chapter 20 electrochemistry. 20.1 oxidation states and oxidation-reduction reactions

Chapter 20Electrochemistry

20.1 Oxidation States and Oxidation-Reduction Reactions

Electrochemical Reactions

In electrochemical reactions, electrons are transferred from one species to another.

Oxidation Numbers

In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

Oxidation and Reduction

A species is oxidized when it loses electrons.Here, zinc loses two electrons to go from neutral zinc

metal to the Zn2+ ion.


A species is reduced when it gains electrons.Here, each of the H+ gains an electron, and they

combine to form H2.


What is reduced is the oxidizing agent.H+ oxidizes Zn by taking electrons from it.

What is oxidized is the reducing agent.Zn reduces H+ by giving it electrons.

Assigning Oxidation Numbers

1. Elements in their elemental form have an oxidation number of 0.

2. The oxidation number of a monatomic ion is the same as its charge.


3. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.

Oxygen has an oxidation number of −2, except in the peroxide ion, which has an oxidation number of −1.

Hydrogen is −1 when bonded to a metal and +1 when bonded to a nonmetal.


3. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.

Fluorine always has an oxidation number of −1.

The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.


4. The sum of the oxidation numbers in a neutral compound is 0.

5. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

p.845 GIST: What are the oxidation numbers of the elements in the nitrite ion, NO2

-?

The nickel-cadmium (nicad) battery, a rechargeable “dry cell” used in battery-operated devices, uses the following redox reaction to generate electricity:

Cd(s) + NiO2(s) + 2 H2O(l) → Cd(OH)2(s) + Ni(OH)2(s)

Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent.

Sample Exercise 20.1

Identify the oxidizing and reducing agents in the oxidation-reduction reaction

2 H2O(l) + Al(s) + MnO4–

(aq) → Al(OH)4–(aq) + MnO2(s)

Practice Exercise

Types of Redox Reactions

1. Hydrogen DisplacementCa(s) + 2H2O(l) --> Ca(OH)2(s) + H2

2. Metal DisplacementZn(s) + CuSO4(aq) ---> ZnSO4(aq) + Cu(s)

3. Halogen DisplacementCl2(g) + KBr(aq) ----> 2KCl(aq) + Br2(l)

4. CombustionCH4(g) + 2O2(g) ---> CO2(g) + H2O(g)

Types of Redox ReactionsDisportionationThis is where one substance both oxidizes

and reduces

Cl2(g) + 2OH-(aq)---> OCl-(aq) + Cl-(aq) + H2O(l)

Reactions involving oxoanions such as Cr2O7

2-

14H+ +Cr2O72- + 6Fe2+ --> Cr3+ +7H2O + 6Fe3+

20.2 Balancing Oxidation-Reduction Reactions

Balancing Oxidation-Reduction Equations

Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.

Balancing Oxidation-Reduction Equations

This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.

The Half-Reaction Method

1. Assign oxidation numbers to determine what is oxidized and what is reduced.

2. Write the oxidation and reduction half-reactions.


3. Balance each half-reaction.a. Balance elements other than H and O.

b. Balance O by adding H2O.c. Balance H by adding H+.d. Balance charge by adding electrons.

4. Multiply the half-reactions by integers so that the electrons gained and lost are the same.


5. Add the half-reactions, subtracting things that appear on both sides.

6. Make sure the equation is balanced according to mass.

7. Make sure the equation is balanced according to charge.


Consider the reaction between MnO4- and C2O4

2- :

MnO4-

(aq) + C2O42-

(aq) Mn2+ (aq) + CO2 (aq)

Practice

Sn2+(aq) + Hg2+

(aq) + Cl-(aq)→ Sn4+(aq) + Hg2Cl2(s)

Hints for balancing in acidic solution

1) Divide into half reactions and balance atoms other than oxygen and hydrogen

2) Balance oxygen by adding water to the side that needs oxygen. Add one H2O for each oxygen needed.

3) Balance the hydrogen by adding H+ to the other side that needs hydrogen. Add one H+ for each hydrogen needed.

4) Balance the charges in the half-reactions and continue as before

Example: Cl- + MnO4- → Cl2 + Mn2+

(acidic solution)

Sample Exercise 20.2 Balancing Redox Equations in Acidic Solution

Complete and balance this equation by the method of half-reactions:

Complete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution.

Practice Exercise

(c) Cr2O72- + H2S → Cr2+ + S

Balancing in Basic Solution

If a reaction occurs in basic solution, one can balance it as if it occurred in acid.

Once the equation is balanced, add OH- to each side to “neutralize” the H+ in the equation and create water in its place.

Cancel any H2O that are the same on both sides.

Basic Solution Half-reaction Example

Pb → PbO

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:

Complete and balance the following equations for oxidation-reduction reactions that occur in basic solution:

Practice Exercise

(c) Pb(OH)3- + OCl- → PbO2 + Cl-

p.848 GIST: If the equation for a redox reaction is balanced, will free electrons appear in the equation as either reactants or products?

20.3 Voltaic Cells

Voltaic Cells

In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

Voltaic Cells

We can use that energy to do work if we make the electrons flow through an external device.

We call such a setup a voltaic cell.

Voltaic Cells

A typical cell looks like this.

The oxidation occurs at the anode.

The reduction occurs at the cathode.

Voltaic Cells

Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

Voltaic Cells

Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.Cations move toward

the cathode.Anions move toward

the anode.

Voltaic CellsIn the cell, then,

electrons leave the anode and flow through the wire to the cathode.

As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

Voltaic CellsAs the electrons

reach the cathode, cations in the cathode are attracted to the now negative cathode.

The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

Sample Exercise 20.4 Describing a Voltaic Cell

The oxidation-reduction reaction

is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution 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 direction of ion migration, and the signs of the electrodes.

The two half-reactions in a voltaic cell are

(a) Indicate which reaction occurs at the anode and which at the cathode. (b) Which electrode is consumed in the cell reaction? (c) Which electrode is positive?

Practice Exercise

20.3 GIST

Why do anions in a salt bridge migrate toward the anode?

What happens to the surface Zn atoms as they lose electrons?

20.4 Cell EMF Under Standard Conditions

Electromotive Force (emf)Water only

spontaneously flows one way in a

waterfall.Likewise, electrons

only spontaneously flow one way in a

redox reaction—from higher to lower

potential energy.

Electromotive Force (emf)

The potential difference between the anode and cathode in a cell is called the electromotive force (emf).

It is also called the cell potential and is designated Ecell.

Cell Potential

Cell potential is measured in volts (V).

1 V = 1 JC

Standard Reduction Potentials

Reduction potentials for

many electrodes have been

measured and tabulated.

Standard Hydrogen Electrode

Their values are referenced to a standard hydrogen electrode (SHE).

By definition, the reduction potential for hydrogen is 0 V:

2 H+ (aq, 1M) + 2 e− H2 (g, 1 atm)

Standard Cell Potentials

The cell potential at standard conditions can be found through this equation:

Ecell = Ered (cathode) − Ered (anode)

Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

Cell Potentials

For the oxidation in this cell,

For the reduction,

Ered = −0.76 V

Ered = +0.34 V

Cell Potentials

Ecell = Ered

(cathode) − Ered (anode)

= +0.34 V − (−0.76 V)

= +1.10 V

Sample Exercise 20.5 Calculating E°red from E°cell

For the Zn–Cu2+ voltaic cell shown in Figure 20.5, we have

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:

A voltaic cell is based on the half-reactions

The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate E°red for the reduction of In3+ to In+.

Practice Exercise

Sample Exercise 20.6 Calculating E°cell from E°red

Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction

Using data in Table 20.1, calculate the standard emf for a cell that employs the following overall cell reaction:

Practice Exercise

20.4 GIST

If a standard cell potential is +0.85 V at 25°C, is the redox reaction of the cell spontaneous?

For the half-reaction Cl2(g) + 2e- → 2Cl-(aq), what are the standard conditions for the reactant and product?

True or false: the smaller the difference is between the standard reduction potentials of the cathode and anode, the smaller the driving force for the overall reaction.

Sample Exercise 20.7 Determining Half-Reactions at electrodes and Calculating Cell EMF

A voltaic cell is based on the following two standard half-reactions:

By using the data in Appendix E, determine (a) the half-reactions that occur at the cathode and the anode, and (b) the standard cell potential.

A voltaic cell is based on a Co2+/Co half-cell and an AgCl/Ag half-cell.

(a) What half-reaction occurs at the anode?

(b) What is the standard cell potential?

Practice Exercise

Oxidizing and Reducing Agents

The strongest oxidizers have the most positive reduction potentials.

The strongest reducers have the most negative reduction potentials.

Oxidizing and Reducing Agents

The greater the difference between the two, the greater the voltage of the cell.

Sample Exercise 20.8 Determining the Relative Strengths of Oxidizing Agents

Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO3

–(aq), Ag+(aq), Cr2O72–

(aq).Using Table 20.1, rank the following species from the strongest to the weakest reducing agent: I–(aq), Fe(s), Al(s).

Practice Exercise

20.5 Free Energy and Redox Reactions

Sample Exercise 20.9 Spontaneous or Not?

Using standard reduction potentials in Table 20.1, determine whether the following reactions are spontaneous under standard conditions.

Using the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions:

.

Practice Exercise

p.863 GIST: Based on Table 4.5, which is the stronger reducing agent: Hg(l) or Pb(s)?

Free Energy

G for a redox reaction can be found by using the equation

G = −nFE

where n is the number of moles of electrons transferred, and F is a constant, the Faraday.

1 F = 96,485 C/mol = 96,485 J/V-mol

Free Energy

Under standard conditions,

G = −nFE

Sample Exercise 20.10 Determining ΔG° and K

(a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, , and the equilibrium constant, K, at 298 K for the reaction

(b) Suppose the reaction in part (a) was written

What are the values of E°, ΔG°, and K when the reaction is written in this way?

For the reaction

(a) What is the value of n? (b) Use the data in Appendix E to calculate ΔG°. (c) Calculate K at T = 298 K.

Practice Exercise

20.6 Cell EMF Under Nonstandard Conditions

Nernst Equation

Remember that

G = G + RT ln Q

This means

−nFE = −nFE + RT ln Q

Nernst Equation

Dividing both sides by −nF, we get the Nernst equation:

E = E −RTnF

ln Q

or, using base-10 logarithms,

E = E −2.303 RTnF

log Q

Nernst Equation

At room temperature (298 K),

Thus the equation becomes

E = E −0.0592n

log Q

2.303 RTF

= 0.0592 V


Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 when [Cr2O7

2-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M and [Cr3+] = 1.0 x 10-5 M.

Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2Cr3+

(aq) + 3I2(s) + 7H2O(l)

Practice Exercise

Calculate the emf generated by the cell described in the practice exercise accompanying Sample Exercise 20.6 when [Al3+] = 4.0 x 10-3 M and [I-] = 0.010 M.


If the voltage of a Zn-H+ cell (like that in Figure 20.11) is 0.45 V at 25°C when [Zn2+] = 1.0 M and PH2 = 1.0 atm, what is the concentration of H+?

Practice Exercise

What is the pH of the solution in the cathode compartment of the cell pictured in Figure 20.11 when PH2 = 1.0 atm, [Zn2+] in the anode compartment is 0.10 M, and cell emf is 0.542 V?

Example

Calculate the emf at 298 K generated by the cell described below:VO2

+ + 2H+ + e- VO2+ + H2O °= 1.00 V

Zn2+ + 2e- Zn °= -0.76VAt 25°C, [VO2

+]=2.0 M, [H+]=0.50 M, [VO2+]=0.010 M, [Zn2+]= 0.10 M

Concentration Cell

If the same half reaction is on each side in the same concentration, the cell =0

Cell where both compartments contain same half-reaction but with different concentrations

Goal: to even out the concentrationsUse up the ion on the more concentrated

side

Concentration Cell

Ag+ is used up on R

and made on L

Ag(s) is used up on L

and created on R

electrons move LR

Example

Determine the direction of e- flow and the anode and cathode

Concentration Cells

Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.

• For such a cell, would be 0, but Q would not.Ecell

• Therefore, as long as the concentrations are different, E will not be 0.

Sample Exercise 20.13A voltaic cell is constructed with two hydrogen

electrodes. Electrode 1 ahs PH2 = 1.00 atm and an unknown concentration of H+(aq). Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, PH2 = 1.00 atm). At 298 K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate [H+] for the solution at electrode 1. What is the pH?

Practice Exercise

A concentration cell is constructed with two Zn(s)-Zn2+(aq) half-cells. The first half-cell has [Zn2+] = 1.35 M, and the second half-cell has [Zn2+] = 3.75 x 10-4 M. A) Which half-cell is the anode of the cell?

B) What is the emf of the cell?

20.9 Electolysis

Electrolysis Rxns or Electrolytic Cells

Use of electrical energy to cause nonspontaneous redox rxns to occur

Consists of 2 electrodes in a molten salt or solution

A battery or other source of electric current acts an electron pump pulling e- from one electrode (anode) and pushing e- to another electrode (cathode)

Electroplating

Uses electrolysis to deposit a thin layer of one metal on another metal to improve beauty or resistance to corrosion

Figure 20.30

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.

(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?

Practice Exercise

Sample Exercise 20.15 Calculating Energy in Kilowatt-hours

Calculate the number of kilowatt-hours of electricity required to produce 1.0 × 103 kg of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V.

Sample Exercise 20.15 Calculating Energy in Kilowatt-hours

Calculate the number of kilowatt-hours of electricity required to produce 1.00 kg of Mg from electrolysis of molten MgCl2 if the applied emf is 5.00 V. Assume that the process is 100% efficient.

Practice Exercise

Sample Integrative Exercise Putting Concepts Together

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 FeF2(s) in water. (b) Use the Ksp value and the standard reduction potential of Fe2+(aq) to calculate the standard reduction potential for the half-reaction in part (a). (c) Rationalize the difference in the reduction potential for the half-reaction in part (a) with that for Fe2+(aq).

chapter 20 electrochemistry. 20.1 oxidation states and oxidation-reduction reactions

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