the Zn–Cu voltaic cell in Figure 20.5, for example, the standard cell potential at 25 °C is +1.10 V:

Zn1s2 + Cu2+1aq, 1 M2 ¡ Zn2+1aq, 1 M2 + Cu1s2 E°cell = +1.10 V

Recall that the superscript ° indicates standard-state conditions. (Section 5.7)

Give it Some thoughtIf a standard cell potential is E °cell = +0.85 V at 25 °C, is the redox reaction of the cell spontaneous?

Standard Reduction PotentialsThe standard cell potential of a voltaic cell, E°cell, depends on the particular cathode and anode half-cells. We could, in principle, tabulate the standard cell potentials for all pos-sible cathode–anode combinations. However, it is not necessary to undertake this ardu-ous task. Rather, we can assign a standard potential to each half-cell and then use these half-cell potentials to determine E°cell. The cell potential is the difference between two half-cell potentials. By convention, the potential associated with each electrode is cho-sen to be the potential for reduction at that electrode. Thus, standard half-cell potentials are tabulated for reduction reactions, which means they are standard reduction poten-tials, denoted E°red. The standard cell potential, E°cell, is the standard reduction potential of the cathode reaction, E°red (cathode), minus the standard reduction potential of the anode reaction, E°red (anode):

E°cell = E°red 1cathode2 - E°red 1anode2 [20.8]

It is not possible to measure the standard reduction potential of a half-reaction directly. If we assign a standard reduction potential to a certain reference half-reaction, however, we can then determine the standard reduction potentials of other half- reactions relative to that reference value. The reference half-reaction is the reduction of H+1aq2 to H21g2 under standard conditions, which is assigned a standard reduction potential of exactly 0 V:

2 H+1aq, 1 M2 + 2 e- ¡ H21g, 1 atm2 E°red = 0 V [20.9]

An electrode designed to produce this half-reaction is called a standard hydrogen electrode (SHE). An SHE consists of a platinum wire connected to a piece of plati-num foil covered with finely divided platinum that serves as an inert surface for the reaction (Figure 20.8). The SHE allows the platinum to be in contact with both 1 M H+1aq2 and a stream of hydrogen gas at 1 atm. The SHE can operate as either the anode or cathode of a cell, depending on the nature of the other electrode.

Figure 20.9 shows a voltaic cell using an SHE. The spontaneous reaction is the one shown in Figure 20.1, namely, oxidation of Zn and reduction of H+:

Zn1s2 + 2 H+1aq2 ¡ Zn2+1aq2 + H21g2


When the cell is operated under standard conditions, the cell potential is +0.76 V. By using the standard cell potential 1E°cell = 0.76 V2, the defined standard reduction potential of H+1E°red = 0 V2 and Equation 20.8, we can determine the standard reduc-tion potential for the Zn2+>Zn half-reaction:

E°cell = E°red 1cathode2 - E°red 1anode2 +0.76 V = 0 V - E°red 1anode2

E°red 1anode2 = -0.76 V

Thus, a standard reduction potential of -0.76 V can be assigned to the reduction of Zn2+ to Zn:

Zn2+1aq, 1 M2 + 2 e- ¡ Zn1s2 E°red = -0.76 V

▲ Figure 20.8 the standard hydrogen electrode (SHe) is used as a reference electrode.

Reduction

Oxidation

e−

e−

e−

e−

H+ ionPt atom H2 molecule

H2 molecule H+ ion

SHE as cathode(H+ reduced to H2)

SHE as anode(H2 oxidized to H+)

0.76 V

Zn2+

Na+

H+NO3−

NO3−

NO3−

NO3−

Voltmeter

−Znanode

2 H+(aq) + 2 e− H2(g)Zn(s) Zn2+(aq) + 2 e−

Cathodehalf-cell(standardhydrogenelectrode, SHE)Anode

half-cell

H2(g)

e− e−

+

▲ Figure 20.9 A voltaic cell using a standard hydrogen electrode (SHe). The anode half-cell is Zn metal in a Zn1NO3221aq2 solution, and the cathode half-cell is the SHE in a HNO31aq2 solution.

GO FiGuREWhy do Na+ ions migrate into the cathode half-cell as the cell reaction proceeds?


We write the reaction as a reduction even though the Zn reaction in Figure 20.9 is an oxidation. Whenever we assign an electrical potential to a half-reaction, we write the reaction as a reduction. Half-reactions, however, are reversible, being able to oper-ate as either reductions or oxidations. Consequently, half-reactions are sometimes written using two arrows 1∆2 between reactants and products, as in equilibrium reactions.

The standard reduction potentials for other half-reactions can be determined in a fashion analogous to that used for the Zn2+>Zn half-reaction. ▼ Table 20.1 lists some standard reduction potentials; a more complete list is found in Appendix E. These stan-dard reduction potentials, often called half-cell potentials, can be combined to calculate E°cell values for a large variety of voltaic cells.

Give it Some thoughtFor the half-reaction Cl21g2 + 2 e- ¡ 2 Cl-1aq2, what are the standard conditions for the reactant and product?

Because electrical potential measures potential energy per electrical charge, stan-dard reduction potentials are intensive properties. (Section 1.3) In other words, if we increase the amount of substances in a redox reaction, we increase both the energy and the charges involved, but the ratio of energy (joules) to electrical charge (coulombs) remains constant 1V = J>C2. Thus, changing the stoichiometric coeffi-cient in a half-reaction does not affect the value of the standard reduction potential.

Table 20.1 standard Reduction Potentials in Water at 25 °c

E°red1V2 Reduction half-Reaction

+2.87 F21g2 + 2 e- ¡ 2 F-1aq2+1.51 MnO4

-1aq2 + 8 H+1aq2 + 5 e- ¡ Mn2+1aq2 + 4 H2O1l2+1.36 Cl21g2 + 2 e- ¡ 2 Cl-1aq2+1.33 Cr2O7

2-1aq2 + 14 H+1aq2 + 6 e- ¡ 2 Cr3+1aq2 + 7 H2O1l2+1.23 O21g2 + 4 H+1aq2 + 4 e- ¡ 2 H2O1l2+1.06 Br21l2 + 2 e- ¡ 2 Br-1aq2+0.96 NO3

-1aq2 + 4 H+1aq2 + 3 e- ¡ NO1g2 + 2 H2O1l2+0.80 Ag+1aq2 + e- ¡ Ag1s2+0.77 Fe3+1aq2 + e- ¡ Fe2+1aq2+0.68 O21g2 + 2 H+1aq2 + 2 e- ¡ H2O21aq2+0.59 MnO4

-1aq2 + 2 H2O1l2 + 3 e- ¡ MnO21s2 + 4 OH-1aq2+0.54 l21s2 + 2 e- ¡ 2 l-1aq2+0.40 O21g2 + 2 H2O1l2 + 4 e- ¡ 4 OH-1aq2+0.34 Cu2+1aq2 + 2 e- ¡ Cu1s20 [defined] 2 H+1aq2 + 2 e- ¡ H21g2-0.28 Ni2+1aq2 + 2 e- ¡ Ni1s2-0.44 Fe2+ 1aq2 + 2 e- ¡ Fe1s2-0.76 Zn2+1aq2 + 2 e- ¡ Zn1s2-0.83 2 H2O1l2 + 2 e- ¡ H21g2 + 2 OH-1aq2-1.66 Al3+1aq2 + 3 e- ¡ Al1s2-2.71 Na+1aq2 + e- ¡ Na1s2-3.05 Li+1aq2 + e- ¡ Li1s2


For example, E°red for the reduction of 10 mol Zn2+ is the same as that for the reduction of 1 mol Zn2+:

10 Zn2+1aq, 1 M2 + 20 e- ¡ 10 Zn1s2 E°red = -0.76 V

sOluTiOnAnalyze We are given E°cell and E°red for Zn2+ and asked to calculate E°red for Cu2+.Plan In the voltaic cell, Zn is oxidized and is therefore the anode. Thus, the given E°red for Zn2+ is E°red (anode). Because Cu2+ is reduced, it is in the cathode half-cell. Thus, the unknown reduction potential for Cu2+ is E°red (cathode). Knowing E°cell and E°red (anode), we can use Equation 20.8 to solve for E °red (cathode).Solve

E°cell = E°red 1cathode2 - E°red 1anode2 1.10 V = E°red 1cathode2 - 1-0.76 V2

E°red 1cathode2 = 1.10 V - 0.76 V = 0.34 V

Check This standard reduction potential agrees with the one listed in Table 20.1.Comment The standard reduction potential for Cu2+ can be repre-sented as E°Cu2+ = 0.34 V and that for Zn2+ as E°Zn2+ = -0.76 V. The subscript identifies the ion that is reduced in the reduction half-reaction.

samPlE ExERcisE 20.5 calculating E°red from E°cell

For the Zn–Cu2+ voltaic cell shown in Figure 20.5, we haveZn1s2 + Cu2+1aq, 1 M2 ¡ Zn2+1aq, 1 M2 + Cu1s2 E°cell = 1.10 V

Given that the standard reduction potential of Zn2+ to Zn(s) is -0.76 V, calculate the E°red for the reduction of Cu2+ to Cu:

Cu2+1aq, 1 M2 + 2 e- ¡ Cu1s2

Practice Exercise 1A voltaic cell based on the reaction 2 Eu2+1aq2 + Ni2+1aq2 ¡ 2 Eu3+1aq2 + Ni1s2 generates E°cell = 0.07 V. Given the standard reduction potential of Ni2+ given in Table 20.1 what is the standard reduction potential for the reaction Eu3+1aq2 + e- ¡ Eu2+1aq2? (a) -0.35 V, (b) 0.35 V, (c) -0.21 V, (d) 0.21 V, (e) 0.07 V.

Practice Exercise 2The standard cell potential is 1.46 V for a voltaic cell based on the following half-reactions:

In+1aq2 ¡ In3+1aq2 + 2 e-

Br21l2+2 e- ¡ 2 Br-1aq2Using Table 20.1, calculate E°red for the reduction of In3+ to In+.

Use Table 20.1 to calculate E °cell for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction

Cr2O72-1aq2 + 14 H+1aq2 + 6 I-1aq2 ¡ 2 Cr3+1aq2 + 3 I21s2 + 7 H2O1l2

sOluTiOnAnalyze We are given the equation for a redox reaction and asked to use data in Table 20.1 to cal-culate the standard cell potential for the associated voltaic cell.Plan Our first step is to identify the half-reactions that occur at the cathode and anode, which we did in Sample Exercise 20.4. Then we use Table 20.1 and Equation 20.8 to calculate the standard cell potential.Solve The half-reactions are

Cathode: Cr2 O72-1aq2 + 14 H+1aq2 + 6 e- ¡ 2 Cr3+1aq2 + 7 H2O1l2

Anode: 6 I-1aq2 ¡ 3 I21s2 + 6 e-

According to Table 20.1, the standard reduction potential for the reduction of Cr2O72- to Cr3+

is +1.33 V and the standard reduction potential for the reduction of I2 to I- (the reverse of the oxidation half-reaction) is +0.54 V. We use these values in Equation 20.8:

E°cell = E°red 1cathode2 - E°red 1anode2 = 1.33 V - 0.54 V = 0.79 V

samPlE ExERcisE 20.6 calculating E°cell from E°red


For each half-cell in a voltaic cell, the standard reduction potential provides a measure of the tendency for reduction to occur: The more positive the value of E°red, the greater the tendency for reduction under standard conditions. In any voltaic cell operat-ing under standard conditions, the E°red value for the reaction at the cathode is more positive than the E°red value for the reaction at the anode. Thus, electrons flow sponta-neously through the external circuit from the electrode with the more negative value of E°red to the electrode with the more positive value of E°red. ▶ Figure 20.10 graphically illustrates the relationship between the standard reduction potentials for the two half-reactions in the Zn–Cu voltaic cell of Figure 20.5.

Give it Some thoughtThe standard reduction potential of Ni2+1aq2 is E °red = -0.28 V and that of Fe2+1aq2 is E °red = -0.44 V. In a Ni–Fe voltaic cell which electrode is the cathode, Ni or Fe?

Although we must multiply the iodide half-reaction by 3 to obtain a balanced equation, we do not multiply the E°red value by 3. As we have noted, the standard reduction potential is an inten-sive property and so is independent of the stoichiometric coefficients.Check The cell potential, 0.79 V, is a positive number. As noted earlier, a voltaic cell must have a positive potential.

Practice Exercise 1Using the data in Table 20.1 what value would you calculate for the standard emf 1E°cell2 for a vol-taic cell that employs the overall cell reaction 2 Ag+1aq2 + Ni1s2 ¡ 2 Ag1s2 + Ni2+1aq2?(a) +0.52 V, (b) -0.52 V, (c) +1.08 V, (d) -1.08 V, (e) +0.80 V.

Practice Exercise 2Using data in Table 20.1, calculate the standard emf for a cell that employs the overall cell reaction 2 Al1s2 + 3 I21s2 ¡ 2 Al3+1aq2 + 6 I-1aq2.

▲ Figure 20.10 Half-cell potentials and standard cell potential for the Zn–cu voltaic cell.

Morepositive

E°cell = (+0.34) − (−0.76) = +1.10 V

+0.34

−0.76

E° re

d (V

)

Zn Zn2+ + 2 e−

Cu2+ + 2 e− Cu

Cathode

Anode

A voltaic cell is based on the two standard half-reactions

Cd2+1aq2 + 2 e- ¡ Cd1s2Sn2+1aq2 + 2 e- ¡ Sn1s2

Use data in Appendix E to determine (a) which half-reaction occurs at the cathode and which occurs at the anode and (b) the standard cell potential.

sOluTiOnAnalyze We have to look up E°red for two half-reactions. We then use these values first to deter-mine the cathode and the anode and then to calculate the standard cell potential, E°cell.Plan The cathode will have the reduction with the more positive E°red value, and the anode will have the less positive E°red. To write the half-reaction at the anode, we reverse the half-reaction written for the reduction, so that the half-reaction is written as an oxidation.Solve

(a) According to Appendix E, E°red1Cd2+>Cd2 = -0.403 V and E°red1Sn2+>Sn2 = -0.136 V. The standard reduction potential for Sn2+ is more positive (less negative) than that for Cd2+. Hence, the reduction of Sn2+ is the reaction that occurs at the cathode:

Cathode: Sn2+1aq2 + 2 e- ¡ Sn1s2 The anode reaction, therefore, is the loss of electrons by Cd:

Anode: Cd1s2 ¡ Cd2+1aq2 + 2 e-

(b) The cell potential is given by the difference in the standard reduction potentials at the cath-ode and anode (Equation 20.8):

E°cell = E°red 1cathode2 - E°red 1anode2 = 1-0.136 V2 - 1-0.403 V2 = 0.267 V

samPlE ExERcisE 20.7 determining half-Reactions at Electrodes

and calculating cell Potentials


Strengths of Oxidizing and Reducing AgentsTable 20.1 lists half-reactions in the order of decreasing tendency to undergo reduction. For example, F2 is located at the top of the table, having the most positive value for E°red. Thus, F2 is the most easily reduced species in Table 20.1 and therefore the strongest oxidizing agent listed.

Among the most frequently used oxidizing agents are the halogens, O2, and oxy-anions such as MnO4

-, Cr2O72-, and NO3

-, whose central atoms have high positive oxi-dation states. As seen in Table 20.1, all these species have large positive values of E°red and therefore easily undergo reduction.

The lower the tendency for a half-reaction to occur in one direction, the greater the tendency for it to occur in the opposite direction. Thus, the half-reaction with the most negative reduction potential in Table 20.1 is the one most easily reversed and run as an oxidation. Being at the bottom of Table 20.1, Li+1aq2 is the most difficult species in the list to reduce and is therefore the poorest oxidizing agent listed. Although Li+1aq2 has little tendency to gain electrons, the reverse reaction, oxidation of Li(s) to Li+1aq2, is highly favorable. Thus, Li is the strongest reducing agent among the substances listed in Table 20.1. (Note that, because Table 20.1 lists half-reactions as reductions, only the substances on the reactant side of these equations can serve as oxidizing agents; only those on the product side can serve as reducing agents.)

Commonly used reducing agents include H2 and the active metals, such as the alkali metals and the alkaline earth metals. Other metals whose cations have negative E°red values—Zn and Fe, for example—are also used as reducing agents. Solutions of reducing agents are difficult to store for extended periods because of the ubiquitous presence of O2, a good oxidizing agent.

The information contained in Table 20.1 is summarized graphically in ▶ Figure 20.11. For the half-reactions at the top of Table 20.1 the substances on the reactant side of the equation are the most readily reduced species in the table and are therefore the stron-gest oxidizing agents. Substances on the product side of these reactions are the most dif-ficult to oxidize and so are the weakest reducing agents in the table. Thus, Figure 20.11 shows F21g2 as the strongest oxidizing agent and F-1aq2 as the weakest reducing agent. Conversely, the reactants in half-reactions at the bottom of Table 20.1, such as Li+1aq2 are the most difficult to reduce and so are the weakest oxidizing agents, while the prod-ucts of these reactions, such as Li(s), are the most readily oxidized species in the table and so are the strongest reducing agents.

Notice that it is unimportant that the E°red values of both half-reactions are negative; the negative values merely indicate how these reductions compare to the reference reaction, the reduction of H+1aq2.

Check The cell potential is positive, as it must be for a voltaic cell.

Practice Exercise 1Consider three voltaic cells, each similar to the one shown in Figure 20.5. In each voltaic cell, one half-cell contains a 1.0 M Fe1NO3221aq2 solution with an Fe electrode. The contents of the other half-cells are as follows:

Cell 1: a 1.0 M CuCl21aq2 solution with a Cu electrodeCell 2: a 1.0 M NiCl21aq2 solution with a Ni electrodeCell 3: a 1.0 M ZnCl21aq2 solution with a Zn electrode

In which voltaic cell(s) does iron act as the anode? (a) Cell 1, (b) Cell 2, (c) Cell 3, (d) Cells 1 and 2, (e) All three cells.

Practice Exercise 2A voltaic cell is based on a Co2+>Co half-cell and a AgCl>Ag half-cell.(a) What half-reaction occurs at the anode? (b) What is the standard cell potential?


This inverse relationship between oxidizing and reducing strength is similar to the inverse relationship between the strengths of conjugate acids and bases. (Section 16.2 and Figure 16.3)

Incr

easi

ng s

tren

gth

of o

xid

izin

g ag

ent

Incr

easi

ng s

tren

gth

of r

educ

ing

agen

t

F2(g) + 2 e− 2 F−(aq)

Cl2(g) + 2 e− 2 Cl−(aq)

Al3+(aq) + 3 e− Al(s)

H2(g)

Li+(aq) + e−

2 H+(aq) + 2 e−

Li(s)

Easiest to reduce,strongest oxidizing agent

Most dif�cult to oxidize,weakest reducing agent

Most positive values of Er̊ed

Most negative values of Er̊ed

Most dif�cult to reduce,weakest oxidizing agent

Easiest to oxidize,strongest reducing agent

▲ Figure 20.11 relative strengths of oxidizing and reducing agents.

GO FiGuRECan an acidic solution oxidize a piece of aluminum?

Using Table 20.1, rank the following ions in the order of increasing strength as oxidizing agents: NO3

-1aq2, Ag+1aq2, Cr2O72-.

sOluTiOnAnalyze We are asked to rank the abilities of several ions to act as oxidizing agents.Plan The more readily an ion is reduced (the more positive its E°red value), the stronger it is as an oxidizing agent.Solve From Table 20.1, we have

NO3-1aq2 + 4 H+1aq2 + 3 e- ¡ NO1g2 + 2 H2O1l2 E°red = +0.96 V

Ag+1aq2 + e- ¡ Ag1s2 E°red = +0.80 V

Cr2O72-1aq2 + 14 H+1aq2 + 6 e- ¡ 2 Cr3+1aq2 + 7 H2O1l2 E°red = +1.33 V

samPlE ExERcisE 20.8 determining Relative strengths

of Oxidizing agents


20.5 | Free Energy and Redox Reactions

We have observed that voltaic cells use spontaneous redox reactions to produce a posi-tive cell potential. Given half-cell potentials, we can determine whether a given redox reaction is spontaneous. In this endeavor, we can use a form of Equation 20.8 that de-scribes redox reactions in general, not just reactions in voltaic cells:

E° = E°red 1reduction process2 - E°red 1oxidation process2 [20.10]

In writing the equation this way, we have dropped the subscript “cell” to indicate that the calculated emf does not necessarily refer to a voltaic cell. Also, we have generalized the stan-dard reduction potentials by using the general terms reduction and oxidation rather than the terms specific to voltaic cells, cathode and anode. We can now make a general statement about the spontaneity of a reaction and its associated emf, E: A positive value of E indicates a spontaneous process; a negative value of E indicates a nonspontaneous process. We use E to represent the emf under nonstandard conditions and E° to indicate the standard emf.

Because the standard reduction potential of Cr2O72- is the most positive, Cr2O7

2- is the strongest oxidizing agent of the three. The rank order is Ag+ 6 NO3

- 6 Cr2O72-.

Practice Exercise 1Based on the data in Table 20.1 which of the following species would you expect to be the strongest oxidizing agent? (a) Cl-1aq2, (b) Cl21g2, (c) O21g2, (d) H+1aq2, (e) Na+1aq2.

Practice Exercise 2Using Table 20.1, rank the following species from the strongest to the weakest reducing agent: I- 1aq), Fe1s2, Al1s2.

Solve

(a) We first must identify the oxidation and reduction half-reactions that when combined give the overall reaction.

sOluTiOnAnalyze We are given two reactions and must determine whether each is spontaneous.

samPlE ExERcisE 20.9 determining spontaneity

Use Table 20.1 to determine whether the following reactions are spontaneous under standard conditions.(a) Cu1s2 + 2 H+1aq2 ¡ Cu2+1aq2 + H21g2(b) Cl21g2 + 2 I-1aq2 ¡ 2 Cl-1aq2 + I21s2

Equation 20.10 to calculate the standard emf, E°, for the reaction. If a reaction is spontaneous, its standard emf must be a positive number.

Plan To determine whether a redox reaction is spontaneous under standard conditions, we first need to write its reduction and oxidation half-reactions. We can then use the standard reduction potentials and

Reduction: 2 H+1aq2 + 2 e- ¡ H21g2 Oxidation: Cu1s2 ¡ Cu2+1aq2 + 2 e-

We look up standard reduction potentials for both half-reactions and use them to calculate E° using Equation 20.10:

E° = E°red 1reduction process2 - E°red 1oxidation process2 = 10 V2 - 10.34 V2 = -0.34 V

Because E° is negative, the reaction is not spontaneous in the direction written. Copper metal does not react with acids as written in Equation (a). The reverse reaction, however, is sponta-neous and has a positive E° value: Cu2+1aq2 + H21g2 ¡ Cu1s2 + 2 H+1aq2 E° = +0.34 V

Thus, Cu2+ can be reduced by H2.(b) We follow a procedure analogous to that in (a): Reduction: Cl21g2 + 2 e- ¡ 2 Cl-1aq2

Oxidation: 2 I-1aq2 ¡ I21s2 + 2 e-

SecTioN 20.5 Free energy and redox reactions 877

We can use standard reduction potentials to understand the activity series of met-als. (Section 4.4) Recall that any metal in the activity series (Table 4.5) is oxidized by the ions of any metal below it. We can now recognize the origin of this rule based on standard reduction potentials. The activity series is based on the oxidation reactions of the metals, ordered from strongest reducing agent at the top to weakest reducing agent at the bottom. (Thus, the ordering is inverted relative to that in Table 20.1.) For example, nickel lies above silver in the activity series, making nickel the stronger reduc-ing agent. Because a reducing agent is oxidized in any redox reaction, nickel is more easily oxidized than silver. In a mixture of nickel metal and silver cations, therefore, we expect a displacement reaction in which the silver ions are displaced in the solution by nickel ions:

Ni1s2 + 2 Ag+1aq2 ¡ Ni2+1aq2 + 2 Ag1s2In this reaction Ni is oxidized and Ag+ is reduced. Therefore, the standard emf for the reaction is

E° = E°red 1Ag+>Ag2 - E°red1Ni2+>Ni2 = 1+0.80 V2 - 1-0.28 V2 = +1.08 V

The positive value of E° indicates that the displacement of silver by nickel resulting from oxidation of Ni metal and reduction of Ag+ is a spontaneous process. Remember that although we multiply the silver half-reaction by 2, the reduction potential is not multiplied.

Give it Some thoughtBased on their relative positions in Table 4.5, which will have a more positive standard reduction potential, Sn2+ or Ni2+?

Emf, Free Energy, and the Equilibrium ConstantThe change in the Gibbs free energy, ∆G, is a measure of the spontaneity of a process that occurs at constant temperature and pressure. (Section 19.5) The emf, E, of a redox reaction also indicates whether the reaction is spontaneous. The relationship be-tween emf and the free-energy change is ∆G = -nFE [20.11]In this equation, n is a positive number without units that represents the number of moles of electrons transferred according to the balanced equation for the reaction, and F is the Faraday constant, named after Michael Faraday (▶ Figure 20.12):

F = 96,485 C>mol = 96,485 J>V@molThe Faraday constant is the quantity of electrical charge on 1 mol of electrons.

In this case

Because the value of E° is positive, this reaction is spontaneous and could be used to build a voltaic cell.

Practice Exercise 1Which of the following elements is capable of oxidizing Fe2+1aq2 ions to Fe3+1aq2 ions: chlorine, bromine, iodine? (a) I2, (b) Cl2, (c) Cl2 and I2, (d) Cl2 and Br2, (e) all three elements.

Practice Exercise 2Using the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions:(a) I21s2 + 5 Cu2+1aq2 + 6 H2O1l2 ¡ 2 IO3

-1aq2 + 5 Cu1s2 + 12 H+1aq2(b) Hg2+1aq2 + 2 I-1aq2 ¡ Hg1l2 + I21s2(c) H2SO31aq2 + 2 Mn1s2 + 4 H+1aq2 ¡ S1s2 + 2 Mn2+1aq2 + 3 H2O1l2

E° = 11.36 V2 - 10.54 V2 = +0.82 V

▲ Figure 20.12 Michael Faraday. Faraday (1791–1867) was born in England, a child of a poor blacksmith. At the age of 14 he was apprenticed to a bookbinder who gave him time to read and to attend lectures. In 1812 he became an assistant in Humphry Davy’s laboratory at the Royal Institution. He succeeded Davy as the most famous and influential scientist in England, making an amazing number of important discoveries, including his formulation of the quantitative relationships between electrical current and the extent of chemical reaction in electrochemical cells.


The units of ∆G calculated with Equation 20.11 are J>mol. As in Equation 19.19, we use “per mole” to mean per mole of reaction as indicated by the coefficients in the balanced equation. (Section 19.7)

Because both n and F are positive numbers, a positive value of E in Equation 20.11 leads to a negative value of ∆G. Remember: A positive value of E and a negative value of ∆G both indicate a spontaneous reaction. When the reactants and products are all in their standard states, Equation 20.11 can be modified to relate ∆G° and E°. ∆G° = -nFE° [20.12]

Because ∆G° is related to the equilibrium constant, K, for a reaction by the expression ∆G° = -RT ln K (Equation 19.20), we can relate E° to K by solving Equation 20.12 for E° and then substituting the Equation 19.20 expression for ∆G°.

E° =∆G°-nF

=-RT ln K

-nF=

RTnF

ln K [20.13]

◀ Figure 20.13 summarizes the relationships among E°, ∆G°, and K.

∆G°∆G° = −nFE°∆G° = − RT ln K

E°

KE° = RT ln KnF

▲ Figure 20.13 relationships of E °, �G, and K. Any one of these important parameters can be used to calculate the other two. The signs of E° and ∆G determine the direction in which the reaction proceeds under standard conditions. The magnitude of K determines the relative amounts of reactants and products in an equilibrium mixture.

GO FiGuREWhat does the variable n represent in the ∆G ° and E ° equations?

(a) Use the standard reduction potentials in Table 20.1 to calculate the standard free-energy change, ∆G°, and the equilibrium constant, K, at 298 K for the reaction

4 Ag1s2 + O21g2 + 4 H+1aq2 ¡ 4 Ag+1aq2 + 2 H2O1l2(b) Suppose the reaction in part (a) is written

2 Ag1s2 +12

O21g2 + 2 H+1aq2 ¡ 2 Ag+1aq2 + H2O1l2What are the values of E°, ∆G°, and K when the reaction is written in this way?

sOluTiOnAnalyze We are asked to determine ∆G° and K for a redox reaction, using standard reduction potentials.Plan We use the data in Table 20.1 and Equation 20.10 to determine E° for the reaction and then use E° in Equation 20.12 to calculate ∆G°. We can then use either Equation 19.20 or Equation 20.13 to calculate K.Solve

(a) We first calculate E° by breaking the equation into two half-reactions and obtaining E°red val-ues from Table 20.1 (or Appendix E):

Reduction: O21g2 + 4 H+1aq2 + 4 e- ¡ 2 H2O1l2 E°red = +1.23 VOxidation: 4 Ag1s2 ¡ 4 Ag+1aq2 + 4 e- E°red = +0.80 V

Even though the second half-reaction has 4 Ag, we use the E°red value directly from Table 20.1 because emf is an intensive property.

Using Equation 20.10, we haveE ° = 11.23 V2 - 10.80 V2 = 0.43 V

The half-reactions show the transfer of four electrons. Thus, for this reaction n = 4. We now use Equation 20.12 to calculate ∆G°:

∆G° = -nFE° = -142196,485 J>V@mol21+0.43 V2 = -1.7 * 105 J>mol = -170 kJ>mol

Now we need to calculate the equilibrium constant, K, using ∆G° = RT ln K. Because ∆G° is a large negative number, which means the reaction is thermodynamically very favorable, we expect K to be large.

∆G° = -RT ln K -1.7 * 105 J>mol = -18.314 J>K mol2 1298 K2 ln K

ln K =-1.7 * 105J>mol

-18.314 J>K mol21298 K2 ln K = 69

K = 9 * 1029

samPlE ExERcisE 20.10 using standard Reduction Potentials

to calculate ∆G° and K

SecTioN 20.5 Free energy and redox reactions 879

(b) The overall equation is the same as that in part (a), multiplied by 12

. The half-reactions are

Reduction: 12 O21g2 + 2 H+1aq2 + 2 e- ¡ H2O1l2 E°red = +1.23 VOxidation: 2 Ag1s2 ¡ 2 Ag+1aq2 + 2 e- E°red = +0.80 V

The values of E°red are the same as they were in part (a); they are not changed by multiplying

the half-reactions by 12. Thus, E° has the same value as in part (a): E° = +0.43 V. Notice, though, that the value of n has changed to n = 2, which is one-half the value in part (a).

Thus, ∆G° is half as large as in part (a):

∆G° = -122196,485 J>V@mol21+0.43 V2 = -83 kJ>mol

The value of ∆G° is half that in part (a) because the coefficients in the chemical equation are half those in (a).

Now we can calculate K as before: -8.3 * 104 J>mol = -18.314 J>K mol21298 K2 ln K

K = 4 * 1014

Comment E° is an intensive quantity, so multiplying a chemical equation by a certain factor will not affect the value of E°. Multiplying an equation will change the value of n, however, and hence the value of ∆G°. The change in free energy, in units of J/mol of reaction as written, is an exten-sive quantity. The equilibrium constant is also an extensive quantity.

Practice Exercise 1For the reaction

3 Ni2+1aq2 + 2 Cr1OH231s2 + 10 OH-1aq2 ¡ 3 Ni1s2 + 2 CrO42-1aq2 + 8 H2O1l2

∆G ° = +87 kJ>mol. Given the standard reduction potential of Ni2+1aq2 in Table 20.1, what value do you calculate for the standard reduction potential of the half-reaction

CrO42-1aq2 + 4 H2O1l2 + 3 e- ¡ Cr1OH231s2 + 5 OH-1aq2?

(a) -0.43 V (b) -0.28 V (c) 0.02 V (d) -0.13 V (e) -0.15 V

Practice Exercise 2Consider the reaction 2 Ag+1aq2 + H21g2 ¡ 2 Ag1s2 + 2 H+1aq2. Calculate ∆G°f for the Ag+1aq2 ion from the standard reduction potentials in Table 20.1 and the fact that ∆G°f for H21g2, Ag1s2, and H+1aq2 are all zero. Compare your answer with the value given in Appendix C.

If a reaction is not spontaneous, ∆G is positive and E is negative. To force a nonspontaneous reaction to occur in an electrochemical cell, we need to apply an external potential, Eext , that exceeds � Ecell � . For example, if a nonspontaneous process has E = -0.9 V, then the external potential Eext must be greater than +0.9 V in order for the process to occur. We will examine such nonspontaneous processes in Section 20.9.

Electrical work can be expressed in energy units of watts times time. The watt (W) is a unit of electrical power (that is, rate of energy expenditure):

1 W = 1 J>s

Thus, a watt-second is a joule. The unit employed by electric util-ities is the kilowatt-hour (kWh), which equals 3.6 * 106 J:

1 kWh = 11000 W211 h2a 3600 s1 h

b a 1 J>s1 W

b = 3.6 * 106 J


a closer look

electrical Work

For any spontaneous process, ∆G is a measure of the maximum use-ful work, wmax , that can be extracted from the process: ∆G = wmax.

(Section 19.5) Because ∆G = -nFE, the maximum useful elec-trical work obtainable from a voltaic cell is

wmax = -nFEcell [20.14]

Because cell emf, Ecell, is always positive for a voltaic cell, wmax is nega-tive, indicating that work is done by a system on its surroundings, as we expect for a voltaic cell. (Section 5.2)

As Equation 20.14 shows, the more charge a voltaic cell moves through a circuit (that is, the larger nF is) and the larger the emf pushing the electrons through the circuit (that is, the larger E°cell is), the more work the cell can accomplish. In Sample Exercise 20.10, we calculated ∆G° = -170 kJ>mol for the reaction 4 Ag1s2 +O21g2 + 4 H+1aq2 ¡ 4 Ag+1aq2 + 2 H2O1l2. Thus, a voltaic cell utilizing this reaction could perform a maximum of 170 kJ of work in consuming 4 mol Ag, 1 mol O2, and 4 mol H+.


20.6 | cell Potentials under nonstandard conditions

We have seen how to calculate the emf of a cell when the reactants and products are under standard conditions. As a voltaic cell is discharged, however, reactants are con-sumed and products are generated, so concentrations change. The emf progressively drops until E = 0, at which point we say the cell is “dead.” In this section, we examine how the emf generated under nonstandard conditions can be calculated by using an equation first derived by Walther Nernst (1864–1941), a German chemist who estab-lished many of the theoretical foundations of electrochemistry.

The Nernst EquationThe effect of concentration on cell emf can be obtained from the effect of concentra-tion on free-energy change. (Section 19.7) Recall that the free-energy change for any chemical reaction, ∆G, is related to the standard free-energy change for the reac-tion, ∆G°:

∆G = ∆G° + RT ln Q [20.15]

The quantity Q is the reaction quotient, which has the form of the equilibrium-constant expression except that the concentrations are those that exist in the reaction mixture at a given moment. (Section 15.6)

Substituting ∆G = -nFE (Equation 20.11) into Equation 20.15 gives

-nFE = -nFE° + RT ln Q

Solving this equation for E gives the Nernst equation:

E = E° -RTnF

ln Q [20.16]

This equation is customarily expressed in terms of the base-10 logarithm:

E = E° -2.303 RT

nF log Q [20.17]

At T = 298 K, the quantity 2.303 RT>F equals 0.0592, with units of volts, and so the Nernst equation simplifies to

E = E° -0.0592 V

n log Q 1T = 298 K2 [20.18]

We can use this equation to find the emf E produced by a cell under nonstandard con-ditions or to determine the concentration of a reactant or product by measuring E for the cell. For example, consider the following reaction:

Zn1s2 + Cu2+1aq2 ¡ Zn2+1aq2 + Cu1s2In this case n = 2 (two electrons are transferred from Zn to Cu2+), and the standard emf is +1.10 V. (Section 20.4) Thus, at 298 K the Nernst equation gives

E = 1.10 V -0.0592 V

2 log

3Zn2+43Cu2+4 [20.19]

Recall that pure solids are excluded from the expression for Q. (Section 15.6) According to Equation 20.19, the emf increases as 3Cu2+4 increases and as 3Zn2+4 decreases. For example, when 3Cu2+4 is 5.0 M and 3Zn2+4 is 0.050 M, we have

E = 1.10 V -0.0592 V

2 log a 0.050

5.0b

= 1.10 V -0.0592 V

21-2.002 = 1.16 V

SecTioN 20.6 cell potentials Under Nonstandard conditions 881

Thus, increasing the concentration of reactant Cu2+ and decreasing the concentration of product Zn2+ relative to standard conditions increases the emf of the cell relative to standard conditions 1E° = +1.10 V2.

The Nernst equation helps us understand why the emf of a voltaic cell drops as the cell discharges. As reactants are converted to products, the value of Q increases, so the value of E decreases, eventually reaching E = 0. Because ∆G = -nFE (Equation 20.11), it follows that ∆G = 0 when E = 0. Recall that a system is at equi-librium when ∆G = 0. (Section 19.7) Thus, when E = 0, the cell reaction has reached equilibrium, and no net reaction occurs.

In general, increasing the concentration of reactants or decreasing the concentra-tion of products increases the driving force for the reaction, resulting in a higher emf. Conversely, decreasing the concentration of reactants or increasing the concentration of products causes the emf to decrease from its value under standard conditions.

Calculate the emf at 298 K generated by a voltaic cell in which the reaction isCr2O7

2-1aq2 + 14 H+1aq2 + 6 I-1aq2 ¡ 2 Cr3+1aq2 + 3 I21s2 + 7 H2O1l2when

3Cr2O72-4 = 2.0 M, 3H+4 = 1.0 M, 3I-4 = 1.0 M, and 3Cr3+4 = 1.0 * 10-5 M

sOluTiOnAnalyze We are given a chemical equation for a voltaic cell and the concentrations of reactants and products under which it operates. We are asked to calculate the emf of the cell under these nonstandard conditions.Plan To calculate the emf of a cell under nonstandard conditions, we use the Nernst equation in the form of Equation 20.18.Solve We calculate E° for the cell from standard reduction potentials (Table 20.1 or Appendix E). The standard emf for this reaction was calculated in Sample Exercise 20.6: E° = 0.79 V. As that exercise shows, six electrons are transferred from reducing agent to oxidizing agent, so n = 6. The reaction quotient, Q, is

Q =3Cr3+42

3Cr2O72-43H+414 3I-46 =

11.0 * 10-522

12.0211.0214 11.026 = 5.0 * 10-11

Using Equation 20.18, we have

E = 0.79 V - a 0.0592 V6

b log15.0 * 10-112

= 0.79 V - a 0.0592 V6

b1-10.302 = 0.79 V + 0.10 V = 0.89 V

Check This result is qualitatively what we expect: Because the concentration of Cr2O72- (a reac-

tant) is greater than 1 M and the concentration of Cr3+ (a product) is less than 1 M, the emf is greater than E°. Because Q is about 10-10, log Q is about -10. Thus, the correction to E° is about 0.06 * 10>6, which is 0.1, in agreement with the more detailed calculation.

Practice Exercise 1Consider a voltaic cell whose overall reaction is Pb2+1aq2 + Zn1s2 ¡ Pb1s2 + Zn2+1aq2. What is the emf generated by this voltaic cell when the ion concentrations are 3Pb2+4 = 1.5 * 10-3 M and 3Zn2+4 = 0.55 M? (a) 0.71 V, (b) 0.56 V, (c) 0.49 V, (d) 0.79 V, (e) 0.64 V.

Practice Exercise 2For the Zn–Cu voltaic cell depicted in Figure 20.5 would the emf increase, decrease, or stay the same, if you increased the Cu2+1aq2 concentration by adding CuSO4 # 5H2O to the cathode compartment?

samPlE ExERcisE 20.11 cell Potential under nonstandard conditions


Q has the form of the equilibrium constant for the reaction:

Concentration CellsIn the voltaic cells we have looked at thus far, the reactive species at the anode has been different from the reactive species at the cathode. Cell emf depends on concentration, however, so a voltaic cell can be constructed using the same species in both half-cells as long as the concentrations are different. A cell based solely on the emf generated be-cause of a difference in a concentration is called a concentration cell.

An example of a concentration cell is diagrammed in ▶ Figure 20.14(a). One half-cell consists of a strip of nickel metal immersed in a 1.00 * 10-3 M solution of Ni2+1aq2. The other half-cell also has an Ni(s) electrode, but it is immersed in a 1.00 M solution of Ni2+1aq2. The two half-cells are connected by a salt bridge and by an

Solve The cell reaction is

sOluTiOnAnalyze We are given a description of a voltaic cell, its emf, the concentration of Zn2+, and the partial pressure of H2 (both products in the cell reaction). We are asked to calculate the pH of the cathode solution, which we can calculate from the concentration of H+, a reactant.Plan We write the equation for the cell reaction and use standard reduction potentials to calculate E° for the reaction. After determining the value of n from our reaction equation, we solve the Nernst equation, Equation 20.18, for Q. We use the equation for the cell reaction to write an expression for Q that contains 3H+4 to deter-mine 3H+4. Finally, we use 3H+4 to calculate pH.

samPlE ExERcisE 20.12 calculating concentrations in a Voltaic cell

If the potential of a Zn–H+ cell (like that in Figure 20.9) is 0.45 V at 25 °C when 3Zn2+4 = 1.0 M and PH2

= 1.0 atm, what is the pH of the cathode solution?

Zn1s2 + 2 H+1aq2 ¡ Zn2+1aq2 + H21g2The standard emf is E° = E°red 1reduction2 - E°red 1oxidation2

= 0 V - 1-0.76 V2 = +0.76 V

Because each Zn atom loses two electrons, n = 2

Using Equation 20.18, we can solve for Q: 0.45 V = 0.76 V -0.0592 V

2 log Q

Q = 1010.5 = 3 * 1010

Q =3Zn2+4PH2

3H+42 =11.0211.023H+42 = 3 * 1010

Solving for 3H+4, we have 3H+42 =1.0

3 * 1010 = 3 * 10-11

3H+4 = 23 * 10-11 = 6 * 10-6 M

Finally, we use 3H+4 to calculate the pH of the cathode solution.

pH = log 3H+4 = - log 16 * 10-62 = 5.2

Comment A voltaic cell whose cell reaction involves H+ can be used to measure 3H+4 or pH. A pH meter is a specially designed voltaic cell with a voltmeter calibrated to read pH directly. (Section 16.4)

Practice Exercise 1Consider a voltaic cell where the anode half-reaction is Zn(s) ¡ Zn2+1aq2 + 2 e- and the cathode half-reaction is Sn2+1aq2 + 2e- ¡ Sn1s2. What is the concentration of Sn2+ if Zn2+ is 2.5 * 10-3 M and the cell emf is 0.660 V? Use the reduction potentials in Appendix E that are reported to three signifi-cant figures. (a) 3.3 * 10-2 M, (b) 1.9 * 10-4 M, (c) 9.0 * 10-3 M, (d) 6.9 * 10-4 M, (e) 7.6 * 10-3 M.

Practice Exercise 2What is the pH of the solution in the cathode half-cell in Figure 20.9 when PH2

= 1.0 atm, 3Zn2+4 in the anode half-cell is 0.10 M, and the cell emf is 0.542 V?


external wire running through a voltmeter. The half-cell reactions are the reverse of each other:

Anode: Ni1s2 ¡ Ni2+1aq2 + 2 e- E°red = -0.28 V

Cathode: Ni2+1aq2 + 2 e- ¡ Ni1s2 E°red = -0.28 V

Although the standard emf for this cell is zero,

E°cell = E°red 1cathode2 - E°red 1anode2 = 1-0.28 V2 - 1-0.28 V2 = 0 V

the cell operates under nonstandard conditions because the concentration of Ni2+1aq2 is not 1 M in both half-cells. In fact, the cell operates until 3Ni2+4anode = 3Ni2+4cathode. Oxidation of Ni(s) occurs in the half-cell containing the more dilute solution, which means this is the anode of the cell. Reduction of Ni2+1aq2 occurs in the half-cell con-taining the more concentrated solution, making it the cathode. The overall cell reaction is therefore

Anode: Ni1s2 ¡ Ni2+1aq, dilute2 + 2 e-

Cathode: Ni2+1aq, concentrated2 + 2 e- ¡ Ni1s2Overall: Ni2+1aq, concentrated2 ¡ Ni2+1aq, dilute2We can calculate the emf of a concentration cell by using the Nernst equation. For

this particular cell, we see that n = 2. The expression for the reaction quotient for the overall reaction is Q = 3Ni2+4dilute>3Ni2+4concentrated. Thus, the emf at 298 K is

E = E° -0.0592 V

n log Q

= 0 -0.0592 V

2 log

3Ni2+4dilute

3Ni2+4concentrated= -

0.0592 V2

log 1.00 * 10-3 M

1.00 M

= +0.0888 V

This concentration cell generates an emf of nearly 0.09 V even though E° = 0. The difference in concentration provides the driving force for the cell. When the concentra-tions in the two half-cells become the same, Q = 1 and E = 0.

The idea of generating a potential by a concentration difference is the basis for the operation of pH meters. It is also a critical aspect in biology. For example, nerve cells in the

0.0888 V 0.0000 V

[Ni2+] = 1.00 M[Ni2+] = 1.00 × 10−3 M

Nicathode

Salt bridge

(a)

Nianode

[Ni2+] = 0.5 M[Ni2+] = 0.5 M

(b)

Discharge

e− e−

▲ Figure 20.14 concentration cell based on Ni2+ -Ni cell reaction. (a) Concentrations of Ni2+ 1aq2 in the two half-cells are unequal, and the cell generates an electrical current. (b) The cell operates until 3Ni2+ 1aq24 is the same in the two half-cells, at which point the cell has reached equilibrium and the emf goes to zero.

GO FiGuREAssuming that the solutions are made from Ni1NO322, how do the ions migrate as the cell operates?


brain generate a potential across the cell membrane by having different concentrations of ions on the two sides of the membrane. Electric eels use cells called electrocytes that are based on a similar principle to generate short, but intense pulses of electricity to stun prey and dissuade predators. (▲ Figure 20.15). The regulation of the heartbeat in mammals, as discussed in the following “Chemistry and Life” box, is another example of the importance of electrochemistry to living organisms.

▲ Figure 20.15 An electric eel. Differences in ion concentrations, mainly Na+ and K+, in special cells called electrocytes produce an emf on the order of 0.1 V. By connecting thousands of these cells in series these South American fish are able to generate short electric pulses as high as 500 V.

chemistry and life

Heartbeats and electrocardiography

The human heart is a marvel of efficiency and dependability. In a typi-cal day an adult’s heart pumps more than 7000 L of blood through the circulatory system, usually with no maintenance required beyond a sensible diet and lifestyle. We generally think of the heart as a me-chanical device, a muscle that circulates blood via regularly spaced muscular contractions. However, more than two centuries ago, two pioneers in electricity, Luigi Galvani (1729–1787) and Alessandro Volta (1745–1827), discovered that the contractions of the heart are controlled by electrical phenomena, as are nerve impulses throughout the body. The pulses of electricity that cause the heart to beat result from a remarkable combination of electrochemistry and the proper-ties of semipermeable membranes. (Section 13.5)

Cell walls are membranes with variable permeability with respect to a number of physiologically important ions (especially Na+, K+, and Ca2+). The concentrations of these ions are different for the fluids in-side the cells (the intracellular fluid, or ICF) and outside the cells (the

extracellular fluid, or ECF). In cardiac muscle cells, for example, the concentrations of K+ in the ICF and ECF are typically about 135 mil-limolar (mM) and 4 mM, respectively. For Na+, however, the concen-tration difference between the ICF and ECF is opposite to that for K+; typically, 3Na+4ICF = 10 mM and 3Na+4ECF = 145 mM.

The cell membrane is initially permeable to K+ions but is much less so to Na+ and Ca2+. The difference in concentration of K+ ions between the ICF and ECF generates a concentration cell. Even though the same ions are present on both sides of the membrane, there is a potential difference between the two fluids that we can calculate using the Nernst equation with E° = 0. At physiological temperature 137 °C2 the potential in millivolts for moving K+ from the ECF to the ICF is

E = E° -2.30 RT

nF log

3K+4ICF

3K+4ECF

= 0 - 161.5 mV2 log a 135 mM4 mM

b = -94 mV


In essence, the interior of the cell and the ECF together serve as a vol-taic cell. The negative sign for the potential indicates that work is re-quired to move K+ into the ICF.

Changes in the relative concentrations of the ions in the ECF and ICF lead to changes in the emf of the voltaic cell. The cells of the heart that govern the rate of heart contraction are called the pacemaker cells. The membranes of the cells regulate the concentrations of ions in the ICF, allowing them to change in a systematic way. The concentra-tion changes cause the emf to change in a cyclic fashion, as shown in ▼ Figure 20.16. The emf cycle determines the rate at which the heart beats. If the pacemaker cells malfunction because of disease or injury, an artificial pacemaker can be surgically implanted. The artificial

pacemaker contains a small battery that generates the electrical pulses needed to trigger the contractions of the heart.

During the late 1800s, scientists discovered that the electrical im-pulses that cause the contraction of the heart muscle are strong enough to be detected at the surface of the body. This observation formed the basis for electrocardiography, noninvasive monitoring of the heart by using a complex array of electrodes on the skin to measure voltage changes during heartbeats. A typical electrocardiogram is shown in ▼ Figure 20.17. It is quite striking that, although the heart’s major function is the mechanical pumping of blood, it is most easily moni-tored by using the electrical impulses generated by tiny voltaic cells.

▲ Figure 20.16 changes in electrical potential in the human heart. Variation of the electrical potential caused by changes of ion concentrations in the pacemaker cells of the heart.

Time

Ele

ctri

cal p

oten

tial

▲ Figure 20.17 A typical electrocardiogram. The printout records the electrical events monitored by electrodes attached to the body surface.

1 second

Time

emf

Solve Using the Nernst equation, we have

sOluTiOnAnalyze We are given the potential of a concentration cell and the direction in which the current flows. We also have the concentrations or partial pressures of all reactants and products except for 3H+4 in half-cell 1, which is our unknown.Plan We can use the Nernst equation to determine Q and then use Q to calculate the unknown concentration. Because this is a concentration cell, E°cell = 0 V.

samPlE ExERcisE 20.13 determining ph using a concentration cell

A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2= 1.00 atm

and an unknown concentration of H+1aq2. Electrode 2 is a standard hydrogen electrode 1PH2

= 1.00 atm, 3H+4 = 1.00 M2. At 298 K the measured cell potential is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate 3H+4 for the solution at electrode 1. What is the pH of the solution?

0.211 V = 0 -0.0592 V

2 log Q

log Q = -10.211 V2a 20.0592 V

b = -7.13

Q = 10-7.13 = 7.4 * 10-8

Because electrons flow from electrode 1 to electrode 2, electrode 1 is the anode of the cell and electrode 2 is the cathode. The elec-trode reactions are therefore as follows, with the concentration of H+1aq2 in electrode 1 represented with the unknown x:

Electrode 1: H21g, 1.00 atm2 ¡ 2 H+1aq, x M2 + 2 e- E°red = 0Electrode 2: 2 H+1aq; 1.00 M2 + 2 e- ¡ H21g, 1.00 atm2 E°red = 0

Overall: 2 H+1aq; 1.00 M2 ¡ 2 H+1aq, x M2

Thus, Q =3H+ 1aq, x M242

3H+ 1aq, 1.00 M242

=x2

11.0022 = x2 = 7.4 * 10-8

x = 3H+4 = 27.4 * 10-8 = 2.7 * 10-4

At electrode 1, therefore, the pH of the solution is pH = - log3H+4 = - log12.7 * 10-42 = 3.57


Comment The concentration of H+ at electrode 1 is lower than that in electrode 2, which is why electrode 1 is the anode of the cell: The oxidation of H2 to H+1aq2 increases 3H+4 at electrode 1.

Practice Exercise 1A concentration cell constructed from two hydrogen electrodes, both with PH2

= 1.00. One electrode is immersed in pure H2O and the other in 6.0 M hydrochloric acid. What is the emf generated by the cell and what is the identity of the electrode that is immersed in hydrochloric acid? (a) -0.23 V, cathode, (b) 0.46 V, anode, (c) 0.023 V, anode, (d) 0.23 V, cathode, (e) 0.23 V, anode.

Practice Exercise 2A concentration cell is constructed with two Zn1s2-Zn2+1aq2 half-cells. In one half-cell 3Zn2+4 = 1.35 M, and in the other 3Zn2+4 = 3.75 * 10-4 M. (a) Which half-cell is the anode? (b) What is the emf of the cell?

20.7 | Batteries and Fuel cellsA battery is a portable, self-contained electrochemical power source that consists of one or more voltaic cells. For example, the 1.5-V batteries used to power flashlights and many consumer electronic devices are single voltaic cells. Greater voltages can be achieved by using multiple cells, as in 12-V automotive batteries. When cells are connected in series (which means the cathode of one attached to the anode of another), the battery produces a voltage that is the sum of the voltages of the individual cells. Higher voltages can also be achieved by using multiple batteries in series (◀ Figure 20.18). Battery electrodes are marked following the convention of Figure 20.6—plus for cathode and minus for anode.

Although any spontaneous redox reaction can serve as the basis for a voltaic cell, making a commercial battery that has specific performance characteristics requires considerable ingenuity. The substances oxidized at the anode and reduced by the cath-ode determine the voltage, and the usable life of the battery depends on the quantities of these substances packaged in the battery. Usually a barrier analogous to the porous barrier of Figure 20.6 separates the anode and cathode half-cells.

Different applications require batteries with different properties. The battery required to start a car, for example, must be capable of delivering a large electrical cur-rent for a short time period, whereas the battery that powers a heart pacemaker must be very small and capable of delivering a small but steady current over an extended time period. Some batteries are primary cells, meaning they cannot be recharged and must be either discarded or recycled after the voltage drops to zero. A secondary cell can be recharged from an external power source after its voltage has dropped.

As we consider some common batteries, notice how the principles we have dis-cussed so far help us understand these important sources of portable electrical energy.

Lead–Acid BatteryA 12-V lead–acid automotive battery consists of six voltaic cells in series, each producing 2 V. The cathode of each cell is lead dioxide 1PbO22 packed on a lead grid (▶ Figure 20.19). The anode of each cell is lead. Both electrodes are immersed in sulfuric acid.

The reactions that occur during discharge are

▲ Figure 20.18 combining batteries. When batteries are connected in series, as in most flashlights, the total voltage is the sum of the individual voltages.

1.5 V

+

−

+

−

1.5 V

3.0 V

Cathode: PbO21s2 + HSO4-1aq2 + 3 H+1aq2 + 2 e- ¡ PbSO41s2 + 2 H2O1l2

Anode: Pb1s2 + HSO4-1aq2 ¡ PbSO41s2 + H+1aq2 + 2 e-

Overall: PbO21s2 + Pb1s2 + 2 HSO4-1aq2 + 2 H+1aq2 ¡ 2 PbSO41s2 + 2 H2O1l2 [20.20]

The standard cell potential can be obtained from the standard reduction potentials in Appendix E:

E°cell = E°red 1cathode2 - E°red 1anode2 = 1+1.685 V2 - 1-0.356 V2 = +2.041 V

The reactants Pb and PbO2 are the electrodes. Because these reactants are solids, there is no need to separate the cell into half-cells; the Pb and PbO2 cannot come into con-tact with each other unless one electrode touches another. To keep the electrodes from touching, wood or glass-fiber spacers are placed between them (Figure 20.19). Using a reaction whose reactants and products are solids has another benefit. Because solids

SecTioN 20.7 Batteries and Fuel cells 887

are excluded from the reaction quotient Q, the relative amounts of Pb1s2, PbO21s2, and PbSO41s2 have no effect on the voltage of the lead storage battery, helping the battery maintain a relatively constant voltage during discharge. The voltage does vary some-what with use because the concentration of H2SO4 varies with the extent of discharge. As Equation 20.20 indicates, H2SO4 is consumed during the discharge.

A major advantage of the lead–acid battery is that it can be recharged. During recharging, an external source of energy is used to reverse the direction of the cell reac-tion, regenerating Pb(s) and PbO21s2:

2 PbSO41s2 + 2 H2O1l2 ¡ PbO21s2 + Pb1s2 + 2 HSO4-1aq2 + 2 H+1aq2

In an automobile, the alternator provides the energy necessary for recharging the bat-tery. Recharging is possible because PbSO4 formed during discharge adheres to the electrodes. As the external source forces electrons from one electrode to the other, the PbSO4 is converted to Pb at one electrode and to PbO2 at the other.

Alkaline BatteryThe most common primary (nonrechargeable) battery is the alkaline battery (▶ Figure 20.20). The anode is powdered zinc metal immobilized in a gel in contact with a concentrated solution of KOH (hence, the name alkaline battery). The cathode is a mixture of MnO21s2 and graphite, separated from the anode by a porous fabric. The bat-tery is sealed in a steel can to reduce the risk of any of the concentrated KOH escaping.

The cell reactions are complex but can be approximately represented as follows:

Cathode: 2 MnO21s2 + 2 H2O1l2 + 2 e- ¡ 2 MnO1OH21s2 + 2 OH-1aq2Anode: Zn1s2 + 2 OH-1aq2 ¡ Zn1OH221s2 + 2 e-

Nickel–Cadmium and Nickel–Metal Hydride BatteriesThe tremendous growth in high-power-demand portable electronic devices in the last decade has increased the demand for lightweight, readily rechargable batteries. One relatively common rechargeable battery is the nickel–cadmium (nicad) battery. Dur-ing discharge, cadmium metal is oxidized at the anode while nickel oxyhydroxide 3NiO1OH21s24 is reduced at the cathode:

Cathode: 2 NiO1OH21s2 + 2 H2O1l2 + 2 e- ¡ 2 Ni1OH221s2 + 2 OH-1aq2Anode: Cd1s2 + 2 OH-1aq2 ¡ Cd1OH221s2 + 2 e-

As in the lead–acid battery, the solid reaction products adhere to the electrodes, which permits the electrode reactions to be reversed during charging. A single nicad voltaic cell has a voltage of 1.30 V. Nicad battery packs typically contain three or more cells in series to produce the higher voltages needed by most electronic devices.

Although nickel–cadmium batteries have a number of attractive characteristics the use of cadmium as the anode introduces significant limitations. Because cadmium is toxic, these batteries must be recycled. The toxicity of cadmium has led to a decline in their popularity from a peak annual production level of approximately 1.5 billion batter-ies in the early 2000s. Cadmium also has a relatively high density, which increases bat-tery weight, an undesirable characteristic for use in portable devices and electric vehicles. These shortcomings have fueled the development of the nickel–metal hydride (NiMH) battery. The cathode reaction is the same as that for nickel–cadmium batteries, but the anode reaction is very different. The anode consists of a metal alloy, typically with AM5 stoichiometry, where A is lanthanum (La) or a mixture of metals from the lanthanide series, and M is mostly nickel alloyed with smaller amounts of other transition metals. On charging, water is reduced at the anode to form hydroxide ions and hydrogen atoms that are absorbed into the AM5 alloy. When the battery is operating (discharging) the hydrogen atoms are oxidized and the resultant H+ ions react with OH- ions to form H2O.

Lithium-Ion BatteriesCurrently most portable electronic devices, including cell phones and laptop com-puters, are powered by rechargeable lithium-ion (Li-ion) batteries. Because lithium

Lead grid filled withspongy lead (anode)

+−

Lead grid filled withPbO2 (cathode)

H2SO4electrolyte

▲ Figure 20.19 A 12-v automotive lead–acid battery. Each anode/cathode pair in this schematic cutaway produces a voltage of about 2 V. Six pairs of electrodes are connected in series, producing 12 V.

GO FiGuREWhat is the oxidation state of lead in the cathode of this battery?

▲ Figure 20.20 cutaway view of a miniature alkaline battery.

GO FiGuREWhat substance is oxidized as the battery discharges?

Cathode (MnO2plus graphite)Anode

(Zn plus KOH)

+

−

Separator


is a very light element, Li-ion batteries achieve a greater specific energy density—the amount of energy stored per unit mass—than nickel-based batteries. Because Li+ has a very large negative standard reduction potential (Table 20.1). Li-ion batteries produce a higher voltage per cell than other batteries. A Li-ion battery produces a maximum voltage of 3.7 V per cell, nearly three times higher than the 1.3 V per cell that nickel–cadmium and nickel–metal hydride batteries generate. As a result, a Li-ion battery can deliver more power than other batteries of comparable size, which leads to a higher volumetric energy density–the amount of energy stored per unit volume.

The technology of Li-ion batteries is based on the ability of Li+ ions to be inserted into and removed from certain layered solids. In most commercial cells, the anode is made of graphite, which contains layers of sp2 bonded carbon atoms (Figure 12.29(b)). The cathode is made of a transition metal oxide that also has a layered structure, typically lithium cobalt oxide 1LiCoO22. The two electrodes are separated by an elec-trolyte, which functions like a salt bridge by allowing Li+ ions to pass through it. When the cell is being charged, cobalt ions are oxidized and Li+ ions migrate out of LiCoO2 and into the graphite. During discharge, when the battery is producing elec-tricity for use, the Li+ ions spontaneously migrate from the graphite anode through the electrolyte to the cathode, enabling electrons to flow through the external cir-cuit. (▼ Figure 20.21)

Current collector Current collector

AnodeElectrolyte

Cathode

Li+

The graphite anode contains layers of carbon atoms (black spheres). Li+ can move in and out of the space between the layers.

The cathode contains cobalt oxide layers (blue spheres = Co, red spheres = O). Li+ can move in and out of the space between the layers.

3.7 Ve− e−

+−

▲ Figure 20.21 Schematic of a Li-ion battery. When the battery is discharging (operating) Li+ ions move out of the anode and migrate through the electrolyte where they enter the spaces between the cobalt oxide layers, reducing the cobalt ions. To recharge the battery, electrical energy is used to drive the Li+ back to the anode, oxidizing the cobalt ions in the cathode.

GO FiGuREWhen a Li-ion battery is fully discharged the cathode has an empirical formula of LiCoO2. What is the oxidation number of cobalt in this state? Does the oxidation number of the cobalt increase or decrease as the battery charges?

SecTioN 20.7 Batteries and Fuel cells 889

Hydrogen Fuel CellsThe thermal energy released by burning fuels can be converted to electrical energy. The thermal energy may convert water to steam, for instance, which drives a turbine that in turn drives an electrical generator. Typically, a maximum of only 40% of the energy from combustion is converted to electricity in this manner; the remainder is lost as heat. The direct production of electricity from fuels by a voltaic cell could, in principle, yield a higher rate of conversion of chemical energy to electrical energy. Voltaic cells that perform this conversion using conventional fuels, such as H2 and CH4, are called fuel cells. Fuel cells are not batteries because they are not self-contained systems—the fuel must be continuously supplied to generate electricity.

chemistry Put to Work

batteries for Hybrid and electric vehicles

There has been a tremendous growth over the last two decades in the development of electric vehicles. This growth has been driven by a de-sire to reduce the use of fossil fuels and lower emissions. Today both hybrid electric vehicles and fully electric vehicles are commercially available. Hybrid electric vehicles can be powered either by electricity from batteries or by a conventional combustion engine, while fully elec-tric vehicles are powered exclusively by the batteries (▼ Figure 20.22). Hybrid electric vehicles can be further divided into plug-in hybrids which require the owner to charge the battery by plugging it into a con-ventional outlet, or regular hybrids which use regenerative braking and power from the combustion engine to charge the batteries.

Among the many technological advances needed to make electric vehicles practical, none is more important than advances in battery technology. Batteries for electric vehicles must have a high specific energy density, to reduce the weight of the car, as well as a high volu-metric energy density, to minimize the space needed for the battery pack. A plot of energy densities for various types of rechargeable bat-teries is shown in ▶ Figure 20.23. The lead–acid batteries used in gasoline-powered automobiles are reliable and inexpensive, but their energy densities are far too low for practical use in an electric vehicle. Nickel–metal hydride batteries offer roughly three times higher en-ergy density and until recently were the batteries of choice for com-mercial hybrid vehicles, such as the Toyota Prius.

Fully electric vehicles and plug-in hybrids use Li-ion batter-ies because they offer the highest energy density of all commercially

available batteries. As Li-ion battery technology has advanced, these batteries have started to displace the nickel–metal hydride batteries used in hybrid electric cars. Concern over safety is one factor that has delayed implementation of Li-ion batteries in commercial automo-biles. In rare cases overheating and/or overcharging can cause Li-ion batteries to combust (the most high profile cases occurring in Boeing’s 787 Dreamliner airplanes, Section 7.3). Most electric vehicles now use Li-ion batteries where the LiCoO2 cathode has been replaced by a cathode made from lithium manganese spinel 1LiMn2O42. Bat-teries made with LiMn2O4 cathodes have several advantages. They are not prone to thermal runaway events that can lead to combus-tion, they tend to have longer lifetimes, and manganese is less expen-sive and more environmentally friendly than cobalt. However, they do have one important shortcoming—the capacity of batteries made from LiMn2O4 is only about two-third of that of batteries with LiCoO2 cathodes. Scientists and engineers are intensively looking for new ma-terials that will lead to further improvements in the energy density, cost, lifetime, and safety of batteries.


▲ Figure 20.22 electric automobile. The Tesla Roadster is a fully electric vehicle powered by Li-ion batteries that can go over 200 miles per charge.

▲ Figure 20.23 energy densities of various types of batteries. The higher the volumetric energy density, the smaller the amount of space needed for the batteries. The higher the specific energy density, the smaller the mass of the batteries.” A Watt-hour (W-h) is equivalent to 3.6 * 103 Joules.

Vol

umet

ric

ener

gy d

ensi

ty (W

-h/

L)

Smal

ler

size

Lighter weight

400

350

300

250

200

150

100

50 100 150 200 250Specific energy density (W-h/kg)

50

0

Leadacid

Nicad

NiMH

Li-ion


The most common fuel-cell systems involve the reaction of H21g2 and O21g2 to form H2O1l2. These cells can generate electricity twice as efficiently as the best internal combustion engine. Under acidic conditions, the reactions are

Cathode: O21g2 + 4 H+ + 4 e- ¡ 2 H2O1l2Anode: 2 H21g2 ¡ 4 H+ + 4 e-

Overall: 2 H21g2 + O21g2 ¡ 2 H2O1l2These cells employ hydrogen gas as the fuel and oxygen gas from air as the oxidant and generate about 1 V.

Fuel cells are often named for either the fuel or the electrolyte used. In the hydrogen-PEM fuel cell (the acronym PEM stands for either proton-exchange mem-brane or polymer-electrolyte membrane), the anode and cathode are separated by a membrane that is permeable to protons but not to electrons (▼ Figure 20.24). The mem-brane therefore acts as the salt bridge. The electrodes are typically made from graphite.

The hydrogen-PEM cell operates at around 80 °C. At this temperature the electro-chemical reactions would normally occur very slowly, and so small islands of platinum are deposited on each electrode to catalyze the reactions. The high cost and relative scarcity of platinum is one factor that limits wider use of hydrogen-PEM fuel cells.

In order to power a vehicle, multiple cells must be assembled into a fuel cell stack. The amount of power generated by a stack depends on the number and size of the fuel cells in the stack and on the surface area of the PEM.

Much fuel cell research today is directed toward improving electrolytes and cat-alysts and toward developing cells that use fuels such as hydrocarbons and alcohols, which are less difficult to handle and distribute than hydrogen gas.

O2, H2O outH2 out

Voltmeter

H+

O2 in

Anode CathodePEM

porous membrane

H2 in

H2 O2H+

H+

e−

Electrons flow through an external circuit to produce electricity

H2(g) is oxidized to form H+

Protons produced at the anode migrate through the PEM

Pt catalysts speed reactions at both electrodes.

O2(g) is reduced and reacts with protons to form H2O(g)

+

+

+

▲ Figure 20.24 A hydrogen-PeM fuel cell. The proton-exchange membrane (PEM) allows H+ ions generated by H2 oxidation at the anode to migrate to the cathode, where H2O is formed.

GO FiGuREWhat half-reaction occurs at the cathode?

SecTioN 20.8 corrosion 891

20.8 | corrosionIn this section, we examine the undesirable redox reactions that lead to corrosion of metals. Corrosion reactions are spontaneous redox reactions in which a metal is at-tacked by some substance in its environment and converted to an unwanted compound.

For nearly all metals, oxidation is thermodynamically favorable in air at room tem-perature. When oxidation of a metal object is not inhibited, it can destroy the object. Oxidation can form an insulating protective oxide layer, however, that prevents further reaction of the underlying metal. Based on the standard reduction potential for Al3+, for example, we expect aluminum metal to be readily oxidized. The many aluminum soft-drink and beer cans that litter the environment are ample evidence, however, that aluminum undergoes only very slow chemical corrosion. The exceptional stability of this active metal in air is due to the formation of a thin protective coat of oxide—a hydrated form of Al2O3—on the metal surface. The oxide coat is impermeable to O2 or H2O and so protects the underlying metal from further corrosion.

Magnesium metal is similarly protected, and some metal alloys, such as stainless steel, likewise form protective impervious oxide coats.

Corrosion of Iron (Rusting)The rusting of iron is a familiar corrosion process that carries a significant economic impact. Up to 20% of the iron produced annually in the United States is used to replace iron objects that have been discarded because of rust damage.

Rusting of iron requires both oxygen and water, and the process can be acceler-ated by other factors such as pH, presence of salts, contact with metals more difficult to oxidize than iron, and stress on the iron. The corrosion process involves oxidation and reduction, and the metal conducts electricity. Thus, electrons can move through the metal from a region where oxidation occurs to a region where reduction occurs, as in voltaic cells. Because the standard reduction potential for reduction of Fe2+1aq2 is less positive than that for reduction of O2, Fe1s2 can be oxidized by O21g2:

Cathode: O21g2 + 4 H+1aq2 + 4 e- ¡ 2 H2O1l2 E°red = 1.23 V

Anode: Fe1s2 ¡ Fe2+1aq2 + 2 e- E°red = -0.44 V

A portion of the iron, often associated with a dent or region of strain, can serve as an anode at which Fe is oxidized to Fe2+ (▼ Figure 20.25). The electrons produced in the

Fe oxidized at anode region of metal

1

Electrons from Fe oxidation migrate to region acting as cathode

2

Fe2+ oxidized to Fe3+, rust (Fe2O3) forms

4

O2 reduced at cathode region

3

O2 + 4 H+ + 4 e− 2 H2O

O2 + 2 H2O + 4 e− 4 OH−

Waterdrop

e−

or

O2Fe2+(aq)

Fe Fe2+ + 2 e−

▲ Figure 20.25 corrosion of iron in contact with water. One region of the iron acts as the cathode and another region acts as the anode.

GO FiGuREWhat is the oxidizing agent in this corrosion reaction?


Zinc coating is oxidized (anode)

Iron is not oxidized (cathode)

Zn Zn2+ + 2 e−

O2 + 4 H+ + 4 e− 2 H2O

Zn2+(aq)

e−

O2

Waterdrop

▲ Figure 20.26 cathodic protection of iron in contact with zinc. The standard reduction potentials are E°red, Fe2+ = -0.440 V, E°red, Zn2+ = -0.763 V, making the zinc more readily oxidized.

oxidation migrate through the metal from this anodic region to another portion of the surface, which serves as the cathode where O2 is reduced. The reduction of O2 requires H+, so lowering the concentration of H+ (increasing the pH) makes O2 reduction less favorable. Iron in contact with a solution whose pH is greater than 9 does not corrode.

The Fe2+ formed at the anode is eventually oxidized to Fe3+, which forms the hydrated iron(III) oxide known as rust:*

4 Fe2+1aq2 + O21g2 + 4 H2O1l2 + 2x H2O1l2 ¡ 2 Fe2O3 # x H2O1s2 + 8 H+1aq2Because the cathode is generally the area having the largest supply of O2, rust often deposits there. If you look closely at a shovel after it has stood outside in the moist air with wet dirt adhered to its blade, you may notice that pitting has occurred under the dirt but that rust has formed elsewhere, where O2 is more readily available. The enhanced corrosion caused by the presence of salts is usually evident on autos in areas where roads are heavily salted during winter. Like a salt bridge in a voltaic cell, the ions of the salt provide the electrolyte necessary to complete the electrical circuit.

Preventing Corrosion of IronObjects made of iron are often covered with a coat of paint or another metal to pro-tect against corrosion. Covering the surface with paint prevents oxygen and water from reaching the iron surface. If the coating is broken, however, and the iron exposed to oxygen and water, corrosion begins as the iron is oxidized.

With galvanized iron, which is iron coated with a thin layer of zinc, the iron is protected from corrosion even after the surface coat is broken. The standard reduction potentials are

Fe2+1aq2 + 2 e- ¡ Fe1s2 E°red = -0.44 V

Zn2+1aq2 + 2 e- ¡ Zn1s2 E°red = -0.76 V

Because E°red for Fe2+ is less negative (more positive) than E°red for Zn2+, Zn1s2 is more readily oxidized than Fe(s). Thus, even if the zinc coating is broken and the galvanized iron is exposed to oxygen and water, as in ▼ Figure 20.26, the zinc serves as the anode and is corroded (oxidized) instead of the iron. The iron serves as the cathode at which O2 is reduced.

*Frequently, metal compounds obtained from aqueous solution have water associated with them. For ex-ample, copper(II) sulfate crystallizes from water with 5 mol of water per mole of CuSO4. We represent this substance by the formula CuSO4 # 5H2O. Such compounds are called hydrates. (Section 13.1) Rust is a hydrate of iron(III) oxide with a variable amount of water of hydration. We represent this variable water content by writing the formula Fe2O3 # xH2O.

SecTioN 20.9 electrolysis 893

Protecting a metal from corrosion by making it the cath-ode in an electrochemical cell is known as cathodic protection. The metal that is oxidized while protecting the cathode is called the sacrificial anode. Underground pipelines and stor-age tanks made of iron are often protected against corrosion by making the iron the cathode of a voltaic cell. For example, pieces of a metal that is more easily oxidized than iron, such as magnesium 1E°red = -2.37 V2, are buried near the pipe or storage tank and connected to it by wire (▶ Figure 20.27). In moist soil, where corrosion can occur, the sacrificial metal serves as the anode, and the pipe or tank experiences cathodic protection.

Give it Some thoughtBased on the values in Table 20.1, which of these metals could provide cathodic protection to iron: Al, Cu, Ni, Zn?

20.9 | ElectrolysisVoltaic cells are based on spontaneous redox reactions. It is also possible for nonspontaneous redox reactions to occur, however, by using electrical energy to drive them. For example, electricity can be used to decompose molten sodium chloride into its component elements Na and Cl2. Such processes driven by an outside source of electrical energy are called electrolysis reactions and take place in electrolytic cells.

An electrolytic cell consists of two electrodes immersed either in a molten salt or in a solution. A battery or some other source of electrical energy acts as an electron pump, pushing electrons into one electrode and pulling them from the other. Just as in voltaic cells, the electrode at which reduction occurs is called the cathode, and the electrode at which oxidation occurs is called the anode.

In the electrolysis of molten NaCl, Na+ ions pick up electrons and are reduced to Na at the cathode, ▶ Figure 20.28. As Na+ ions near the cathode are depleted, additional Na+ ions migrate in. Similarly, there is net movement of Cl- ions to the anode where they are oxidized. The electrode reactions for the electrolysis are

Cathode: 2 Na+1l2 + 2 e- ¡ 2 Na1l2Anode: 2 Cl-1l2 ¡ Cl21g2 + 2 e-

Overall: 2 Na+1l2 + 2 Cl-1l2 ¡ 2 Na1l2 + Cl21g2Notice how the energy source is connected to the electrodes in Figure 20.28. The posi-tive terminal is connected to the anode and the negative terminal is connected to the cathode, which forces electrons to move from the anode to the cathode.

Because of the high melting points of ionic substances, the electrolysis of molten salts requires very high temperatures. Do we obtain the same products if we electrolyze the aqueous solution of a salt instead of the molten salt? Frequently the answer is no because water itself might be oxidized to form O2 or reduced to form H2 rather than the ions of the salt.

In our examples of the electrolysis of NaCl, the electrodes are inert; they do not react but merely serve as the surface where oxidation and reduction occur. Several practical applications of electrochemistry, however, are based on active electrodes—electrodes that participate in the electrolysis process. Electroplating, for example, uses electrolysis to deposit a thin layer of one metal on another metal to improve beauty or resistance to corrosion. Examples include electroplating nickel or chromium onto steel and electroplating a precious metal like silver onto a less expensive one.

Figure 20.29 illustrates an electrolytic cell for electroplating nickel onto a piece of steel. The anode is a strip of nickel metal, and the cathode is the steel. The electrodes are

▲ Figure 20.27 cathodic protection of an iron pipe. A mixture of gypsum, sodium sulfate, and clay surrounds the sacrificial magnesium anode to promote conductivity of ions.

Insulatedcopper wire

Solderedconnection

Soilelectrolyte

Iron pipe,cathode

Groundlevel

Sacrificialmagnesium anode

30 cm

Voltage sourceCathodeAnode

2 Cl− Cl2(g) + 2 e−

e−e− + −

2 Na++ 2 e− 2 Na(l)

Na(l)

Molten NaCl

Cl2(g)

Na+Cl−

▲ Figure 20.28 electrolysis of molten sodium chloride. Pure NaCl melts at 801 °C.


immersed in a solution of NiSO41aq2. When an external voltage is applied, reduction occurs at the cathode. The standard reduction potential of Ni2+ 1E°red = -0.28 V2 is

less negative than that of H2O1E°red = -0.83 V2, so Ni2+ is pref-erentially reduced, depositing a layer of nickel metal on the steel cathode.

At the anode, the nickel metal is oxidized. To explain this behavior, we need to compare the substances in contact with the anode, H2O and NiSO41aq2, with the anode material, Ni. For the NiSO41aq2 solution, neither Ni2+ nor SO4

2- can be oxidized (because both already have their elements in their highest possible oxidation state). Both, the H2O solvent and the Ni atoms in the anode, however, can undergo oxidation:

2 H2O1l2 ¡ O21g2 + 4 H+1aq2 + 4 e- E°red = +1.23 VNi1s2 ¡ Ni2+1aq2 + 2 e- E°red = -0.28 V

We saw in Section 20.4 that the half-reaction with the more negative E°red undergoes oxidation more readily. (Remember Fig-ure 20.11: The strongest reducing agents, which are the substances oxidized most readily, have the most negative E°red values.) Thus, it

is the Ni(s), with its E°red = -0.28 V, that is oxidized at the anode rather than the H2O. If we look at the overall reaction, it appears as if nothing has been accomplished. How-ever, this is not true because Ni atoms are transferred from the Ni anode to the steel cathode, plating the steel with a thin layer of nickel atoms.

The standard emf for the overall reaction is

E°cell = E°red 1cathode2 - E°red 1anode2 = 1-0.28 V2 - 1-0.28 V2 = 0

Because the standard emf is zero, only a small emf is needed to cause the transfer of nickel atoms from one electrode to the other.

Quantitative Aspects of ElectrolysisThe stoichiometry of a half-reaction shows how many electrons are needed to achieve an electrolytic process. For example, the reduction of Na+ to Na is a one-electron process:

Na+ + e- ¡ Na

Thus, 1 mol of electrons plates out 1 mol of Na metal, 2 mol of electrons plates out 2 mol of Na metal, and so forth. Similarly, 2 mol of electrons are required to produce 1 mol of Cu from Cu2+, and 3 mol of electrons are required to produce 1 mol of Al from Al3+:

Cu2+ + 2 e- ¡ CuAl3+ + 3 e- ¡ Al

For any half-reaction, the amount of substance reduced or oxidized in an electrolytic cell is directly proportional to the number of electrons passed into the cell.

The quantity of charge passing through an electrical circuit, such as that in an elec-trolytic cell, is generally measured in coulombs. As noted in Section 20.5, the charge on 1 mol of electrons is 96,485 C. A coulomb is the quantity of charge passing a point in a circuit in 1 s when the current is 1 ampere (A). Therefore, the number of coulombs passing through a cell can be obtained by multiplying the current in amperes by the elapsed time in seconds.

coulombs = amperes * seconds [20.21]

◀ Figure 20.30 shows how the quantities of substances produced or consumed in electrolysis are related to the quantity of electrical charge used. The same relation-ships can also be applied to voltaic cells. In other words, electrons can be thought of as “reagents” in electrolysis reactions.

Steelobject

Nickelanode

Ni2+

Ni(s) Ni2+(aq) + 2 e− Ni2+(aq) + 2 e− Ni(s)Anode Cathode

e− e−Voltage source

−+

Ni2+

▲ Figure 20.29 electrolytic cell with an active metal electrode. Nickel dissolves from the anode to form Ni2+1aq2. At the cathode Ni2+1aq2 is reduced and forms a nickel “plate” on the steel cathode.

GO FiGuREWhat E° for this cell?

▲ Figure 20.30 relationship between charge and amount of reactant and product in electrolysis reactions.

Quantity of charge (coulombs) =

current (amperes)× time (seconds)

Moles of electrons

Faradayconstant

Balancedhalf-reaction

Moles ofsubstance oxidized

or reduced

Grams ofsubstance oxidized

or reduced

Formulaweight

SecTioN 20.9 electrolysis 895

Solve First, we calculate the coulombs of electrical charge passed into the electrolytic cell:

samPlE ExERcisE 20.14 Relating Electrical charge and Quantity of Electrolysis

Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.

sOluTiOnAnalyze We are told that AlCl3 is electrolyzed to form Al and asked to calculate the number of grams of Al produced in 1.00 h with 10.0 A.Plan Figure 20.30 provides a roadmap for this problem. Using the current, time, a balanced half-reaction, and the atomic weight of aluminum we can calculate the mass of Al produced.

Coulombs = amperes * seconds = 110.0 A211.00 h2a 3600 sh

b = 3.60 * 104 C

Second, we calculate the number of moles of electrons that pass into the cell: Moles e- = (3.60 * 104 C2a 1 mol e-

96,485 Cb = 0.373 mol e-

Third, we relate number of moles of electrons to number of moles of aluminum formed, using the half-reaction for the reduction of Al3+: Al3+ + 3 e- ¡ Al

Thus, 3 mol of electrons are required to form 1 mol of Al: Moles Al = (0.373 mol e-2a 1 mol Al

3 mol e- b = 0.124 mol Al

Finally, we convert moles to grams:

Because each step involves multiplica-tion by a new factor, we can combine all the steps:

Grams Al = 10.124 mol Al2 a 27.0 g Al1 mol Al

b = 3.36 g Al

Grams Al = 13.60 * 104 C2a 1 mole e-

96,485 Cb a 1 mol Al

3 mol e- b a27.0 g Al1 mol Al

b = 3.36 g Al

Practice Exercise 1How much time is needed to deposit 1.0 g of chromium metal from an aqueous solution of CrCl3 using a current of 1.5 A? (a) 3.8 * 10-2 s, (b) 21 min, (c) 62 min, (d) 139 min, (e) 3.2 * 103 min.

Practice Exercise 2(a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2 e- ¡ Mg. Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 * 103 s. (b) How many seconds would be required to produce 50.0 g of Mg from MgCl2 if the current is 100.0 A?

chemistry Put to Work

electrometallurgy of Aluminum

Many processes used to produce or refine metals are based on elec-trolysis. Collectively these processes are referred to as electrometal-lurgy. Electrometallurgical procedures can be broadly differentiated according to whether they involve electrolysis of a molten salt or of an aqueous solution.

Electrolytic methods using molten salts are important for ob-taining the more active metals, such as sodium, magnesium, and alu-minum. These metals cannot be obtained from aqueous solution because water is more easily reduced than the metal ions. The stan-dard reduction potentials of water under both acidic 1E°red = 0.00 V2 and basic 1E°red = -0.83 V2 conditions are more positive than

those of Na+1E°red = -2.71 V2, Mg2+1E°red = -2.37 V2, and Al3+

1E°red = -1.66 V2.Historically, obtaining aluminum metal has been a challenge. It

is obtained from bauxite ore, which is chemically treated to concen-trate aluminum oxide 1Al2O32. The melting point of aluminum oxide is above 2000 °C, which is too high to permit its use as a molten me-dium for electrolysis.

The electrolytic process used commercially to produce alumi-num is the Hall–Héroult process, named after its inventors, Charles M. Hall and Paul Héroult. Hall (Figure 20.31) began working on the problem of reducing aluminum in about 1885 after he had learned from a professor of the difficulty of reducing ores of very active metals. Before the development of an electrolytic process, aluminum was


obtained by a chemical reduction using sodium or potassium as the reducing agent, a costly procedure that made aluminum metal expen-sive. As late as 1852, the cost of aluminum was $545 per pound, far greater than the cost of gold. During the Paris Exposition in 1855, alu-minum was exhibited as a rare metal, even though it is the third most abundant element in Earth’s crust.

Hall, who was 21 years old when he began his research, utilized handmade and borrowed equipment in his studies and used a wood-shed near his Ohio home as his laboratory. In about a year’s time, he developed an electrolytic procedure using an ionic compound that melts to form a conducting medium that dissolves Al2O3 but does not interfere with the electrolysis reactions. The ionic compound he selected was the relatively rare mineral cryolite 1Na3AlF62. Héroult, who was the same age as Hall, independently made the same discov-ery in France at about the same time. Because of the research of these two unknown young scientists, large-scale production of aluminum became commercially feasible, and aluminum became a common and familiar metal. Indeed, the factory that Hall subsequently built to pro-duce aluminum evolved into Alcoa Corporation.

In the Hall–Héroult process, Al2O3 is dissolved in molten cryo-lite, which melts at 1012 °C and is an effective electrical conductor (▼ Figure 20.32). Graphite rods are employed as anodes and are consumed in the electrolysis:

Anode: C1s2 + 2 O2 - 1l2 ¡ CO21g2 + 4 e-

Cathode: 3e- + Al3+1l2 ¡ Al1l2A large amount of electrical energy is needed in the Hall–Héroult

process with the result that the aluminum industry consumes about 2% of the electrical energy generated in the United States. Because re-cycled aluminum requires only 5% of the energy needed to produce “new” aluminum, considerable energy savings can be realized by in-creasing the amount of aluminum recycled. Approximately 65% of aluminum beverage containers are recycled in the United States.▲ Figure 20.31 charles M. Hall (1863–1914) as a young man.

▲ Figure 20.32 the Hall–Héroult process. Because molten aluminum is denser than the mixture of cryolite 1Na3AlF62 and Al2O3, the metal collects at the bottom of the cell.

Moltenaluminummetal

Carbon-linediron cathode

Al2O3dissolvedin moltencryolite

Graphiteanodes

+

–

inTEGRaTiVE ExERcisE samPlE

The Ksp at 298 K for iron(II) fluoride is 2.4 * 10-6. (a) Write a half-reaction that gives the likely products of the two-electron reduction of FeF21s2 in water. (b) Use the Ksp value and the stan-dard reduction potential of Fe2+1aq2 to calculate the standard reduction potential for the half-reaction in part (a). (c) Rationalize the difference between the reduction potential in part (a) and the reduction potential for Fe2+1aq2.

sOluTiOnAnalyze We are going to combine what we know about equilibrium constants and electrochemis-try to obtain reduction potentials.Plan For (a) we need to determine which ion, Fe2+ or F -, is more likely to be reduced by two electrons and complete the overall reaction FeF2+2 e- ¡ ?. For (b) we need to write the chemical equation associated with the Ksp and see how it relates to E ° for the reduction half-reaction in (a). For (c) we need to compare E° from (b) with the value for the reduc-tion of Fe2+.Solve

(a) Iron(II) fluoride is an ionic substance that consists of Fe2+ and F- ions. We are asked to pre-dict where two electrons could be added to FeF2. We cannot envision adding the electrons to the F- ions to form F2 - , so it seems likely that we could reduce the Fe2+ ions to Fe(s). We therefore predict the half-reaction

FeF21s2 + 2 e- ¡ Fe1s2 + 2 F -1aq2(b) The Ksp for FeF2 refers to the following equilibrium: (Section 17.4)

FeF21s2 ∆ Fe2+1aq2 + 2 F-1aq2 Ksp = 3Fe2+43F-42 = 2.4 * 10-6

Putting concepts Together

chapter Summary and Key Terms 897

chapter summary and Key TermsoxidAtion StAteS And oxidAtion reduction reActionS (introduction And Section 20.1) In this chapter, we have focused on electrochemistry, the branch of chemistry that relates elec-tricity and chemical reactions. Electrochemistry involves oxidation– reduction reactions, also called redox reactions. These reactions involve a change in the oxidation state of one or more elements. In every oxidation–reduction reaction one substance is oxidized (its oxidation state, or number, increases) and one substance is reduced (its oxidation state, or number, decreases). The substance that is oxi-dized is referred to as a reducing agent, or reductant, because it causes the reduction of some other substance. Similarly, the substance that is reduced is referred to as an oxidizing agent, or oxidant, because it causes the oxidation of some other substance.

bALAncinG redox equAtionS (Section 20.2) An oxidization–reduction reaction can be balanced by dividing the reaction into two half-reactions, one for oxidation and one for reduction. A half-reaction is a balanced chemical equation that includes electrons. In oxidation half-reactions the electrons are on the product (right) side of the equa-tion. In reduction half-reactions the electrons are on the reactant (left)

side of the equation. Each half-reaction is balanced separately, and the two are brought together with proper coefficients to balance the elec-trons on each side of the equation, so the electrons cancel when the half-reactions are added.

voLtAic ceLLS (Section 20.3) A voltaic (or galvanic) cell uses a spontaneous oxidation–reduction reaction to generate electricity. In a voltaic cell the oxidation and reduction half-reactions often occur in separate half-cells. Each half-cell has a solid surface called an elec-trode, where the half-reaction occurs. The electrode where oxidation occurs is called the anode, and the electrode where reduction occurs is called the cathode. The electrons released at the anode flow through an external circuit (where they do electrical work) to the cathode. Electri-cal neutrality in the solution is maintained by the migration of ions between the two half-cells through a device such as a salt bridge.

ceLL PotentiALS under StAndArd conditionS (Section 20.4) A voltaic cell generates an electromotive force (emf) that moves the electrons from the anode to the cathode through the external circuit. The origin of emf is a difference in the electrical potential energy of

We were also asked to use the standard reduction potential of Fe2+, whose half-reaction and standard reduction potentials are listed in Appendix E:

Fe2+1aq2 + 2 e- ¡ Fe1s2 E = -0.440 V

According to Hess’s law, if we can add chemical equations to get a desired equation, then we can add their associated thermodynamic state functions, like ∆H or ∆G, to determine the thermodynamic quantity for the desired reaction. (Section 5.6) So we need to consider whether the three equations we are working with can be combined in a similar fashion. No-tice that if we add the Ksp reaction to the standard reduction half-reaction for Fe2+, we get the half-reaction we want: 1. FeF21s2 ¡ Fe2+1aq2 + 2 F-1aq2 2. Fe2+1aq2 + 2 e- ¡ Fe1s2

Overall: 3. FeF21s2 + 2 e- ¡ Fe1s2 + 2 F-1aq2 Reaction 3 is still a half-reaction, so we do see the free electrons. If we knew ∆G° for reactions 1 and 2, we could add them to get ∆G° for reaction 3.

We can relate ∆G° to E° by ∆G° = -nFE° (Equation 20.12) and to K by ∆G° = -RT ln K (Equation 19.20, see also Figure 20.13). Furthermore, we know that K for reaction 1 is the Ksp of FeF2, and we know E° for reaction 2. Therefore, we can calculate ∆G° for reactions 1 and 2:

Reaction 1:∆G° = -RT ln K = -18.314 J>K mol21298 K2 ln12.4 * 10-62 = 3.21 * 104 J>mol

Reaction 2:∆G° = -nFE° = -122196,485 C>mol21-0.440 J>C2 = 8.49 * 104 J>mol

(Recall that 1 volt is 1 joule per coulomb.) Then ∆G° for reaction 3, the one we want, is the sum of the ∆G° values for reactions 1

and 2:3.21 * 104 J>mol + 8.49 * 104 J>mol = 1.17 * 105 J>mol

We can convert this to E° from the relationship ∆G° = -nFE°: 1.17 * 105 J>mol = -122196,485 C>mol2E°

E° =1.17 * 105 J>mol

-122196,485 C>mol2 = -0.606 J>C = -0.606 V

(c) The standard reduction potential for FeF21-0.606 V2 is more negative than that for Fe2+1-0.440 V2, telling us that the reduction of FeF2 is the less favorable process. When FeF2 is reduced, we not only reduce the Fe2+ but also break up the ionic solid. Because this additional energy must be overcome, the reduction of FeF2 is less favorable than the reduc-tion of Fe2+.


the two electrodes in the cell. The emf of a cell is called its cell poten-tial, Ecell, and is measured in volts 11 V = 1 J>C2. The cell potential under standard conditions is called the standard emf, or the standard cell potential, and is denoted E°cell.

A standard reduction potential, E°red, can be assigned for an indi-vidual half-reaction. This is achieved by comparing the potential of the half-reaction to that of the standard hydrogen electrode (SHE), which is defined to have E°red = 0 V and is based on the following half-reaction:

2 H+1aq, 1 M2 + 2 e- ¡ H21g, 1 atm2 E°red = 0 V

The standard cell potential of a voltaic cell is the difference between the standard reduction potentials of the half-reactions that occur at the cathode and the anode:

E°cell = E°red 1cathode2 - E°red 1anode2.

The value of E°cell is positive for a voltaic cell.For a reduction half-reaction, E°red is a measure of the tendency of

the reduction to occur; the more positive the value for E°red, the greater the tendency of the substance to be reduced. Substances that are easily reduced act as strong oxidizing agents; thus, E°red provides a measure of the oxidizing strength of a substance. Substances that are strong oxidizing agents produce products that are weak reducing agents and vice versa.

Free enerGy And redox reActionS (Section 20.5) The emf, E, is related to the change in the Gibbs free energy, ∆G = -nFE, where n is the number of moles of electrons transferred during the redox process and F is the Faraday constant, defined as the quantity of electrical charge on one mole of electrons: F = 96,485 C>mol. Because E is related to ∆G, the sign of E indicates whether a redox process is spontaneous: E 7 0 indicates a spontaneous process, and E 6 0 indicates a nonspontaneous one. Because ∆G is also related to the equilibrium constant for a reaction 1∆G ° = -RT ln K2, we can relate E to K as well.

The maximum amount of electrical work produced by a voltaic cell is given by the product of the total charge delivered, nF, and the emf, E: wmax = -nFE. The watt is a unit of power: 1 W = 1 J>s. Elec-trical work is often measured in kilowatt-hours.

ceLL PotentiALS under nonStAndArd conditionS (Sec-tion 20.6) The emf of a redox reaction varies with temperature and with the concentrations of reactants and products. The nernst equation

relates the emf under nonstandard conditions to the standard emf and the reaction quotient Q:

E = E° - 1RT>nF2 ln Q = E° - 10.0592>n2 log Q

The factor 0.0592 is valid when T = 298 K. A concentration cell is a voltaic cell in which the same half-reaction occurs at both the anode and the cathode but with different concentrations of reactants in each half-cell. At equilibrium, Q = K and E = 0.

bAtterieS And FueL ceLLS (Section 20.7) A battery is a self-contained electrochemical power source that contains one or more voltaic cells. Batteries are based on a variety of different redox reac-tions. Batteries that cannot be recharged are called primary cells, while those that can be recharged are called secondary cells. The common alkaline dry cell battery is an example of a primary cell battery. Lead–acid, nickel–cadmium, nickel–metal hydride, and lithium-ion bat-teries are examples of secondary cells. Fuel cells are voltaic cells that utilize redox reactions in which reactants such as H2 have to be con-tinuously supplied to the cell to generate voltage.

corroSion (Section 20.8) Electrochemical principles help us understand corrosion, undesirable redox reactions in which a metal is attacked by some substance in its environment. The corrosion of iron into rust is caused by the presence of water and oxygen, and it is acceler-ated by the presence of electrolytes, such as road salt. The protection of a metal by putting it in contact with another metal that more readily under-goes oxidation is called cathodic protection. Galvanized iron, for example, is coated with a thin layer of zinc; because zinc is oxidized more readily than iron, the zinc serves as a sacrificial anode in the redox reaction.

eLectroLySiS (Section 20.9) An electrolysis reaction, which is carried out in an electrolytic cell, employs an external source of electric-ity to drive a nonspontaneous electrochemical reaction. The current-carrying medium within an electrolytic cell may be either a molten salt or an electrolyte solution. The products of electrolysis can generally be predicted by comparing the reduction potentials associated with possible oxidation and reduction processes. The electrodes in an elec-trolytic cell can be inert or active, meaning that the electrode can be involved in the electrolysis reaction. Active electrodes are important in electroplating and in metallurgical processes.

The quantity of substances formed during electrolysis can be cal-culated by considering the number of electrons involved in the redox reaction and the amount of electrical charge that passes into the cell. The amount of electrical charge is measured in coulombs and is related to the magnitude of the current and the time it flows 11 C = 1 A@s2.

learning Outcomes after studying this chapter, you should be able to:

• Identify oxidation, reduction, oxidizing agent, and reducing agent in a chemical equation. (Section 20.1)

• Complete and balance redox equations using the method of half-reactions. (Section 20.2)

• Sketch a voltaic cell and identify its cathode, anode, and the direc-tions in which electrons and ions move. (Section 20.3)

• Calculate standard emfs (cell potentials), E°cell, from standard reduction potentials. (Section 20.4)

• Use reduction potentials to predict whether a redox reaction is spontaneous. (Section 20.4)

• Relate E°cell, to ∆G° and equilibrium constants. (Section 20.5)

• Calculate emf under nonstandard conditions. (Section 20.6)

• Identify the components of common batteries. (Section 20.7)

• Describe the construction of a lithium-ion battery and explain how it works (Section 20.7)

• Describe the construction of a fuel cell and explain how it gener-ates electrical energy. (Section 20.7)

• Explain how corrosion occurs and how it is prevented by cathodic protection. (Section 20.8)

• Describe the reactions in electrolytic cells. (Section 20.9)

• Relate amounts of products and reactants in redox reactions to electrical charge. (Section 20.9)

exercises 899

Key Equations • E°cell = E°red 1cathode2 - E°red 1anode2 [20.8] Relating standard emf to standard reduction potentials of the re-

duction (cathode) and oxidation (anode) half-reactions

• ∆G = -nFE [20.11] Relating free-energy change and emf

• E = E° -0.0592 V

n log Q 1at 298 K2 [20.18] The Nernst equation, expressing the effect of concentration on cell

potential

ExercisesVisualizing concepts

20.1 In the Brønsted–Lowry concept of acids and bases, acid–base reactions are viewed as proton-transfer reactions. The stron-ger the acid, the weaker is its conjugate base. If we were to think of redox reactions in a similar way, what particle would be analogous to the proton? Would strong oxidizing agents be analogous to strong acids or strong bases? [Sections 20.1 and 20.2]

20.2 You may have heard that “antioxidants” are good for your health. Based on what you have learned in this chapter, what do you deduce an “antioxidant” is? [Sections 20.1 and 20.2]

20.3 The diagram that follows represents a molecular view of a process occurring at an electrode in a voltaic cell.

(a) Does the process represent oxidation or reduction? (b) Is the electrode the anode or cathode? (c) Why are the atoms in the electrode represented by larger spheres than the ions in the solution? [Section 20.3]

20.4 Assume that you want to construct a voltaic cell that uses the following half-reactions:

A2+1aq2 + 2 e- ¡ A1s2 E°red = -0.10 VB2+1aq2 + 2 e- ¡ B1s2 E°red = -1.10 V

You begin with the incomplete cell pictured here in which the electrodes are immersed in water.

VoltmeterA B

(a) What additions must you make to the cell for it to gen-erate a standard emf? (b) Which electrode functions as the cathode? (c) Which direction do electrons move through the

external circuit? (d) What voltage will the cell generate under standard conditions? [Sections 20.3 and 20.4]

20.5 For a spontaneous reaction A1aq2 + B1aq2 ¡ A-1aq2 +B+1aq2, answer the following questions:(a) If you made a voltaic cell out of this reaction, what half-

reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode?

(b) Which half-reaction from (a) is higher in potential energy?(c) What is the sign of E°cell? [Section 20.3]

20.6 Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution:

Reduction half-Reaction E°(V)

A+1aq2 + e- ¡ A1s2 1.33

B2+1aq2 + 2 e- ¡ B1s2 0.87

C3+1aq2 + e- ¡ C2+1aq2 -0.12

D3+1aq2 + 3 e- ¡ D1s2 -1.59

(a) Which substance is the strongest oxidizing agent? Which is weakest?

(b) Which substance is the strongest reducing agent? Which is weakest?

(c) Which substance(s) can oxidize C2+? [Sections 20.4 and 20.5] 20.7 Consider a redox reaction for which E° is a negative number.

(a) What is the sign of ∆G° for the reaction?(b) Will the equilibrium constant for the reaction be larger or

smaller than 1?(c) Can an electrochemical cell based on this reaction accom-

plish work on its surroundings? [Section 20.5] 20.8 Consider the following voltaic cell:

e−

Ag(s)Fe(s)

1 M Fe2+

1 M Ag+

Salt bridge

Voltmeter

(a) Which electrode is the cathode?(b) What is the standard emf generated by this cell?


Oxidation–Reduction Reactions (section 20.1)

20.13 (a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons appear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent?

20.14 (a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

20.15 Indicate whether each of the following statements is true or false:(a) If something is oxidized, it is formally losing electrons.(b) For the reaction Fe3+1aq2 + Co2+1aq2 ¡ Fe2+1aq2 +

Co3+1aq2, Fe3+1aq2 is the reducing agent and Co2+1aq2 is the oxidizing agent.

(c) If there are no changes in the oxidation state of the reac-tants or products of a particular reaction, that reaction is not a redox reaction.

20.16 Indicate whether each of the following statements is true or false:(a) If something is reduced, it is formally losing electrons.(b) A reducing agent gets oxidized as it reacts.(c) An oxidizing agent is needed to convert CO into CO2.

20.17 In each of the following balanced oxidation–reduction equa-tions, identify those elements that undergo changes in oxi-dation number and indicate the magnitude of the change in each case.(a) I2O51s2 + 5 CO1g2 ¡ I21s2 + 5 CO21g2(b) 2 Hg2+1aq2 + N2H41aq2 ¡ 2 Hg1l2 + N21g2 + 4 H+1aq2(c) 3 H2S1aq2 + 2 H+1aq2 + 2 NO3

-1aq2 ¡ 3 S1s2 +2 NO1g2 + 4 H2O1l2

20.18 In each of the following balanced oxidation–reduction equa-tions, identify those elements that undergo changes in oxi-dation number and indicate the magnitude of the change in each case.(a) 2 MnO4

-1aq2 + 3 S2 - 1aq2 + 4 H2O1l2 ¡ 3 S1s2 +2 MnO21s2 + 8 OH-1aq2

(b) 4 H2O21aq2 + Cl2O71g2 + 2 OH-1aq2 ¡ 2 ClO2-1aq2 +

5 H2O1l2 + 4 O21g2(c) Ba2+1aq2 + 2 OH-1aq2 + H2O21aq2 + 2 ClO21aq2 ¡

Ba1ClO2221s2 + 2 H2O1l2 + O21g2 20.19 Indicate whether the following balanced equations involve

oxidation–reduction. If they do, identify the elements that undergo changes in oxidation number.(a) PBr31l2 + 3 H2O1l2 ¡ H3PO31aq2 + 3 HBr1aq2(b) NaI1aq2 + 3 HOCl1aq2 ¡ NaIO31aq2 + 3 HCl1aq2(c) 3 SO21g2 + 2 HNO31aq2 + 2 H2O1l2 ¡

3 H2SO41aq2 + 2 NO1g2 20.20 Indicate whether the following balanced equations involve

oxidation–reduction. If they do, identify the elements that undergo changes in oxidation number.(a) 2 AgNO31aq2 + CoCl21aq2 ¡ 2 AgCl1s2 +

Co1NO3221aq2(b) 2 PbO21s2 ¡ 2 PbO1s2 + O21g2(c) 2 H2SO41aq2 + 2 NaBr1s2 ¡ Br21l2 + SO21g2 +

Na2SO41aq2 + 2 H2O1l2

(c) What is the change in the cell voltage when the ion concentra-tions in the cathode half-cell are increased by a factor of 10?

(d) What is the change in the cell voltage when the ion concentrations in the anode half-cell are increased by a factor of 10? [Sections 20.4 and 20.6]

20.9 Consider the half-reaction Ag+1aq2 + e- ¡ Ag1s2. (a) Which of the lines in the following diagram indicates how the reduction potential varies with the concentration of Ag+? (b) What is the value of E°red when log3Ag+4 = 0? [Section 20.6]

1

2

3

log[Ag+]

Ered

20.10 The electrodes in a silver oxide battery are silver oxide 1Ag2O2 and zinc. (a) Which electrode acts as the anode? (b) Which battery do you think has an energy density most similar to the silver oxide battery: a Li-ion battery, a nickel–cadmium bat-tery, or a lead–acid battery? [Section 20.7]

20.11 Bars of iron are put into each of the three beakers as shown here. In which beaker—A, B, or C—would you expect the iron to show the most corrosion ? [Section 20.8]

Beaker APure waterpH = 7.0

Beaker BDilute HCl (aq)

solutionpH = 4.0

Beaker CDilute NaOH

solutionpH = 10.0

20.12 Magnesium is produced commercially by electrolysis from a mol-ten salt using a cell similar to the one shown here. (a) What salt is used as the electrolyte? (b) Which electrode is the anode, and which one is the cathode? (c) Write the overall cell reaction and individual half-reactions. (d) What precautions would need to be taken with respect to the magnesium formed? [Section 20.9]

Cl2(g)out

LiquidMg

Steelelectrode

Carbonelectrode

Electrolyte

+−Voltage source

exercises 901

(c) Cr2O72 - 1aq2 + CH3OH1aq2 ¡ HCO2H1aq2 +

Cr3+1aq2 (acidic solution)(d) BrO3

-1aq2 + N2H41g2 ¡ Br-1aq2 + N21g2 (acidic solution)

(e) NO2-1aq2 + Al1s2 ¡ NH4

+1aq2 + AlO2-1aq2

(basic solution)(f) H2O21aq2 + ClO21aq2 ¡ ClO2

-1aq2 + O21g2 (basic solution)

Voltaic cells (section 20.3)

20.27 (a) What are the similarities and differences between Figure 20.3 and Figure 20.4? (b) Why are Na+ ions drawn into the cathode half-cell as the voltaic cell shown in Figure 20.5 operates?

20.28 (a) What is the role of the porous glass disc shown in Figure 20.4? (b) Why do NO3

- ions migrate into the anode half-cell as the voltaic cell shown in Figure 20.5 operates?

20.29 A voltaic cell similar to that shown in Figure 20.5 is con-structed. One electrode half-cell consists of a silver strip placed in a solution of AgNO3, and the other has an iron strip placed in a solution of FeCl2. The overall cell reaction is

Fe1s2 + 2 Ag+1aq2 ¡ Fe2+1aq2 + 2 Ag1s2 (a) What is being oxidized, and what is being reduced?

(b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

20.30 A voltaic cell similar to that shown in Figure 20.5 is con-structed. One half-cell consists of an aluminum strip placed in a solution of A11NO323, and the other has a nickel strip placed in a solution of NiSO4. The overall cell reaction is

2 Al1s2 + 3 Ni2+1aq2 ¡ 2 Al3+1aq2 + 3 Ni1s2 (a) What is being oxidized, and what is being reduced? (b) Write

the half-reactions that occur in the two half-cells. (c) Which elec-trode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode or from the nickel to the alumi-num? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.

cell Potentials under standard conditions (section 20.4)

20.31 (a) What does the term electromotive force mean? (b) What is the definition of the volt? (c) What does the term cell potential mean?

20.32 (a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?

20.33 (a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell. (b) Write the half-reaction that occurs at a hydro-gen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard hydrogen electrode?

Balancing Oxidation–Reduction Reactions (section 20.2)

20.21 At 900 °C titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chloride. (a) Write a balanced equation for this re-action. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant?

20.22 Hydrazine 1N2H42 and dinitrogen tetroxide 1N2O42 form a self-igniting mixture that has been used as a rocket propel-lant. The reaction products are N2 and H2O. (a) Write a bal-anced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent and which as the oxidizing agent?

20.23 Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction.(a) Sn2+1aq2 ¡ Sn4+1aq2 (acidic solution)(b) TiO21s2 ¡ Ti2+1aq2 (acidic solution)(c) ClO3

-1aq2 ¡ Cl-1aq2 (acidic solution)(d) N21g2 ¡ NH4

+1aq2 (acidic solution)(e) OH-1aq2 ¡ O21g2 (basic solution)(f) SO3

2-1aq2 ¡ SO42-1aq2 (basic solution)

(g) N21g2 ¡ NH31g2 (basic solution) 20.24 Complete and balance the following half-reactions. In each

case indicate whether the half-reaction is an oxidation or a reduction.(a) Mo3+1aq2 ¡ Mo1s2 (acidic solution)(b) H2SO31aq2 ¡ SO4

2-1aq2 (acidic solution)(c) NO3

-1aq2 ¡ NO1g2 (acidic solution)(d) O21g2 ¡ H2O1l2 (acidic solution)(e) O21g2 ¡ H2O1l2 (basic solution)(f) Mn2+1aq2 ¡ MnO21s2 (basic solution)(g) Cr1OH231s2 ¡ CrO4

2-1aq2 (basic solution) 20.25 Complete and balance the following equations, and identify

the oxidizing and reducing agents:(a) Cr2O7

2-1aq2 + I-1aq2 ¡ Cr3+1aq2 + IO3-1aq2

(acidic solution)(b) MnO4

-1aq2 + CH3OH1aq2 ¡ Mn2+1aq2 +HCO2H1aq2 (acidic solution)

(c) I21s2 + OCl-1aq2 ¡ IO3-1aq2 + Cl-1aq2 (acidic

solution)(d) As2O31s2 + NO3

-1aq2 ¡ H3AsO41aq2 + N2O31aq2 (acidic solution)

(e) MnO4-1aq2 + Br-1aq2 ¡ MnO21s2 + BrO3

-1aq2 (basic solution)

(f) Pb1OH242-1aq2 + ClO-1aq2 ¡ PbO21s2 + Cl-1aq2

(basic solution) 20.26 Complete and balance the following equations, and identify

the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.)(a) NO2

-1aq2 + Cr2O72-1aq2 ¡ Cr3+1aq2 + NO3

-1aq2 (acidic solution)

(b) S1s2 + HNO31aq2 ¡ H2SO31aq2 + N2O1g2 (acidic solution)


the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calcu-late that value.

20.41 A 1 M solution of Cu1NO322 is placed in a beaker with a strip of Cu metal. A 1 M solution of SnSO4 is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal elec-trodes. (a) Which electrode serves as the anode and which as the cathode? (b) Which electrode gains mass and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

20.42 A voltaic cell consists of a strip of cadmium metal in a solu-tion of Cd1NO322 in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with Cl2 gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

strengths of Oxidizing and Reducing agents (section 20.4)

20.43 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger reducing agent:(a) Fe(s) or Mg(s)(b) Ca(s) or Al(s)(c) H2 (g, acidic solution) or H2S1g2(d) BrO3

-1aq2 or IO3-1aq2

20.44 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent:(a) Cl21g2 or Br21l2(b) Zn2+1aq2 or Cd2+1aq2(c) Cl-1aq2 or ClO3

-1aq2(d) H2O21aq2 or O31g2

20.45 By using the data in Appendix E, determine whether each of the following substances is likely to serve as an oxidant or a re-ductant: (a) Cl21g2, (b) MnO4

- (aq, acidic solution), (c) Ba(s), (d) Zn(s).

20.46 Is each of the following substances likely to serve as an oxi-dant or a reductant: (a) Ce3+1aq2, (b) Ca(s), (c) ClO3

-1aq2, (d) N2O51g2?

20.47 (a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solu-tion: Cr2O7

2-, H2O2, Cu2+, Cl2, O2. (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: Zn, I-, Sn2+, H2O2, Al.

20.48 Based on the data in Appendix E, (a) which of the following is the strongest oxidizing agent and which is the weakest in

acidic solution: Br2, H2O2, Zn, Cr2O72 - ? (b) Which of the fol-

lowing is the strongest reducing agent, and which is the weak-est in acidic solution: F-, Zn, N2H5

+, I2, NO?

20.34 (a) What conditions must be met for a reduction potential to be a standard reduction potential? (b) What is the stan-dard reduction potential of a standard hydrogen electrode? (c) Why is it impossible to measure the standard reduction potential of a single half-reaction?

20.35 A voltaic cell that uses the reaction

T13+1aq2 + 2 Cr2+1aq2 ¡ T1+1aq2 + 2 Cr3+1aq2 has a measured standard cell potential of +1.19 V. (a) Write

the two half-cell reactions. (b) By using data from Appendix E, determine E°red for the reduction of T13+1aq2 to T1+1aq2. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

20.36 A voltaic cell that uses the reaction

PdCl42-1aq2 + Cd1s2 ¡ Pd1s2 + 4 Cl-1aq2 + Cd2+1aq2

has a measured standard cell potential of +1.03 V. (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine E°red for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

20.37 Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions:(a) Cl21g2 + 2 I-1aq2 ¡ 2 Cl-1aq2 + I21s2(b) Ni1s2 + 2 Ce4+1aq2 ¡ Ni2+1aq2 + 2 Ce3+1aq2(c) Fe1s2 + 2 Fe3+1aq2 ¡ 3 Fe2+1aq2(d) 2 NO3

-1aq2 + 8 H+1aq2 + 3 Cu1s2 ¡ 2 NO1g2 +4 H2O1l2 + 3 Cu2+1aq2

20.38 Using data in Appendix E, calculate the standard emf for each of the following reactions:(a) H21g2 + F21g2 ¡ 2 H+1aq2 + 2 F-1aq2(b) Cu2+1aq2 + Ca1s2 ¡ Cu1s2 + Ca2+1aq2(c) 3 Fe2+1aq2 ¡ Fe1s2 + 2 Fe3+1aq2(d) 2 ClO3

-1aq2 + 10 Br-1aq2 + 12 H+1aq2 ¡ Cl21g2 +5 Br21l2 + 6 H2O1l2

20.39 The standard reduction potentials of the following half- reactions are given in Appendix E:

Ag+1aq2 + e- ¡ Ag1s2Cu2+1aq2 + 2 e- ¡ Cu1s2Ni2+1aq2 + 2 e- ¡ Ni1s2Cr3+1aq2 + 3 e- ¡ Cr1s2

(a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell poten-tial and calculate the value. (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell potential and calculate the value.

20.40 Given the following half-reactions and associated standard reduction potentials:

AuBr4-1aq2 + 3 e- ¡ Au1s2 + 4 Br-1aq2

E°red = -0.858 VEu3+1aq2 + e- ¡ Eu2+1aq2

E°red = -0.43 VIO-1aq2 + H2O1l2 + 2 e- ¡ I-1aq2 + 2 OH-1aq2

E°red = +0.49 V (a) Write the equation for the combination of these half-cell

reactions that leads to the largest positive emf and calculate

exercises 903

20.59 A voltaic cell is based on the reaction

Sn1s2 + I21s2 ¡ Sn2+1aq2 + 2I-1aq2 Under standard conditions, what is the maximum electrical

work, in joules, that the cell can accomplish if 75.0 g of Sn is consumed?

20.60 Consider the voltaic cell illustrated in Figure 20.5, which is based on the cell reaction

Zn1s2 + Cu2+1aq2 ¡ Zn2+1aq2 + Cu1s2 Under standard conditions, what is the maximum electrical

work, in joules, that the cell can accomplish if 50.0 g of copper is formed?

cell EmF under nonstandard conditions (section 20.6)

20.61 (a) In the Nernst equation what is the numerical value of the reaction quotient, Q, under standard conditions? (b) Can the Nernst equation be used at temperatures other than room temperature?

20.62 (a) A voltaic cell is constructed with all reactants and prod-ucts in their standard states. Will the concentration of the reactants increase, decrease, or remain the same as the cell operates? (b) What happens to the emf of a cell if the concen-trations of the products are increased?

20.63 What is the effect on the emf of the cell shown in Figure 20.9, which has the overall reaction Zn1s2 + 2 H+1aq2¡ Zn2+1aq2 + H21g2, for each of the following changes? (a) The pressure of the H2 gas is increased in the cathode half-cell. (b) Zinc nitrate is added to the anode half-cell. (c) So-dium hydroxide is added to the cathode half-cell, decreasing 3H+4. (d) The surface area of the anode is doubled.

20.64 A voltaic cell utilizes the following reaction:

Al1s2 + 3 Ag+1aq2 ¡ Al3+1aq2 + 3 Ag1s2 What is the effect on the cell emf of each of the following

changes? (a) Water is added to the anode half-cell, dilut-ing the solution. (b) The size of the aluminum electrode is increased. (c) A solution of AgNO3 is added to the cathode half-cell, increasing the quantity of Ag+ but not changing its concentration. (d) HCl is added to the AgNO3 solution, pre-cipitating some of the Ag+ as AgCl.

20.65 A voltaic cell is constructed that uses the following reaction and operates at 298 K:

Zn1s2 + Ni2+1aq2 ¡ Zn2+1aq2 + Ni1s2 (a) What is the emf of this cell under standard conditions?

(b) What is the emf of this cell when 3Ni2+4 = 3.00 M? and 3Zn2+4 = 0.100 M? (c) What is the emf of the cell when 3Ni2+4 = 0.200 M and 3Zn2+4 = 0.900 M?

20.66 A voltaic cell utilizes the following reaction and operates at 298 K:

3 Ce4+1aq2 + Cr1s2 ¡ 3 Ce3+1aq2 + Cr3+1aq2 (a) What is the emf of this cell under standard conditions?

(b) What is the emf of this cell when 3Ce4+4 = 3.0 M,3Ce3+4 = 0.10 M, and 3Cr3+4 = 0.010 M? (c) What is the emf of the cell when 3Ce4+4 = 0.010 M, 3Ce3+4 = 2.0 M, and

3Cr3+4 = 1.5 M?

20.49 The standard reduction potential for the reduction of Eu3+1aq2 to Eu2+1aq2 is -0.43 V. Using Appendix E, which of the fol-lowing substances is capable of reducing Eu3+1aq2 to Eu2+1aq2 under standard conditions: Al, Co, H2O2, N2H5

+, H2C2O4? 20.50 The standard reduction potential for the reduction of

RuO4-1aq2 to RuO4

2-1aq2 is +0.59 V. By using Appendix E, which of the following substances can oxidize RuO4

2-1aq2 to RuO4

-1aq2 under standard conditions: Br21l2, BrO3-1aq2,

Mn2+1aq2, O21g2, Sn2+1aq2?

Free Energy and Redox Reactions (section 20.5)

20.51 Given the following reduction half-reactions:

Fe3+1aq2 + e- ¡ Fe2+1aq2 E°red = +0.77 VS2O6

2 - 1aq2 + 4 H+1aq2 + 2 e- ¡ 2 H2SO31aq2 E°red = +0.60 VN2O1g2 + 2 H+1aq2 + 2 e- ¡ N21g2 + H2O1l2 E°red = -1.77 VVO2

+1aq2 + 2 H+1aq2 + e- ¡ VO2+ + H2O1l2 E°red = +1.00 V

(a) Write balanced chemical equations for the oxidation of Fe2+1aq2 by S2O6

2-1aq2, by N2O1aq2, and by VO2+1aq2.

(b) Calculate ∆G° for each reaction at 298 K. (c) Calculate the equilibrium constant K for each reaction at 298 K.

20.52 For each of the following reactions, write a balanced equation, calculate the standard emf, calculate ∆G° at 298 K, and calcu-late the equilibrium constant K at 298 K. (a) Aqueous iodide ion is oxidized to I21s2 by Hg2

2+1aq2. (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, Cr1OH231s2 is oxidized to CrO4

2-1aq2 by ClO-1aq2.

20.53 If the equilibrium constant for a two-electron redox reaction at 298 K is 1.5 * 10-4, calculate the corresponding ∆G° and E°red.

20.54 If the equilibrium constant for a one-electron redox reaction at 298 K is 8.7 * 104, calculate the corresponding ∆G° and E°red.

20.55 Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following re-actions at 298 K:(a) Fe1s2 + Ni2+1aq2 ¡ Fe2+1aq2 + Ni1s2(b) Co1s2 + 2 H+1aq2 ¡ Co2+1aq2 + H21g2(c) 10 Br-1aq2 + 2 MnO4

-1aq2 + 16 H+1aq2 ¡2 Mn2+1aq2 + 8 H2O1l2 + 5 Br21l2

20.56 Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K:(a) Cu1s2 + 2 Ag+1aq2 ¡ Cu2+1aq2 + 2 Ag1s2(b) 3 Ce4+1aq2 + Bi1s2 + H2O1l2 ¡ 3 Ce3+1aq2 +

BiO+1aq2 + 2H+1aq2(c) N2H5

+1aq2 + 4 Fe1CN263 - 1aq2 ¡ N21g2 +

5 H+1aq2 + 4 Fe1CN264-1aq2

20.57 A cell has a standard cell potential of +0.177 V at 298 K. What is the value of the equilibrium constant for the reaction (a) if n = 1? (b) if n = 2? (c) if n = 3?

20.58 At 298 K a cell reaction has a standard cell potential of +0.17 V. The equilibrium constant for the reaction is

5.5 * 105. What is the value of n for the reaction?


20.75 Heart pacemakers are often powered by lithium–silver chro-mate “button” batteries. The overall cell reaction is

2 Li1s2 + Ag2CrO41s2 ¡ Li2CrO41s2 + 2 Ag1s2 (a) Lithium metal is the reactant at one of the electrodes of

the battery. Is it the anode or the cathode? (b) Choose the two half-reactions from Appendix E that most closely approxi-mate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half- reactions? (c) The battery generates an emf of +3.5 V. How close is this value to the one calculated in part (b)? (d) Calculate the emf that would be generated at body temperature, 37 °C. How does this compare to the emf you calculated in part (b)?

20.76 Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are

HgO1s2 + H2O1l2 + 2 e- ¡ Hg1l2 + 2 OH-1aq2 Zn1s2 + 2 OH-1aq2 ¡ ZnO1s2 + H2O1l2 + 2 e-

(a) Write the overall cell reaction. (b) The value of E°red for the cathode reaction is +0.098 V. The overall cell potential is +1.35 V. Assuming that both half-cells operate under stan-dard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction oc-curred in an acidic medium?

20.77 (a) Suppose that an alkaline battery was manufactured us-ing cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel–metal hydride batteries over nickel–cadmium batteries?

20.78 In some applications nickel–cadmium batteries have been replaced by nickel–zinc batteries.The overall cell reaction for this relatively new battery is:

2 H2O1l2 + 2 NiO1OH21s2 + Zn1s2¡ 2 Ni1OH221s2 + Zn1OH221s2

(a)What is the cathode half-reaction? (b)What is the anode half-reaction? (c) A single nickel–cadmium cell has a voltage of 1.30 V. Based on the difference in the standard reduction potentials of Cd2+ and Zn2+, what voltage would you estimate a nickel–zinc battery will produce? (d) Would you expect the specific energy density of a nickel–zinc battery to be higher or lower than that of a nickel–cadmium battery?

20.79 In a Li-ion battery the composition of the cathode is LiCoO2 when completely discharged. On charging approximately 50% of the Li+ ions can be extracted from the cathode and trans-ported to the graphite anode where they intercalate between the layers. (a) What is the composition of the cathode when the battery is fully charged? (b) If the LiCoO2 cathode has a mass of 10 g (when fully discharged), how many coulombs of electricity can be delivered on completely discharging a fully charged battery?

20.80 Li-ion batteries used in automobiles typically use a LiMn2O4 cathode in place of the LiCoO2 cathode found in most Li-ion batteries. (a) Calculate the mass percent lithium in each elec-trode material? (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with


4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (a) What is the emf of this cell under standard conditions?

(b) What is the emf of this cell when 3Fe2+4 = 1.3 M, 3Fe3+4 =0.010 M, Po2

= 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50?


2 Fe3+1aq2 + H21g2 ¡ 2 Fe2+1aq2 + 2 H+1aq2 (a) What is the emf of this cell under standard conditions?

(b) What is the emf for this cell when 3Fe3+4 = 3.50 M, PH2=

0.95 atm, 3Fe2+4 = 0.0010 M, and the pH in both half-cells is 4.00?

20.69 A voltaic cell is constructed with two Zn2+ - Zn elec-trodes. The two half-cel ls have 3Zn2+4 = 1.8 M and 3Zn2+4 = 1.00 * 10-2 M, respectively. (a) Which electrode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether 3Zn2+4 will increase, decrease, or stay the same as the cell operates.

20.70 A voltaic cell is constructed with two silver–silver chlo-ride electrodes, each of which is based on the following half-reaction:

AgCl1s2 + e- ¡ Ag1s2 + Cl-1aq2 The two half-cells have 3Cl-4 = 0.0150 M and 3Cl-4 = 2.55 M,

respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, pre-dict whether 3Cl-4 will increase, decrease, or stay the same as the cell operates.

20.71 The cell in Figure 20.9 could be used to provide a measure of the pH in the cathode half-cell. Calculate the pH of the cath-ode half-cell solution if the cell emf at 298 K is measured to be +0.684 V when 3Zn2+4 = 0.30 M and PH2

= 0.90 atm. 20.72 A voltaic cell is constructed that is based on the following

reaction:

Sn2+1aq2 + Pb1s2 ¡ Sn1s2 + Pb2+1aq2 (a) If the concentration of Sn2+ in the cathode half-cell is

1.00 M and the cell generates an emf of +0.22 V, what is the concentration of Pb2+ in the anode half-cell? (b) If the an-ode half-cell contains 3SO4

2-4 = 1.00 M in equilibrium with PbSO41s2, what is the Ksp of PbSO4?

Batteries and Fuel cells (section 20.7)

20.73 During a period of discharge of a lead–acid battery, 402 g of Pb from the anode is converted into PbSO41s2. (a) What mass of PbO21s2 is reduced at the cathode during this same period? (b) How many coulombs of electrical charge are transferred from Pb to PbO2?

20.74 During the discharge of an alkaline battery, 4.50 g of Zn is consumed at the anode of the battery. (a) What mass of MnO2 is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred

from Zn to MnO2?

additional exercises 905

Electrolysis (section 20.9)

20.89 (a) What is electrolysis? (b) Are electrolysis reactions thermo-dynamically spontaneous? Explain. (c) What process occurs at the anode in the electrolysis of molten NaCl? (d) Why is so-dium metal not obtained when an aqueous solution of NaCl undergoes electrolysis?

20.90 (a) What is an electrolytic cell? (b) The negative terminal of a voltage source is connected to an electrode of an electrolytic cell. Is the electrode the anode or the cathode of the cell? Ex-plain. (c) The electrolysis of water is often done with a small amount of sulfuric acid added to the water. What is the role of the sulfuric acid? (d) Why are active metals such as Al ob-tained by electrolysis using molten salts rather than aqueous solutions?

20.91 (a) A Cr3+1aq2 solution is electrolyzed, using a current of 7.60 A. What mass of Cr(s) is plated out after 2.00 days? (b) What amperage is required to plate out 0.250 mol Cr from a Cr3+ solution in a period of 8.00 h?

20.92 Metallic magnesium can be made by the electrolysis of molten MgCl2. (a) What mass of Mg is formed by passing a current of 4.55 A through molten MgCl2, for 4.50 days? (b) How many minutes are needed to plate out 25.00 g Mg from molten MgCl2 using 3.50 A of current?

20.93 (a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of 7.5 * 104 A flowing for a period of 24 h. Assume the electrolytic cell is 85% efficient. (b) What is the minimum voltage required to drive the reaction?

20.94 Elemental calcium is produced by the electrolysis of molten CaCl2. (a) What mass of calcium can be produced by this pro-cess if a current of 7.5 * 103 A is applied for 48 h? Assume that the electrolytic cell is 68% efficient. (b) What is the mini-mum voltage needed to cause the electrolysis?

20.95 Metallic gold is collected from below the anode when crude copper metal is refined by electrolysis. Explain this behavior.

20.96 The crude copper that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, Te4+, is

Te4+1aq2 + 4 e- ¡ Te1s2 E°red = 0.57 V

Given this information, describe the probable fate of tellu-rium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If they go into solution, do they plate out on the cathode?

LiMn2O4 cathodes deliver less power on discharging? (c) In a battery that uses a LiCoO2 cathode approximately 50% of the lithium migrates from the cathode to the anode on charging. In a battery that uses a LiMn2O4 cathode what fraction of the lithium in LiMn2O4 would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

20.81 The hydrogen–oxygen fuel cell has a standard emf of 1.23 V. What advantages and disadvantages are there to using this device as a source of power compared to a 1.55-V alkaline battery?

20.82 (a) What is the difference between a battery and a fuel cell? (b) Can the “fuel” of a fuel cell be a solid? Explain.

corrosion (section 20.8)

20.83 (a) Write the anode and cathode reactions that cause the cor-rosion of iron metal to aqueous iron(II). (b) Write the bal-anced half-reactions involved in the air oxidation of Fe2+1aq2 to Fe2O3 # 3 H2O.

20.84 (a) Based on standard reduction potentials, would you ex-pect copper metal to oxidize under standard conditions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed be-tween the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play?

20.85 (a) Magnesium metal is used as a sacrificial anode to protect underground pipes from corrosion. Why is the magnesium re-ferred to as a “sacrificial anode”? (b) Looking in Appendix E, suggest what metal the underground pipes could be made from in order for magnesium to be successful as a sacrificial anode.

20.86 An iron object is plated with a coating of cobalt to protect against corrosion. Does the cobalt protect iron by cathodic protection? Explain.

20.87 A plumber’s handbook states that you should not connect a brass pipe directly to a galvanized steel pipe because electro-chemical reactions between the two metals will cause corro-sion. The handbook recommends you use instead an insulating fitting to connect them. Brass is a mixture of copper and zinc. What spontaneous redox reaction(s) might cause the corro-sion? Justify your answer with standard emf calculations.

20.88 A plumber’s handbook states that you should not connect a cop-per pipe directly to a steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook rec-ommends you use instead an insulating fitting to connect them. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

additional Exercises 20.97 A disproportionation reaction is an oxidation–reduction re-

action in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions:(a) Ni+1aq2 ¡ Ni2+1aq2 + Ni1s2 (acidic solution)(b) MnO4

2-1aq2 ¡ MnO4-1aq2 + MnO21s2

(acidic solution)

(c) H2SO31aq2 ¡ S1s2 + HSO4-1aq2 (acidic solution)

(d) Cl21aq2 ¡ Cl-1aq2 + ClO-1aq2 (basic solution) 20.98 A common shorthand way to represent a voltaic cell is

anode � anode solution � � cathode solution �cathode A double vertical line represents a salt bridge or a porous bar-

rier. A single vertical line represents a change in phase, such as


20.104 (a) Write the reactions for the discharge and charge of a nickel–cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell:

Cd1OH221s2 + 2 e- ¡ Cd1s2 + 2 OH-1aq2 E°red = -0.76 V

NiO1OH21s2 + H2O1l2 + e- ¡ Ni1OH221s2 + OH-1aq2E°red = +0.49 V

(c) A typical nicad voltaic cell generates an emf of +1.30 V. Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

20.105 The capacity of batteries such as the typical AA alkaline bat-tery is expressed in units of milliamp-hours (mAh). An AA alkaline battery yields a nominal capacity of 2850 mAh. (a) What quantity of interest to the consumer is being ex-pressed by the units of mAh? (b) The starting voltage of a fresh alkaline battery is 1.55 V. The voltage decreases dur-ing discharge and is 0.80 V when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electri-cal work the battery could perform during discharge.

20.106 If you were going to apply a small potential to a steel ship rest-ing in the water as a means of inhibiting corrosion, would you apply a negative or a positive charge? Explain.

[20.107] (a) How many coulombs are required to plate a layer of chro-mium metal 0.25 mm thick on an auto bumper with a total area of 0.32 m2 from a solution containing CrO4

2-? The den-sity of chromium metal is 7.20 g>cm3. (b) What current flow is required for this electroplating if the bumper is to be plated in 10.0 s? (c) If the external source has an emf of +6.0 V and the electrolytic cell is 65% efficient, how much electrical power is expended to electroplate the bumper?

20.108 Magnesium is obtained by electrolysis of molten MgCl2. (a) Why is an aqueous solution of MgCl2 not used in the elec-trolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are 96% efficient in producing the desired prod-ucts in electrolysis, what mass of Mg is formed by passing a current of 97,000 A for a period of 24 h?

20.109 Calculate the number of kilowatt-hours of electricity required to produce 1.0 * 103 kg (1 metric ton) of aluminum by elec-trolysis of Al3+ if the applied voltage is 4.50 V and the process is 45% efficient.

20.110 Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship us-ing a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about 7 * 108 mol of H2 to provide the buoyancy to lift the ship (J. Chem. Educ., 1973, Vol. 50, 61). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate H2 and O2 if the pressure on the gases at the depth of the wreckage (2 mi) is 300 atm? (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary H2 if the electricity costs 85 cents per kilowatt-hour to generate at the site?

from solid to solution. (a) Write the half-reactions and over-all cell reaction represented by Fe � Fe2+ � � Ag+ �Ag; sketch the cell. (b) Write the half-reactions and overall cell reaction rep-resented by Zn � Zn2+ � � H+ �H2; sketch the cell. (c) Using the notation just described, represent a cell based on the follow-ing reaction:ClO3

-1aq2 + 3 Cu1s2 + 6 H+1aq2¡ Cl-1aq2 + 3 Cu2+1aq2 + 3 H2O1l2

Pt is used as an inert electrode in contact with the ClO3- and

Cl-. Sketch the cell. 20.99 Predict whether the following reactions will be spontaneous

in acidic solution under standard conditions: (a) oxidation of Sn to Sn2+ by I2 (to form I-), (b) reduction of Ni2+ to Ni by I- (to form I2), (c) reduction of Ce4+ to Ce3+ by H2O2, (d) reduc-tion of Cu2+ to Cu by Sn2+ (to form Sn4+).

[20.100] Gold exists in two common positive oxidation states, +1 and +3. The standard reduction potentials for these oxida-tion states are

Au+1aq2 + e- ¡ Au1s2 E°red = +1.69 V

Au3+1aq2 + 3 e- ¡ Au1s2 E°red = +1.50 V (a) Can you use these data to explain why gold does not tar-

nish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble com-plex ion of gold forms in the aqueous solution because of the redox reaction

4 Au1s2 + 8 NaCN1aq2 + 2 H2O1l2 + O21g2¡ 4 Na3Au1CN2241aq2 + 4 NaOH1aq2

What is being oxidized and what is being reduced in this reac-tion? (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metal. Write a balanced redox reaction for this process. What is being oxi-dized, and what is being reduced?

20.101 A voltaic cell is constructed from an Ni2+1aq2 - Ni1s2 half-cell and an Ag+1aq2 - Ag1s2 half-cell. The initial concentration of Ni2+1aq2 in the Ni2+ - Ni half-cell is 3Ni2+4 = 0.0100 M. The initial cell voltage is +1.12 V. (a) By using data in Table 20.1, calculate the standard emf of this voltaic cell. (b) Will the concentration of Ni2+1aq2 increase or decrease as the cell operates? (c) What is the initial concen-tration of Ag+1aq2 in the Ag+ -Ag half-cell?

[20.102] A voltaic cell is constructed that uses the following half-cell reactions:

Cu+1aq2 + e- ¡ Cu1s2I21s2 + 2 e- ¡ 2 I-1aq2

The cell is operated at 298 K with 3Cu+4 = 0.25 M and 3I-4 = 3.5 M. (a) Determine E for the cell at these concentra-tions. (b) Which electrode is the anode of the cell? (c) Is the an-swer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If 3Cu+4 were equal to 0.15 M, at what concentration of I- would the cell have zero potential?

20.103 Using data from Appendix E, calculate the equilibrium con-stant for the disproportionation of the copper(I) ion at room temperature:

2 Cu+1aq2 ¡ Cu2+1aq2 + Cu1s2.

Design an experiment 907

integrative Exercises 20.111 The Haber process is the principal industrial route for con-

verting nitrogen into ammonia:

N21g2 + 3 H21g2 ¡ 2 NH31g2 (a) What is being oxidized, and what is being reduced? (b) Using

the thermodynamic data in Appendix C, calculate the equilib-rium constant for the process at room temperature. (c) Calculate the standard emf of the Haber process at room temperature.

[20.112] In a galvanic cell the cathode is an Ag+11.00 M2>Ag1s2 half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing 0.10 M benzoic acid 1C6H5COOH2 and 0.050 M sodium benzoate 1C6H5COO-Na+2. The mea-sured cell voltage is 1.030 V. What is the pKa of benzoic acid?

20.113 Consider the general oxidation of a species A in solution: A ¡ A+ + e-. The term oxidation potential is sometimes used to describe the ease with which species A is oxidized—the easier a species is to oxidize, the greater its oxidation potential. (a) What is the relationship between the standard oxidation potential of A and the standard reduction poten-tial of A+? (b) Which of the metals listed in Table 4.5 has the highest standard oxidation potential? Which has the lowest? (c) For a series of substances, the trend in oxidation poten-tial is often related to the trend in the first ionization energy. Explain why this relationship makes sense.

20.114 A voltaic cell is based on Ag+1aq2>Ag1s2 and Fe3+1aq2>Fe2+1aq2 half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use S° values in Appendix C and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or de-creases when the temperature is raised above 25 °C.

20.115 Hydrogen gas has the potential for use as a clean fuel in reac-tion with oxygen. The relevant reaction is

2 H21g2 + O21g2 ¡ 2 H2O1l2 Consider two possible ways of utilizing this reaction as an

electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is cur-rently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at 85 °C. (a) Use data in Appendix C

to calculate ∆H° and ∆S° for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of ∆G for the reaction as the tem-perature increases? (c) What is the significance of the change in the magnitude of ∆G with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

20.116 Cytochrome, a complicated molecule that we will represent as CyFe2+, reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions. (Section 19.7) At pH 7.0 the following reduction potentials pertain to this oxidation of CyFe2+:

O21g2 + 4 H+1aq2 + 4 e- ¡ 2 H2 O1l2 E°red = +0.82 V

CyFe3+1aq2 + e- ¡ CyFe2+1aq2 E°red = +0.22 V

(a) What is ∆G for the oxidation of CyFe2+ by air? (b) If the synthesis of 1.00 mol of ATP from adenosine diphosphate (ADP) requires a ∆G of 37.7 kJ, how many moles of ATP are synthesized per mole of O2?

[20.117] The standard potential for the reduction of AgSCN(s) is +0.0895 V.

AgSCN1s2 + e- ¡ Ag1s2 + SCN-1aq2 Using this value and the electrode potential for Ag+1aq2, cal-

culate the Ksp for AgSCN. [20.118] The Ksp value for PbS(s) is 8.0 * 10-28. By using this value to-

gether with an electrode potential from Appendix E, determine the value of the standard reduction potential for the reaction

PbS1s2 + 2 e- ¡ Pb1s2 + S2-1aq2 20.119 A student designs an ammeter (a device that measures electri-

cal current) that is based on the electrolysis of water into hy-drogen and oxygen gases. When electrical current of unknown magnitude is run through the device for 2.00 min, 12.3 mL of water-saturated H21g2 is collected. The temperature of the sys-tem is 25.5 °C, and the atmospheric pressure is 768 torr. What is the magnitude of the current in amperes?

design an ExperimentYou are asked to construct a voltaic cell that would simulate an alkaline battery by providing an elec-trical output of 1.50 V at the beginning of its discharge. After you complete it your voltaic cell will be used to power an external device that draws a constant current of 0.50 amperes for 2.0 hours. You are given the following supplies: electrodes of each transition metal from manganese to zinc, the chlo-ride salts of the +2 transition metal ions from Mn2+ to Zn2+ (MnCl2, FeCl2, CoCl2, NiCl2, CuCl2 and ZnCl2), two 100 mL beakers, a salt bridge, a voltmeter, and wires to make electrical connections between the electrodes and the voltmeter. (a) Sketch out your voltaic cell labeling the metal used for each electrode and the type and concentration of the solutions in which each electrode is immersed. Be sure to describe how many grams of the salt are dissolved and the total volume of solution in each beaker. (b) What will be the concentrations of the transition metal ion in each solution at the end of the 2-h discharge? (c) What voltage will the cell register at the end of the discharge? (d) How long would your cell run before it died because the reactant in one of the half-cells was completely consumed? Assume the current stays constant throughout the discharge.

21 Nuclear ChemistryThe chemical energy we have discussed thus far originates in the making and breaking of chemical bonds, which result from the interaction of electrons between atoms. When chemical bonds are made or broken, the manner in which atoms are connected changes, but the number of atoms of each type is the same on both sides of the chemical equation. Indeed, for all the chemical reactions we have studied thus far, atoms are neither created nor destroyed. In this chapter, however, we examine a very different type of chemical transformation: reactions in which the nuclei of atoms undergo change, thereby changing the very identities of the atoms involved.

21.3 Nuclear TraNsmuTaTioNs We explore nuclear transmutations, which are nuclear reactions induced by bombardment of a nucleus by a neutron or an accelerated charged particle.

21.4 raTes of radioacTive decay We learn that radioisotope decays are first-order kinetic processes with characteristic half-lives. Decay rates can be used to determine the age of ancient artifacts and geological formations.

21.5 deTecTioN of radioacTiviTy We see that the radiation emitted by a radioactive substance can be detected by a variety of devices, including dosimeters, Geiger counters, and scintillation counters.

21.1 radioacTiviTy aNd Nuclear equaTioNs We begin by learning how to describe nuclear reactions using equations analogous to chemical equations, in which the nuclear charges and masses of reactants and products are in balance. We see that radioactive nuclei most commonly decay by emission of alpha, beta, or gamma radiation.

21.2 PaTTerNs of Nuclear sTabiliTy We see that nuclear stability is determined largely by the neutron-to-proton ratio. For stable nuclei, this ratio increases with increasing atomic number. All nuclei with 84 or more protons are radioactive. Heavy nuclei gain stability by a series of nuclear disintegrations leading to stable nuclei.

WhaT’s ahead

▶ A PELLET OF PLUTONIUM-238 DIOXIDE, which generates heat and light from its radioactive decay. Pellets such as these are used in radioisotope thermoelectric generators (RTGs) to generate electricity in space vehicles.

Transformations of atomic nuclei, which are called nuclear reactions, can involve enor-mous changes in energy—far greater than that involved in bond making and breaking. The energy of stars, such as our Sun, and the energy created by nuclear power plants offer examples of the tremendous energy released in nuclear reactions. Our chapter-opening photo shows a pellet of the dioxide of plutonium-238, an unstable isotope that undergoes a spontaneous process called nuclear decay. The pellet’s glow is a consequence of the significant amount of heat generated by the nuclear decay of plutonium-238. This heat is utilized to generate electricity in space vehicles using a device called a radioisotope thermoelectric generator (RTG). For example, all of the electricity used in the Curiosity rover vehicle currently exploring the surface of Mars is generated using an RTG that contains 4.8 kg of plutonium-238 dioxide. After its nuclear decay, the atoms of plutonium have changed into a different element—because the number of protons and neutrons in the nucleus generally changes during a nuclear decay, the identity of the atom changes.

21.8 Nuclear Power: fusioN We learn that in nuclear fusion two light nuclei are fused together to form a more stable, heavier nucleus.

21.9 radiaTioN iN The eNviroNmeNT aNd liviNg sysTems We discover that naturally occurring radioisotopes bathe our planet—and us—with low levels of radiation. The radiation emitted in nuclear reactions can damage living cells but also has diagnostic and therapeutic applications.

21.6 eNergy chaNges iN Nuclear reacTioNs We learn that energy changes in nuclear reactions are related to mass changes via Einstein’s equation, E = mc2. The nuclear binding energy of a nucleus is the difference between the mass of the nucleus and the sum of the masses of its nucleons.

21.7 Nuclear Power: fissioN We explore nuclear fission, in which a heavy nucleus splits to form two or more product nuclei. Nuclear fission is the energy source for nuclear power plants, and we look at the operating principles of these plants.

910 cHApTEr 21 Nuclear chemistry

Nuclear reactions are used to generate electricity on Earth as well as in space. Roughly 13% of the electricity generated worldwide comes from nuclear power plants, though the percentage varies from one country to the next, as ▲ Figure 21.1 shows.

The use of nuclear energy for power generation and the disposal of nuclear wastes from power plants, as well as concerns about nuclear weaponry, are controversial social and political issues. It is imperative, therefore, that as a citizen with a stake in these matters, you have some understanding of nuclear reactions and the uses of radioactive substances.

Nuclear chemistry is the study of nuclear reactions, with an emphasis on their uses and their effects on biological systems. Nuclear chemistry affects our lives in many ways, particularly in energy and medical applications. Radioactivity is also used to help determine the mechanisms of chemical reactions, to trace the movement of atoms in biological systems and the environment, and to date historical artifacts. Different iso-topes of the same element can undergo very different nuclear reactions, and one of our goals in this chapter is to gain a deeper appreciation for the differences among differ-ent radioactive isotopes and the ways in which they undergo decay and other nuclear transformations.

21.1 | Radioactivity and Nuclear equations

To understand nuclear reactions, we must review and develop some ideas introduced in Section 2.3. First, recall that two types of subatomic particles reside in the nucleus: protons and neutrons. We will refer to these particles as nucleons. Recall also that all

▲ Figure 21.1 Sources of electricity generation, worldwide and for select countries (Sources: U.S. Energy Information Administration and the International Energy Agency, 2010–2012 data).

Coal41%

Coal79%

Gas22%

Hydroelectric16%

France United States

Worldwide

China

Nuclear13%

Nuclear77%

Nuclear19%

Nuclear2%

Coal5%

Coal37%

Other2%

Other6%

Gas4%

Gas30%

Gasandoil2%

Hydroelectric11%

Hydroelectric7%

Hydroelectric17%

Oil1%

Oil1%

Oil5%

Other3%

sEcTioN 21.1 radioactivity and Nuclear Equations 911

atoms of a given element have the same number of protons; this number is the ele-ment’s atomic number. The atoms of a given element can have different numbers of neutrons, however, so they can have different mass numbers; the mass number is the total number of nucleons in the nucleus. Atoms with the same atomic number but dif-ferent mass numbers are known as isotopes.

The different isotopes of an element are distinguished by their mass numbers. For example, the three naturally occurring isotopes of uranium are uranium-234, ura-nium-235, and uranium-238, where the numerical suffixes represent the mass num-bers. These isotopes are also written 234

92U, 23592U, and 238

92U, where the superscript is the mass number and the subscript is the atomic number.*

Different isotopes of an element have different natural abundances. For example, 99.3% of naturally occurring uranium is uranium-238, 0.7% is uranium-235, and only a trace is uranium-234. Different isotopes of an element also have different stabilities. Indeed, the nuclear properties of any given isotope depend on the number of protons and neutrons in its nucleus.

A nuclide is a nucleus containing a specified number of protons and neutrons. Nuclides that are radioactive are called radionuclides, and atoms containing these nuclei are called radioisotopes.

Nuclear EquationsMost nuclei in nature are stable and remain intact indefinitely. Radionuclides, however, are unstable and spontaneously emit particles and electromagnetic radiation. Emission of radiation is one of the ways in which an unstable nucleus is transformed into a more stable one that has less energy. The emitted radiation is the carrier of the excess energy. Uranium-238, for example, is radioactive, undergoing a nuclear reaction emitting helium-4 nuclei. The helium-4 particles are known as alpha (A) particles, and a stream of them is called alpha radiation. When a 238

92U nucleus loses an alpha particle, the remain-ing fragment has an atomic number of 90 and a mass number of 234. The element with atomic number 90 is Th, thorium. Therefore, the products of uranium-238 decomposi-tion are an alpha particle and a thorium-234 nucleus. We represent this reaction by the nuclear equation

23892U ¡ 234

90Th + 42He [21.1]

When a nucleus spontaneously decomposes in this way, it is said either to have decayed or to have undergone radioactive decay. Because an alpha particle is involved in this reaction, scientists also describe the process as alpha decay.

Give It Some ThoughtWhat change in the mass number of a nucleus occurs when the nucleus emits an alpha particle?

In Equation 21.1 the sum of the mass numbers is the same on both sides of the equation 1238 = 234 + 42. Likewise, the sum of the atomic numbers on both sides of the equation is equal 192 = 90 + 22. Mass numbers and atomic numbers must be balanced in all nuclear equations.

The radioactive properties of the nucleus in an atom are independent of the chemi-cal state of the atom. In writing nuclear equations, therefore, we are not concerned with the chemical form (element or compound) of the atom in which the nucleus resides.

*As noted in Section 2.3, we often do not explicitly write the atomic number of an isotope because the element symbol is specific to the atomic number. In studying nuclear chemistry, however, it is often use-ful to include the atomic number in order to help us keep track of changes in the nuclei.


Types of Radioactive DecayThe three most common kinds of radiation given off when a radionuclide decays are alpha 1a2, beta 1b2, and gamma 1g2 radiation. (Section 2.2) ▼ Table 21.1 sum-marizes some of the important properties of these types of radiation. As just described, alpha radiation consists of a stream of helium-4 nuclei known as alpha particles, which we denote as 42He or simply a.

Beta radiation consists of streams of beta (B) particles, which are high-speed elec-trons emitted by an unstable nucleus. Beta particles are represented in nuclear equa-tions by 0

-1e or more commonly by b-. The superscript 0 indicates that the mass of the electron is exceedingly small relative to the mass of a nucleon. The subscript -1 represents the negative charge of the beta particle, which is opposite that of the proton.

Iodine-131 is an isotope that undergoes decay by beta emission:

13153I ¡ 131

54Xe + 0-1e [21.2]

You can see from this equation that beta decay causes the atomic number of the reac-tant to increase from 53 to 54, which means a proton was created. Therefore, beta emis-sion is equivalent to the conversion of a neutron (1

0n or simply n) to a proton (11H or

simply p):

10n ¡ 1

1H + 0-1e or n ¡ p + b- [21.3]

Just because an electron is emitted from a nucleus in beta decay, we should not think that the nucleus is composed of these particles any more than we consider a match to

soluTioNAnalyze We are asked to determine the nucleus that results when radium-226 loses an alpha particle.Plan We can best do this by writing a balanced nuclear reaction for the process.Solve The periodic table shows that radium has an atomic number of 88. The complete chemical symbol for radium-226 is therefore 226

88Ra. An alpha particle is a helium-4 nucleus, and so its symbol is 42He. The alpha particle is a product of the nuclear reaction, and so the equation is of the form

22688Ra ¡ A

ZX + 42He

where A is the mass number of the product nucleus and Z is its atomic number. Mass numbers and atomic numbers must bal-ance, so

226 = A + 4and

88 = Z + 2

sample exeRCise 21.1 predicting the product of a Nuclear Reaction

What product is formed when radium-226 undergoes alpha decay?

Hence,A = 222 and Z = 86

Again, from the periodic table, the element with Z = 86 is radon (Rn). The product, therefore, is 222

86Rn, and the nuclear equation is226

88Ra ¡ 22286Rn + 4

2He

practice exercise 1The plutonium-238 that is shown in the chapter-opening pho-tograph undergoes alpha decay. What product forms when this radionuclide decays?(a) Plutonium-234 (b) Uranium-234 (c) Uranium-238(d) Thorium-236 (e) Neptunium-237

practice exercise 2Which element undergoes alpha decay to form lead-208?

Table 21.1 properties of alpha, Beta, and Gamma Radiation

Type of Radiation

property A B G

Charge 2+ 1- 0Mass 6.64 * 10-24g 9.11 * 10-28g 0

Relative penetrating power 1 100 10,000Nature of radiation 4

2He nuclei Electrons High-energy photons

sEcTioN 21.1 radioactivity and Nuclear Equations 913

be composed of sparks simply because it gives them off when struck. The beta-particle electron comes into being only when the nucleus undergoes a nuclear reaction. Fur-thermore, the speed of the beta particle is sufficiently high that it does not end up in an orbital of the decaying atom.

Gamma 1G2 radiation (or gamma rays) consists of high-energy photons (that is, electromagnetic radiation of very short wavelength). It changes neither the atomic number nor the mass number of a nucleus and is represented as either 0

0g or simply g. Gamma radiation usually accompanies other radioactive emission because it repre-sents the energy lost when the nucleons in a nuclear reaction reorganize into more stable arrangements. Often gamma rays are not explicitly shown when writing nuclear equations.

Two other types of radioactive decay are positron emission and electron capture. A positron, 0

+1e, or, often simply as b+, is a particle that has the same mass as an elec-tron (thus, we use the letter e and superscript 0 for the mass) but the opposite charge (represented by the +1 subscript).*

The isotope carbon-11 decays by positron emission:

116C ¡ 11

5B + 0+1e [21.4]

Positron emission causes the atomic number of the reactant in this equation to decrease from 6 to 5. In general, positron emission has the effect of converting a proton to a neu-tron, thereby decreasing the atomic number of the nucleus by 1 while not changing the mass number:

11p ¡ 1

0n + 0+1e or p ¡ n + b+ [21.5]

Electron capture is the capture by the nucleus of an electron from the electron cloud surrounding the nucleus, as in this rubidium-81 decay:

8137Rb + 0

-1e 1orbital electron2 ¡ 8136Kr [21.6]

Because the electron is consumed rather than formed in the process, it is shown on the reactant side of the equation. Electron capture, like positron emission, has the effect of converting a proton to a neutron:

11p + 0

-1e ¡ 10n [21.7]

▶ Table 21.2 summarizes the symbols used to represent the particles commonly encountered in nuclear reactions. The various types of radioactive decay are summa-rized in ▼ Table 21.3.

Give It Some ThoughtWhich particles in Table 21.2 result in no change in nuclear charge when emitted in nuclear decay?

Table 21.2 particles Found in Nuclear Reactions

particle symbol

Neutron 10n or n

Proton 11H or p

Electron 0-1e

Alpha particle 42He or a

Beta particle 0-1e or b-

Positron 0+1e or b+

*The positron has a very short life because it is annihilated when it collides with an electron, producing gamma rays: 0

+1e + 0-1e ¡ 20

0g.

Table 21.3 Types of Radioactive decay

Change in Change in Type Nuclear equation atomic Number mass Number

Alpha decay AZX ¡ A-4

Z-2Y + 42He -2 -4

Beta emission AZX ¡ A

Z+1Y + 0-1e +1 Unchanged

Positron emission AZX ¡ A

Z-1Y + 0+1e -1 Unchanged

Electron capture* AZX + 0

-1e ¡ AZ-1Y -1 Unchanged

*The electron captured comes from the electron cloud surrounding the nucleus.


21.2 | patterns of Nuclear stabilitySome nuclides, such as 12

6C and 136C are stable, whereas others, such as 14

6C are unsta-ble and undergo fission. Why is it that some nuclides are stable but others that may have only one more or one fewer neutron are not? No single rule allows us to predict whether a particular nucleus is radioactive and, if it is, how it might decay. However, several empirical observations can help us predict the stability of a nucleus.

Neutron-to-Proton RatioBecause like charges repel each other, it may seem surprising that a large number of protons can reside within the small volume of the nucleus. At close distances, however, a strong force of attraction, called the strong nuclear force, exists between nucleons. Neutrons are intimately involved in this attractive force. All nuclei other than 1

1H contain neutrons. As the number of protons in a nucleus increases, there is an ever greater need for neutrons to counteract the proton–proton repulsions. Stable nuclei with atomic numbers up to about 20 have approximately equal numbers of neutrons and protons. For nuclei with atomic number above 20, the number of neutrons exceeds the number of protons. Indeed, the number of neutrons necessary to create a stable nucleus increases more rapidly than the number of protons. Thus, the neutron-to-proton ratios of stable nuclei increase with increasing atomic number, as illustrated by the most common isotopes of carbon, 126C 1n/p = 12, manganese, 55

25Mn 1n/p = 1.202, and gold, 19779Au 1n/p = 1.492.

▶ Figure 21.2 shows all known isotopes of the elements through Z = 100 plotted according to their numbers of protons and neutrons. Notice how the plot goes above the line for 1:1 neutron-to-proton for heavier elements. The dark blue dots in the figure represent stable (nonradioactive) isotopes. The region of the graph covered by these dark blue dots is known as the belt of stability. The belt of stability ends at element 83 (bismuth), which means that all nuclei with 84 or more protons are radioactive. For example, all isotopes of uranium, Z = 92, are radioactive.

soluTioNAnalyze We must write balanced nuclear equations in which the masses and charges of reactants and products are equal.Plan We can begin by writing the complete chemical symbols for the nuclei and decay particles that are given in the problem.Solve

(a) The information given in the question can be summarized as20180Hg + 0

-1e ¡ AZX

The mass numbers must have the same sum on both sides of the equation:

201 + 0 = AThus, the product nucleus must have a mass number of 201. Simi-larly, balancing the atomic numbers gives

80 - 1 = ZThus, the atomic number of the product nucleus must be 79, which identifies it as gold (Au):

20180Hg + 0

-1e ¡ 20179Au

(b) In this case we must determine what type of particle is emitted in the course of the radioactive decay:

23190Th ¡ 231

91Pa + AZX

sample exeRCise 21.2 Writing Nuclear equations

Write nuclear equations for (a) mercury-201 undergoing electron capture; (b) thorium-231 decaying to protactinium-231.

From 231 = 231 + A and 90 = 91 + Z, we deduce A = 0 and Z = -1. According to Table 21.2, the particle with these characteris-tics is the beta particle (electron). We therefore write

23190Th ¡ 231

91Pa + 0-1e or 231

90Th ¡ 23191Pa + b-

practice exercise 1The radioactive decay of thorium-232 occurs in multiple steps, called a radioactive decay chain. The second product produced in this chain is actinium-228. Which of the following processes could lead to this product starting with thorium-232?(a) Alpha decay followed by beta emission(b) Beta emission followed by electron capture(c) Positron emission followed by alpha decay(d) Electron capture followed by positron emission(e) More than one of the above is consistent with the observed

transformation

practice exercise 2Write a balanced nuclear equation for the reaction in which oxygen-15 undergoes positron emission.

sEcTioN 21.2 patterns of Nuclear stability 915

The type of radioactive decay that a particular radionuclide undergoes depends largely on how its neutron-to-proton ratio compares with those of nearby nuclei that lie within the belt of stability. We can envision three general situations:

1. Nuclei above the belt of stability (high neutron-to-proton ratios). These neutron-rich nuclei can lower their ratio and thereby move toward the belt of stability by emitting a beta particle because beta emission decreases the number of neutrons and increases the number of protons (Equation 21.3).

2. Nuclei below the belt of stability (low neutron-to-proton ratios). These proton-rich nuclei can increase their ratio and so move closer to the belt of stabil-ity by either positron emission or electron capture because both decays increase the number of neutrons and decrease the number of protons (Equations 21.5 and 21.7). Positron emission is more common among lighter nuclei. Electron capture becomes increasingly common as the nuclear charge increases.

3. Nuclei with atomic numbers #84. These heavy nuclei tend to undergo alpha emission, which decreases both the number of neutrons and the number of pro-tons by two, moving the nucleus diagonally toward the belt of stability.

▲ Figure 21.2 Stable and radioactive isotopes as a function of numbers of neutrons and protons in a nucleus. The stable nuclei (dark blue dots) define a region known as the belt of stability.

Go FiGuReEstimate the optimal number of neutrons for a nucleus containing 70 protons.

0 10 20 30 40 50 60 70 80 90 100Number of protons, p

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

160

Num

ber

of n

eutr

ons,

n

Nuclei below belt of stability, dominant decay mode = positron emission or electron capture

Nuclei with Z ≥ 84, dominant decay mode = alpha emission

Bi83209

(n/p = 1.52)

I53127

(n/p = 1.40)

26

816

(n/p = 1.00)O

Fe56

1:1 neutron-to-proton ratio

0

Nuclei above belt of stability, dominant decay mode = beta emission

(n/p = 1.15)


Radioactive Decay ChainsSome nuclei cannot gain stability by a single emission. Consequently, a series of succes-sive emissions occurs as shown for uranium-238 in ▶ Figure 21.3. Decay continues until a stable nucleus—lead-206 in this case—is formed. A series of nuclear reactions that be-gins with an unstable nucleus and terminates with a stable one is known as a radioactive decay chain or a nuclear disintegration series. Three such series occur in nature: uranium-238 to lead-206, uranium-235 to lead-207, and thorium-232 to lead-208. All of the decay processes in these series are either alpha emissions or beta emissions.

Further ObservationsTwo further observations can help us to predict stable nuclei: • Nucleiwiththemagic numbers of 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50,

82, or 126 neutrons are generally more stable than nuclei that do not contain these numbers of nucleons.

• Nucleiwithevennumbersofprotons,neutrons,orbotharemorelikelytobesta-ble than those with odd numbers of protons and/or neutrons. Approximately 60% of stable nuclei have an even number of both protons and neutrons, whereas less than 2% have odd numbers of both (◀ Table 21.4).These observations can be understood in terms of the shell model of the nucleus, in

which nucleons are described as residing in shells analogous to the shell structure for electrons in atoms. Just as certain numbers of electrons correspond to stable filled-shell electron configurations, so also the magic numbers of nucleons represent filled shells in nuclei.

There are several examples of the stability of nuclei with magic numbers of nucle-ons. For example, the radioactive series in Figure 21.3 ends with the stable 203

82Pb nucleus, which has a magic number of protons (82). Another example is the observa-tion that tin, which has a magic number of protons (50), has ten stable isotopes, more than any other element.

soluTioNAnalyze We are asked to predict the modes of decay of two nuclei.

Plan To do this, we must locate the respective nuclei in Figure 21.2 and determine their positions with respect to the belt of stability in order to predict the most likely mode of decay.

Solve

(a) Carbon is element 6. Thus, carbon-14 has 6 protons and 14 - 6 = 8 neutrons, giving it a neutron-to-proton ratio of 1.25. Elements with Z 6 20 normally have stable nuclei that contain approximately equal numbers of neutrons and protons 1n/p = 12. Thus, carbon-14 is located above the belt of stability and we expect it to decay by emitting a beta particle to decrease the n/p ratio:

146C ¡ 14

7N + 0-1e

This is indeed the mode of decay observed for carbon-14, a reaction that lowers the n/p ratio from 1.25 to 1.0.(b) Xenon is element 54. Thus, xenon-118 has 54 protons and

118 - 54 = 64 neutrons, giving it an n/p ratio of 1.18. Ac-cording to Figure 21.2, stable nuclei in this region of the belt of stability have higher neutron-to-proton ratios than xenon-118.

sample exeRCise 21.3 predicting modes of Nuclear decay

Predict the mode of decay of (a) carbon-14, (b) xenon-118.

The nucleus can increase this ratio by either positron emission or electron capture:

11854 Xe ¡ 118

53I + 0+1e

11854Xe + 0

-1e ¡ 11853I

In this case both modes of decay are observed.Comment Keep in mind that our guidelines do not always work. For example, thorium-233, which we might expect to undergo alpha de-cay, actually undergoes beta emission. Furthermore, a few radioactive nuclei lie within the belt of stability. Both 146

60Nd and 14860Nd, for ex-

ample, are stable and lie in the belt of stability. 14760 Nd, however, which

lies between them, is radioactive.

practice exercise 1Which of the following radioactive nuclei is most likely to decay via emission of a b- particle?(a) nitrogen-13 (b) magnesium-23 (c) rubidium-83(d) iodine-131 (e) neptunium-237

practice exercise 2Predict the mode of decay of (a) plutonium-239, (b) indium-120.

Table 21.4 Number of stable isotopes with even and odd Numbers of protons and Neutrons

Number of stable isotopes

proton Number

Neutron Number

157 Even Even 53 Even Odd 50 Odd Even 5 Odd Odd

sEcTioN 21.2 patterns of Nuclear stability 917

Evidence also suggests that pairs of protons and pairs of neutrons have a special stability, analogous to the pairs of electrons in molecules. This evidence accounts for the second observation noted earlier, that stable nuclei with an even number of protons and/or neutrons are far more numerous than those with odd numbers. The preference for even numbers of protons is illustrated in ▼ Figure 21.4, which shows the number

▲ Figure 21.3 Nuclear decay chain for uranium-238. The decay continues until the stable nucleus 206Pb is formed.

Each blue arrow represents decay by alpha emission

Each red arrow represents decay by beta emission

81Tl

82Pb

83Bi

84Po

85At

86Rn

87Fr

88Ra

89Ac

90Th

91Pa

92U

93Np

204

206

208

210

212

214

216

218

220

222

224

226

228

230

232

234

238

236

Atomic number

U

UPaTh

PoBiPb

BiPb

Pb

Po

Mas

s nu

mbe

r

Th

Ra

Rn

Po

1H(2)

4Be(1)

3Li(2)11Na(1)

19K(2)

21Sc(1)

37Rb(1)

12Mg(3)

20Ca(5)

38Sr(3)

23V(2)

41Nb(1)

39Y(1)

22Ti(5)

40Zr(4)

24Cr(4)

42Mo(6)

25Mn(1)

43Tc(0)

5B(2)

13Al(1)

31Ga(2)

49In(1)

6C(2)

14Si(3)

32Ge(4)

50Sn(10)

7N(2)

15P

(1)

33As(1)

51Sb(2)

8O(3)

16S

(4)

34Se(5)

52Te(6)

9F

(1)

17Cl(2)

35Br(2)

53I

(1)

2He(2)10Ne(3)

18Ar(3)

36Kr(6)

54Xe(9)

26Fe(4)

44Ru(7)

28Ni(5)

46Pd(6)

29Cu(2)

47Ag(2)

30Zn(5)

48Cd(6)

27Co(1)

45Rh(1)

Elements with twoor fewer stable isotopes

Elements with threeor more stable isotopes

Number ofstable isotopes

▲ Figure 21.4 Number of stable isotopes for elements 1–54.

Go FiGuReAmong the elements shown here, how many have an even number of protons and fewer than three stable isotopes? How many have an odd number of protons and more than two stable isotopes?


of stable isotopes for all elements up to Xe. Notice that once we move past nitrogen, the elements with an odd number of protons invariably have fewer stable isotopes than their neighbors with an even number of protons.

Give It Some ThoughtWhat can you say about the number of neutrons in the stable isotopes of fluorine, sodium, aluminum, and phosphorus?

21.3 | Nuclear TransmutationsThus far we have examined nuclear reactions in which a nucleus decays spontaneously. A nucleus can also change identity if it is struck by a neutron or by another nucleus. Nuclear reactions induced in this way are known as nuclear transmutations.

In 1919, Ernest Rutherford performed the first conversion of one nucleus into another, using alpha particles emitted by radium to convert nitrogen-14 into oxygen-17:

147N + 4

2He ¡ 178O + 1

1H or 147N + a ¡ 17

8O + p [21.8]

Such reactions have allowed scientists to synthesize hundreds of radioisotopes in the laboratory.

A shorthand notation often used to represent nuclear transmutations lists the tar-get nucleus, the bombarding particle and the ejected particle in parentheses, followed by the product nucleus. Using this condensed notation, Equation 21.8 becomes

N (α, p) O147

178

Targetnucleus

Productnucleus

Bombardingparticle

Ejectedparticle

soluTioNAnalyze We must go from the condensed descriptive form of the reac-tion to the balanced nuclear equation.Plan We arrive at the balanced equation by writing n and a, each with its associated subscripts and superscripts.Solve The n is the abbreviation for a neutron 11

0n2 and a represents an alpha particle 14

2He2. The neutron is the bombarding particle, and the alpha particle is a product. Therefore, the nuclear equation is

2713Al + 1

0n ¡ 2411Na + 4

2He or 2713Al + n ¡ 24

11Na + a

sample exeRCise 21.4 Writing a Balanced Nuclear equation

Write the balanced nuclear equation for the process summarized as 2713Al1n, a224

11Na.

practice exercise 1Consider the following nuclear transmutation: 238

92U1n, b-2X. What is the identity of nucleus X?(a) 238

93Np (b) 23992U (c) 239

92U+ (d) 23590Th (e) 239

93Np

practice exercise 2Write the condensed version of the nuclear reaction

168O + 1

1H ¡ 137N + 4

2He

Accelerating Charged ParticlesAlpha particles and other positively charged particles must move very fast to overcome the electrostatic repulsion between them and the target nucleus. The higher the nuclear charge on either the bombarding particle or the target nucleus, the faster the bombarding particle must move to bring about a nuclear reaction. Many methods have been devised to accelerate charged particles, using strong magnetic and electrostatic fields. These particle accelerators, popularly called “atom smashers,” bear such names as cyclotron and synchrotron.

sEcTioN 21.3 Nuclear Transmutations 919

A common theme in all particle accelerators is the need to create charged parti-cles so that they can be manipulated by electric and magnetic fields. The tubes through which the particles move must be kept at high vacuum so that the particles do not inad-vertently collide with any gas-phase molecules.

▲ Figure 21.5 shows the Relativistic Heavy Ion Collider (RHIC) located at Brookhaven National Laboratory. This facility and the Large Hadron Collider (LHC) at CERN (Conseil Européen pour la Recherche Nucléaire) near Geneva are two of the largest particle accelerators in the world. Both LHC and RHIC are capable of acceler-ating protons, as well as heavy ions such as gold and lead, to speeds approaching the speed of light. Scientists study the outcomes of collisions involving these ultra-high-energy particles. These experiments are used to investigate the fundamental structure of matter and ultimately answer questions about the beginning of the universe. In 2013, the existence of an important fundamental particle in particle physics, called the Higgs boson, was experimentally confirmed at the LHC.

Reactions Involving NeutronsMost synthetic isotopes used in medicine and scientific research are made using neu-trons as the bombarding particles. Because neutrons are neutral, they are not repelled by the nucleus. Consequently, they do not need to be accelerated to cause nuclear reactions. The neutrons are produced in nuclear reactors. For example, cobalt-60, which is used in cancer radiation therapy, is produced by neutron capture. Iron-58 is placed in a nuclear reactor and bombarded by neutrons to trigger the following sequence of reactions:

5826Fe + 1

0n ¡ 5926Fe [21.9]

5926Fe ¡ 59

27Co + 0-1e [21.10]

5927Co + 1

0n ¡ 6027Co [21.11]

▲ Figure 21.5 The Relativistic Heavy Ion Collider. This particle accelerator is located at Brookhaven National Laboratory on Long Island, New York.

Gold atoms are ionized, creating ions that are accelerated in a Tandem van de Graaff accelerator

1

The booster synchrotron and alternate gradient synchrotron (AGS) further accelerate the ions to 99.7% of the speed of light

3

Finally the ions are transferred to RHIC, which has a circumfer-ence of 3.8 km. Ions moving in opposite directions can collide at one of six points on the ring, marked with white rectangles

4

If needed, beams of H+ ions can be generated in the Linac

2


Give It Some ThoughtCan an electrostatic or magnetic field be used to accelerate neutrons in a particle accelerator? Why or why not?

Transuranium ElementsNuclear transmutations have been used to produce the elements with atomic number above 92, collectively known as the transuranium elements because they follow ura-nium in the periodic table. Elements 93 (neptunium, Np) and 94 (plutonium, Pu) were produced in 1940 by bombarding uranium-238 with neutrons:

23892U + 1

0n ¡ 23992U ¡ 239

93Np + 0-1e [21.12]

23993Np ¡ 239

94Pu + 0-1e [21.13]

Elements with still larger atomic numbers are normally formed in small quantities in particle accelerators. Curium-242, for example, is formed when a plutonium-239 tar-get is bombarded with accelerated alpha particles:

23994Pu + 4

2He ¡ 24296Cm + 1

0n [21.14]

New advances in the detection of the decay patterns of single atoms have led to recent additions to the periodic table. Between 1994 and 2010, elements 110 through 118 were discovered via the nuclear reactions that occur when nuclei of much lighter elements collide with high energy. For example, in 1996 a team of European scientists based in Germany synthesized element 112, copernicium, Cn, by bombarding a lead target con-tinuously for three weeks with a beam of zinc atoms:

20882Pb + 70

30Zn ¡ 277112Cn + 1

0n [21.15]

Amazingly, their discovery was based on the detection of only one atom of the new element, which decays after roughly 100 ms by alpha decay to form darmstadtium-273 (element 110). Within one minute, another five alpha decays take place producing fermium-253 (element 100). The finding has been verified in both Japan and Russia.

Because experiments to create new elements are very complicated and produce only a very small number of atoms of the new elements, they need to be carefully evaluated and reproduced before the new element is made an official part of the periodic table. The International Union for Pure and Applied Chemistry (IUPAC) is the international body that authorizes names of new elements after their experimental discovery and confirmation. In 2012, IUPAC officially approved names for the two latest elements added to the periodic table, which are flerovium (element 114) and livermorium (element 116).

21.4 | Rates of Radioactive decaySome radioisotopes, such as uranium-238, are found in nature even though they are not stable. Other radioisotopes do not exist in nature but can be synthesized in nuclear reactions. To understand this distinction, we must realize that different nuclei undergo radioactive decay at different rates. Many radioisotopes decay essentially completely in fractions of a second, so we do not find them in nature. Uranium-238, on the other hand, decays very slowly. Therefore, despite its instability, we can still observe what re-mains from its formation in the early history of the universe.

Radioactive decay is a first-order kinetic process. Recall that a first-order process has a characteristic half-life, which is the time required for half of any given quantity of a substance to react. (Section 14.4) Nuclear decay rates are commonly expressed in terms of half-lives. Each radioisotope has its own characteristic half-life. For example, the half-life of strontium-90 is 28.8 yr (◀ Figure 21.6). If we start with 10.0 g of strontium-90, only

▲ Figure 21.6 Decay of a 10.0-g sample of strontium-90 1 t1/2 = 28.8 yr 2 . The 10 * 10 grids show how much of the radioactive isotope remains after various amounts of time.

Go FiGuReIf we start with a 50.0-g sample, how much of it remains after three half-lives have passed?

0 20 40 60 80 100 120

2.0

4.0

6.0

8.0

10.0

Time (yr)

Mas

s of

90Sr

(g)

sEcTioN 21.4 rates of radioactive Decay 921

5.0 g of that isotope remains after 28.8 yr, 2.5 g remains after another 28.8 yr, and so on. Strontium-90 decays to yttrium-90 via beta emission:

9038Sr ¡ 90

39Y + 0-1e [21.16]

Half-lives as short as millionths of a second and as long as billions of years are known. The half-lives of some radioisotopes are listed in ▲ Table 21.5. One important feature of half-lives for nuclear decay is that they are unaffected by external conditions such as temperature, pressure, or state of chemical combination. Unlike toxic chemicals, therefore, radioactive atoms cannot be rendered harmless by chemical reaction or by any other practical treatment. At this point, we can do nothing but allow these nuclei to lose radioactivity at their characteristic rates. In the meantime, we must take precautions to prevent radioisotopes, such as those produced in nuclear power plants (Section 21.7), from entering the environment because of the damage radiation can cause.

Table 21.5 The half-lives and Type of decay for several Radioisotopes

isotope half-life (yr) Type of decay

Natural radioisotopes 23892U 4.5 * 109 Alpha

23592U 7.0 * 108 Alpha

23290Th 1.4 * 1010 Alpha

4019K 1.3 * 109 Beta146C 5700 Beta

Synthetic radioisotopes 23994Pu 24,000 Alpha

13755Cs 30.2 Beta

9038Sr 28.8 Beta13153I 0.022 Beta

soluTioNAnalyze We are given the half-life for cobalt-60 and asked to calculate the amount of cobalt-60 remaining from an initial 1.000-mg sample after 15.81 yr.Plan We will use the fact that the amount of a radioactive substance decreases by 50% for every half-life that passes.Solve Because 5.27 * 3 = 15.81, 15.81 yr is three half-lives for cobalt-60. At the end of one half-life, 0.500 mg of cobalt-60 remains, 0.250 mg at the end of two half-lives, and 0.125 mg at the end of three half-lives.

practice exercise 1A radioisotope of technetium is useful in medical imaging tech-niques. A sample initially contains 80.0 mg of this isotope. After

sample exeRCise 21.5 Calculation involving half-lives

The half-life of cobalt-60 is 5.27 yr. How much of a 1.000-mg sample of cobalt-60 is left after 15.81 yr?

24.0 h, only 5.0 mg of the technetium isotope remains. What is the half-life of the isotope?(a) 3.0 h (b) 6.0 h (c) 12.0 h (d) 16.0 h (e) 24.0 h

practice exercise 2Carbon-11, used in medical imaging, has a half-life of 20.4 min. The carbon-11 nuclides are formed, and the carbon atoms are then incorporated into an appropriate compound. The resulting sample is injected into a patient, and the medical image is obtained. If the entire process takes five half-lives, what percentage of the original carbon-11 remains at this time?

Radiometric DatingBecause the half-life of any particular nuclide is constant, the half-life can serve as a “nuclear clock” to determine the age of objects. The method of dating objects based on their isotopes and isotope abundances is called radiometric dating.

When carbon-14 is used in radiometric dating, the technique is known as radiocar-bon dating. The procedure is based on the formation of carbon-14 as neutrons created by


cosmic rays in the upper atmosphere convert nitrogen-14 into carbon-14 (▲ Figure 21.7). The 14C reacts with oxygen to form 14CO2 in the atmosphere, and this “labeled” CO2 is taken up by plants and introduced into the food chain through photosynthesis. This pro-cess provides a small but reasonably constant source of carbon-14, which is radioactive and undergoes beta decay with a half-life of 5700 yr (to two significant figures):

146C ¡ 14

7N + 0-1e [21.17]

Because a living plant or animal has a constant intake of carbon compounds, it is able to maintain a ratio of carbon-14 to carbon-12 that is nearly identical with that of the atmosphere. Once the organism dies, however, it no longer ingests carbon compounds to replenish the carbon-14 lost through radioactive decay. The ratio of carbon-14 to carbon-12 therefore decreases. By measuring this ratio and comparing it with that of the atmosphere, we can estimate the age of an object. For example, if the ratio diminishes to half that of the atmosphere, we can conclude that the object is one half-life, or 5700 yr old.

This method cannot be used to date objects older than about 50,000 yr because after this length of time the radioactivity is too low to be measured accurately.

In radiocarbon dating, a reasonable assumption is that the ratio of carbon-14 to carbon-12 in the atmosphere has been relatively constant for the past 50,000 yr. How-ever, because variations in solar activity control the amount of carbon-14 produced in the atmosphere, that ratio can fluctuate. We can correct for this effect by using other kinds of data. Recently scientists have compared carbon-14 data with data from tree rings, corals, lake sediments, ice cores, and other natural sources to correct variations in the carbon-14 “clock” back to 26,000 yr.

Cosmic rays (largely protons) enter the atmosphere and collide with atoms, creating neutrons

1

Nitrogen atoms capture a neutron and emit a proton, forming 14C

2

14C atoms are incorporated in CO2, which is taken up by plants and made into more complex molecules through photosynthesis

3

Animals and people take in 14C by eating plants

4

Once an organism dies, intake of 14C ceases and its concentration decreases through beta emission to form 14N

5

1n0

1n014N + 7

1p114C + 6

14C 614N + 7

0e−1

14C + O214CO2

▲ Figure 21.7 Creation and distribution of carbon-14. The ratio of carbon-14 to carbon-12 in a dead animal or plant is related to the time since death occurred.

Go FiGuReHow does 14CO2 become incorporated into the mammalian food chain?


Other isotopes can be similarly used to date other types of objects. For example, it takes 4.5 * 109 yr for half of a sample of uranium-238 to decay to lead-206. The age of rocks containing uranium can therefore be determined by measuring the ratio of lead-206 to uranium-238. If the lead-206 had somehow become incorporated into the rock by normal chemical processes instead of by radioactive decay, the rock would also contain large amounts of the more abundant isotope lead-208. In the absence of large amounts of this “geonormal” isotope of lead, it is assumed that all of the lead-206 was at one time uranium-238.

The oldest rocks found on Earth are approximately 3 * 109 yr old. This age indi-cates that Earth’s crust has been solid for at least this length of time. Scientists estimate that it required 1 * 109 to 1.5 * 109 yr for Earth to cool and its surface to become solid, making the age of Earth 4.0 to 4.5 * 109 yr.

Calculations Based on Half-LifeSo far, our discussion has been mainly qualitative. We now consider the topic of half-lives from a more quantitative point of view. This approach enables us to determine the half-life of a radioisotope or the age of an object.

As noted earlier, radioactive decay is a first-order kinetic process. Its rate, there-fore, is proportional to the number of radioactive nuclei N in a sample:

Rate = kN [21.18]

The first-order rate constant, k, is called the decay constant.The rate at which a sample decays is called its activity, and it is often expressed as

number of disintegrations per unit time. The becquerel (Bq) is the SI unit for express-ing activity. A becquerel is defined as one nuclear disintegration per second. An older, but still widely used, unit of activity is the curie (Ci), defined as 3.7 * 1010 disintegra-tions per second, which is the rate of decay of 1 g of radium. Thus, a 4.0-mCi sample of cobalt-60 undergoes

4.0 * 10-3 Ci *3.7 * 1010 disintegrations>s

1 Ci= 1.5 * 108 disintegrations>s

and so has an activity of 1.5 * 108 Bq.As a radioactive sample decays, the amount of radiation emanating from the sam-

ple decays as well. For example, the half-life of cobalt-60 is 5.27 yr. The 4.0-mCi sample of cobalt-60 would, after 5.27 yr, have a radiation activity of 2.0 mCi, or 7.5 * 107 Bq.

Give It Some ThoughtIf the size of a radioactive sample is doubled, what happens to the activity of the sample in Bq?

As we saw in Section 14.4, a first-order rate law can be transformed into the equation

ln Nt

N0= -kt [21.19]

In this equation t is the time interval of decay, k is the decay constant, N0 is the initial number of nuclei (at time zero), and Nt is the number remaining after the time interval. Both the mass of a particular radioisotope and its activity are proportional to the num-ber of radioactive nuclei. Thus, either the ratio of the mass at any time t to the mass at time t = 0 or the ratio of the activities at time t and t = 0 can be substituted for Nt>N0 in Equation 21.19.

From Equation 21.19 we can obtain the relationship between the decay constant, k, and half-life, t1>2: (Section 14.4)

k =0.693t1>2

[21.20]


where we have used the value ln1Nt>N02 = ln10.52 = -0.693 for one half-life. Thus, if we know the value of either the decay constant or the half-life, we can calculate the value of the other.

Give It Some ThoughtWould doubling the mass of a radioactive sample change the half-life for the radioactive decay?

soluTioNAnalyze We are told that a rock sample has a certain amount of lead-206 for every unit mass of uranium-238 and asked to estimate the age of the rock.Plan Lead-206 is the product of the radioactive decay of uranium-238. We will assume that the only source of lead-206 in the rock is from the decay of uranium-238, with a known half-life. To apply first-order kinetics expressions (Equations 21.19 and 21.20) to calculate the time elapsed since the rock was formed, we first need to calculate how much initial uranium-238 there was for every 1 mg that remains today.Solve Let’s assume that the rock currently contains 1.000 mg of uranium-238 and therefore 0.257 mg of lead-206. The amount of uranium-238 in the rock when it was first formed therefore equals 1.000 mg plus the quantity that has decayed to lead-206. Because the mass of lead atoms is not the same as the mass of uranium atoms, we cannot just add 1.000 mg and 0.257 mg. We have to multiply the present mass of lead-206 (0.257 mg) by the ratio of the mass number of uranium to that of lead, into which it has decayed. Therefore, the original mass of 238

92U was

Original 23892U = 1.000 mg +

238206

10.257 mg2 = 1.297 mg

Using Equation 21.20, we can calculate the decay constant for the pro-cess from its half-life:

k =0.693

4.5 * 109 yr= 1.5 * 10-10 yr-1

sample exeRCise 21.6 Calculating the age of objects using Radioactive decay

A rock contains 0.257 mg of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is 4.5 * 109 yr. How old is the rock?

Rearranging Equation 21.19 to solve for time, t, and substituting known quantities gives

t = -1k

lnNt

N0= -

11.5 * 10-10 yr-1 ln

1.0001.297

= 1.7 * 109 yr

Comment To check this result, you could use the fact that the decay of uranium-235 to lead-207 has a half-life of 7 * 108 yr and measure the relative amounts of uranium-235 and lead-207 in the rock.

practice exercise 1Cesium-137, which has a half-life of 30.2 yr, is a component of the radioactive waste from nuclear power plants. If the activity due to cesium-137 in a sample of radioactive waste has decreased to 35.2% of its initial value, how old is the sample?(a) 1.04 yr (b) 15.4 yr (c) 31.5 yr (d) 45.5 yr (e) 156 yr

practice exercise 2A wooden object from an archeological site is subjected to radiocarbon dating. The activity due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5700 yr. What is the age of the archeologi-cal sample?

If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 yr. (a) What is the half-life of strontium-90? (b) How much strontium-90 will remain after 5.00 yr? (c) What is the initial activity of the sample in becquerels and curies?

soluTioN(a) Analyze We are asked to calculate a half-life, t1>2, based on data that tell us how much of a radioactive nucleus has decayed in a time interval t = 2.00 yr and the information N0 = 1.000 g, Nt = 0.953 g.Plan We first calculate the rate constant for the decay, k, and then use that to compute t1>2.

sample exeRCise 21.7 Calculations involving Radioactive

decay and Time


Solve Equation 21.19 is solved for the decay constant, k, and then Equation 21.20 is used to cal-culate half-life, t1>2:

k = -1t

lnNt

N0= -

12.00 yr

ln0.953 g1.000 g

= -1

2.00 yr 1-0.04812 = 0.0241 yr-1

t1>2 =0.693

k=

0.6930.0241 yr-1 = 28.8 yr

(b) Analyze We are asked to calculate the amount of a radionuclide remaining after a given period of time.

Plan We need to calculate Nt, the amount of strontium present at time t, using the initial quan-tity, N0, and the rate constant for decay, k, calculated in part (a).Solve Again using Equation 21.19, with k = 0.0241 yr-1, we have

ln Nt

N0= -kt = -10.0241 yr-1215.00 yr2 = -0.120

Nt>N0 is calculated from ln1Nt>N02 = -0.120 using the ex or INV LN function of a calculator:

Nt

N0= e-0.120 = 0.887

Because N0 = 1.000 g, we have

Nt = 10.8872N0 = 10.887211.000 g2 = 0.887 g

(c) Analyze We are asked to calculate the activity of the sample in becquerels and curies.Plan We must calculate the number of disintegrations per atom per second and then multiply by the number of atoms in the sample.Solve The number of disintegrations per atom per second is given by the decay constant, k:

k = a 0.0241yr

b a 1 yr365 days

b a 1 day24 h

b a 1 h3600 s

b = 7.64 * 10-10 s-1

To obtain the total number of disintegrations per second, we calculate the number of atoms in the sample. We multiply this quantity by k, where we express k as the number of disintegrations per atom per second, to obtain the number of disintegrations per second:

11.000 g 90Sr2a 1 mol 90Sr90 g 90Sr

b a 6.022 * 1023 atoms Sr1 mol 90Sr

b = 6.7 * 1021 atoms 90Sr

Total disintegrations/s = a 7.64 * 10-10 disintegrationsatom # s

b16.7 * 1021 atoms2= 5.1 * 1012 disintegrations/s

Because 1 Bq is one disintegration per second, the activity is 5.1 * 1012 Bq. The activity in curies is given by

15.1 * 1012 disintegrations/s2a 1 Ci3.7 * 1010 disintegrations/s

b = 1.4 * 102 Ci

We have used only two significant figures in products of these calculations because we do not know the atomic weight of 90Sr to more than two significant figures without looking it up in a special source.

practice exercise 1As mentioned in the previous Practice Exercise 1, cesium-137, a component of radioactive waste, has a half-life of 30.2 yr. If a sample of waste has an initial activity of 15.0 Ci due to cesium-137, how long will it take for the activity due to cesium-137 to drop to 0.250 Ci?(a) 0.728 yr (b) 60.4 yr (c) 78.2 yr (d) 124 yr (e) 178 yr

practice exercise 2A sample to be used for medical imaging is labeled with 18F, which has a half-life of 110 min. What percentage of the original activity in the sample remains after 300 min?


21.5 | detection of RadioactivityA variety of methods have been devised to detect emissions from radioactive sub-stances. Henri Becquerel discovered radioactivity because radiation caused fogging of photographic plates, and since that time photographic plates and film have been used to detect radioactivity. The radiation affects photographic film in much the same way as X-rays do. The greater the extent of exposure to radiation, the darker the area of the developed negative. People who work with radioactive substances carry film badges to record the extent of their exposure to radiation (▼ Figure 21.8).

Radioactivity can also be detected and measured by a Geiger counter. The operation of this device is based on the fact that radiation is able to ionize matter. The ions and electrons produced by the ionizing radiation permit conduction of an electrical current. The basic design of a Geiger counter is shown in ▶ Figure 21.9. A current pulse between the anode and the metal cylinder occurs whenever enter-ing radiation produces ions. Each pulse is counted in order to estimate the amount of radiation.

Give It Some ThoughtWhich type of radiation—alpha, beta, or gamma—is most likely to be stopped by the window of a Geiger counter?

Substances that are electronically excited by radiation can also be used to detect and measure radiation. For example, some substances excited by radiation give off light as electrons return to their lower-energy states. These substances are called phosphors. Different substances respond to different particles. Zinc sulfide, for example, responds to alpha particles. An instrument called a scintillation counter is used to detect and measure radiation, based on the tiny flashes of light produced when radiation strikes a suitable phosphor. The flashes of light are magnified electronically and counted to measure the amount of radiation.

The film strip is white before exposure to radiation

The film strip is darkened on exposure to radiation

▲ Figure 21.8 Badge dosimeters monitor the extent to which the individual has been exposed to high-energy radiation. The radiation dose is determined from the extent of darkening of the film in the dosimeter.

Go FiGuReWhich type of radiation—alpha, beta, or gamma—is likely to fog a film that is sensitive to X rays?

sEcTioN 21.5 Detection of radioactivity 927

RadiotracersBecause radioisotopes can be detected readily, they can be used to follow an element through its chemical reactions. The incorporation of carbon atoms from CO2 into glucose during photosynthesis, for example, has been studied using CO2 enriched in carbon-14:

6 14CO2 + 6 H2O ¡Sunlight

Chlorophyll 14C6H12O6 + 6 O2 [21.21]

The use of the carbon-14 label provides direct experimental evidence that carbon diox-ide in the environment is chemically converted to glucose in plants. Analogous labeling experiments using oxygen-18 show that the O2 produced during photosynthesis comes from water, not carbon dioxide. When it is possible to isolate and purify intermediates and products from reactions, detection devices such as scintillation counters can be used to “follow” the radioisotope as it moves from starting material through intermedi-ates to final product. These types of experiments are useful for identifying elementary steps in a reaction mechanism. (Section 14.6)

The use of radioisotopes is possible because all isotopes of an element have essen-tially identical chemical properties. When a small quantity of a radioisotope is mixed with the naturally occurring stable isotopes of the same element, all the isotopes go through the same reactions together. The element’s path is revealed by the radioactivity of the radioisotope. Because the radioisotope can be used to trace the path of the ele-ment, it is called a radiotracer.

Give It Some ThoughtCan you think of a process not involving radioactive decay for which 14CO2 would behave differently from 12CO2?

Highvoltage

Amplifierand counter

Thin windowpenetrated by

radiation

γ-ray

Argon gas

Metal cylinderacting as

cathode (−)Anode (+)

Radiation (α-, β-, or γ- rays) penetrates thin window

Current is amplified and measured as series of pulses, with each pulse signaling detection of a radioactive particle or ray.

Radiation ionizes gaseous atoms (usually Ar or He), creating positively charged ions and electrons

Charged particles moving between anode and cathode create electric current.

1

4

23

Ar+

e−

3 9 8 0 0 0

▲ Figure 21.9 Schematic drawing of a Geiger counter.

Go FiGuReWhich property of the atoms of gas inside a Geiger counter is most relevant to the operation of the device?


Chemistry and life

Medical Applications of Radiotracers

Radiotracers have found wide use as diagnostic tools in medicine. ▼ Table 21.6 lists some radiotracers and their uses. These radioiso-topes are incorporated into a compound that is administered to the patient, usually intravenously. The diagnostic use of these isotopes is based on the ability of the radioactive compound to localize and con-centrate in the organ or tissue under investigation. Iodine-131, for example, has been used to test the activity of the thyroid gland. This gland is the only place in which iodine is incorporated significantly in the body. The patient drinks a solution of NaI containing iodine-131. Only a very small amount is used so that the patient does not receive a harmful dose of radioactivity. A Geiger counter placed close to the thyroid, in the neck region, determines the ability of the thyroid to take up the iodine. A normal thyroid will absorb about 12% of the io-dine within a few hours.

The medical applications of radiotracers are further illustrated by positron emission tomography (PET). PET is used for clinical

diagnosis of many diseases. In this method, compounds contain-ing radionuclides that decay by positron emission are injected into a patient. These compounds are chosen to enable researchers to monitor blood flow, oxygen and glucose metabolic rates, and other biological functions. Some of the most interesting work involves the study of the brain, which depends on glucose for most of its en-ergy. Changes in how this sugar is metabolized or used by the brain may signal a disease such as cancer, epilepsy, Parkinson’s disease, or schizophrenia.

The compound to be detected in the patient must be labeled with a radionuclide that is a positron emitter. The most widely used nuclides are carbon-11 1t1>2 = 20.4 min2, fluorine-18 1t1>2 = 110 min2, oxy-gen-15 1t1>2 = 2 min2, and nitrogen-13 1t1>2 = 10 min2. Glucose, for example, can be labeled with carbon@11. Because the half-lives of positron emitters are so short, they must be generated on site us-ing a cyclotron and the chemist must quickly incorporate the radio-nuclide into the sugar (or other appropriate) molecule and inject the compound immediately. The patient is placed in an instrument that measures the positron emission and constructs a computer-based im-age of the organ in which the emitting compound is localized. When the element decays, the emitted positron quickly collides with an electron. The positron and electron are annihilated in the collision, producing two gamma rays that move in opposite directions. The gamma rays are detected by an encircling ring of scintillation counters (▼ Figure 21.10). Because the rays move in opposite directions but were created in the same place at the same time, it is possible to ac-curately locate the point in the body where the radioactive isotope decayed. The nature of this image provides clues to the presence of disease or other abnormality and helps medical researchers under-stand how a particular disease affects the functioning of the brain. For example, the images shown in ▼ Figure 21.11 reveal that levels of ac-tivity in brains of patients with Alzheimer’s disease are different from the levels in those without the disease.

Related Exercises: 21.55, 21.56, 21.82, 21.83

Table 21.6 some Radionuclides used as Radiotracers

Nuclide half-life area of the Body studied

Iodine-131 8.04 days ThyroidIron-59 44.5 days Red blood cellsPhosphorus-32 14.3 days Eyes, liver, tumorsTechnetium-99a 6.0 hours Heart, bones, liver, and

lungsThallium-201 73 hours Heart, arteriesSodium-24 14.8 hours Circulatory systemaThe isotope of technetium is actually a special isotope of Tc-99 called Tc-99m, where the m indicates a so-called metastable isotope.

▲ Figure 21.11 Positron emission tomography (PET) scans showing glucose metabolism levels in the brain. Red and yellow colors show higher levels of glucose metabolism.

Normal Mild cognitiveimpairment

Alzheimer’sdisease

▲ Figure 21.10 Schematic representation of a positron emission tomography (PET) scanner.

Scintillation countersdetect gamma rays

Radioactive isotope emits a positron

Gamma rays produced when positron and electron collide

sEcTioN 21.6 Energy changes in Nuclear reactions 929

21.6 | energy Changes in Nuclear Reactions

Why are the energies associated with nuclear reactions so large, in many cases orders of magnitude larger than those associated with nonnuclear chemical reactions? The answer to this question begins with Einstein’s celebrated equation from the theory of relativity that relates mass and energy:

E = mc2 [21.22]

In this equation E stands for energy, m for mass, and c for the speed of light, 2.9979 * 108 m/s. This equation states that mass and energy are equivalent and can be converted into one another. If a system loses mass, it loses energy; if it gains mass, it gains energy. Because the proportionality constant between energy and mass, c2, is such a large number, even small changes in mass are accompanied by large changes in energy.

The mass changes in chemical reactions are too small to detect. For example, the mass change associated with the combustion of 1 mol of CH4 (an exothermic process) is -9.9 * 10-9 g. Because the mass change is so small, it is possible to treat chemical reactions as though mass is conserved. (Section 2.1)

The mass changes and the associated energy changes in nuclear reactions are much greater than those in chemical reactions. The mass change accompanying the radioac-tive decay of 1 mol of uranium-238, for example, is 50,000 times greater than that for the combustion of 1 mol of CH4. Let’s examine the energy change for the nuclear reaction

23892U ¡ 234

90Th + 42He

The masses of the nuclei are 23892U, 238.0003 amu; 234

90Th, 233.9942 amu; and 42He, 4.0015 amu. The mass change, ∆m, is the total mass of the products minus the total mass of the reactants. The mass change for the decay of 1 mol of uranium-238 can then be expressed in grams:

233.9942 g + 4.0015 g - 238.0003 g = -0.0046 g

The fact that the system has lost mass indicates that the process is exothermic. All spon-taneous nuclear reactions are exothermic.

The energy change per mole associated with this reaction is

∆E = ∆1mc22 = c2∆m

= 12.9979 * 108 m/s221-0.0046 g2a 1 kg1000 g

b

= -4.1 * 1011kg@m2

s2 = -4.1 * 1011J

Notice that ∆m must be converted to kilograms, the SI unit of mass, to obtain ∆E in joules, the SI unit of energy. The negative sign for the energy change indicates that energy is released in the reaction—in this case, over 400 billion joules per mole of uranium!

How much energy is lost or gained when 1 mol of cobalt-60 undergoes beta decay, 6027Co ¡ 60

28Ni + 0-1e? The mass of a 60

27Co atom is 59.933819 amu, and that of a 6028Ni atom is

59.930788 amu.

soluTioNAnalyze We are asked to calculate the energy change in a nuclear reaction.Plan We must first calculate the mass change in the process. We are given atomic masses, but we need the masses of the nuclei in the reaction. We calculate these by taking account of the masses of the electrons that contribute to the atomic masses.

sample exeRCise 21.8 Calculating mass Change in a Nuclear Reaction


Nuclear Binding EnergiesScientists discovered in the 1930s that the masses of nuclei are always less than the masses of the individual nucleons of which they are composed. For example, the helium-4 nucleus (an alpha particle) has a mass of 4.00150 amu. The mass of a proton is 1.00728 amu and that of a neutron is 1.00866 amu. Consequently, two protons and two neutrons have a total mass of 4.03188 amu:

Mass of two protons = 211.00728 amu2 = 2.01456 amu Mass of two neutrons = 211.00866 amu2 = 2.01732 amu

Total mass = 4.03188 amu

The mass of the individual nucleons is 0.03038 amu greater than that of the helium-4 nucleus:

Mass of two protons and two neutrons = 4.03188 amu Mass of 42He nucleus = 4.00150 amu Mass difference ∆m = 0.03038 amu

The mass difference between a nucleus and its constituent nucleons is called the mass defect. The origin of the mass defect is readily understood if we consider that energy must be added to a nucleus to break it into separated protons and neutrons:

Energy + 42He ¡ 2 11H + 2 10n [21.23]

Solve A 6027Co atom has 27 electrons. The mass of an electron is 5.4858 * 10-4 amu. (See the

list of fundamental constants in the back inside cover.) We subtract the mass of the 27 electrons from the mass of the 60

27Co atom to find the mass of the 6027Co nucleus:

59.933819 amu - 127215.4858 * 10-4 amu2 = 59.919007 amu 1or 59.919007 g>mol2 Likewise, for 60

28Ni, the mass of the nucleus is59.930788 amu - 128215.4858 * 10-4 amu2 = 59.915428 amu 1or 59.915428 g>mol2

The mass change in the nuclear reaction is the total mass of the products minus the mass of the reactant:

∆m = mass of electron + mass 6028Ni nucleus - mass of 60

27Co nucleus = 0.00054858 amu + 59.915428 amu - 59.919007 amu = -0.003030 amu

Thus, when a mole of cobalt-60 decays,∆m = -0.003030 g

Because the mass decreases 1∆m 6 02, energy is released 1∆E 6 02. The quantity of energy released per mole of cobalt-60 is calculated using Equation 21.22:

∆E = c2 ∆m

= 12.9979 * 108 m/s221-0.003030 g2a 1 kg1000 g

b

= -2.723 * 1011kg@m2

s2 = -2.723 * 1011 J

practice exercise 1The nuclear reaction that powers the radioisotope thermoelectric generator shown in the chapter-opening photograph is 238

94Pu ¡ 23492U + 4

2He. The atomic masses of plutonium-238 and uranium-234 are 238.049554 amu and 234.040946 amu, respectively. The mass of an alpha particle is 4.001506 amu. How much energy in kJ is released when 1.00 g of plutonium-238 decays to uranium-234?(a) 2.27 * 106 kJ (b) 2.68 * 106 kJ (c) 3.10 * 106 kJ (d) 3.15 * 106 kJ(e) 7.37 * 108 kJ

practice exercise 2Positron emission from 11C, 11

6C ¡ 115B + 0

+ 1e, occurs with release of 2.87 * 1011 J per mole of 11C. What is the mass change per mole of 11C in this nuclear reaction? The masses of 11B and 11C are 11.009305 and 11.011434 amu, respectively.

sEcTioN 21.6 Energy changes in Nuclear reactions 931

By Einstein’s relation, the addition of energy to a system must be accompanied by a proportional increase in mass. The mass change we just calculated for the conversion of helium-4 into separated nucleons is ∆m = 0.03038 amu. Therefore, the energy required for this process is

∆E = c2 ∆m

= 12.9979 * 108 m>s2210.03038 amu2a 1 g6.022 * 1023 amu

b a 1 kg1000 g

b

= 4.534 * 10-12 J

The energy required to separate a nucleus into its individual nucleons is called the nuclear binding energy. The mass defect and nuclear binding energy for three ele-ments are compared in ▲ Table 21.7.

Give It Some ThoughtThe mass of a single atom of iron-56 is 55.93494 amu. Why is this number different from the mass of the nucleus given in Table 21.7?

Values of binding energies per nucleon can be used to compare the stabilities of different combinations of nucleons (such as two protons and two neutrons arranged either as 4

2He or as 2 21H). ▶ Figure 21.12 shows average binding energy per nucleon plotted against mass number. Binding energy per nucleon at first increases in magni-tude as mass number increases, reaching about 1.4 * 10-12 J for nuclei whose mass numbers are in the vicinity of iron-56. It then decreases slowly to about 1.2 * 10-12 J for very heavy nuclei. This trend indicates that nuclei of intermediate mass numbers are more tightly bound (and therefore more stable) than those with either smaller or larger mass numbers.

This trend has two significant consequences: First, heavy nuclei gain stability and therefore give off energy if they are fragmented into two midsized nuclei. This process, known as fission, is used to generate energy in nuclear power plants. Sec-ond, because of the sharp increase in the graph for small mass numbers, even greater amounts of energy are released if very light nuclei are combined, or fused together, to give more mas-sive nuclei. This fusion process is the essential energy-produc-ing process in the Sun and other stars.

Give It Some ThoughtWould fusing two stable nuclei that have mass numbers in the vicinity of 100 be an energy-releasing process?

Table 21.7 mass defects and Binding energies for Three Nuclei

Nucleusmass of Nucleus (amu)

mass of individual Nucleons (amu)

mass defect (amu)

Binding energy (J)

Binding energy per Nucleon (J)

42He 4.00150 4.03188 0.03038 4.53 * 10-12 1.13 * 10-12

5626Fe 55.92068 56.44914 0.52846 7.90 * 10-11 1.41 * 10-12

23892U 238.00031 239.93451 1.93420 2.89 * 10-10 1.21 * 10-12

▲ Figure 21.12 Nuclear binding energies. The average binding energy per nucleon increases initially as the mass number increases and then decreases slowly. Because of these trends, fusion of light nuclei and fission of heavy nuclei are exothermic processes.

Splitting nuclei(fission)

releases energy

Combining nuclei(fusion)

releases energy

0 50

0

1

100 150 200 250

Bin

din

g en

ergy

per

nuc

leon

(10−

12 J)

Incr

ease

d s

tabi

lity

Mass number

56Fe

235U4He


21.7 | Nuclear power: FissionCommercial nuclear power plants and most forms of nuclear weaponry depend on nu-clear fission for their operation. The first nuclear fission reaction to be discovered was that of uranium-235. This nucleus, as well as those of uranium-233 and plutonium-239, undergoes fission when struck by a slow-moving neutron (▼ Figure 21.13).*

A heavy nucleus can split in many ways. Two ways that the uranium-235 nucleus splits, for instance, are

0n10n

1

392U

235

36Kr91

56Ba142

0n1240Zr

9752Te

137

++ +

+ +

[21.24]

[21.25]

The nuclei produced in equations 21.24 and 21.25—called the fission products—are themselves radioactive and undergo further nuclear decay. More than 200 isotopes of 35 elements have been found among the fission products of uranium-235. Most of them are radioactive.

Slow-moving neutrons are required in fission because the process involves initial absorption of the neutron by the nucleus. The resulting more massive nucleus is often unstable and spontaneously undergoes fission. Fast neutrons tend to bounce off the nucleus, and little fission occurs.

Note that the coefficients of the neutrons produced in Equations 21.24 and 21.25 are 2 and 3. On average, 2.4 neutrons are produced by every fission of a uranium-235 nucleus. If one fission produces two neutrons, the two neutrons can cause two addi-tional fissions, each producing two neutrons. The four neutrons thereby released can produce four fissions, and so forth, as shown in ▶ Figure 21.14. The number of fis-sions and the energy released quickly escalate, and if the process is unchecked, the result is a violent explosion. Reactions that multiply in this fashion are called chain reactions.

For a fission chain reaction to occur, the sample of fissionable material must have a certain minimum mass. Otherwise, neutrons escape from the sample before they have the opportunity to strike other nuclei and cause additional fission. The amount of fis-sionable material large enough to maintain a chain reaction with a constant rate of fis-sion is called the critical mass. When a critical mass of material is present, one neutron on average from each fission is subsequently effective in producing another fission and

*Other heavy nuclei can be induced to undergo fission. However, these three are the only ones of practical importance.

▲ Figure 21.13 Uranium-235 fission. This is just one of many fission patterns. In this reaction, 3.5 * 10-11 J of energy is released per 235U nucleus that is split.

Go FiGuReWhat is the relationship between the sum of the mass numbers on the two sides of this reaction?

0n1

0n1

92U235

36Kr91

56Ba142

sEcTioN 21.7 Nuclear power: Fission 933

the fission continues at a constant, controllable rate. The critical mass of uranium-235 is about 50 kg for a bare sphere of the metal.*

If more than a critical mass of fissionable material is present, very few neutrons escape. The chain reaction thus multiplies the number of fissions, which can lead to a nuclear explosion. A mass in excess of a critical mass is referred to as a supercritical mass. The effect of mass on a fission reaction is illustrated in ▼ Figure 21.15.

▼ Figure 21.16 shows a schematic diagram of the first atomic bomb used in warfare, the bomb, code-named “Little Boy,” that was dropped on Hiroshima, Japan, on August 6, 1945. The bomb contained about 64 kg of uranium-235, which had been separated from the nonfissionable uranium-238 primarily by gaseous diffu-sion of uranium hexafluoride, UF6. (Section 10.8) To trigger the fission reaction, two subcritical masses of uranium-235 were slammed together using chemical explosives. The combined masses of the uranium formed a supercritical mass, which led to a rapid, uncontrolled chain reaction and, ultimately, a nuclear explosion. The energy released by the bomb dropped on Hiroshima was equiva-lent to that of 16,000 tons of TNT (it therefore is called a 16-kiloton bomb). Unfortunately, the basic design of a fission-based atomic bomb is quite simple, and the fissionable materials are potentially available to any nation with a nuclear reactor. The combination of design simplicity and materials availability has generated international concerns about the proliferation of atomic weapons.

Neutron in

Nucleus

2 neutronsfrom fission

▲ Figure 21.14 Fission chain reaction.

Go FiGuReIf this figure were extended one more “generation” down, how many neutrons would be produced?

*The exact value of the critical mass depends on the shape of the radioactive substance. The critical mass can be reduced if the radioisotope is surrounded by a material that reflects some neutrons.

Supercritical massCritical massSubcritical massRate of neutron loss> rate of neutroncreation by fission

Rate of neutron loss= rate of neutroncreation by fission

Rate of neutron loss< rate of neutroncreation by fission

▲ Figure 21.15 Subcritical, critical, and supercritical nuclear fission.

Go FiGuReWhich of these criticality scenarios—subcritical, critical, or supercritical—is desirable in a nuclear power plant that generates electricity?

Subcriticaluranium-235wedge

Subcriticaluranium-235target

Chemicalexplosive

▲ Figure 21.16 Schematic drawing of an atomic bomb. A conventional explosive is used to bring two subcritical masses together to form a supercritical mass.


Nuclear ReactorsNuclear power plants use nuclear fission to generate energy. The core of a typical nu-clear reactor consists of four principal components: fuel elements, control rods, a mod-erator, and a primary coolant (◀ Figure 21.18). The fuel is a fissionable substance, such as uranium-235. The natural isotopic abundance of uranium-235 is only 0.7, too low to sustain a chain reaction in most reactors. Therefore, the 235U content of the fuel must be enriched to 3–5% for use in a reactor. The fuel elements contain enriched uranium in the form of UO2 pellets encased in zirconium or stainless steel tubes.

The control rods are composed of materials that absorb neutrons, such as boron-10 or an alloy of silver, indium, and cadmium. These rods regulate the flux of neutrons to keep the reaction chain self-sustaining and also prevent the reactor core from overheating.*

The probability that a neutron will trigger fission of a 235U nucleus depends on the speed of the neutron. The neutrons produced by fission have high speeds (typically in excess of 10,000 km/s). The function of the moderator is to slow down the neutrons (to speeds of a few kilometers per second) so that they can be captured more readily by the fissionable nuclei. The moderator is typically either water or graphite.

The primary coolant is a substance that transports the heat generated by the nuclear chain reaction away from the reactor core. In a pressurized water reactor, which is the most

on the fission reaction. An enormous research project, known as the Manhattan Project, began.

On December 2, 1942, the first artificial self-sustaining nuclear fis-sion chain reaction was achieved in an abandoned squash court at the University of Chicago. This accomplishment led to the development of the first atomic bomb, at Los Alamos National Laboratory in New Mex-ico in July 1945 (▼ Figure 21.17). In August 1945 the United States dropped atomic bombs on two Japanese cities, Hiroshima and Nagasaki. The nuclear age had arrived, albeit in a sadly destructive fashion. Hu-manity has struggled with the conflict between the positive potential of nuclear energy and its terrifying potential as a weapon ever since.

a Closer look

The Dawning of the Nuclear Age

Uranium-235 fission was first achieved during the late 1930s by Enrico Fermi and coworkers in Rome and shortly thereafter by Otto Hahn and coworkers in Berlin. Both groups were trying to produce transuranium elements. In 1938, Hahn identified barium among his reaction products. He was puzzled by this observation and questioned the identification because the presence of barium was so unexpected. He sent a letter describing his experiments to Lise Meitner, a former coworker who had been forced to leave Germany because of the anti-Semitism of the Third Reich and had settled in Sweden. She surmised that Hahn’s experiment indicated a nuclear process was occurring in which the uranium-235 split. She called this process nuclear fission.

Meitner passed word of this discovery to her nephew, Otto Frisch, a physicist working at Niels Bohr’s institute in Copenhagen. Frisch repeated the experiment, verifying Hahn’s observations, and found that tremendous energies were involved. In January 1939, Meitner and Frisch published a short article describing the reaction. In March 1939, Leo Szilard and Walter Zinn at Columbia University discovered that more neutrons are produced than are used in each fis-sion. As we have seen, this result allows a chain reaction to occur.

News of these discoveries and an awareness of their potential use in explosive devices spread rapidly within the scientific community. Several scientists finally persuaded Albert Einstein, the most famous physicist of the time, to write a letter to President Franklin D. Roos-evelt explaining the implications of these discoveries. Einstein’s letter, written in August 1939, outlined the possible military applications of nuclear fission and emphasized the danger that weapons based on fis-sion would pose if they were developed by the Nazis. Roosevelt judged it imperative that the United States investigate the possibility of such weapons. Late in 1941, the decision was made to build a bomb based

▲ Figure 21.17 The Trinity test for the atom bomb developed during World War II. The first human-made nuclear explosion took place on July 16, 1945, on the Alamogordo test range in New Mexico.

▲ Figure 21.18 Schematic diagram of a pressurized water reactor core.

Water acts as bothmoderator and coolant

Fuel elements

Controlrods

Control-roddrive

*The reactor core cannot reach supercritical levels and explode with the violence of an atomic bomb because the concentration of uranium-235 is too low. However, if the core overheats, sufficient damage can lead to release of radioactive materials into the environment.

sEcTioN 21.7 Nuclear power: Fission 935

common commercial reactor design, water acts as both the moderator and the primary coolant.

The design of a nuclear power plant is basically the same as that of a power plant that burns fossil fuel (except that the burner is replaced by a reactor core). The nuclear power plant design shown in ▲ Figure 21.19, a pressurized water reactor, is currently the most popular. The primary coolant passes through the core in a closed system, which lessens the chance that radioactive products could escape the core. As an added safety precaution, the reactor is surrounded by a reinforced concrete containment shell to shield personnel and nearby residents from radiation and to protect the reactor from external forces. After passing through the reactor core, the very hot primary coolant passes through a heat exchanger where much of its heat is transferred to a secondary coolant, converting the latter to high-pressure steam that is used to drive a turbine. The secondary coolant is then condensed by transferring heat to an external source of water, such as a river or lake.

Approximately two-thirds of all commercial reactors are pressurized water reac-tors, but there are several variations on this basic design, each with advantages and disadvantages. A boiling water reactor generates steam by boiling the primary coolant; thus, no secondary coolant is needed. Pressurized water reactors and boiling water reactors are collectively referred to as light water reactors because they use H2O as mod-erator and primary coolant. A heavy water reactor uses D2O (D = deuterium, 2H2 as moderator and primary coolant, and a gas-cooled reactor uses a gas, typically CO2, as primary coolant and graphite as the moderator. Use of either D2O or graphite as the moderator has the advantage that both substances absorb fewer neutrons than H2O. Consequently, the uranium fuel does not need to be as enriched.

Electricgenerator

Condensor

PumpHeatexchanger

PumpPump

River27 °C 38 °C

TurbineContainmentshell

Reactorcore

Primarycoolant (H2O)

Steam

Secondarycoolant (H2O)

Pressurized water is heated in the reactor core

1

Heat is transferred to the secondary coolant in the heat exchanger, generating steam

2

The steam drives an electric generator, creating electricity

3

Heat is transferred to an external source of water, condensing the secondary coolant, which is pumped back to the heat exchanger

4

▲ Figure 21.19 Basic design of a pressurized water reactor nuclear power plant.

Go FiGuReWhy are nuclear power plants usually located near a large body of water?


In a high-temperature pebble-bed reactor, the fuel elements are spheres (“pebbles”) roughly the size of an orange (▲ Figure 21.20). The spheres are made of graphite, which acts as the moderator, and thousands of tiny fuel particles are embedded in the interior of each sphere. Each fuel particle is a kernel of fissionable material, typically 235U in the form of UO2, surrounded by carbon and a coating of a ceramic material, such as SiC. Hundreds of thousands of these spheres are loosely packed in the reactor core, and helium gas, which acts as the primary coolant, flows up through the packed spheres. The reactor core operates at temperatures considerably higher than those in a light water reactor, approaching 950 °C. A pebble-bed reactor is not subject to steam explosions and does not need to be shut down to refuel. Engineers can remove spent spheres from the bottom of the reactor core and add fresh ones to the top. This design is relatively new and is not yet in commercial use.

Nuclear WasteThe fission products that accumulate as a reactor operates decrease the efficiency of the reactor by capturing neutrons. For this reason, commercial reactors must be stopped periodically to either replace or reprocess the nuclear fuel. When the fuel elements are removed from the reactor, they are initially very radioactive. It was originally intended that they be stored for several months in pools at the reactor site to allow decay of short-lived radioactive nuclei. They were then to be transported in shielded contain-ers to reprocessing plants where the unspent fuel would be separated from the fission products. Reprocessing plants have been plagued with operational difficulties, however, and there is intense opposition in the United States to the transport of nuclear wastes on the nation’s roads and rails.

Even if the transportation difficulties could be overcome, the high level of radio-activity of the spent fuel makes reprocessing a hazardous operation. At present in the United States spent fuel elements are kept in storage at reactor sites. Spent fuel is repro-cessed, however, in France, Russia, the United Kingdom, India, and Japan.

Storage of spent nuclear fuel poses a major problem because the fission products are extremely radioactive. It is estimated that 10 half-lives are required for their radio-activity to reach levels acceptable for biological exposure. Based on the 28.8-yr half-life of strontium-90, one of the longer-lived and most dangerous of the products, the wastes must be stored for nearly 300 yr. Plutonium-239 is one of the by-products present in spent fuel elements. It is formed by absorption of a neutron by uranium-238, followed by two successive beta emissions. (Remember that most of the uranium in the fuel ele-ments is uranium-238.) If the elements are reprocessed, the plutonium-239 is largely recovered because it can be used as a nuclear fuel. However, if the plutonium is not removed, spent elements must be stored for a very long time because plutonium-239 has a half-life of 24,000 yr.

▲ Figure 21.20 Fuel spheres used in a high-temperature pebble-bed reactor. The image on the right is an optical microscope image of a fuel particle.

Graphite shell

Embedded fuelparticles

Cross-sectional view Coated fuel particle

6 cm

~1 mm

Graphite Siliconcarbide

Porouscarbonbuffer

UO2 fuelkernel

sEcTioN 21.8 Nuclear power: Fusion 937

A fast breeder reactor offers one approach to getting more power out of existing uranium sources and potentially reducing radioactive waste. This type of reactor is so named because it creates (“breeds”) more fissionable material than it consumes. The reactor operates without a moderator, which means the neutrons used are not slowed down. In order to capture the fast neutrons, the fuel must be highly enriched with both uranium-235 and plutonium-239. Water cannot be used as a primary coolant because it would moderate the neutrons, and so a liquid metal, usually sodium, is used. The core is surrounded by a blanket of uranium-238 that captures neutrons that escape the core, producing plutonium-239 in the process. The plutonium can later be sepa-rated by reprocessing and used as fuel in a future cycle.

Because fast neutrons are more effective at decaying many radioactive nuclides, the material separated from the uranium and plutonium during reprocessing is less radio-active than waste from other reactors. However, generation of relatively high levels of plutonium coupled with the need for reprocessing is problematic in terms of nuclear nonproliferation. Thus, political factors coupled with increased safety concerns and higher operational costs make fast breeder reactors quite rare.

A considerable amount of research is being devoted to disposal of radioactive wastes. At present, the most attractive possibilities appear to be formation of glass, ceramic, or synthetic rock from the wastes, as a means of immobilizing them. These solid materials would then be placed in containers of high corrosion resistance and durability and buried deep underground. The U.S. Department of Energy (DOE) had designated Yucca Mountain in Nevada as a disposal site, and extensive construction has been done there. However, in 2010 this project was discontinued because of tech-nological and political concerns. The long-term solution to nuclear waste storage in the United States remains unclear. Whatever solution is finally decided on, there must be assurances that the solids and their containers will not crack from the heat gener-ated by nuclear decay, allowing radioactivity to find its way into underground water supplies.

In spite of all these difficulties, nuclear power is making a modest comeback as an energy source. Concerns about climate change caused by escalating atmospheric CO2 levels (Section 18.2) have increased support for nuclear power as a major energy source in the future. Increasing demand for power in rapidly developing countries, par-ticularly China, has sparked a rise in construction of new nuclear power plants in those parts of the world.

21.8 | Nuclear power: FusionEnergy is produced when light nuclei fuse into heavier ones. Reactions of this type are responsible for the energy produced by the Sun. Spectroscopic studies indicate that the mass composition of the Sun is 73% H, 26% He, and only 1% all other elements. The following reactions are among the numerous fusion processes believed to occur in the Sun:

11H + 1

1H ¡ 21H + 0

+1e [21.26] 1

1H + 21H ¡ 3

2He [21.27] 3

2He + 32He ¡ 4

2He + 2 11H [21.28] 3

2He + 11H ¡ 4

2He + 0+1e [21.29]

Fusion is appealing as an energy source because of the availability of light isotopes on Earth and because fusion products are generally not radioactive. Despite this fact, fusion is not presently used to generate energy. The problem is that, in order for two nuclei to fuse, extremely high temperatures and pressures are needed to overcome the electrostatic repulsion between nuclei in order to fuse them. Fusion reactions are there-fore also known as thermonuclear reactions. The lowest temperature required for any fusion is about 40,000,000 K, the temperature needed to fuse deuterium and tritium:

21H + 3

1H ¡ 42He + 1

0n [21.30]


Such high temperatures have been achieved by using an atomic bomb to initi-ate fusion. This is the operating principle behind a thermonuclear, or hydrogen, bomb. This approach is obviously unacceptable, however, for a power generation plant.*

Numerous problems must be overcome before fusion becomes a practical energy source. In addition to the high temperatures necessary to initiate the reaction, there is the problem of confining the reaction. No known structural material is able to with-stand the enormous temperatures necessary for fusion. Research has centered on the use of an apparatus called a tokamak, which uses strong magnetic fields to contain and to heat the reaction. Temperatures of over 100,000,000 K have been achieved in a toka-mak. Unfortunately, scientists have not yet been able to generate more power than is consumed over a sustained period of time.

21.9 | Radiation in the environment and living systems

We are continuously bombarded by radiation from both natural and artificial sources. We are exposed to infrared, ultraviolet, and visible radiation from the Sun; radio waves from radio and television stations; microwaves from microwave ovens; X rays from medical procedures; and radioactivity from natural materials (▼ Table 21.8). Under-standing the different energies of these various kinds of radiation is necessary in order to understand their different effects on matter.

When matter absorbs radiation, the radiation energy can cause atoms in the matter to be either excited or ionized. In general, radiation that causes ionization, called ion-izing radiation, is far more harmful to biological systems than radiation that does not cause ionization. The latter, called nonionizing radiation, is generally of lower energy, such as radiofrequency electromagnetic radiation (Section 6.7) or slow-moving neutrons.

Most living tissue contains at least 70% water by mass. When living tissue is irradi-ated, water molecules absorb most of the energy of the radiation. Thus, it is common to define ionizing radiation as radiation that can ionize water, a process requiring a minimum energy of 1216 kJ/mol. Alpha, beta, and gamma rays (as well as X rays and higher-energy ultraviolet radiation) possess energies in excess of this quantity and are therefore forms of ionizing radiation.

Table 21.8 average abundances and activities of Natural Radionuclides†

potassium-40 Rubidium-87 Thorium-232 uranium-238

Land elemental abundance (ppm) 28,000 112 10.7 2.8Land activity (Bq/kg) 870 102 43 35Ocean elemental concentration (mg/L) 339 0.12 1 * 10-7 0.0032

Ocean activity (Bq/L) 12 0.11 4 * 10-7 0.040

Ocean sediments elemental abundance (ppm) 17,000 — 5.0 1.0Ocean sediments activity (Bq/kg) 500 — 20 12Human body activity (Bq) 4000 600 0.08 0.4‡

†Data from “Ionizing Radiation Exposure of the Population of the United States,” Report 93, 1987, and Report 160, 2009, National Council on Radiation Protection.‡Includes lead-210 and polonium-210, daughter nuclei of uranium-238.

*Historically a nuclear weapon that relies solely on a fission process to release energy is called an atomic bomb, whereas one that also releases energy via a fusion reaction is called a hydrogen bomb.

sEcTioN 21.9 radiation in the Environment and Living systems 939

life on Earth. Let’s look at the factors responsible for the relatively high abundance of carbon and oxygen in the universe.

A star is born from a cloud of gas and dust called a nebula. When conditions are right, gravitational forces collapse the cloud, and its core density and temperature rise until nuclear fusion com-mences. Hydrogen nuclei fuse to form deuterium, 2

1H, and eventually 42He through the reactions shown in Equations 21.26 through 21.29. Because 4

2He has a larger binding energy than any of its immediate neighbors (Figure 21.12), these reactions release an enormous amount of energy. This process, called hydrogen burning, is the dominant pro-cess for most of a star’s lifetime.

Once a star’s supply of hydrogen is nearly exhausted, several im-portant changes occur as the star enters next phase of its life, and is transformed into a red giant. The decrease in nuclear fusion causes the core to contract, triggering an increase in core temperature and pres-sure. At the same time, the outer regions expand and cool enough to make the star emit red light (thus, the name red giant). The star now must use 42He nuclei as its fuel. The simplest reaction that can occur in the He-rich core, fusion of two alpha particles to form a 8

4Be nucleus, does occur. The binding energy per nucleon for 8

4Be is very slightly smaller than that for 4

2He, so this fusion process is very slightly endo-thermic. The 8

4Be nucleus is highly unstable (half-life of 7 * 10-17 s) and so falls apart almost immediately. In a tiny fraction of cases, how-ever, a third 42He collides with a 84Be nucleus before it decays, forming carbon-12 through the triple-alpha process:

42He + 4

2He ¡ 84Be

84Be + 4

2He ¡ 126C

Some of the 126C nuclei go on to react with alpha particles to form

oxygen-16:126C + 4

2He ¡ 168O

This stage of nuclear fusion is called helium burning. Notice that carbon, element 6, is formed without prior formation of elements 3, 4, and 5, explaining in part their unusually low abundance. Nitrogen is relatively abundant because it can be produced from carbon through a series of re-actions involving proton capture and positron emission.

Most stars gradually cool and dim as the helium is con-verted to carbon and oxygen, ending their lives as white dwarfs, a phase in which stars become incredibly dense—generally about one million times denser than the Sun. The extreme density of white dwarfs is accompanied by much higher temperatures and pressures at the core, where a variety of fusion processes lead to synthesis of the elements from neon to sulfur. These fusion reac-tions are collectively called advanced burning.

Eventually progressively heavier elements form at the core until it becomes predominantly 56Fe as shown in ◀ Figure 21.22. Because this is such a stable nucleus, fur-ther fusion to heavier nuclei consumes energy rather than re-leasing it. When this happens, the fusion reactions that power the star diminish, and immense gravitational forces lead to a dramatic collapse called a supernova explosion. Neutron cap-ture coupled with subsequent radioactive decays in the dying moments of such a star are responsible for the presence of all elements heavier than iron and nickel.

Without these dramatic supernova events in the past his-tory of the universe, heavier elements that are so familiar to us, such as silver, gold, iodine, lead, and uranium, would not exist.


a Closer look

Nuclear Synthesis of the Elements

The lightest elements—hydrogen and helium along with very small amounts of lithium and beryllium—were formed as the universe expanded in the moments following the Big Bang. All the heavier elements owe their existence to nuclear reactions that occur in stars. These heavier elements are not all created equally, however. In our solar system, for example, carbon and oxygen are a million times more abundant than lithium and boron, for instance, and over 100 million times more abundant that beryllium (▼ Figure 21.21)! In fact, of the elements heavier than helium, carbon and oxygen are the most abundant. This is more than an academic curiosity given the fact that these ele-ments, together with hydrogen, are the most important elements for

▲ Figure 21.21 Relative abundance of elements 1–10 in the solar system. Note the logarithmic scale used for the y-axis.

0 2 4 6 8 101 × 10−12

1 × 10−10

1 × 10−8

1 × 10−6

1 × 10−4

1 × 10−2

1 × 100

Atomic number

He

Li

Be

B

C

N

O

F

Ne

H

Rel

ativ

e ab

und

ance

in s

olar

sys

tem

▲ Figure 21.22 Fusion processes going on in a red giant just prior to a supernova explosion.

56Fe core26

Hydrogen burning(H He)

Helium burning(He C, O)

Advanced burning(Ne through S)


When ionizing radiation passes through living tissue, electrons are removed from water molecules, forming highly reactive H2O+ ions. An H2O+ ion can react with another water molecule to form an H3O+ ion and a neutral OH molecule:

H2O+ + H2O ¡ H3O+ + OH [21.31]

The unstable and highly reactive OH molecule is a free radical, a substance with one or more unpaired electrons, as seen in the Lewis structure shown in the margin. The OH molecule is also called the hydroxyl radical, and the presence of the unpaired elec-tron is often emphasized by writing the species with a single dot, # OH. In cells and tissues, hydroxyl radicals can attack biomolecules to produce new free radicals, which in turn attack yet other biomolecules. Thus, the formation of a single hydroxyl radical via Equation 21.31 can initiate a large number of chemical reactions that are ultimately able to disrupt the normal operations of cells.

The damage produced by radiation depends on the activity and energy of the radi-ation, the length of exposure, and whether the source is inside or outside the body. Gamma rays are particularly harmful outside the body because they penetrate human tissue very effectively, just as X rays do. Consequently, their damage is not limited to the skin. In contrast, most alpha rays are stopped by skin, and beta rays are able to pen-etrate only about 1 cm beyond the skin surface (◀ Figure 21.23). Neither alpha rays nor beta rays are as dangerous as gamma rays, therefore, unless the radiation source some-how enters the body. Within the body, alpha rays are particularly dangerous because they transfer their energy efficiently to the surrounding tissue, causing considerable damage.

In general, the tissues damaged most by radiation are those that reproduce rapidly, such as bone marrow, blood-forming tissues, and lymph nodes. The principal effect of extended exposure to low doses of radiation is to cause cancer. Cancer is caused by damage to the growth-regulation mechanism of cells, inducing the cells to reproduce uncontrollably. Leukemia, which is characterized by excessive growth of white blood cells, is probably the major type of radiation-caused cancer.

In light of the biological effects of radiation, it is important to determine whether any levels of exposure are safe. Unfortunately, we are hampered in our attempts to set realistic standards because we do not fully understand the effects of long-term expo-sure. Scientists concerned with setting health standards have used the hypothesis that the effects of radiation are proportional to exposure. Any amount of radiation is assumed to cause some finite risk of injury, and the effects of high dosage rates are extrapolated to those of lower ones. Other scientists believe, however, that there is a threshold below which there are no radiation risks. Until scientific evidence enables us to settle the matter with some confidence, it is safer to assume that even low levels of radiation present some danger.

Radiation DosesTwo units are commonly used to measure exposure to radiation. The gray (Gy), the SI unit of absorbed dose, corresponds to the absorption of 1 J of energy per kilogram of tissue. The rad (radiation absorbed dose) corresponds to the absorption of 1 * 10-2 J of energy per kilogram of tissue. Thus, 1 Gy = 100 rad. The rad is the unit most often used in medicine.

Not all forms of radiation harm biological materials to the same extent even at the same level of exposure. For example, 1 rad of alpha radiation can produce more damage than 1 rad of beta radiation. To correct for these differences, the radiation dose is multiplied by a factor that measures the relative damage caused by the radiation. This multiplication factor is known as the relative biological effectiveness, RBE. The RBE is approximately 1 for gamma and beta radiation, and 10 for alpha radiation.

Bone Organs

SkinTissue

αβ−

γ

▲ Figure 21.23 Relative penetrating abilities of alpha, beta, and gamma radiation.

Go FiGuReWhy are alpha rays much more dangerous when the source of radiation is located inside the body?

O H


The exact value of the RBE varies with dose rate, total dose, and type of tissue affected. The product of the radiation dose in rads and the RBE of the radiation give the effective dosage in rem (roentgen equivalent for man):

Number of rem = 1number of rad21RBE2 [21.32]

The SI unit for effective dose is the sievert (Sv), obtained by multiplying the RBE times the SI unit for radiation dose, the gray; because a gray is 100 times larger than a rad, 1 Sv = 100 rem. The rem is the unit of radiation damage usually used in medicine.

Give It Some ThoughtIf a 50-kg person is uniformly irradiated by 0.10-J alpha radiation, what is the absorbed dosage in rad and the effective dosage in rem?

The effects of short-term exposure to radiation appear in ▼ Table 21.9. An expo-sure of 600 rem is fatal to most humans. To put this number in perspective, a typical dental X-ray entails an exposure of about 0.5 mrem. The average exposure for a person in 1 yr due to all natural sources of ionizing radiation (called background radiation) is about 360 mrem (▼ Figure 21.24).

Radioisotopes in the body (40 mrem)

Naturalsources(82%)

Human-madesources (18%)

50 100 150 200 250

Average annual exposure (mrem)

Radon (200 mrem)

Rocks and soil (28 mrem)

Cosmic rays (27 mrem)

Medical X rays (39 mrem)

Nuclear medicine (14 mrem)

Consumer products (11 mrem)

0

Table 21.9 effects of short-Term exposures to Radiation

dose (rem) effect

0–25 No detectable clinical effects25–50 Slight, temporary decrease in white blood cell counts100–200 Nausea; marked decrease in white blood cell counts500 Death of half the exposed population within 30 days

▲ Figure 21.24 Sources of U.S. average annual exposure to high-energy radiation. The total average annual exposure is 360 mrem.Data from “Ionizing Radiation Exposure of the Population of the United States,” Report 93, 1987 and Report 160, 2009, National Council on Radiation Protection.


RadonRadon-222 is a product of the nuclear disintegration series of uranium-238 (Figure 21.3) and is continuously generated as uranium in rocks and soil decays. As Figure 21.24 indi-cates, radon exposure is estimated to account for more than half the 360-mrem average annual exposure to ionizing radiation.

The interplay between the chemical and nuclear properties of radon makes it a health hazard. Because radon is a noble gas, it is extremely unreactive and is therefore free to escape from the ground without chemically reacting along the way. It is readily inhaled and exhaled with no direct chemical effects. Its half-life, however, is only 3.82 days. It decays, by losing an alpha particle, into a radioisotope of polonium:

22286Rn ¡ 218

84Po + 42He [21.33]

Because radon has such a short half-life and because alpha particles have a high RBE, inhaled radon is considered a probable cause of lung cancer. Even worse than the radon, however, is the decay product because polonium-218 is an alpha-emitting chemically active element that has an even shorter half-life (3.11 min) than radon-222:

21884Po ¡ 214

82Pb + 42He [21.34]

When a person inhales radon, therefore, atoms of polonium-218 can become trapped in the lungs, where they bathe the delicate tissue with harmful alpha radiation. The resulting damage is estimated to contribute to 10% of all lung cancer deaths in the United States.

The U.S. Environmental Protection Agency (EPA) has recommended that radon-222 levels not exceed 4 pCi per liter of air in homes. Homes located in areas where the natural uranium content of the soil is high often have levels much greater than that (▼ Figure 21.25). Because of public awareness, radon-testing kits are readily available in many parts of the country.

▲ Figure 21.25 EPA map of radon zones in the United States.* The color coding shows average indoor radon levels as a function of geographic location.

Zone 1 Predicted average indoor radon level greater than 4 pCi/LZone 2 Predicted average indoor radon level between 2 and 4 pCi/LZone 3 Predicted average indoor radon level less than 2 pCi/L

*Data from “Ionizing Radiation Exposure of the Population of the United States,” Report 93, 1987, National Council on Radiation Protection and Measurements.


Chemistry and life

Radiation Therapy

Healthy cells are either destroyed or damaged by high-energy radia-tion, leading to physiological disorders. This radiation can also de-stroy unhealthy cells, however, including cancerous cells. All cancers are characterized by runaway cell growth that can produce malignant tumors. These tumors can be caused by the exposure of healthy cells to high-energy radiation. Paradoxically, however, they can be destroyed by the same radiation that caused them because the rapidly reproduc-ing cells of the tumors are very susceptible to radiation damage. Thus, cancerous cells are more susceptible to destruction by radiation than healthy ones, allowing radiation to be used effectively in the treatment of cancer. As early as 1904, physicians used the radiation emitted by radioactive substances to treat tumors by destroying the mass of un-healthy tissue. The treatment of disease by high-energy radiation is called radiation therapy.

Many radionuclides are currently used in radiation therapy. Some of the more commonly used ones are listed in ▼ Table 21.10. Most of them have short half-lives, meaning that they emit a great deal of radiation in a short period of time.

The radiation source used in radiation therapy may be inside or outside the body. In almost all cases, radiation therapy uses gamma

radiation emitted by radioisotopes. Any alpha or beta radiation that is emitted concurrently can be blocked by appropriate packaging. For example, 192Ir is often administered as “seeds” consisting of a core of radioactive isotope coated with 0.1 mm of platinum metal. The plati-num coating stops the alpha and beta rays, but the gamma rays pen-etrate it readily. The radioactive seeds can be surgically implanted in a tumor.

In some cases, human physiology allows a radioisotope to be ingested. For example, most of the iodine in the human body ends up in the thyroid gland, so thyroid cancer can be treated by using large doses of 131I. Radiation therapy on deep organs, where a surgi-cal implant is impractical, often uses a 60Co “gun” outside the body to shoot a beam of gamma rays at the tumor. Particle accelerators are also used as an external source of high-energy radiation for radiation therapy.

Because gamma radiation is so strongly penetrating, it is nearly impossible to avoid damaging healthy cells during radiation therapy. Many cancer patients undergoing radiation treatment experience un-pleasant and dangerous side effects such as fatigue, nausea, hair loss, a weakened immune system, and occasionally even death. However, if other treatments such as chemotherapy (the use of drugs to combat cancer) fail, radiation therapy can be a good option.

Much current research in radiation therapy is engaged in de-veloping new drugs that specifically target tumors using a method called neutron capture therapy. In this technique, a nonradioactive isotope, usually boron-10, is concentrated in the tumor by using specific tumor-seeking reagents. The boron-10 is then irradiated with neutrons, where it undergoes the following nuclear reaction:

105B + 10n ¡ 7

3Li + 42He

Tumor cells are killed or damaged by exposure to the alpha particles. Healthy tissue farther away from the tumor is unaffected because of the short-range penetrating power of alpha particles. Thus, neutron-capture therapy has the promise to be a “silver bullet” that specifically targets unhealthy cells for exposure to radiation.


Table 21.10 some Radioisotopes used in Radiation Therapy

isotope half-life isotope half-life

32P 14.3 days 137Cs 30 yr60Co 5.27 yr 192Ir 74.2 days90Sr 28.8 yr 198Au 2.7 days125I 60.25 days 222Rn 3.82 days131I 8.04 days 226Ra 1600 yr

Potassium ion is present in foods and is an essential nutrient in the human body. One of the naturally occurring isotopes of potassium, potassium-40, is radioactive. Potassium-40 has a nat-ural abundance of 0.0117% and a half-life t1>2 = 1.28 * 109 yr. It undergoes radioactive decay in three ways: 98.2% is by electron capture, 1.35% is by beta emission, and 0.49% is by positron emission. (a) Why should we expect 40K to be radioactive? (b) Write the nuclear equations for the three modes by which 40K decays. (c) How many 40K+ ions are present in 1.00 g of KCl? (d) How long does it take for 1.00% of the 40K in a sample to undergo radioactive decay?

soluTioN(a) The 40K nucleus contains 19 protons and 21 neutrons. There are very few stable nuclei with

odd numbers of both protons and neutrons (Section 21.2).(b) Electron capture is capture of an inner-shell electron by the nucleus:

4019K + 0

-1e ¡ 4018Ar

sample iNTeGRaTive exeRCise putting Concepts Together


Chapter summary and Key TermsINTRODUCTION TO RADIOACTIvITy AND NUCLEAR EqUATIONS (SECTION 21.1) The nucleus of an atom contains protons and neutrons, both of which are called nucleons. Reactions that involve changes in atomic nuclei are called nuclear reactions. Nuclei that spon-taneously change by emitting radiation are said to be radioactive. Radioactive nuclei are called radionuclides, and the atoms containing them are called radioisotopes. Radionuclides spontaneously change through a process called radioactive decay. The three most important types of radiation given off as a result of radioactive decay are alpha 1A 2 particles 14

2He or a2, beta 1B 2 particles 1 0-1e or b-2, and gamma 1G 2

radiation 100g or g2. Positrons 1 0

+1e or b+2, which are particles with the same mass as an electron but the opposite charge, can also be pro-duced when a radioisotope decays.

In nuclear equations, reactant and product nuclei are represented by giving their mass numbers and atomic numbers, as well as their chemi-cal symbol. The totals of the mass numbers on both sides of the equation are equal; the totals of the atomic numbers on both sides are also equal. There are four common modes of radioactive decay: alpha decay, which reduces the atomic number by 2 and the mass number by 4, beta emis-sion, which increases the atomic number by 1 and leaves the mass number unchanged, positron emission and electron capture, both of which reduce the atomic number by 1 and leave the mass number unchanged.

PATTERNS OF NUCLEAR STABILITy (SECTION 21.2) The neutron-to-proton ratio is an important factor determining nuclear stability. By comparing a nuclide’s neutron-to-proton ratio with those of stable nuclei, we can predict the mode of radioactive decay. In general, neutron-rich nuclei tend to emit beta particles; proton-rich nuclei tend to either emit positrons or undergo elec-tron capture; and heavy nuclei tend to emit alpha particles. The presence of magic numbers of nucleons and an even number of pro-tons and neutrons also help determine the stability of a nucleus. A nuclide may undergo a series of decay steps before a stable nuclide forms. This series of steps is called a radioactive decay chain or a nuclear disintegration series.

NUCLEAR TRANSMUTATIONS (SECTION 21.3) Nuclear transmuta-tions, induced conversions of one nucleus into another, can be brought about by bombarding nuclei with either charged particles or neutrons. Particle accelerators increase the kinetic energies of positively charged particles, allowing these particles to overcome their electrostatic repul-sion by the nucleus. Nuclear transmutations are used to produce the transuranium elements, those elements with atomic numbers greater than that of uranium.

Beta emission is loss of a beta particle 1 0-1e2 by the nucleus:

4019K ¡ 40

20Ca + 0-1e

Positron emission is loss of a positron 1 0+1e2 by the nucleus:

4019K ¡ 40

18Ar + 0+ 1e

(c) The total number of K+ ions in the sample is

11.00 g KCl2a 1 mol KCl74.55 g KCl

b a 1 mol K+

1 mol KClb a 6.022 * 1023 K+

1 mol K+ b = 8.08 * 1021 K+ ions

Of these, 0.0117% are 40K+ ions:

18.08 * 1021 K+ ions2a 0.0117 40 K+ ions100+ ions

b = 9.45 * 1017 potassium@40 ions

(d) The decay constant (the rate constant) for the radioactive decay can be calculated from the half-life, using Equation 21.20:

k =0.693t1>2

=0.693

1.28 * 109 yr= 15.41 * 10-102/yr

The rate equation, Equation 21.19, then allows us to calculate the time required:

lnNt

N0= -kt

ln99

100= -315.41 * 10-102/yr4t

-0.01005 = -315.41 * 10-102/yr4t t =

-0.010051-5.41 * 10-102/yr

= 1.86 * 107 yr

That is, it would take 18.6 million years for just 1.00% of the 40K in a sample to decay.

Key Equations 945

RADIOACTIvE DECAy RATES AND DETECTION OF RADIOACTIvITy (SECTIONS 21.4 AND 21.5) The SI unit for the activity of a radioac-tive source is the becquerel (Bq), defined as one nuclear disintegration per second. A related unit, the curie (Ci), corresponds to 3.7 * 1010 disintegrations per second. Nuclear decay is a first-order process. The decay rate (activity) is therefore directly proportional to the number of radioactive nuclei. The half-life of a radionuclide, which is a constant independent of temperature, is the time needed for one-half of the nuclei to decay. Some radioisotopes can be used to date objects; 14C, for example, is used to date organic objects. Geiger counters and scin-tillation counters count the emissions from radioactive samples. The ease of detection of radioisotopes also permits their use as radiotracers to follow elements through reactions.

ENERGy CHANGES IN NUCLEAR REACTIONS (SECTION 21.6) The energy produced in nuclear reactions is accompanied by mea-surable changes of mass in accordance with Einstein’s relationship, ∆E = c2 ∆m. The difference in mass between nuclei and the nucleons of which they are composed is known as the mass defect. The mass defect of a nuclide makes it possible to calculate its nuclear binding energy, the energy required to separate the nucleus into individual nucleons. Because of trends in the nuclear binding energy with atomic number, energy is produced when heavy nuclei split (fission) and when light nuclei fuse (fusion).

NUCLEAR FISSION AND FUSION (SECTIONS 21.7 AND 21.8) Uranium-235, uranium-233, and plutonium-239 undergo fission when they capture a neutron, splitting into lighter nuclei and releas-ing more neutrons. The neutrons produced in one fission can cause further fission reactions, which can lead to a nuclear chain reaction. A reaction that maintains a constant rate is said to be critical, and the mass necessary to maintain this constant rate is called

a critical mass. A mass in excess of the critical mass is termed a supercritical mass.

In nuclear reactors the fission rate is controlled to generate a constant power. The reactor core consists of fuel elements containing fissionable nuclei, control rods, a moderator, and a primary coolant. A nuclear power plant resembles a conventional power plant except that the reactor core replaces the fuel burner. There is concern about the disposal of highly radioactive nuclear wastes that are generated in nuclear power plants.

Nuclear fusion requires high temperatures because nuclei must have large kinetic energies to overcome their mutual repulsions. Fusion reactions are therefore called thermonuclear reactions. It is not yet possible to generate power on Earth through a controlled fusion process.

NUCLEAR CHEMISTRy AND LIvING SySTEMS (SECTION 21.9) Ionizing radiation is energetic enough to remove an electron from a water molecule; radiation with less energy is called nonionizing radia-tion. Ionizing radiation generates free radicals, reactive substances with one or more unpaired electrons. The effects of long-term exposure to low levels of radiation are not completely understood, but there is evi-dence that the extent of biological damage varies in direct proportion to the level of exposure.

The amount of energy deposited in biological tissue by radia-tion is called the radiation dose and is measured in units of gray or rad. One gray (Gy) corresponds to a dose of 1 J/kg of tissue. It is the SI unit of radiation dose. The rad is a smaller unit; 100 rad = 1 Gy. The effective dose, which measures the biological damage created by the deposited energy, is measured in units of rem or sievert (Sv). The rem is obtained by multiplying the number of rad by the relative biological effectiveness (RBE); 100 rem = 1 Sv.

learning outcomes after studying this chapter, you should be able to:

• Write balanced nuclear equations. (Section 21.1)

• Predict nuclear stability and expected type of nuclear decay from the neutron-to-proton ratio of an isotope. (Section 21.2)

• Write balanced nuclear equations for nuclear transmutations. (Section 21.3)

• Calculate ages of objects and/or the amount of a radionuclide remain-ing after a given period of time using the half-life of the radionuclide in question. (Section 21.4)

• Calculate mass and energy changes for nuclear reactions. (Section 21.6)

• Calculate the binding energies for nuclei. (Section 21.6)

• Describe the difference between fission and fusion. (Sections 21.7 and 21.8)

• Explain how a nuclear power plant operates and know the differ-ences among various types of nuclear power plants. (Section 21.7)

• Compare different measurements and units of radiation dosage. (Section 21.9)

• Describe the biological effects of different kinds of radiation. (Section 21.9)

Key equations

• lnNt

N0 = -kt [21.19] First-order rate law for nuclear decay

• k =0.693t1>2

[21.20] Relationship between nuclear decay constant and half-life; this is derived from the previous equation at Nt =

12N0

• E = mc2 [21.22] Einstein’s equation that relates mass and energy


exercisesvisualizing Concepts

21.1 Indicate whether each of the following nuclides lies within the belt of stability in Figure 21.2: (a) neon-24, (b) chlorine-32, (c) tin-108, (d) polonium-216. For any that do not, describe a nuclear decay process that would alter the neutron-to-proton ratio in the direction of increased stability. [Section 21.2]

21.2 Write the balanced nuclear equation for the reaction repre-sented by the diagram shown here. [Section 21.2]

45 46 47 4861

62

63

64

Number of protons

Num

ber

of n

eutr

ons

21.3 Draw a diagram similar to that shown in Exercise 21.2 that illustrates the nuclear reaction 211

83Bi ¡ 42He + 207

81Tl. [Section 21.2]

21.4 In the sketch below, the red spheres represent protons and the gray spheres represent neutrons. (a) What are the identities of the four particles involved in the reaction depicted? (b) Write the transformation represented below using condensed no-tation. (c) Based on its atomic number and mass number, do you think the product nucleus is stable or radioactive? [Section 21.3]

+

+ ++

++

+

+ +

++

+

21.5 The steps below show three of the steps in the radioactive decay chain for 232

90Th. The half-life of each isotope is shown below the symbol of the isotope. (a) Identify the type of radio-active decay for each of the steps (i), (ii), and (iii). (b) Which of the isotopes shown has the highest activity? (c) Which of the isotopes shown has the lowest activity? (d) The next step in the decay chain is an alpha emission. What is the next iso-tope in the chain? [Sections 21.2 and 21.4]

1.41 × 109 yr

(i) (ii) (iii)Th23290

5.7 yr

Rn22888

6.1 min

Ac22889

1.9 yr

Th22890

21.6 The accompanying graph illustrates the decay of 8842Mo, which

decays via positron emission. (a) What is the half-life of the decay? (b) What is the rate constant for the decay? (c) What fraction of the original sample of 88

42Mo remains after 12 min? (d) What is the product of the decay process? [Section 21.4]

6420 8Time (min)

Mas

s of

1.0

0.9

0.7

0.8

0.6

0.5

0.4

0.3

0.2

0.1

10 12 14 16

88M

o(g)

42

21.7 All the stable isotopes of boron, carbon, nitrogen, oxygen, and fluorine are shown in the accompanying chart (in red), along with their radioactive isotopes with t1>2 7 1 min (in blue). (a) Write the chemical symbols, including mass and atomic numbers, for all of the stable isotopes. (b) Which ra-dioactive isotopes are most likely to decay by beta emission? (c) Some of the isotopes shown are used in positron emission tomography. Which ones would you expect to be most useful for this application? (d) Which isotope would decay to 12.5% of its original concentration after 1 hour? [Sections 21.2, 21.4, and 21.5]

5700 yr

110 min

2 min

71 s

10 min

20 min

5 6 7 8 9

5

6

7

8

9

10

Number of protons

Num

ber

of n

eutr

ons

64 s

21.8 The diagram shown here illustrates a fission process. (a) What is the unidentified product of the fission? (b) Use Figure 21.2 to predict whether the nuclear products of this fission reac-tion are stable. [Section 21.7]

0n1 0n

1

0n1

94Pu239 40Zr

95

?

+

Exercises 947

56 58 60 62 64 66 68 70 722

3

4

5

Atomic number

Log

of a

bund

ance

in E

arth

’s c

rust

(par

ts p

er b

illio

n by

mas

s)

Lu

Yb

Tm

Er

Ho

Dy

Tb

Gd

Eu

Sm

Nd

Pr

Ce

La

21.25 Using the concept of magic numbers, explain why alpha emis-sion is relatively common, but proton emission is nonexistent.

21.26 Which of the following nuclides would you expect to be ra-dioactive: 58

26Fe, 6027Co, 92

41Nb, mercury-202, radium-226? Justify your choices.

Nuclear Transmutations (section 21.3)

21.27 Why are nuclear transmutations involving neutrons generally easier to accomplish than those involving protons or alpha particles?

21.28 In 1930 the American physicist Ernest Lawrence designed the first cyclotron in Berkeley, California. In 1937 Lawrence bom-barded a molybdenum target with deuterium ions, producing for the first time an element not found in nature. What was this element? Starting with molybdenum-96 as your reactant, write a nuclear equation to represent this process.

21.29 Complete and balance the following nuclear equations by sup-plying the missing particle:(a) 252

98Cf + 105B ¡ 3 10n + ?

(b) 21H + 3

2He ¡ 42He + ?

(c) 11H + 11

5B ¡ 3?(d) 122

53I ¡ 12254Xe + ?

(e) 5926Fe ¡ 0

-1e + ? 21.30 Complete and balance the following nuclear equations by

supplying the missing particle:(a) 14

7N + 42He ¡ ? + 1

1H(b) 40

19K + 0-1e 1orbital electron2 ¡ ?

(c) ? + 42He ¡ 30

14Si + 11H

(d) 5826Fe + 2 10n ¡ 60

27Co + ?(e) 235

92U + 10n ¡ 135

54Xe + 2 10n + ? 21.31 Wr it e b a l an c e d e qu at i ons for ( a ) 238

92U1a, n224194Pu,

(b) 147N1a, p217

8O, (c) 5626Fe1a, b-260

29Cu. 21.32 Write balanced equations for each of the following nuclear reac-

tions: (a) 23892U1n, g2239

92U, (b) 168O1p, a213

7N, (c) 188O1n, b-219

9F.

Rates of Radioactive decay (section 21.4)

21.33 Each statement that follows refers to a comparison between two radioisotopes, A and X. Indicate whether each of the fol-lowing statements is true or false, and why.(a) If the half-life for A is shorter than the half-life for X, A

has a larger decay rate constant.

Radioactivity and Nuclear equations (section 21.1)

21.9 Indicate the number of protons and neutrons in the following nuclei: (a) 56

24Cr, (b) 193Tl, (c) argon-38. 21.10 Indicate the number of protons and neutrons in the following

nuclei: (a) 12953I, (b) 138Ba, (c) neptunium-237.

21.11 Give the symbol for (a) a neutron, (b) an alpha particle, (c) gamma radiation.

21.12 Give the symbol for (a) a proton, (b) a beta particle, (c) a positron.

21.13 Write balanced nuclear equations for the following processes: (a) rubidium-90 undergoes beta emission; (b) selenium-72 undergoes electron capture; (c) krypton-76 undergoes posi-tron emission; (d) radium-226 emits alpha radiation.

21.14 Write balanced nuclear equations for the following trans-formations: (a) bismuth-213 undergoes alpha decay; (b) nitrogen-13 undergoes electron capture; (c) technicium-98 undergoes electron capture; (d) gold-188 decays by positron emission.

21.15 Decay of which nucleus will lead to the following prod-ucts: (a) bismuth-211 by beta decay; (b) chromium-50 by positron emission; (c) tantalum-179 by electron capture; (d) radium-226 by alpha decay?

21.16 What particle is produced during the following decay pro-cesses: (a) sodium-24 decays to magnesium-24; (b) mercury-188 decays to gold-188; (c) iodine-122 decays to xenon-122; (d) plutonium-242 decays to uranium-238?

21.17 The naturally occurring radioactive decay series that begins with 235

92U stops with formation of the stable 20782Pb nucleus.

The decays proceed through a series of alpha-particle and beta-particle emissions. How many of each type of emission are involved in this series?

21.18 A radioactive decay series that begins with 23290Th ends with

formation of the stable nuclide 20882Pb. How many alpha-

particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

patterns of Nuclear stability (section 21.2)

21.19 Predict the type of radioactive decay process for the following radionuclides: (a) 8

5B, (b) 6829Cu, (c) phosphorus-32,

(d) chlorine-39. 21.20 Each of the following nuclei undergoes either beta decay

or positron emission. Predict the type of emission for each: (a) tritium, 31H, (b) 89

38Sr, (c) iodine-120, (d) silver-102. 21.21 One of the nuclides in each of the following pairs is radioac-

tive. Predict which is radioactive and which is stable: (a) 3919K

and 4019K, (b) 209Bi and 208Bi, (c) nickel-58 and nickel-65.

21.22 One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: (a) 40

20Ca and 4520Ca,

(b) 12C and 14C, (c) lead-206 and thorium-230. Explain your choice in each case.

21.23 Which of the following nuclides have magic numbers of both protons and neutrons: (a) helium-4, (b) oxygen-18, (c) cal-cium-40, (d) zinc-66, (e) lead-208?

21.24 Despite the similarities in the chemical reactivity of elements in the lanthanide series, their abundances in Earth’s crust vary by two orders of magnitude. This graph shows the rela-tive abundance as a function of atomic number. How do you explain the sawtooth variation across the series?


21.46 An analytical laboratory balance typically measures mass to the nearest 0.1 mg. What energy change would accompany the loss of 0.1 mg in mass?

21.47 How much energy must be supplied to break a single alu-minum-27 nucleus into separated protons and neutrons if an aluminum-27 atom has a mass of 26.9815386 amu? How much energy is required for 100.0 grams of aluminum-27? (The mass of an electron is given on the inside back cover.)

21.48 How much energy must be supplied to break a single 21Ne nucleus into separated protons and neutrons if the nucleus has a mass of 20.98846 amu? What is the nuclear binding en-ergy for 1 mol of 21Ne?

21.49 The atomic masses of hydrogen-2 (deuterium), helium-4, and lithium-6 are 2.014102 amu, 4.002602 amu, and 6.0151228 amu, respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon. (d) Which of these three isotopes has the largest nuclear binding energy per nucleon? Does this agree with the trends plotted in Figure 21.12?

21.50 The atomic masses of nitrogen-14, titanium-48, and xe-non-129 are 13.999234 amu, 47.935878 amu, and 128.904779 amu, respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon.

21.51 The energy from solar radiation falling on Earth is 1.07 * 1016 kJ>min. (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction

235U + 10n ¡ 141

56Ba + 9236Kr + 31

0n

(235U nuclear mass, 234.9935 amu; 141Ba nuclear mass, 140.8833 amu; 92Kr nuclear mass, 91.9021 amu) is taken as typical of that occurring in a nuclear reactor, what mass of uranium-235 is required to equal 0.10% of the solar energy that falls on Earth in 1.0 day?

21.52 Based on the following atomic mass values:1H, 1.00782 amu; 2H, 2.01410 amu; 3H, 3.01605 amu; 3He, 3.01603 amu; 4He, 4.00260 amu—and the mass of the neutron given in the text, calculate the energy released per mole in each of the fol-lowing nuclear reactions, all of which are possibilities for a controlled fusion process:(a) 2

1H + 31H ¡ 4

2He + 10n

(b) 21H + 2

1H ¡ 32He + 1

0n(c) 2

1H + 32He ¡ 4

2He + 11H

21.53 Which of the following nuclei is likely to have the largest mass defect per nucleon: (a) 59Co, (b) 11B, (c) 118Sn, (d) 243Cm? Ex-plain your answer.

21.54 The isotope 6228Ni has the largest binding energy per nucleon

of any isotope. Calculate this value from the atomic mass of nickel-62 (61.928345 amu) and compare it with the value given for iron-56 in Table 21.7.

Nuclear power and Radioisotopes (sections 21.7–21.9)

21.55 Iodine-131 is a convenient radioisotope to monitor thyroid ac-tivity in humans. It is a beta emitter with a half-life of 8.02 days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of NaI, in which only a small fraction of the iodide is radioactive.

(b) If X is “not radioactive,” its half-life is essentially zero.(c) If A has a half-life of 10 yr, and X has a half-life of 10,000 yr,

A would be a more suitable radioisotope to measure processes occurring on the 40-yr time scale.

21.34 It has been suggested that strontium-90 (generated by nuclear testing) deposited in the hot desert will undergo radioactive de-cay more rapidly because it will be exposed to much higher av-erage temperatures. (a) Is this a reasonable suggestion? (b) Does the process of radioactive decay have an activation energy, like the Arrhenius behavior of many chemical reactions (Section 14.5)?

21.35 Some watch dials are coated with a phosphor, like ZnS, and a polymer in which some of the 1H atoms have been replaced by 3H atoms, tritium. The phosphor emits light when struck by the beta particle from the tritium decay, causing the dials to glow in the dark. The half-life of tritium is 12.3 yr. If the light given off is assumed to be directly proportional to the amount of tritium, by how much will a dial be dimmed in a watch that is 50 yr old?

21.36 It takes 4 h 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. What is the half-life of radium-230?

21.37 Cobalt-60 is a strong gamma emitter that has a half-life of 5.26 yr. The cobalt-60 in a radiotherapy unit must be replaced when its radioactivity falls to 75% of the original sample. If an original sample was purchased in June 2013, when will it be necessary to replace the cobalt-60?

21.38 How much time is required for a 6.25-mg sample of 51Cr to decay to 0.75 mg if it has a half-life of 27.8 days?

[21.39] Radium-226, which undergoes alpha decay, has a half-life of 1600 yr. (a) How many alpha particles are emitted in 5.0 min by a 10.0-mg sample of 226Ra? (b) What is the activity of the sample in mCi?

21.40 Cobalt-60, which undergoes beta decay, has a half-life of 5.26 yr. (a) How many beta particles are emitted in 600 s by a 3.75-mg sample of 60Co? (b) What is the activity of the sample in Bq?

21.41 The cloth shroud from around a mummy is found to have a 14C activity of 9.7 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. From the half-life for 14C decay, 5715 yr, calculate the age of the shroud.

21.42 A wooden artifact from a Chinese temple has a 14C activity of 38.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. From the half-life for 14C decay, 5715 yr, determine the age of the artifact.

21.43 Potassium-40 decays to argon-40 with a half-life of 1.27 * 109 yr. What is the age of a rock in which the mass ratio of 40Ar to 40K is 4.2?

21.44 The half-life for the process 238U ¡ 206Pb is 4.5 * 109 yr. A mineral sample contains 75.0 mg of 238U and 18.0 mg of 206Pb. What is the age of the mineral?

energy Changes in Nuclear Reactions (section 21.6)

21.45 The thermite reaction, Fe2O31s2 + 2 Al1s2 ¡ 2 Fe1s2 +Al2O31s2, ∆H° = -851.5 kJ>mol, is one of the most exother-mic reactions known. Because the heat released is sufficient to melt the iron product, the reaction is used to weld metal un-der the ocean. How much heat is released per mole of Fe2O3 produced? How does this amount of thermal energy compare with the energy released when 2 mol of protons and 2 mol of neutrons combine to form 1 mol of alpha particles?

Additional Exercises 949

which requires a temperature of 106 to 107 K. (a) Use the mass of the helium-4 nucleus given in Table 21.7 to determine how much energy is released when the reaction is run with 1 mol of hydrogen atoms. (b) Why is such a high temperature required?

21.64 The spent fuel elements from a fission reactor are much more intensely radioactive than the original fuel elements. (a) What does this tell you about the products of the fission process in re-lationship to the belt of stability, Figure 21.2? (b) Given that only two or three neutrons are released per fission event and know-ing that the nucleus undergoing fission has a neutron-to-proton ratio characteristic of a heavy nucleus, what sorts of decay would you expect to be dominant among the fission products?

21.65 Which type or types of nuclear reactors have these characteristics?(a) Does not use a secondary coolant(b) Creates more fissionable material than it consumes(c) Uses a gas, such as He or CO2, as the primary coolant

21.66 Which type or types of nuclear reactors have these characteristics?(a) Can use natural uranium as a fuel(b) Does not use a moderator(c) Can be refueled without shutting down

21.67 Hydroxyl radicals can pluck hydrogen atoms from molecules (“hydrogen abstraction”), and hydroxide ions can pluck pro-tons from molecules (“deprotonation”). Write the reaction equations and Lewis dot structures for the hydrogen abstrac-tion and deprotonation reactions for the generic carboxylic acid R ¬ COOH with hydroxyl radical and hydroxide ion, respectively. Why is hydroxyl radical more toxic to living sys-tems than hydroxide ion?

21.68 Which are classified as ionizing radiation: X rays, alpha par-ticles, microwaves from a cell phone, and gamma rays?

21.69 A laboratory rat is exposed to an alpha-radiation source whose activity is 14.3 mCi. (a) What is the activity of the radi-ation in disintegrations per second? In becquerels? (b) The rat has a mass of 385 g and is exposed to the radiation for 14.0 s, absorbing 35% of the emitted alpha particles, each having an energy of 9.12 * 10-13 J. Calculate the absorbed dose in mil-lirads and grays. (c) If the RBE of the radiation is 9.5, calculate the effective absorbed dose in mrem and Sv.

21.70 A 65-kg person is accidentally exposed for 240 s to a 15-mCi source of beta radiation coming from a sample of 90Sr. (a) What is the activity of the radiation source in disintegrations per second? In becquerels? (b) Each beta particle has an energy of 8.75 * 10-14 J. and 7.5% of the radiation is absorbed by the person. Assuming that the absorbed radiation is spread over the person’s entire body, calculate the absorbed dose in rads and in grays. (c) If the RBE of the beta particles is 1.0, what is the ef-fective dose in mrem and in sieverts? (d) Is the radiation dose equal to, greater than, or less than that for a typical mammogram (300 mrem)?

(a) Why is NaI a good choice for the source of iodine? (b) If a Geiger counter is placed near the person’s thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thy-roid will take up about 12% of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to 0.01% of the original amount?

21.56 Why is it important that radioisotopes used as diagnostic tools in nuclear medicine produce gamma radiation when they de-cay? Why are alpha emitters not used as diagnostic tools?

21.57 (a) Which of the following are required characteristics of an isotope to be used as a fuel in a nuclear power reactor? (i) It must emit gamma radiation. (ii) On decay, it must release two or more neutrons. (iii) It must have a half-life less than one hour. (iv) It must undergo fission upon the absorption of a neutron. (b) What is the most common fissionable isotope in a commercial nuclear power reactor?

21.58 (a) Which of the following statements about the uranium used in nuclear reactors is or are true? (i) Natural uranium has too little 235U to be used as a fuel. (ii) 238U cannot be used as a fuel because it forms a supercritical mass too easily. (iii) To be used as fuel, uranium must be enriched so that it is more than 50% 235U in composition. (iv) The neutron-induced fission of 235U releases more neutrons per nucleus than fission of 238U. (b) Which of the following statements about the plutonium shown in the chapter-opening photograph explains why it cannot be used for nuclear power plants or nuclear weapons? (i) None of the isotopes of Pu possess the characteristics needed to support nuclear fission chain reactions. (ii) The orange glow indicates that the only radioactive decay products are heat and visible light. (iii) The particular isotope of plutonium used for RTGs is incapable of sustaining a chain reaction. (iv) Pluto-nium can be used as a fuel, but only after it decays to uranium.

21.59 What is the function of the control rods in a nuclear reactor? What substances are used to construct control rods? Why are these substances chosen?

21.60 (a) What is the function of the moderator in a nuclear reac-tor? (b) What substance acts as the moderator in a pressur-ized water generator? (c) What other substances are used as a moderator in nuclear reactor designs?

21.61 Complete and balance the nuclear equations for the following fission or fusion reactions:(a) 2

1H + 21H ¡ 3

2He + :

(b) 23992U + 1

0n ¡ 13351Sb + 98

41Nb + : 10n

21.62 Complete and balance the nuclear equations for the following fission reactions:(a) 235

92U + 10n ¡ 160

62Sm + 7230Zn + : 10n

(b) 23994 Pu + 1

0n ¡ 14458 Ce + : + 2 10n

21.63 A portion of the Sun’s energy comes from the reaction4 11H ¡ 4

2He + 2 01e

additional exercises 21.71 The table to the right gives the number of protons (p) and

neutrons (n) for four isotopes. (a) Write the symbol for each of the isotopes. (b) Which of the isotopes is most likely to be unstable? (c) Which of the isotopes involves a magic num-ber of protons and/or neutrons? (d) Which isotope will yield potassium-39 following positron emission?

(i) (ii) (iii) (iv)

p 19 19 20 20n 19 21 19 20


atoms of strontium-90 to which this dose corresponds. To what mass of strontium-90 does this correspond? The half-life for strontium-90 is 28.8 yr.

[21.80] Suppose you had a detection device that could count every decay event from a radioactive sample of plutonium-239 (t1>2 is 24,000 yr). How many counts per second would you obtain from a sample containing 0.385 g of plutonium-239?

21.81 Methyl acetate 1CH3COOCH32 is formed by the reaction of acetic acid with methyl alcohol. If the methyl alcohol is labeled with oxygen-18, the oxygen-18 ends up in the methyl acetate:

CH3COH H18OCH3+

O

CH3C18OCH3 H2O+

O

(a) Do the C ¬ OH bond of the acid and the O ¬ H bond of the alcohol break in the reaction, or do the O ¬ H bond of the acid and the C ¬ OH bond of the alcohol break? (b) Imagine a similar experiment using the radioisotope 3H, which is called tritium and is usually denoted T. Would the reaction between CH3COOH and TOCH3 provide the same information about which bond is broken as does the above experiment with H18OCH3?

21.82 An experiment was designed to determine whether an aquatic plant absorbed iodide ion from water. Iodine-131 (t1>2 = 8.02 days) was added as a tracer, in the form of iodide ion, to a tank containing the plants. The initial activity of a 1.00@mL sample of the water was 214 counts per minute. After 30 days the level of activity in a 1.00@mL sample was 15.7 counts per minute. Did the plants absorb iodide from the water?

21.83 Each of the following transmutations produces a radionuclide used in positron emission tomography (PET). (a) In equa-tions (i) and (ii), identify the species signified as “X.” (b) In equation (iii), one of the species is indicated as “d.” What do you think it represents?(i) 14N1p, a2X (ii) 18O1p, X218F (iii) 14N1d, n215O

21.84 The nuclear masses of 7Be, 9Be, and 10Be are 7.0147, 9.0100, and 10.0113 amu, respectively. Which of these nuclei has the largest binding energy per nucleon?

21.85 A 26.00-g sample of water containing tritium, 31H, emits

1.50 * 103 beta particles per second. Tritium is a weak beta emitter with a half-life of 12.3 yr. What fraction of all the hy-drogen in the water sample is tritium?

21.86 The Sun radiates energy into space at the rate of 3.9 * 1026 J/s. (a) Calculate the rate of mass loss from the Sun in kg/s. (b) How does this mass loss arise? (c) It is estimated that the Sun contains 9 * 1056 free protons. How many protons per second are consumed in nuclear reactions in the Sun?

21.87 The average energy released in the fission of a single ura-nium-235 nucleus is about 3 * 10-11 J. If the conversion of this energy to electricity in a nuclear power plant is 40% effi-cient, what mass of uranium-235 undergoes fission in a year in a plant that produces 1000 megawatts? Recall that a watt is 1 J/s.

21.88 Tests on human subjects in Boston in 1965 and 1966, follow-ing the era of atomic bomb testing, revealed average quanti-ties of about 2 pCi of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits 8 * 10-13 J of energy and if the average person weighs 75 kg, calculate the number of rads and rems of radiation in 1 yr from such a level of plutonium.

21.72 Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

21.73 Equation 21.28 is the nuclear reaction responsible for much of the helium-4 production in our Sun. How much energy is released in this reaction?

21.74 Chlorine has two stable nuclides, 35Cl and 37Cl. In contrast, 36Cl is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of 36Cl? (b) Based on the em-pirical rules about nuclear stability, explain why the nucleus of 36Cl is less stable than either 35Cl or 37Cl.

21.75 When two protons fuse in a star, the product is 2H plus a pos-itron (Equation 21.26). Why do you think the more obvious product of the reaction, 2He, is unstable?

21.76 Nuclear scientists have synthesized approximately 1600 nu-clei not known in nature. More might be discovered with heavy-ion bombardment using high-energy particle accelera-tors. Complete and balance the following reactions, which in-volve heavy-ion bombardments:(a) 6

3Li + 5628Ni ¡ ?

(b) 4020Ca + 248

96Cm ¡ 14762Sm + ?

(c) 8838Sr + 84

36Kr ¡ 11646Pd + ?

(d) 4020Ca + 238

92U ¡ 7030Zn + 4 10n + 2?

21.77 In 2010, a team of scientists from Russia and the U.S. reported creation of the first atom of element 117, which is not yet named and is denoted [117]. The synthesis involved the col-lision of a target of 249

97Bk with accelerated ions of an isotope which we will denote Q. The product atom, which we will call Z, immediately releases neutrons and forms 294

11731174:249

97Bk + Q ¡ Z ¡ 294117[117] + 3 10n

(a) What are the identities of isotopes Q and Z? (b) Isotope Q is unusual in that it is very long-lived (its half-life is on the order of 1019 yr) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.2). Can you propose a reason for its un-usual stability? (c) Collision of ions of isotope Q with a target was also used to produce the first atoms of livermorium, Lv. The initial product of this collision was 296

116Lv. What was the target isotope with which Q collided in this experiment?

21.78 The synthetic radioisotope technetium-99, which decays by beta emission, is the most widely used isotope in nuclear med-icine. The following data were collected on a sample of 99Tc:

disintegrations per minute Time (h)

180 0130 2.5104 5.0 77 7.5 59 10.0 46 12.5 24 17.5

Using these data, make an appropriate graph and curve fit to determine the half-life.

[21.79] According to current regulations, the maximum permissible dose of strontium-90 in the body of an adult is 1 mCi11 * 10-6 Ci2. Using the relationship rate = kN, calculate the number of

Design an Experiment 951

integrative exercises 21.89 A 53.8-mg sample of sodium perchlorate contains radioactive

chlorine-36 (whose atomic mass is 36.0 amu). If 29.6% of the chlorine atoms in the sample are chlorine-36 and the remain-der are naturally occurring nonradioactive chlorine atoms, how many disintegrations per second are produced by this sample? The half-life of chlorine-36 is 3.0 * 105 yr.

21.90 Calculate the mass of octane, C8H181l2, that must be burned in air to evolve the same quantity of energy as produced by the fusion of 1.0 g of hydrogen in the following fusion reaction:

4 11H ¡ 42He + 2 01e

Assume that all the products of the combustion of octane are in their gas phases. Use data from Exercise 21.50, Appendix C, and the inside covers of the text. The standard enthalpy of formation of octane is -250.1 kJ>mol.

21.91 Natural ly found uranium consists of 99.274% 238U, 0.720% 235U, and 0.006% 233U. As we have seen, 235U is the isotope that can undergo a nuclear chain reaction. Most of the 235U used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, UF61g2. (a) What is the mass of UF6 in a 30.0-L vessel of UF6 at a pressure of 695 torr at 350 K? (b) What is the mass of 235U in the sample described in part (a)? (c) Now suppose that the UF6 is diffused through a porous barrier and that the change in the ratio of 238U and 235U in the diffused gas can be described by Equation 10.23. What

is the mass of 235U in a sample of the diffused gas analogous to that in part (a)? (d) After one more cycle of gaseous diffusion, what is the percentage of 235UF6 in the sample?

21.92 A sample of an alpha emitter having an activity of 0.18 Ci is stored in a 25.0-mL sealed container at 22 °C for 245 days. (a) How many alpha particles are formed during this time? (b) Assuming that each alpha particle is converted to a helium atom, what is the partial pressure of helium gas in the con-tainer after this 245-day period?

[21.93] Charcoal samples from Stonehenge in England were burned in O2, and the resultant CO2 gas bubbled into a solution of Ca1OH22 (limewater), resulting in the precipitation of CaCO3. The CaCO3 was removed by filtration and dried. A 788-mg sample of the CaCO3 had a radioactivity of 1.5 * 10-2 Bq due to carbon-14. By comparison, living organisms undergo 15.3 dis-integrations per minute per gram of carbon. Using the half-life of carbon-14, 5700 yr, calculate the age of the charcoal sample.

21.94 A 25.0-mL sample of 0.050 M barium nitrate solution was mixed with 25.0 mL of 0.050 M sodium sulfate solution labeled with ra-dioactive sulfur-35. The activity of the initial sodium sulfate solu-tion was 1.22 * 106 Bq>mL. After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 250 Bq/mL. (a) Write a balanced chemical equa-tion for the reaction that occurred. (b) Calculate the Ksp for the precipitate under the conditions of the experiment.

design an experimentThis chapter has focused on the properties of elements that exhibit radioactivity. Because radio-activity can have harmful effects on human health, very stringent experimental procedures and precautions are required when undertaking experiments on radioactive materials. As such, we typically do not have experiments involving radioactive substances in general chemistry labora-tories. We can nevertheless ponder the design of some hypothetical experiments that would allow us to explore some of the properties of radium, which was discovered by Marie and Pierre Curie in 1898.(a) A key aspect of the discovery of radium was Marie Curie’s observation that pitchblende, a

natural ore of uranium, had greater radioactivity than pure uranium metal. Design an experiment to reproduce this observation and to obtain a ratio of the activity of pitchblende relative to that of pure uranium.

(b) Radium was first isolated as halide salts. Suppose you had pure samples of radium metal and radium bromide. The sample sizes are on the order of milligrams and are not amenable to the usual forms of elemental analysis. Could you use a device that measures radioactivity quantitatively to determine the empirical formula of radium bromide? What information must you use that the Curies may not have had at the time of their discovery?

(c) Suppose you had a 1-yr time period in order to measure the half-life of radium and related elements. You have some pure samples and a device that measures radioactivity quantitatively. Could you determine the half-life of the elements in the samples? Would you have different experimental constraints depending on whether the half-life were 10 yr or 1000 yr?

(d) Before its negative health effects were better understood, small amounts of radium salts were used in “glow in the dark” watches, such as the one shown here. The glow is not due to the radioactivity of radium directly; rather, the radium is combined with a luminescent sub-stance, such as zinc sulfide, which glows when it is exposed to radiation. Suppose you had pure samples of radium and zinc sulfide. How could you determine whether the glow of zinc sulfide is due to alpha, beta, or gamma radiation? What type of device could you design to use the glow as a quantitative measure of the amount of radioactivity in a sample?

supercritical solvents · 2018-08-30 · learning outcomes 803 chapter summary and key terms...

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