chapter 17 | electrochemistry: the quest for …postonp/ch223/pdf/chemat_ism_ch17.pdf292 chapter 17...

292

CHAPTER 17 | Electrochemistry: The Quest for Clean Energy

17.1. Collect and Organize For the voltaic cell shown in Figure P17.1, we are to explain why a porous separator is not required.

Analyze The porous separator serves to keep the reduction and oxidation half-reactions separate so that electrons are passed through the external circuit. Solve Because of the careful layering, each half-cell has its metal in contact with its cation solution. The solutions are not mixing, but, nevertheless, the layers allow the ions needed to balance the charge in each half-cell to pass.

Think about It The half-reactions and overall reaction for this voltaic cell are

Zn(s) → Zn2+(aq) + 2 e– Eanode = – 0.762 V

Cu2+(aq) + 2 e– → Cu(s) Ecathode = 0.342 V

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Ecell = Ecathode

– Eanode

= 0.342 – (–0.762) = 1.104 V

17.2. Collect and Organize

From the standard reduction potentials of Cu2+ and Cd2+ in Appendix 6, we can identify the cathode and anode and indicate the direction of flow of electrons in the external circuit in Figure P17.2.

Analyze A voltaic cell has a positive Ecell° indicating a spontaneous reaction. At one electrode (the anode) oxidation occurs, and at the other electrode (the cathode) reduction occurs. From Appendix 6 in the textbook, Ecell° (Cu2+) = 0.342 V and Ecell° (Cd2+) = –0.403 V. Solve To give a positive Ecell° , Cu2+ is reduced and Cd is oxidized:

Cu2+ (aq)+ 2 e– → Cu(s) Ecathode = 0.342 V

Cd(s)→ Cd2+ (aq)+ 2 e– Eanode = –0.403 V

Thus, Cu is the cathode, Cd is the anode, and electrons flow from Cd to Cu (to the left in the circuit shown in Figure P17.2).

Electrochemistry | 293

Think about It The overall reaction and potential for this voltaic cell is

Cu2+(aq) + Cd(s) → Cu(s) + Cd2+(aq) Ecell = Ecathode

– Eanode = 0.342 – (–0.403) = 0.745 V


For the voltaic cell shown in Figure P17.3, in which an Ag+/Ag cell is connected to a standard hydrogen electrode (SHE), we are to determine which electrode is the anode and which is the cathode and indicate in which direction the electrons flow in the outside circuit.

Analyze A voltaic cell runs spontaneously when Ecell is positive. By comparing the reduction potentials of each half-cell, we can write the reaction that is spontaneous for the cell. The half-cell where reduction (gain of electrons) occurs contains the cathode, and the half-cell where oxidation occurs contains the anode. Electrons flow from the anode, where they are produced by oxidation, toward the cathode, where they are required for reduction. Solve The spontaneous reaction for this cell is

2 × (Ag+ + e– → Ag) Ecathode = 0.7996 V

H2 → 2 H+ + 2 e– Eanode = 0.000 V

2 Ag+ + H2 → 2 H+ + Ag Ecell = Ecathode

– Eanode

= 0.7996 – 0.000 = 0.7996 V Thus, Ag is the cathode, Pt in the SHE is the anode, and electrons flow from the SHE to Ag (to the left in the circuit shown in Figure P17.3).

Think about It The shorthand notation for this cell would be

Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s) where Pt is an inert electrode used in the SHE.

17.4. Collect and Organize From the highlighted elements in the periodic table in Figure P17.4, we are to choose the group of elements that make the best inert electrodes.

Analyze Inert metal electrodes would have very negative (not favorable) oxidation potentials. These metals include those from which durable materials are made (jewelry, electrical contacts).

294 | Chapter 17

Solve The elements in blue (Pt, Au, and Hg) are the best metals to use as inert electrodes.

Think about It Of the three, platinum is often used in electrochemistry as an inert electrode for the construction of the standard hydrogen electrode (SHE). Mercury, although a liquid, is also used as an electrode. It is used in polarography as a dropping mercury electrode.

17.5. Collect and Organize The graph of cell potential versus [H2SO4] (Figure P17.5) shows four lines, some curved, some linear, some increasing, and some decreasing, as the concentration of H2SO4 decreases. From the shape of the curves and their trends we are to choose the line that best represents the trend in potential versus [H2SO4] in a lead–acid battery.

Analyze The scale for [H2SO4] is logarithmic, and voltage in the battery varies with the [H2SO4] according to the Nernst equation:

[ ]cell 22 4

0.0592 12.04 V – log 2 H SO

E =

Solve From the Nernst equation, we see that the cell potential drops as the log 1/[H2SO4]2 decreases. So as [H2SO4] decreases, the cell potential also decreases. The red line on the graph shows the opposite trend: the voltage increases as [H2SO4] decreases. In considering which of the remaining lines might describe the lead–acid battery, we must consider that because the cell voltage drops as log 1/[H2SO4]2 we expect that the decrease in potential is linear. Therefore, the blue line best describes the potential as a function of [H2SO4] concentration. Think about It Another characteristic of lead–acid batteries is that their cell voltage does not drop substantially until more than 90% of the battery has been discharged (see Figure 17.12).


From among the measures of battery performance, we are to choose which differ among different sizes of batteries.

Analyze The voltage describes the electrochemical potential of the battery and is dependent on the chemicals used in the battery. All the batteries shown in Figure P17.6 are alkaline batteries, so the voltage does not differ among them. A watt-hour describes the energy delivered in 1 hour and a milliampere-hour describes the charge delivered in 1 hour. These are both dependent on the amount of battery material, which differs among the sizes of the batteries. Solve Both (b) watt-hour and (c) milliampere-hour differ among the batteries because of their different sizes and different amounts of battery material.

Think about It A larger battery of the same voltage can be used to provide more power to an electric motor or other such electrical device or to power the device for a longer time. For example, the Chevy Volt has a total battery capacity of 16 kWh. This battery can produce 16,000 W for 1 hour or 1 W for 16,000 hours or anything in between. To increase the life of the lithium−ion battery in the Volt, however, the capacity used is a little over 10 kWh.


For the electrolysis of water in which the two product gases, H2 and O2, are collected in burets (Figure P17.7),


we are to write the half-reactions occurring at each electrode and discuss why a small amount of acid was added to the water to speed up the reaction.

Analyze In the electrolysis of water, electricity is supplied to make the nonspontaneous oxidation and reduction reactions occur. From Figure P17.7, we notice that the left buret has collected twice the volume of gas as the right buret. From the overall balanced equation

2 H2O()→ 2 H2(g)+O2(g)

we can identify the left buret as containing H2 and the right buret as containing O2. Ered° values are given in Appendix 6. Solve (a) The left electrode is the cathode where reduction is occurring:

2 H2O()+ 2 e– → H2(g)+ 2 OH– (aq) Ecathode = –0.828 V

The right electrode is the anode where oxidation is occurring:

2 H2O()→ O2(g)+ 4 H+ (aq)+ 4 e– Eanode = 1.229 V

(b) A small amount of H2SO4 is added to the water to increase the conductivity of the solution.

Think about It The electrochemical potential for the overall process is

Ecell = Ecathode

− Eanode = –0.828 V –1.229 V = –2.057 V


When the hydrolysis of water shown in Figure P17.7 is conducted with added Na2CO3, we are to write the half-reactions for the electrodes and explain why Na2CO3 makes the reaction go more quickly.

296 | Chapter 17

Analyze The added ions, Na+ and CO3

2–, do not get involved in the electrolysis reaction. The reduction of Na+ is too negative (–2.71 V) and Appendix 6 does not give any reduction reaction involving CO3

2–. Solve (a) The half-reactions are the same as Problem 17.7. The left electrode is the cathode where reduction is occurring:

2 H2O()+ 2 e– → H2(g)+ 2 OH– (aq) Ecathode = –0.828 V

The right electrode is the anode where oxidation is occurring:

2 H2O()→ O2(g)+ 4 H+ (aq)+ 4 e– Eanode = 1.229 V

(b) Na2CO3 speeds up the reaction by increasing the conductivity of the solution. Think about It The balanced equation for the overall process is

2 H2O()→ 2 H2(g)+O2(g) Ecell = Ecathode

− Eanode = –2.057 V


We are to explain what is meant by a half-reaction. Analyze A half-reaction is a part of how we describe oxidation–reduction reactions. Solve A half-reaction is one half of an oxidation–reduction (redox) reaction. It can be either the reduction reaction or the oxidation reaction. Think about It A half-reaction is used in balancing electrochemical equations and in designing batteries.

17.10. Collect and Organize In comparing the reactions between a Cu2+ solution and Zn metal in Figures 17.1 and 17.2, we are to explain why one generates electricity and the other does not. Analyze In Figure 17.1, the Cu2+ solution is in direct contact with the Zn metal and the spontaneous reaction occurs at the surface of the Zn metal. In Figure 17.2, the oxidation and reduction half-reactions are separated from each other. Solve When the half-reactions are in direct contact with each other, the electrons are transferred at the surface of the metal and cannot be “captured” to produce electricity. If the half-reactions are placed into separate cells, the electrons produced by the oxidation reaction travel through the external circuit to the reduction half-cell; this generates electricity. Think about It Any spontaneous redox reaction, when the oxidation and reduction half-cells are separated, can be used in a battery.

17.11. Collect and Organize We are to explain why a metal wire cannot function as a porous separator in an electrochemical cell. Analyze The porous separator allows charged ions to pass through to maintain electrical neutrality in each cell as the redox reaction progresses.


Solve A wire can pass only electrons through it, not ions, so it cannot function as a porous separator. Think about It A wire is used as the conduit for electrons in the external circuit of the voltaic cell.


We are asked why the cathode is labeled (+) and the anode (–) in a voltaic cell.

Analyze The site at which oxidation occurs is called the anode and generates electrons from each metal atom, allowing anions to flow toward the anode. Reduction occurs at the cathode where electrons are consumed, allowing cations to flow toward the cathode. Solve The electrons move from the negative terminal, where electrons are generated at the anode, to the positive terminal, where electrons are used at the cathode.

Think about It In an electrolytic cell the labels are opposite because we use an external current to run the spontaneous redox reaction in reverse.


For each half-reaction, we are to balance it by adding the appropriate number of electrons to the reactant or product side and then identify each reaction as either a reduction or an oxidation reaction.

Analyze All the reactions in this question are balanced for atoms, so we need only add electrons to balance the total charge between the reactants side and the products side. A reduction reaction will have electrons on the reactants side; an oxidation reaction will have electrons on the products side. Solve (a) Br2(ℓ) + 2 e– → 2 Br–(aq); this is a reduction reaction. (b) Pb(s) + 2 Cl–(aq) → PbCl2(s) + 2 e–; this is an oxidation reaction. (c) O3(g) + 2 H+(aq) + 2 e– → O2(g) + H2O(ℓ); this is a reduction reaction. (d) H2S(g) → S(s) + 2 H+(aq) + 2 e–; this is an oxidation reaction.

Think about It In real chemical systems, oxidation and reduction reactions are always paired.


For each half-reaction, we are to balance it by adding the appropriate number of electrons to the reactants or products side and then identify each reaction as either a reduction or an oxidation reaction.

Analyze All the reactions in this question are balanced for atoms, so we need only add electrons to balance the total charge between the reactants side and the products side. A reduction reaction will have electrons on the reactants side; an oxidation reaction will have electrons on the products side. Solve (a) Fe2+(aq) → Fe3+(aq) + e–; this is an oxidation reaction. (b) AgI(s) + e– → Ag(s) + I–(aq); this is a reduction reaction. (c) VO2

+(aq) + 2 H+(aq) + e– → VO2+(aq) + H2O(ℓ); this is a reduction reaction. (d) I2(s) + 6 H2O(ℓ) → 2 IO3

–(aq) + 12 H+(aq) +10 e–; this is an oxidation reaction.

Think about It In real chemical systems, oxidation and reduction reactions are always paired.

298 | Chapter 17

17.15. Collect and Organize For the oxidation reaction of Fe3O4 to Fe2O3 in an acidic water environment, we are to balance the half-reaction describing this process.

Analyze First, we will need to balance the iron atoms in the half-reaction and then the oxygen atoms by adding H2O. Next, we balance hydrogen by adding an appropriate number of H+ ions, followed by adding electrons (we expect that they will be added to the products side because this is an oxidation reaction) to balance the charge in the half-reaction. Solve Balancing the iron atoms:

2 Fe3O4(s) → 3 Fe2O3(s) Balancing the oxygen atoms:

2 Fe3O4(s) + H2O(ℓ) → 3 Fe2O3(s) Balancing the hydrogen atoms:

2 Fe3O4(s) + H2O(ℓ) → 3 Fe2O3(s) + 2 H+(aq) Balancing the charge:

2 Fe3O4(s) + H2O(ℓ) → 3 Fe2O3(s) + 2 H+(aq) + 2 e–

Think about It Indeed, this reaction does balance so that electrons are on the products side, consistent with our expectation that this is an oxidation reaction.


For the oxidation reaction of MnCO3 to MnO2 in an acidic water environment where a principal carbonate species is HCO3

–, we are to balance the half-reaction describing this process.

Analyze First, we will need to balance the manganese and carbon atoms in the half-reaction and then the oxygen atoms by adding H2O. Next, we balance hydrogen by adding an appropriate number of H+ ions, followed by adding electrons (we expect that they will be added to the products side because this is an oxidation reaction) to balance the charge in the half-reaction. Solve Balancing the manganese and carbon atoms:

MnCO3(s) → MnO2(s) + HCO3–(aq)

Balancing the oxygen atoms: MnCO3(s) + 2 H2O(ℓ) → MnO2(s) + HCO3

–(aq) Balancing the hydrogen atoms:

MnCO3(s) + 2 H2O(ℓ) → MnO2(s) + HCO3–(aq) + 3 H+(aq)

Balancing the charge: MnCO3(s) + 2 H2O(ℓ) → MnO2(s) + HCO3

–(aq) + 3 H+(aq) + 3 e–

Think about It Indeed, this reaction does balance so that electrons are on the products side, consistent with our expectation that this is an oxidation reaction.

17.17. Collect and Organize For the Pb2+/Pb and Zn2+/Zn voltaic cell in which we are told that the products are Pb(s) and Zn2+(aq), we are to write the appropriate half-reactions, write the balanced overall cell reaction, and diagram the cell.

Analyze Because we know that Pb and Zn2+ are produced, Pb2+ is reduced and Zn is oxidized in this process. In balancing the overall reaction, we must be sure to cancel all the electrons produced by oxidation with those used


in reduction. To do so, we might have to multiply either half-reaction, or both, by some factor. Finally, to diagram the cell we use the following convention:

anode | oxidation half-reaction species || reduction half-reaction species | cathode making sure to indicate the phases of the species involved and to use vertical lines to separate phases. Solve (a) Pb2+(aq) + 2 e– → Pb(s) cathode, reduction Zn(s) → Zn2+(aq) + 2 e– anode, oxidation (b) We simply add the above reactions to obtain the overall reaction:

Pb2+(aq) + Zn(s) → Zn2+(aq) + Pb(s) (c) Zn(s) | Zn2+(aq) || Pb2+(aq) | Pb(s)

Think about It As this reaction proceeds, the lead electrode gets heavier and the zinc electrode loses mass.

17.18. Collect and Organize For the Ag+/Ag and Ni2+/Ni voltaic cell in which we are told that the products are Ag(s) and Ni2+(aq), we are to write the appropriate half-reactions, write the balanced overall cell reaction, and diagram the cell.

Analyze Because we know that Ag and Ni2+ are produced, Ag+ is reduced and Ni is oxidized in this process. In balancing the overall reaction, we must be sure to cancel all the electrons produced by oxidation with those used in reduction. To do so, we might have to multiply either half-reaction, or both, by some factor. Finally, to diagram the cell we use the following convention:

anode | oxidation half-reaction species || reduction half-reaction species | cathode making sure to indicate the phases of the species involved. Solve (a) Ag+(aq) + e– → Ag(s) cathode, reduction

Ni(s) → Ni2+(aq) + 2 e– anode, oxidation (b) To cancel the electrons, we multiply the cathode half-reaction by 2:

+ –2 Ag ( ) 2 e 2 Ag( )+ →aq s – 2+Ni( ) 2 Ni ( )e→ +s aq

+ 2+2 Ag ( ) Ni( ) 2 Ag( ) Ni ( )+ → +aq s s aq (c) Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)

Think about It As this reaction proceeds, the silver electrode gets heavier and the nickel electrode loses mass.

17.19. Collect and Organize For the Cd/Cd(OH)2 and MnO4

–/MnO2 voltaic cell in which we are told that the products are Cd(OH)2(s) and MnO2(s), we are to write the appropriate half-reactions, write the balanced overall cell reaction, and diagram the cell. The electrolyte is an alkaline aqueous solution.

Analyze Because we know that Cd(OH)2 and MnO2 are produced, MnO4

– is reduced and Cd is oxidized in this process. In balancing the overall reaction, we must be sure to cancel all the electrons produced by oxidation with those used in reduction. To do so, we might have to multiply either half-reaction, or both, by some factor. Finally, to diagram the cell we use the following convention:

anode | oxidation half-reaction species || reduction half-reaction species | cathode making sure to indicate the phases of the species involved and to use vertical lines to separate phases.

300 | Chapter 17

Solve (a) MnO4

–(aq) + 2 H2O(ℓ) + 3 e– → MnO2(s) + 4 OH–(aq) cathode, reduction Cd(s) + 2 OH–(aq) → Cd(OH)2(s) + 2 e– anode, oxidation (b) To cancel the electrons, we multiply the cathode reaction by 2 and the anode reaction by 3:

2 MnO4–(aq) + 4 H2O() + 6 e– → 2 MnO2(s) + 8 OH–(aq)

3 Cd(s) + 6 OH–(aq) → 3 Cd(OH)2(s) + 6 e–

2 MnO4–(aq) + 4 H2O(ℓ) + 3 Cd(s) → 2 MnO2(s) + 3 Cd(OH)2(s) + 2 OH–

(aq) (c) Cd(s) | Cd(OH)2(s) || MnO4

–(aq) | MnO2(s) | Pt(s)

Think about It Because we would find it difficult to make an electrode out of MnO2(s), this cell requires the use of an inert electrode such as Pt.

17.20. Collect and Organize For the Ag+/Ag and Sn2+/Sn voltaic cell in which we are told that the products are Ag(s) and Sn2+(aq), we are to write the appropriate half-reactions, write the balanced overall cell reaction, and diagram the cell.

Analyze Because we know that Ag and Sn2+ are produced, Ag+ is reduced and Sn is oxidized in this process. In balancing the overall reaction, we must be sure to cancel all the electrons produced by oxidation with those used in reduction. To do so, we might have to multiply either half-reaction, or both, by some factor. Finally, to diagram the cell we use the usual convention:

anode | oxidation half-reaction species || reduction half-reaction species | cathode making sure to indicate the phases of the species involved. Solve (a) Ag+(aq) + e– → Ag(s) cathode, reduction Sn(s) → Sn2+(aq) + 2 e– anode, oxidation (b) To cancel the electrons, we multiply the cathode reaction by 2:

2 Ag+(aq) + 2 e– → 2 Ag(s) Sn(s) → 2 e–+ Sn2+(aq) 2 Ag+(aq) + Sn(s) → 2 Ag(s) + Sn2+(aq)

(c) Sn(s) | Sn2+(aq) || Ag+(aq) | Ag(s)

Think about It As this reaction proceeds, the silver electrode gets heavier and the tin electrode loses mass.

17.21. Collect and Organize For the “super iron” battery, we are to determine the number of electrons transferred in the spontaneous reaction and the oxidation states for Fe and Zn in the reactants and products. Finally, we are to diagram the cell.

Analyze (a) By writing the half-reactions involved in the redox reaction and balancing them, we can determine the number of electrons transferred. (b) Knowing the typical oxidation states of oxygen (–2) and potassium (+1) helps us determine the oxidation states of Fe in K2FeO4 and Zn in ZnO and K2ZnO2. (c) To diagram the cell we use the following cell notation:

anode | oxidation half-reaction species || reduction half-reaction species | cathode making sure to indicate the phases of the species involved and to use vertical lines to separate phases.


Solve (a) The half-reactions are as follows (K+ is a spectator ion): 5 H2O(ℓ) + 3 Zn(s) → ZnO(s) + 2 ZnO2

2–(aq) + 10 H+(aq) + 6 e– 6 e– + 10 H+(aq) + 2 FeO4

2–(aq) → Fe2O3(s) + 5 H2O(ℓ) Six electrons are transferred in this reaction. (b) FeO4

2– has Fe6+ Fe2O3 has Fe3+ Zn has Zn0 ZnO and ZnO2

2– have Zn2+ (c) Zn(s) | ZnO(s) | ZnO2

2–(aq) || FeO42–(aq) | Fe2O3(s) | Pt(s)

Think about It Because we would find it difficult to make an electrode out of Fe2O3(s), this cell requires the use of an inert electrode such as Pt.


For the aluminum–air battery described, we are to write the two half-reactions and diagram the cell. Analyze The oxidation reaction forms Al(OH)3 from Al in OH–(aq). The reduction reaction forms OH– from O2 with an inert M(s) electrode. Solve The half-reactions in the aluminum–air battery are Al(s) + 3 OH–(aq) → Al(OH)3(s) + 3 e– 4 e– + 2 H2O(ℓ) + O2(g) → 4 OH–(aq) The cell diagram is

Al(s) | Al(OH)3(s) || OH–(aq) | O2(g) | M(s)

Think about It To balance the redox reaction, we would multiply the oxidation half-reaction by 4 and the reduction half-reaction by 3. The number of electrons transferred, then, is 12.


We are asked to equate two different ways to calculate Ecell° .

Analyze Equation 17.1 is

Ecell = Ered

(cathode)− Ered (anode)

Other textbooks express Ecell° in terms of Ered° and Eox° :

Ecell = Ered

(cathode)+ Eox (anode)

Solve Because Ered

(anode) = –Eox (anode), we can substitute –Eox

(anode) for Ered (anode) in the expression

Ecell = Ered

(cathode)+ Eox (anode)

to obtain

Ecell = Ered

(cathode)+ –Ered (anode)( )

= Ered (cathode) – Ered

(anode)

This is equal to the expression for Ecell° in Equation 17.1.

Think about It Either expression to calculate Ecell° is valid.

302 | Chapter 17

17.24. Collect and Organize Using Ered

° values from Appendix 6, we can explain why O2 is a stronger oxidizing agent in acid or in base.

Analyze An oxidizing agent causes another substance to be oxidized and therefore it is itself reduced. The more positive the Ered

° for O2 in base or in acid, the better its ability to oxidize another substance. Solve O2 is a stronger oxidizing agent in acid because Ered

° (acid, 1.229 V) is greater than Ered ° (base, 0.401 V).

Think about It This also means that OH– (with Eanode

° = –0.0401 V) is more easily oxidized than H2O (with Eanode ° = –1.229 V).

17.25. Collect and Organize From the standard reduction potentials for Cu2+, Co2+, and Hg2+ we are to find a combination of half-reactions that would give the highest positive and lowest positive value of Ecell° . Analyze The standard potentials for the half-reactions from Appendix 6 are

Cu2+(aq) + 2 e– → Cu(s) Ecell ° = 0.342 V

Co2+(aq) + 2 e– → Co(s) Ecell ° = –0.277 V

Hg2+(aq) + 2 e– → Hg(ℓ) Ecell ° = 0.851 V

Solve (a) The largest value of Ecell

° is for the reduction of Hg2+ by Co. Hg2+(aq) + 2 e– → Hg(ℓ) Ecathode

° = 0.851 V Co(s) → Co2+(aq) + 2 e– Eanode

° = –0.277 V Ecell ° = Ecathode

° – Eanode ° = 0.851 V – (–0.277 V) = 1.128 V

(b) The smallest positive value of E°cell is for the reduction of Hg2+ by Cu. Hg2+(aq) + 2 e– → Hg(ℓ) Ecathode

° = 0.851 V Cu(s) → Cu2+(aq) + 2 e– Eanode


° – Eanode ° = 0.851 V – (0.342 V) = 0.509 V

Think about It Another positive cell potential would be for the reduction of Cu2+ by Co with a standard cell potential of 0.619 V.


Using the standard reduction potentials in Appendix 6, we are to identify an oxidizing agent for Cr(s) but not for Cd(s) and a reducing agent for Br2(ℓ) but not for I2(s). Analyze The standard potentials for the half-reactions for Cr3+, Cd,2+ Br2, and I2 from Appendix 6 are

Cr3+(aq) + 3 e– → Cr(s) Ecell° = –0.74 V Cd2+(aq) + 2 e– → Cd(S) Ecell° = –0.403 V Br2(ℓ) + 2 e– → 2 Br–(aq) Ecell° = 1.066 V I2(s) + 2 e– → 2 I–(aq) Ecell° = 0.5355 V

Solve (a) To oxidize Cr but not Cd, we need an oxidizing agent with a standard reduction potential between –0.403 and –0.74 V. The reduction of Fe2+, with a reduction potential of –0.447 V, will oxidize Cr but not Cd.


Fe2+(aq) + 2 e– → Fe(s) Ecathode ° = –0.447 V

Cr(s) → Cr3+(aq) + 3 e– Eanode ° = –0.74 V

Ecell ° = Ecathode

° – Eanode ° = –0.447 V – (–0.74 V) = 0.293 V

Fe2+(aq) + 2 e– → Fe(s) Ecathode

° = –0.447 V Cd(s) → Cd2+(aq) + 2 e– Eanode


° – Eanode ° = –0.447 V – (–0.403 V) = –0.044 V

(b) To reduce Br2 but not I2 we need a reducing agent with a standard reduction potential between 0.5355 and 1.066 V. The oxidation of Hg to Hg2+, with a reduction potential of 0.851 V, will reduce Br2 but not I2.

Br2(ℓ) + 2 e– → 2 Br–(aq) Ecathode ° = 1.066 V

Hg(ℓ) → Hg2+(aq) + 2 e– Eanode ° = 0.851 V

Ecell ° = Ecathode

° – Eanode ° = 1.066 V – (0.851 V) = 0.215 V

I2(s) + 2 e– → 2 I–(aq) Ecathode

° = 0.5355 V Hg(ℓ) → Hg2+(aq) + 2 e– Eanode

° = 0.851 V Ecell ° = Ecathode

° – Eanode ° = 0.5355 V – (0.851 V) = –0.3155 V

Think about It Your answers here may vary, but be sure that you have chosen one in the ranges indicated.


By calculating Ecell° for the possible redox reaction between Ag and Cu2+, we can determine whether the reaction is spontaneous.

Analyze Because we are told that [Ag+] = [Cu2+] = 1.00 M, this reaction occurs under standard conditions, and we can calculate Ecell° from values of Ecell° in Appendix 6. If Ecell° is calculated as positive, the reaction is spontaneous. Solve The half-reactions, Ecathode° = and Eanode° and the overall reaction and cell potential are

2 Ag(s) → 2 Ag+(aq) + 2 e– Eanode° = 0.7996 V Cu2+(aq) + 2 e– → Cu(s) Ecathode° = 0.342 V 2 Ag(s) + Cu2+(aq) → 2 Ag+(aq) + Cu(s) Ecell

° = Ecathode ° – Eanode

° = – 0.458 V No, the reaction is not spontaneous.

Think about It The reverse reaction, Cu placed into Ag+ to dissolve copper and deposit silver, is spontaneous.


By calculating Ecell ° for the possible redox reaction between Cd and Sn2+, we can determine whether the reaction

is spontaneous.

Analyze Because we are told that [Cd2+] = [Sn2+] = 1.00 M, this reaction occurs under standard conditions, and we can calculate Ecell

° from values of Ered ° in Appendix 6. If Ecell

° is calculated as positive, the reaction is spontaneous. Solve The half-reactions, Ecathode° = and Eanode° and the overall reaction and cell potential are

Cd(s) → Cd2+(aq) + 2 e– Eanode° = – 0.403 V

Sn2+(aq) + 2 e– → Sn(s) Ecathode° = – 0.136 V

Cd(s) + Sn2+(aq) → Cd2+(aq) + Sn(s) Ecell ° = Ecathode

° – Eanode °

= 0.267 V

The reaction is spontaneous.

304 | Chapter 17

Think about It The reverse reaction, Sn placed into Cd2+ to dissolve tin and deposit cadmium, is not spontaneous.


For a Ni–Zn cell, we are asked to compare its Ecell ° with that of a Cu–Zn cell that has Ecell

° = 1.10 V.

Analyze Using Ered

° values in Appendix 6, we can identify the spontaneous reaction (having positive Ecell° ) and calculate Ecell° . Solve The spontaneous reaction in the Ni–Zn cell is

Ni2+(aq) + 2 e– → Ni(s) Ecathode ° = – 0.257 V

Zn(s) → Zn2+(aq) + 2 e– Eanode° = – 0.762 V Ecell


° = 0.505 V

The Ecell ° for a Ni–Zn cell is less than Ecell

° for a Cu–Zn cell.

Think about It As we will see later in the chapter, because Ecell

° for the Ni–Zn cell is less positive than that of a Cu–Zn cell, we also know that the cell reaction is less spontaneous.

17.30. Collect and Organize For an Ag–Zn cell we are to calculate Ecell

° and determine whether Zn or Ag functions as the anode.

Analyze Using Ered

° values in Appendix 6, we can identify the spontaneous reaction (having positive Ecell° ) and calculate Ecell° . The anode is the electrode in the oxidation half-cell. Solve (a)

2 × [Ag+(aq) + e– → Ag(s)] Ecathode ° = 0.7996 V

Zn(s) → Zn2+(aq) + 2 e– Eanode° = – 0.762 V 2 Ag+(aq) + Zn(s) → 2 Ag(s) + Zn2+(aq) Ecell


° = 1.562 V (b) Zinc is the anode in this cell.

Think about It As we will see later in this chapter, because this cell has a positive potential, it is a spontaneous reaction.


For each pair of reduction reactions, we can use the reduction potentials in Appendix 6 to write the equation for the spontaneous cell reaction, identifying the reactions occurring at the anode and cathode.

Analyze The overall cell potential, Ecell° , must be positive for the spontaneous voltaic cell reaction. To calculate Ecell

° we use

Ecell ° = Ecathode

° – Eanode °

The oxidation reaction occurs at the anode, and the reduction reaction occurs at the cathode. Solve (a)

Anode Zn(s) → Zn2+(aq) + 2 e– Eanode° = – 0.762 V Cathode Hg2+(aq) + 2 e–

→ Hg(ℓ) Ecathode ° = 0.851 V

Zn(s) + Hg2+(aq) → Zn2+(aq) + Hg(ℓ) Ecell ° = Ecathode

° – Eanode ° = 1.613 V


(b) Anode Zn(s) + 2 OH–(aq) → ZnO(s) + H2O(ℓ) + 2 e– Eanode

° = –1.25 V Cathode Ag2O(s) + H2O(ℓ) + 2 e– → 2 Ag(s) + 2 OH–(aq) Ecathode

° = 0.342 V Zn(s) + Ag2O(s) → ZnO(s) + 2 Ag(s) Ecell


° = 1.59 V (c)

Anode 2 × [Ni(s) + 2 OH–(aq) → Ni(OH)2(s) + 2 e–] Eanode ° = – 0.72 V

Cathode O2(g) + 2 H2O(ℓ) + 4 e– → 4 OH–(aq) Ecathode ° = 0.401 V

2 Ni(s) + O2(g) +2 H2O(ℓ) → 2 Ni(OH)2(s) Ecell ° = Ecathode

° – Eanode ° = 1.12 V

Think about It When we multiply a half-reaction to obtain the balanced cell reaction, the oxidation or reduction potential does not change. This is because cell potential is an intensive, not extensive, property.

17.32. Collect and Organize For each pair of reduction reactions, we will use the reduction potentials in Appendix 6 to write the equation for the spontaneous cell reaction, identifying the reactions occurring at the anode and cathode.

Analyze The overall cell potential, Ecell

° must be positive for the spontaneous voltaic cell reaction. To calculate Ecell ° we

use

Ecell ° = Ecathode

° – Eanode °

The oxidation reaction occurs at the anode, and the reduction reaction occurs at the cathode. Solve (a)

Anode Cd(s) → Cd2+(aq) + 2 e– Eanode ° = – 0.403 V

Cathode [Ag+(aq) + e– → Ag(s)] × 2 Ecathode

° = 0.7996 V Cd(s) + 2 Ag+(aq) → Cd2+(aq) + 2 Ag(s) Ecell


° = 1.203 V

(b) Anode [Ag(s) + Br–(aq) → AgBr(s) + e–] × 2 Eanode

° = 0.095 V Cathode MnO2(s) + 4 H+(aq) + 2 e– → Mn2+(aq) + 2 H2O(ℓ) Ecathode

° = 1.23 V 2 Ag(s) + 2 Br–(aq) + MnO2(s) + 4 H+(aq) →

2 AgBr(s) + Mn2+(aq) + 2 H2O(ℓ) Ecell ° = Ecathode

° – Eanode ° = 1.14 V

(c) Anode [Ag(s) + Cl–(aq) → AgCl(s) + e–] × 2 Eanode

° = 0.2223 V Cathode PtCl4

2–(aq) + 2 e– → Pt(s) + 4 Cl–(aq) Ecathode ° = 0.73 V

2 Ag(s) + PtCl42–(aq) → 2 AgCl(s) + Pt(s) + 2 Cl–(aq) Ecell


° = 0.51 V Think about It When we multiply a half-reaction to obtain the balanced cell reaction, the oxidation or reduction potential does not change. This is because cell potential is an intensive, not extensive, property.

17.33. Collect and Organize / Analyze

We are to explain how the negative sign in welec = –QEcell

is consistent with a positive cell voltage doing negative electrical work. Solve The sign of electrical work is negative because in passing voltage through a cell (or outside circuit), the redox reaction is doing work on the environment; therefore, the sign of work from the perspective of the system (the redox reaction) is negative.

Think about It An example of positive electrical work would be charging a battery.

306 | Chapter 17

17.34. Collect and Organize / Analyze We are to relate physical work (moving an object) to electrical work.

Solve From the equations given w = F × d

welec = C × E we see that the force in an electrochemical reaction is the electrical potential, E, and what is moved is the charge, C. A joule is a coulomb ⋅ volt, which is the energy needed to move a unit of charge across a potential of 1 V.

Think about It In a simple electric circuit containing an electrochemical cell (the internal circuit) and an external circuit, the electrons travel through the external circuit from a region of high potential to low potential.


For each redox reaction given, we are to calculate the cell potential after calculating the value of free energy.

Analyze We calculate ∆G° for each reaction by using free energy of formation values for the products and reactants from Appendix 4. The cell potential can then be calculated by using

ΔGcell = –nFEcell

Ecell = –ΔG

nF

Solve (a)

ΔG = (1 mol Cu2+ × 65.5 kJ mol) + 1 mol Cu × 0.0 kJ mol( )⎡⎣ ⎤⎦ – (2 mol Cu+ ×50.0 kJ mol) = –34.5 kJ

Ecell = +34,500 J

1 mol× 9.65×104 C/mol= +0.358 V

(b)

ΔG = (1 mol Ag+ × 77.1 kJ/mol)+ (1 mol Fe2+ × −78.9 kJ/mol)⎡⎣ ⎤⎦ −

(1 mol Ag × 0.0 kJ/mol)+ (1 mol Fe3+ × −4.7 kJ/mol)⎡⎣ ⎤⎦ = 2.9 kJ

Ecell = −2900 J

1 mol× (9.65×104 C/mol)= −0.030 V

Think about It Reaction (a) is spontaneous and reaction (b) is nonspontaneous.

17.36. Collect and Organize For each redox reaction given, we are to calculate the cell potential after calculating the value of free energy.

Analyze We calculate ∆G° for each reaction by using ∆Gf

° values for the products and reactants from Appendix 4. The cell potential can then be calculated by using

ΔGcell = –nFEcell

Ecell

o = –ΔG

nF


Solve (a)

ΔG = (1 mol Fe × 0 kJ/mol)+ (1 mol H2O × –237.2 kJ/mol⎡⎣ ⎤⎦ −

(1 mol FeO × −255.2 kJ/mol)+ (2 mol H2 × 0 kJ/mol)⎡⎣ ⎤⎦ = 18.0 kJ

Ecell =−18,000 J

2 mol× 9.65×104 C/mol= −0.0933 V

(b)

ΔG = (2 mol PbSO4 × −813.0 kJ/mol)+ (2 mol H2O × –237.2 kJ/mol⎡⎣ ⎤⎦ −

(2 mol Pb× 0 kJ/mol)+ (1 mol O2 × 0 kJ/mol)+ (2 mol H2SO4 × −744.5 kJ/mol)⎡⎣ ⎤⎦ = −611.4 kJ

Ecell =611,400 J

4 mol× 9.65×104 C/mol= +1.58 V

Think about It Reaction (a) is nonspontaneous and reaction (b) is spontaneous.

17.37. Collect and Organize Using the relationship between free energy and cell potential

∆Gcell = –nFEcell we can calculate ∆Gcell for a Zn–MnO2 battery generating 1.50 V.

Analyze To use the equation, we need the value of n, the number of electrons transferred in the overall balanced equation. From the cell reaction provided, we see that Zn is oxidized to Zn2+ and 2 mol of Mn4+ (in MnO2) is reduced to Mn3+ (in Mn2O3), so 2 mol of electrons is transferred in the reaction. We should also be aware of the units for the quantities in this equation. The value of F is 9.65 × 104 C/mol and E is in volts, which, being equivalent to joules per coulomb (J/C), means that ∆G is calculated in joules, which we can convert to kilojoules, the usual units for free energy. Solve

4

cell9.65 10 C 1.50 J 1 kJ–2 mol –290 kJ

mol C 1000 J×

Δ = × × × =G

Think about It This reaction is spontaneous because the cell potential is positive, making ∆Gcell negative.


∆Gcell = –nFEcell we can calculate ∆Gcell for a nicad battery generating 1.20 V.

Analyze To use the equation, we need the value of n, the number of electrons transferred in the overall balanced equation. From the cell reaction provided, we see that Cd is oxidized to Cd2+ [in Cd(OH)2] and 2 mol of Ni3+ [in NiO(OH)] is reduced to 2 mol of Ni2+ [in Ni(OH)2], so 2 mol of electrons is transferred in the reaction. We should also be aware of the units for the quantities in this equation. The value of F is 9.65 × 104 C/mol and E is in volts, which, being equivalent to joules per coulomb (J/C), means that ∆G is calculated in joules, which we can convert to kilojoules, the usual units for free energy.

308 | Chapter 17

Solve 4

cell9.65 10 C 1.20 J 1 kJ–2 mol –232 kJ

mol C 1000 J×

Δ = × × × =G



∆Gcell = –nFEcell we can calculate ∆Gcell for a nickel–metal hydride battery cell generating 1.20 V.

Analyze To use the equation we need the value of n, the number of electrons transferred in the overall balanced equation. From the cell reaction provided we see that Ni3+ [in NiO(OH)] is reduced to Ni2+ [in Ni(OH)2] and M– (in MH) is oxidized to M, so 1 mol of electrons is transferred in the reaction. We should also be aware of the units for the quantities in this equation. The value of F is 9.65 × 104 C/mol and E is in volts, which, being equivalent to joules per coulomb (J/C), means that ∆G is calculated in joules, which we can convert to kilojoules, the usual units for free energy. Solve

4

cell9.65 10 C 1.20 J 1 kJ–1 mol –116 kJ

mol C 1000 J×Δ = × × × =G



∆Gcell = –nFEcell we can calculate ∆Gcell for a lead–acid battery cell generating 2.00 V.

Analyze To use the equation, we need the value of n, the number of electrons transferred in the overall balanced equation. From the cell reaction provided, we see that Pb is oxidized to Pb2+ (in PbSO4) and PbO2 (where Pb is +4) is reduced to Pb2+ (also in PbSO4), so 2 mol of electrons is transferred in the reaction. We should also be aware of the units for the quantities in this equation. The value of F is 9.65 × 104 C/mol and E is in volts, which, being equivalent to joules per coulomb, means that ∆G is calculated in joules, which we can convert to kilojoules, the usual units for free energy. Solve

4

cell9.65 10 C 2.00 J 1 kJ–2 mol –386 kJ

mol C 1000 JG ×

Δ = × × × =

Think about It A redox reaction, like this one, in which two species transfer electrons to become the same species (PbO2 + Pb → PbSO4) is called comproportionation.

17.41. Collect and Organize For each reaction, we can break it up into its appropriate half-reactions and, using Ered ° values from Appendix 6, calculate Ecell° . From this cell potential, we can then calculate the free energy for each reaction.


Analyze To calculate the cell potential, we use

Ecell ° = Ecathode

° – Eanode °

Once we have calculated Ecell° , we use

ΔG = –nFEcell

to calculate the free energy of the reaction. Here, we have to remember that n is the number of moles of electrons transferred in the overall balanced equation.

Solve (a)

Cu(s) → Cu2+(aq) + 2 e– Eanode °

= 0.342 V Sn2+(aq) + 2 e– → Sn(s) Ecathode

° = – 0.136 V Cu(s) + Sn2+(aq) → Cu2+(aq) + Sn(s) Ecell


° = – 0.478 V 49.65 10 C –0.478 J 1 kJ–2 mol 92.2 kJ

mol C 1000 JG ×

Δ = × × × =o

(b) Zn(s) → Zn2+(aq) + 2 e– Eanode

° = – 0.762 V Ni2+(aq) + 2 e– → Ni(s) Ecathode

° = – 0.257 V Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) Ecell


° = 0.505 V

ΔG = –2 mol× 9.65×104 C

mol× 0.505 J

C× 1 kJ

1000 J= –97.5 kJ

Think about It Reaction (b), with its positive cell potential, is spontaneous. Reaction (a) is nonspontaneous, so copper metal in contact with Sn2+ solution does not oxidize.

17.42. Collect and Organize For each reaction we can break it up into its appropriate half-reactions and, using 𝐸𝐸red ° values from Appendix 6, calculate 𝐸𝐸cell° . From this cell potential, we can then calculate the free energy for each reaction.

Analyze To calculate the cell potential, we use

Ecell ° = Ecathode

° – Eanode °

Once we have calculated Ecell° , we use

ΔG = –nFEcell

to calculate the free energy of the reaction. Here, we have to remember that n is the number of moles of electrons transferred in the overall balanced equation. Solve (a)

Fe(s) → Fe2+(aq) + 2 e– Eanode ° = – 0.447 V

Cu2+(aq) + 2 e– → Cu(s) Ecathode ° = 0.342 V

Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) Ecell ° = Ecathode

° – Eanode ° = 0.789 V

ΔG = –2 mol× 9.65×104 C

mol× 0.7889 J

C× 1 kJ

1000 J= –152 kJ

(b) Ag(s) → Ag+(aq) + e– Eanode

° = 0.7996 V Fe3+(aq) + e– → Fe2+(aq) Ecathode

° = 0.770 V Ag(s) + Fe3+(aq) → Ag+(aq) + Fe2+(aq) Ecell


° = – 0.0296 V

ΔG = –1 mol× 9.65×104 C

mol× –0.0296 J

C× 1 kJ

1000 J= 2.8 kJ

310 | Chapter 17

Think about It Reaction (a), with its positive cell potential, is spontaneous. Reaction (b) is nonspontaneous, so silver metal in contact with a Fe3+ solution is not expected to corrode.


We are to describe the function of platinum in the standard hydrogen electrode. Solve The platinum electrode transfers electrons to the half-cell; because it is inert, it is not involved in the reaction.

Think about It The inert electrode simply serves as a place for the oxidation or reduction reaction to occur, and as a conduit for electrons for the reaction.


We consider whether we can use a half-reaction for a battery in which the anode has neither reactant nor product as a solid conductor. Solve Yes, we can use these reactions, but we must use an inert electrode such as Pt for the battery.

Think about It For the half-reaction provided as an example in this problem, we could not use Fe(s) as the electrode because it would not be inert in this solution.


We are to explain why the voltage of most batteries changes little until the battery is almost discharged, where the voltage drops significantly.

Analyze Voltage of a battery (a voltaic cell) is governed by the Nernst equation:

Ecell = Ecell

– RTnF

lnQ

As a battery discharges, the value of Q, the reaction quotient, changes: [products][reactants]

x

yQ =

Solve At the start of the reaction, Q is very small because [reactants] >> [products]. As the reaction proceeds [products] grows and Q increases but does not increase significantly until significant amounts of products form, that is, when the battery is nearly discharged.

Think about It As an example, Figure 17.11shows the cell potential of a lead–acid battery as a function of discharge. Notice that the voltage is relatively constant until the battery is approximately 90% discharged.


Using the Nernst equation we can determine how Ecell for the Cu–Zn cell differs, if at all, from 1.10 V (Ecell° ) when both [Cu2+] and [Zn2+] = 0.25 M.

Analyze The Nernst equation expression for this reaction is

Ecell = Ecell

– RTnF

ln [Zn2+ ][Cu2+ ]


Solve If we assume T = 298 K and from the balanced equation we know n = 2, we can calculate Ecell for [Zn2+] = [Cu2+] = 0.25 M:

( )( )( )cell 4

cell

8.314 J/mol K 298 K 0.25 1.10 V – ln0.25 2 9.65 10 C/mol

1.10 V 0 1.10 V

MEM

E

⋅= ×

× ×

= − =

If both concentrations are changed from 1.00 M to 0.25 M, the cell voltage is unchanged because ln(1) = 0.

Think about It For this reaction, whenever [Zn2+] = [Cu2+] at 298 K then Ecell = Ecell° .

17.47. Collect and Organize We can use the Nernst equation to calculate the cell potential when Fe3+ is combined with Cr2+ at nonstandard conditions. Analyze Because T = 298 K, we can use the following form of the Nernst equation:

Ecell = Ecell

–0.0592

n log Q

where Ecell° is the potential of the cell under standard conditions (calculated from tabulated Ered° values), n is the moles of electrons transferred in the overall balanced redox equation, and Q is the reaction quotient. Solve First, we need to calculate Ecell° and determine the value of n. The half-reactions and overall cell reaction are

Fe3+(aq) + e– → Fe2+(aq) Ecathode °

= 0.770 V Cr2+(aq) → Cr3+(aq) + e–

Eanode °

= –0.41 V Fe3+(aq) + Cr2+(aq) → Fe2+(aq) + Cr3+(aq) Ecell


° = 1.18 V We see that n = 1. Now we can use the Nernst equation to calculate Ecell when [Fe3+] = [Cr2+] = 1.50 × 10–3 M and [Fe2+] = [Cr3+] 2.5 × 10–4 M.

( )( )

24

cell 23

2.5 100.05921.18 V – log 1.27 V1.50 10

−

−

×= =

×E

n

Think about It This reaction became more spontaneous (higher cell potential) under these conditions.

17.48. Collect and Organize We can use the Nernst equation to calculate the cell potential when Cu is combined with Ag+ at nonstandard conditions.

Analyze Because T = 298 K, we can use the following form of the Nernst equation:

Ecell = Ecell

–0.0592

n log Q

where Ecell ° is the potential of the cell under standard conditions (calculated from tabulated Ered° values), n is the

moles of electrons transferred in the overall balanced redox equation, and Q is the reaction quotient.

312 | Chapter 17

Solve First, we need to calculate Ecell

° and determine the value of n. The half-reactions and overall cell reaction are Cu(s) → Cu2+(aq) + 2 e– Eanode

° = 0.3419 V 2 Ag+(aq) + 2 e– → 2 Ag(s) Ecathode

° = 0.7996 V Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) Ecell


° = 0.4577 V We see that n = 2. Now we can use the Nernst equation to calculate Ecell when [Ag+] = 2.56 × 10–3 M and [Cu2+] = 8.25 × 10–4 M. Remember that the reaction quotient for this reaction takes the form

2+

+ 2

[Cu ][Ag ]

Q =

because pure solids do not appear in the expression for Q. Thus,

( )– 4

cell 2–3

0.0592 8.25 100.4577 V – log 0.396 V2 2.56 10

×= =×

E

Think about It This reaction became less spontaneous (lower cell potential) under these conditions.


We can use the following equation to calculate the equilibrium constant for the given redox reaction:

log K =

nEcell

0.0592

Analyze First, we have to determine the standard cell potential, Ecell

° , using the Ered° values in Appendix 6, and determine the value of n, the moles of electrons transferred in the overall balanced equation. Solve

Fe3+(aq) + e– → Fe2+(aq) Ecathode ° = 0.770 V

Cr2+(aq) → Cr3+(aq) + e– Eanode

° = –0.41 V Fe3+(aq) + Cr2+(aq) → Fe2+(aq) + Cr3+(aq) Ecell


° = 1.18 V = 1n

1 1.18 Vlog 19.93240.0592

K ×= =

19.9324 191 10 8.56 10K = × = × Think about It Because K for this reaction is very large, the reaction goes very far to the right.

17.50. Collect and Organize We can use the following equation to calculate the equilibrium constant for the given redox reaction:

log K =

nEcell

0.0592

Analyze First, we have to determine the standard cell potential, Ecell

° using the Ered° values in Appendix 6, and determine the value of n, the moles of electrons transferred in the overall balanced equation.


Solve 2 Fe2+(aq) → 2 Fe3+(aq) + 2 e– Ecathode

° = 0.770 V 2 e– + 4 H+(aq) + MnO2(s) → Mn2+(aq) + 2 H2O(ℓ) Eanode

° = 1.23 V 4 H+(aq) + MnO2(s) + 2 Fe2+(aq) → Mn2+(aq) + Fe3+(aq) + 2 H2O(ℓ) Ecell


° = –0.46 V

log K = 2× –0.46 V

0.0592= –15.54

K = 1×10–15.54 = 2.9×10–16

Think about It Because K for this reaction is very small, the equilibrium lies very far to the left.

17.51. Collect and Organize We can use the Nernst equation to calculate the potential of the hydrogen electrode at pH = 7.00.

Analyze We are reminded that under standard conditions [1 atm H2(g) and 1.00 M H+ = pH = 0.00] the voltage of the hydrogen cell is zero. This is Ecell° The overall reaction for the cell (against SHE) is

2 H+(aq) + 2 e– → H2(g) ([H+] = 1 × 10–7 M)

H2(g) → 2 H+(aq) + 2 e– (SHE) 2 H+(aq) (1 × 10–7 M) + H2(g) (1 atm) → 2 H+(aq) (1.00 M) + H2(g) (1 atm)

The form of Q for the Nernst equation is

( ) ( )( ) ( ) ( )

2

2 2–7 –7

1.00 1 atm 1

1 10 M 1 atm 1 10

MQ

×= =

× × ×

Solve

( )cell 2–7

0.0592 10.000 V – log –0.414 V2 1 10

E = =×

Think about It The spontaneous reaction actually is the reverse reaction:

2 H+(aq) (1.00 M) + H2(g) (1 atm) → 2 H+(aq) (1 × 10–7 M) + H2(g) (1 atm) Ecell° = 0.414 V In this redox cell, acid in the SHE will be reduced and H2 in the cell where pH = 7.00 will be oxidized.

17.52. Collect and Organize Given the standard reduction potentials, we can add the oxidation potential of NADH to the reduction potential of oxaloacetate to determine the potential and then calculate the value of K for the reaction. In this problem, we can assume that the reaction is occurring in the biochemical standard state (pH 7).

Analyze The Eanode° for NADH is 0.320 V,− so we can calculate Ecell° from

Ecell = Ecathode

(oxaloacetate)− Eanode (NADH)

The value of the equilibrium constant at 298 K is given by

log K =

nEcell

0.0592

The process is a two-electron redox reaction, so n = 2.

314 | Chapter 17

Solve (a) Ecell

= –0.166 V − (−0.320 V) = 0.154 V

(b)

5.203 5

2 mol 0.154 Vlog 5.20270.0592

1 10 1.59 10

K

K

×= =

= × = ×

Think about It In this reaction, oxaloacetate functions as the oxidizing agent to NADH.

17.53. Collect and Organize We can use the Nernst equation to calculate the potential for the reduction of MnO4

– to MnO2 in the presence of SO3

2– when [MnO4

–] = 0.150 M, [SO32–] = 0.256 M, [SO4

2–] = 0.178 M, and [OH–] = 0.0100 M. We are also to assess whether the potential increases or decreases as reactants become products in the reaction.

Analyze To use the Nernst equation, we need to know the value of Ecell° and n. We can determine both of these by writing out the half-reactions and balancing the redox reaction. Solve

2 MnO4–(aq) + 4 H2O() + 6 e– → 2 MnO2(s) + 8 OH–(aq) Ecathode

° = 0.59 V

3 SO32–(aq) + 6 OH–(aq) → 3 SO4

2–(aq) + 3 H2O()+ 6 e– Eanode ° = –0.92 V

2 MnO4–(aq) + 3 SO3

2–(aq) + H2O()→ 2 MnO2(s)+ 3 SO42– (aq) 2 OH– (aq) Ecell


° = 1.51 V n = 6

( ) ( )( ) ( )

3 2

cell 2 3

cell

0.178 0.0100 0.05921.51 V – log 6 0.150 0.256

1.54 V

M ME

M ME

=

=

As the reaction proceeds, the concentrations of the reactants decrease and the concentrations of the products increase, so Q increases and log Q becomes more positive. When a more positive log Q is multiplied by 0.0592/6 and then subtracted from the Ecell° , Ecell decreases.

Think about It Be sure to use half-reactions to determine the correct value of n.

17.54. Collect and Organize We can use the Nernst equation to calculate the cell potential for the reduction of MnO2 to Mn2+ in the presence of I–

when [I–] = 0.225 M, [H+] = 0.900 M, [Mn2+] = 0.100 M, and [I2] = 0.00114 M. We are also to assess whether the cell potential increases or decreases as reactants become products in the reaction.

Analyze To use the Nernst equation, we need to know the value of Ecell

° and n. We can determine both of these by writing out the half-reactions and balancing the redox reaction. Solve

MnO2(s) + 4 H+(aq) + 2 e– → 2 Mn2+(aq) + 2 H2O(ℓ) Ecathode ° = 1.23 V

2 I–(aq) → I2(aq) + 2 e– Eanode ° = 0.5355 V

MnO2–(s) + 2 I–(aq) + 4 H+(aq) → 2 Mn2+(aq) + I2(aq) + 2 H2O(ℓ) Ecell


° = 0.6945 V ( ) ( )( ) ( )

2

cell 4 2

cell

0.100 0.00114 0.05920.6945 V – log 2 0.900 0.225

= 0.797 V

M ME

M ME

=


As the reaction proceeds, more and more I2 is put into solution until its concentration is saturated at 0.114 M. In the Nernst equation, this means that Q becomes larger, so log Q gets larger. Because we subtract (0.0592/2)log Q from the cell potential, Ecell decreases. Think about It In using the Nernst equation, remember that the concentrations in Q, the reaction quotient, must be raised to their stoichiometric coefficients.


For the reaction of copper pennies with nitric acid, we are to calculate Ecell ° and then Ecell when [H+] = 0.100 M,

[NO3–] = 0.0250 M, [Cu2+] = 0.0375 M, and PNO = 0.00150 atm.

Analyze Ecell ° is calculated by adding Eanode

° and Ecathode ° . The reaction is spontaneous, so our calculated Ecell

° must be positive. In balancing the reaction, we can also determine the value of n for the Nernst equation Solve (a)

3 Cu(s) → 3 Cu2+(aq) + 6 e– Eanode ° = 0.3419 V

2 NO3– + 8 H+(aq) + 6 e–

→ 2 NO(g) + 4 H2O(ℓ) Ecathode ° = 0.96 V

3 Cu(s) + 2 NO3– + 8 H+(aq) → 3 Cu2+(aq) + 2 NO(g) + 4 H2O(ℓ) Ecell


° = 0.62 V

(b) ( ) ( )( ) ( )

3 2

cell 2 8

0.0375 0.00150 atm0.05920.6181 V – log 0.61 V6 0.0250 0.100

ME

M M= =

Think about It We may mix concentration units of atmospheres and molarity, as we do in this calculation of Q.

17.56. Collect and Organize For the reaction of ClO3

– with Cl– in an acidic solution, we are to calculate 𝐸𝐸cell° and then calculate [ClO3–] when

the system has reached equilibrium where 2ClOP = 2.0 atm,

2ClP = 1.00 atm, and [H+] = [Cl–] = 10.0 M.

Analyze Ecell ° is calculated by adding Eanode

° and Ecathode° . The reaction is spontaneous, so our calculated Ecell ° must be

positive. In balancing the reaction we can determine the value of n. Because the system is at equilibrium in part b, we know that Erxn = 0. Therefore, [ClO3

–] can be determined through the equation

0 = Ecell

– 0.0592n

logPClO2( )2

× PCl2

[ClO3– ]2[Cl2]2[H+ ]4

Solve (a)

2 e– + 2 ClO3–(aq) + 4 H+(aq) → 2 ClO2(g) + 2 H2O(ℓ) Ecathode

° = 1.152 V 2 Cl–(aq) → Cl2(g) + 2 e– Eanode

° = 1.3583 V 2 ClO3

–(aq) + 2 Cl–(aq) + 4 H+(aq) → 2 ClO2(g) + Cl2(g) + 2 H2O(ℓ) Ecell ° = Ecathode

° – Eanode ° = –0.2063 V

(b)

0 = –0.2063 V – 0.05922

log2.0 atm( )2

1.00 atm( )[ClO3

− ]2 10.0 M( )210.0 M( )4

0.2063 V = – 0.05922

log2.0 atm( )2

1.00 atm( )[ClO3

− ]2 10.0 M( )210.0 M( )4

316 | Chapter 17

−6.970 = log 4.0×10– 6

[ClO3− ]2

1×10– 6.970 = 1.072×10–7 = 4.0×10– 6

[ClO3− ]2

[ClO3− ] = 6.1 M

Think about It The Nernst equation is quite versatile. From it we can calculate Ecell; Ecell° ; the equilibrium constant, K; and even the value of n.


For the reaction of NH4+ with O2 in water, we are to calculate Ecell° . We can then use the Nernst equation to

determine [NO3–]/[NH4

+] for PO2 = 0.21 atm and pH = 5.60 at 298 K. Analyze Ecell ° is calculated by adding Eanode

° and Ecathode° . The reaction is spontaneous, so our calculated Ecell ° must be

positive. In balancing the reaction, we can determine the value of n. Because the system is at equilibrium in part b, we know that Ecell = 0. Therefore, [NO3

–]/[NH4+] can be determined through the equation

0 = Ecell – 0.0592

nlog

[NO3– ][H+ ]2

[NH4+ ] PO2( )2

Solve (a)

NH4+(aq) + 3 H2O(ℓ) → NO3

–(aq) + 10 H+ + 8 e– Eanode ° = 0.88 V

2 O2(g) + 8 H+(aq) + 8 e– → 4 H2O(ℓ) Ecathode ° = 1.229 V

NH4+(aq) + 2 O2(g) → NO3

–(aq) + H2O(ℓ) + 2 H+(aq) Ecell ° = Ecathode

° – Eanode ° = 0.349 V

(b)

( )( )

( )( )

( )

2– 63

24

2– 63

24

–10 3

4

47.16 47 –10 3

4

3

[NO ] 2.512 10 0.05920 0.349 V – log8 [NH ] 0.21 atm

[NO ] 2.512 10 0.05920.349 V – log8 [NH ] 0.21 atm

[NO ] 47.16 log 1.431 10

[NH ]

[NO ]1 10 1.453 10 1.43 10

[NH ]

[NO ]

−

+

−

+

−

+

−

+

−

×=

×− =

⎛ ⎞= × ×⎜ ⎟⎜ ⎟⎝ ⎠

× = × = × ×

M

M

57

4

1.02 10[NH ]+ = ×

Think about It This ratio is consistent with a spontaneous reaction as indicated by the positive Ecell° .


For the reaction of silver chloride with hydrogen gas to give silver and hydrochloric acid, we are to determine the value of 𝐸𝐸cell° .


Analyze From the described reaction, we first identify the half-cell reactions and obtain their standard potentials from Appendix 6. We then compute the cell potential by using

Ecell ° = Ecathode

° – Eanode °

Solve

2 × (AgCl(aq) + e– → Ag(s) + Cl–(aq)) Ecathode ° = 0.2223 V

H2(g) → 2 H+(aq) + 2 e– Eanode ° = 0.000 V

2 AgCl(aq) + H2(g) → Ag(s) + Cl–(aq) + 2 H+(aq) Ecell ° = Ecathode

° – Eanode ° = 0.2223 V

Think about It The oxidation half-reaction in this redox reaction is the standard hydrogen electrode oxidation reaction.

17.59. Collect and Organize We are to compare a 12 V lead–acid battery with one that has a lower ampere-hour rating. Analyze An ampere-hour is a unit of electrical charge and is defined as the electric charge transferred by 1 A of current for 1 hr. It is used to describe the life of a battery. Solve The total masses of the electrode materials (c) and the combined surface areas of the electrodes (f) are likely to be different.

Think about It Both batteries use the same components (b and e) and have the same voltage (a and d).

17.60. Collect and Organize Considering the two voltaic cells (Zn–Cu and Cd–Cu) in which all soluble species are present as 1 mol, we are to choose the true statement describing these cells.

Analyze We would expect that the Ered° values for Zn and Cd are different from each other. A quick look at Appendix 6 shows that this is true (Ered° for Zn = –0.7618 V; Ered° for Cd = –0.403 V). Therefore, the potentials of the two cells are different. Furthermore, the mass of the solid electrode does not change the cell potential, nor can we tell from the information given whether the electrodes in the cells are the same mass. We are left to determine whether the quantity of electrical charge or energy that the two cells can produce is the same. The amount of electrical energy depends on the voltage of the cell as well as the cell’s current flow rate. If the two cells have different E° values, the quantity of electrical energy each can produce must be different. However, because each redox reaction is a 2 e– process, the two cells will generate the same amount of electrical charge. Solve (c) The quantities of electrical charge that they can produce are the same.

Think about It In designing batteries, it is important to consider not only the amount of electrical charge the battery could produce, but also the current flow rate (expressed in amperes).


We are to compare two voltaic cells to determine which produces more charge per gram of anode material.

Analyze For each cell we must first identify which species is the anode and the number of electrons transferred when 1 mol of anode is consumed in the reaction. The charge generated by the reaction is

C = nF

318 | Chapter 17

where C = charge in coulombs, n = moles of electrons transferred in the balanced equation, and F = 9.65 × 104 C/mol. This gives the charge per mole of anode. To convert this into charge per gram, we use the molar mass:

charge coulombs/molgram molar mass of anode material

=

Solve For the Ni–Cd voltaic cell, Cd is the anode material:

Cg= 2 mol e– × 9.65×104C / mol

112.41 g/mol= 1.72×103 C/g Cd

For the Al–O2 voltaic cell, Al is the anode material:

Cg= 12 mol e– × 9.65×104C / mol

26.98 g/mol= 4.29×104 C/g Al

Therefore, the Al–O2 cell produces a greater charge per gram. Think about It The number electrons transferred in the oxidation of Al to Al3+ is 4 mol × 3 e–/mol = 12 e–.

17.62. Collect and Organize We are to compare two voltaic cells to determine which produces more charge per gram of anode material.

Analyze For each cell we must first identify which species is the anode and the number of electrons transferred when 1 mol of anode is consumed in the reaction. The charge generated by the reaction is

C = nF where C = charge in coulombs, n = moles of electrons transferred in the balanced equation, and F = 9.65 × 104 C/mol. This gives the charge per mole of anode. To convert this into charge per gram we use the molar mass:

charge coulombs/molgram molar mass of anode material

=

Solve For the Zn–MnO2 voltaic cell, Zn is the anode material:

Cg= 2 mol e− × 9.65×104C/mol

65.38 g/mol= 2.95×103 C/g

For the Li–MnO2 voltaic cell, Li is the anode material:

Cg= 1 mol e– × 9.65×104C / mol

6.941 g/mol= 1.39×104 C/g Li

Therefore, the Li–MnO2 cell produces a greater charge per gram.

Think about It Even though the electrochemical process for the Li–MnO2 cell produces only 1 mol of e–, the light mass of Li makes this cell higher in charge per gram than the Zn–MnO2 cell.


We are to compare two voltaic cells to determine which produces more energy per gram of anode material. Analyze The energy of a voltaic cell is the force to move electrons from the anode to the cathode. The unit of volts is energy per unit charge, so

Energy = volts × charge (in units of V⋅ C)


where 1 V = 1 J/C. The charge in the cell generated by 1 g of anode material is – 4mol e 9.65 10 CCharge 1 g molar mass of anode material

mol anode mol×

= × × ×

Solve For the Zn–Ni(OH)2 cell, Zn is the anode material:

431 mol 2 mol e 9.65 10 CCharge = 1 g 2.952 10 C

65.38 g 1 mol Zn mol e

−

−

×× × × = ×

3 3JEnergy 1.20 2.952 10 C = 3.54 10 JC

= × × ×

For the Li–MnO2 cell, Li is the anode material: – 4

4–

1 mol 1 mol e 9.65 10 CCharge 1 g 1.390 10 C6.941 g 1 mol Li mol e

×= × × × = ×

4 4JEnergy 3.15 1.390 10 C 4.38 10 JC

= × × = ×

Therefore, the Li–MnO2 cell generates more energy per gram of anode. Think about It Although the charge generated per mole of Li versus that of a mole of Zn in these cells is lower, the high voltage of the Li–MnO2 cell means that this cell generates more energy.


We are to compare two voltaic cells to determine which produces more energy per gram of anode material.

Analyze The energy of a voltaic cell is the force to move electrons from the anode to the cathode. The unit of volts is energy per unit charge, so

Energy = volts × charge (in units of V⋅ C) where 1 V = 1 J/C. The charge in the cell generated by 1 g of anode material is

– 4mol e 9.65 10 CCharge 1 g molar mass of anode materialmol anode mol

×= × × ×

Solve For the Zn–Ni(OH)2 cell, Zn is the anode material:



−

−

×× × × = ×

3 3JEnergy 1.50 2.952 10 C = 4.43 10 JC

= × × ×

For the Zn– O2 cell, Zn is the anode material:



−

−

×× × × = ×

3 3JEnergy 2.08 2.952 10 C = 6.14 10 J

C= × × ×

Therefore, the Zn– O2 cell generates more energy per gram of anode. Think about It The higher voltage of the Zn– O2 cell is responsible for its higher energy output.

320 | Chapter 17

17.65. Collect and Organize We are to explain the differences in the signs of the cathode in a voltaic versus an electrolytic cell.

Analyze The signs of the electrodes in a cell indicate the direction of electron flow. Solve In a voltaic cell, the electrons are produced at the anode, so a negative (–) charge builds up there; in an electrolytic cell, electrons are being forced onto the cathode so that it builds up negative (–) charge. The flow of electrons in the outside circuit is reversed in an electrolytic cell compared to the flow in a voltaic cell.

Think about It An electrolytic cell uses an outside source of electrical energy to cause a nonspontaneous reaction to occur.


We are to explain the differences in the signs of the anode in a voltaic versus an electrolytic cell.

Analyze The signs of the electrodes in a cell indicate the direction of electron flow. Solve In a voltaic cell, the electrons are produced at the anode, so a negative (–) charge builds up there; in an electrolytic cell, electrons are being forced onto the cathode so that it builds up a negative (–) charge, with the anode having a positive (+) charge. The flow of electrons in the outside circuit is reversed in an electrolytic cell compared to the flow in a voltaic cell.

Think about It An electrolytic cell uses an outside source of electrical energy to cause a nonspontaneous reaction to occur.


In a mixture of molten Br– and Cl– salts, we are to predict which product, Br2 or Cl2, forms first in an electrolytic cell as the voltage is increased.


Analyze The oxidations of Br– and Cl– are expressed as

2 Br–(ℓ) → Br2(ℓ) + 2 e– 2 Cl–(ℓ) → Cl2(ℓ) + 2 e–

Solve The halide that is first to be oxidized is the one with the lowest ionization energy. Br–, being larger and less electronegative than Cl–, loses its electron more readily; therefore, Br2 forms first in the cell as the voltage is increased.

Think about It If the molten salt also contains F–, F2 would form after Br2 and Cl2.


For the electrolytic cell in Problem 17.67 that uses molten halide salts, we are to explain why molten salts must be used rather than an aqueous solution of the salts (seawater).

Analyze The possible oxidation reactions of the electrolysis of seawater containing Cl– and Br– and H2O are

2 Cl–(aq) → Cl2(g) + 2 e– Eanode° = 1.3583 V 2 Br–(aq) → Br2(ℓ) + 2 e– Eanode° = 1.066 V 2 H2O(ℓ) → O2(g) + 4 H+(aq) + 4 e– Eanode° = 1.229 V

Solve As seen by the Eanode° for water, Cl–, and Br–, the electrolysis of seawater produces Br2 first because Br– is the most easily oxidized (Br2 is the hardest to reduce) but then produces O2 before Cl2. Therefore, the oxidation products of the electrolysis are O2, and Br2, not Cl2. Therefore, the process cannot be a source of Cl2.

Think about It The salts must be molten, not solids, in the electrolysis described in Problem 17.67 because the ions must be able to migrate to the electrodes.


For the electrolysis of a 1.0 M Cu2+ solution, we are to determine whether the potential at the cathode where the reduction of Cu2+ occurs needs to be more negative or less negative than 0.34 V to quantitatively reduce the Cu2+ in solution to Cu.

Analyze We are given that Ered° for Cu2+ is 0.34 V. This is under standard conditions when [Cu2+] = 1.0 M, the concentration of the solution at the start of the electrolysis. As [Cu2+] decreases as Cu is deposited on the electrode, Ecell can be calculated by using the Nernst equation:

cell 2+

0.0592 10.34 V – log2 [Cu ]

E =

Solve As the reaction proceeds and [Cu2+] decreases, the value of log (1/[Cu2+]) becomes more positive. As a result, Ecell decreases, so the cathode potential must be more negative than 0.34 V to complete the reduction.

Think about It A slight overpotential might be required to accomplish the electrolysis, however, because of a kinetic barrier to the reduction reaction.

17.70. Collect and Organize To increase the rate of electrolysis of water, we are to choose the best electrolyte from 2.00 M solutions of H2SO4, HBr, NaI, Na2SO4, and Na2CO3.

322 | Chapter 17

Analyze When electrolytes are added to water, the rate of electrolysis increases because the conductivity of the water increases. The greater the concentration of ions in solution, the greater is the rate of electrolysis. In addition, we have to make sure that the water undergoes electrolysis preferentially to the cations or anions of the salts. We can determine this by comparing the required cell potential for the hydrolysis to that of the reduction of oxidation potentials for the cations and anions, respectively, in these species.

Solve First, we consider the effect of the concentration of ions in solution. The more ions in solution, the more enhanced the electrolysis. When 2.00 M H2SO4 dissolves in water, it completely dissociates into 2.00 M H+ and 2.00 M HSO4

– ions. The HSO4– ion, in turn, partially dissociates to H+ and SO4

2– ions. So, for H2SO4 the concentration of ions in a 2.00 M solution is greater than 4.00 M. When 2.00 M HBr dissolves in water, it completely dissociates to give 4.00 M ions (H+ and Br–). Likewise, 2.00 M NaI also gives 4.00 M ions (Na+ and I–). When 2.00 M Na2SO4 dissolves in water, initially 6.00 M ions (2 Na+ and SO4

2–) form. The SO42– ion

partially hydrolyzes (Kb = 8.33 × 10–13) to give HSO4– and H+ ions, so the concentration of ions in this solution

is greater than 6.00 M. When 2.00 M Na2CO3 dissolves in water, initially 6.00 M ions (2 Na+ and CO32–) form.

The CO32– ions partially hydrolyze (Kb = 2.13 × 10–4) to give HCO3

– and H+ ions, so the concentration of ions in this solution is also greater than 6.00 M. Because the Kb for the hydrolysis of CO3

2– is greater than that of SO42–,

the concentration of ions in 2.00 M Na2CO3 is greater than in 2.00 M Na2SO4 and the electrolysis is most rapid for the solution with dissolved Na2CO3. Therefore, in order of worst to best for enhancing electrolysis on the basis of the number of electrolyte ions: NaI = HBr < H2SO4 < Na2SO4 < Na2CO3. Second, we have to consider that we will be supplying +1.229 V to the solution to electrolyze the water to oxygen and hydrogen. Using data from Appendix 6, we see that we should not use either NaBr or NaI because the halide ion will oxidize (their reduction potentials are 1.066 V and 0.5355 V, respectively). HSO4

– has a high reduction potential (0.17 V), so this would be stable under the electrolysis conditions. Unfortunately, the carbonate ion potential is not included in Appendix 6. This might indicate that it is stable to redox behavior. The student’s best choice would be Na2CO3 both for concentration of electrolyte and for stability under the electrolysis conditions.

Think about It Pure water has a slow rate of hydrolysis because it has only 2 × 10–7 M ions from the dissociation of water into H+ and OH– ions.

17.71. Collect and Organize In an electroplating process, we are to calculate the mass of silver deposited on an object when 1.7 A ⋅ hr of charge is delivered from a battery. To determine this, we need to relate the ampere-hours to the total number of electrons generated. Then we can relate the number of electrons to the mass of Ag deposited from a solution of Ag+.

Analyze We can convert ampere-hours to coulombs:

1 C 3600 sA hrA s hr

⋅ × ×⋅

We can calculate the moles of electrons used in the process from this result, knowing that 1 mol e– = 9.65 × 104 C. To calculate the mass of Ag deposited we also need to know that the reduction of Ag+ to Ag is a one-electron process. Solve

1 C 3600 s1.7 A hr 6120 CA s hr

⋅ × × =⋅

––2 –

4

1 mol e6120 C 6.342 10 mol e9.65 10 C

× = ××


–2 ––

1 mol Ag 107.87 g Ag6.342 10 mol e 6.8 gmol1 mol e

× × × =

Think about It The higher the amps for the battery, the faster an object can be electroplated.


Knowing that a particular NiMH battery requires 100 min to recharge, we are to calculate the mass in grams of Ni(OH)2 that is oxidized in this 0.50 A battery to NiO(OH).

Analyze We first need to determine the amount of charge transferred in this process:

Amperes time in seconds coulombs× =

From the calculated charge, we can find the mass of Ni(OH)2 oxidized: –

224 –

1 mol Ni(OH)1 mol eMass (g) coulombs molar mass of Ni(OH)9.65 10 1 mol e

= × × ××

We know that the oxidation of 1 mol of Ni(OH)2 to NiO(OH) is a one-electron process because Ni2+ is being oxidized to Ni3+. Solve

60 s 1 C0.50 A 100 min 3000 Cmin A s

× × × =⋅

–2 2

4 –

1 mol Ni(OH) 92.71 g Ni(OH)mol e3000 C 2.88 gmol9.65 10 1 mol e

× × × =×

Think about It To speed up the recharging process further, we could increase the amperage of the battery charger.

17.73. Collect and Organize / Analyze From knowing that 0.446 g of gold is produced when the same amount of charge is passed through the solution as when 0.732 g of silver is deposited from AgNO3(aq), we are to determine the oxidation state of the gold. Analyze From the mass of silver deposited, we can calculate the moles of electrons needed to deposit that amount of silver, as 1 mol of electrons would be needed to reduce 1 mol of Ag+ ions. We can then calculate the moles of Au deposited and divide the number of electrons used to deposit that gold by the moles of Au deposited to give the charge on the gold in the solution. Solve Moles of electrons used to deposit 0.732 g of silver:

0.732 g × 1 mol Ag107.87

× 1 mol e–

1 mol Ag= 6.79 ×10−3 mol

Moles of gold deposited:

0.466 g × 1 mol196.97 g

= 2.26 ×10−3 mol

Charge of the gold ions in the solution: 6.79 ×10−3 mol e–

2.26 ×10−3 mol Au= 3

Think about It For the same amount of charge delivered to the silver nitrate solution, one-third the molar amount of gold is

324 | Chapter 17

deposited because of the difference in the charge of their ions (Au3+ versus Ag+).

17.74. Collect and Organize / Analyze We are to calculate the current (in amps) needed for 2.5 hr to deposit 0.750 g of Pt(s) from a solution of [PtCl4]2–.

Analyze We will first need to know the number of moles of electrons needed to deposit the platinum. From the mass of platinum deposited, we can find the moles of platinum from the molar mass. From the balanced half-reaction, we can find the number of moles of electrons required per mole of platinum deposited.

[PtCl4]2–(aq) + 2 e– → Pt(s) + 4 Cl–(aq) Knowing then the number of moles of electrons required, we then convert into coulombs (1 mol e–/9.65 × 104 C) and then into amps (1 C/A ⋅ s) for the time of the deposit (2.5 hr). Solve Moles of electrons required for 0.750 g Pt:

0.750 g × 1 mol Pt195.08 g

× 2 mol e–

1 mole Pt= 7.689 ×10−4 mol e−

Amps required:

7.689 ×10−4 mol e– × 9.65 ×104 Cmol e− × A ⋅s

C× 1

2.5 h× 1 h

3600 s= 0.082 A

Think about It To deposit the same molar amount of silver over the same amount of time would require only half the current because only 1 mol of electrons per mole of Ag+ would be required.


We are to calculate how long it will take to recharge a battery that contains 4.10 g of NiO(OH) and is 50% discharged. This means that 2.05 g of NiO(OH) has been depleted from the battery. The charger for the battery operates at 2.00 A and 1.3 V.

Analyze We first need to calculate the moles of electrons needed to recover 2.05 g of NiO(OH), which in the NiMH battery forms Ni(OH)2 in a one-electron process. Next, we will convert the moles of electrons to coulombs. Because 1 C = 1 A ⋅ s and we know the amperes at which the charger operates, we can then calculate the time it takes the charger to deliver the electrons to recharge the battery. Solve

––

4–

–

1 mol 1 mol e2.05 g NiO(OH) 0.0224 mol e91.70 g 1 mol NiO(OH)

9.65 10 C A s 10.0224 mol e 1080 sC 2.00 Amol e

× × =

× ⋅× × × =

In minutes this is 1080 s × 1 min/60 s = 18.0 min.

Think about It The larger the battery, the more of the reactant is needed to be regenerated and the longer it takes to recharge it.

17.76. Collect and Organize We are to calculate the time it takes to deposit a thin layer (1.00 µm) of gold on a medallion that is 4.0 cm in diameter and 2.0 mm thick. The current for the deposition is 85 A.


Analyze We first need to calculate the mass of gold required to electroplate the disk. We need to include in the surface area (SA) not only the two sides of the medallion, but also the edge:

SA = (2πr 2 )+ (πdw)

where r = radius, d = diameter, and w = thickness of the medallion. The volume of gold needed for the deposition is SA × depth of the deposit (1.00 µm). The mass of gold required to plate the medallion then is volume × density (19.3 g/cm3). The time it would take for the gold to deposit onto the medallion from a solution of Au3+ is

– 4

–

1 3 mol e 9.65 10 C 1 A s 1Time mass Au neededAu molar mass mol Au C amperesmol e

× ⋅= × × × × ×

Solve

SA = 2× π × (2.0 cm)2⎡⎣ ⎤⎦ + π × 4.0 cm × 0.20 cm( ) = 27.646 cm2

Volume of gold deposited = 27.646 cm2 ×1.00×10– 4cm = 2.7646×10–3cm3

Mass of Au = 2.7646×10–3cm3 × 19.3 gcm3 = 0.05336 g

Time to deposit Au = 0.05336 g × 1 mol196.97 g

× 3 mol e–

mol Au× 9.65×104C

mol e– × 1 A ⋅sC

× 185 A

= 0.92 s

Think about It If a solution of Au+ were used for this process, it would take one-third of the time.

17.77. Collect and Organize We are to calculate the amount of O2 that could be generated in 1 hr on a submarine by using electrolysis and then consider the practicality of using seawater as the source of oxygen for the submarine.

Analyze We can calculate the moles of O2 produced by the electrolytic cell through

–2

2 4 –

1 mol O1 C mol eMoles O time in seconds amperesA s 9.65 10 C 4 mol e

= × × × ×⋅ ×

Notice that the oxidation of water to O2 is a 4 e– process. We can then use the ideal gas law to calculate the volume of O2 produced. Solve (a)

–

– 422 24 –

1 mol O3600 s 1 C 1 mol eMoles O 1 hr 0.025 A 2.332 10 mol O1 hr A s 9.65 10 C 4 mol e

= × × × × × = ×⋅ ×

V = 2.332×10–4 mol × 0.08206 L ⋅atm / mol ⋅K × 298 K0.98692 atm/1 bar

= 5.78×10–3 L, or 5.8 mL

(b) Seawater contains a fairly high concentration of Cl– and Br– that can be oxidized, so the direct electrolysis of seawater would not be useful as an oxygen source.

Think about It Submarines probably purify their water, perhaps through a reverse osmosis process, to remove the chloride and bromide and other ions before the electrolysis process.

17.78. Collect and Organize For an electrolytic cell that has a current of 52 mA, we are to calculate how long it would take for the cell to reduce water so that 125 L of H2 is produced at 293 K and 750 torr.

Analyze Using the ideal gas law, we first calculate the moles of H2 in 125 L. We know that the reduction of H2O to

326 | Chapter 17

produce H2 is a 2 e– process, so we can calculate the time to produce the H2 through the equation – 4

2 –2

2 mol e 9.65 10 C 1 A s 1Time in seconds mol Hmol H C amperesmol e

× ⋅= × × × ×

Solve

n =

750 torr × 1 atm760 torr

×125 L

0.08206 L ⋅atm / mol ⋅K × 293 K= 5.13 mol

5.13 mol H2 ×

2 mol e–

mol H2

× 9.65×104Cmol e– × 1 A ⋅s

C× 1

0.052 A= 1.9×107 s

1.9×107 s × 1 hr

3600 s× 1 d

24 hr= 220 d

Think about It The amps used in this electrolysis are too small to generate this amount of H2 in a reasonable time frame. By increasing the amps 1000-fold, we can reduce the time to 5.3 hr.

17.79. Collect and Organize For the process that electroplates nickel, we are to calculate the lowest potential required to deposit Ni onto a piece of iron by using a 0.35 M Ni2+ solution.

Analyze To solve this problem we need to use the Nernst equation:

Ecell = Ecell

− 0.0592n

logQ

where Ecell° is the reduction potential of Ni2+ versus the SHE (– 0.257 V), n is the number of electrons needed to reduce Ni2+ to Ni+0, and Q is 1/[Ni2+] on the basis of the reduction reaction

Ni2+(aq) + 2 e– → Ni(s) Solve

cell0.0592 1–0.257 V log –0.270 V

2 0.35 = − =E

M

Think about It For this electrolysis reaction, using a more dilute solution of the metal cation necessitates an increase in the potential needed to cause Ni2+ to deposit on the iron.

17.80. Collect and Organize For the process that electroplates silver onto cutlery, we are to calculate the lowest potential required by using a 3.50 × 10–5 M Ag+ solution.

Analyze To solve this problem we need to use the Nernst equation:

Ecell = Ecell

− 0.0592n

logQ

where Ecell° is the reduction potential of Ag+ versus the SHE (0.7996 V), n is the number of electrons needed to reduce Ag+ to Ag+0, and Q is 1/[Ag+] on the basis of the reduction reaction

Ag+(aq) + e– → Ag(s)


Solve

cell –5

0.0592 10.7996 V log 0.536 V1 3.50 10

EM

= − =×

Think about It The ammonia in the solution serves to maintain the [Ag+] at a low level for control of the electroplating process.

17.81. Collect and Organize We are to consider the advantages and disadvantages of hybrid power systems versus all-electric fuel-cell systems.

Analyze A parallel hybrid power system uses traditional petroleum-based fuel for high power demands and an electric motor for lower power demands. An all-electric fuel cell system uses only electrochemical power based on combustion half-reactions to supply power. Solve A hybrid vehicle uses a relatively inexpensive fuel (gasoline) in the internal combustion engine and has good fuel economy but still gives off emissions. A fuel-cell vehicle does not give off emissions (the reaction produces H2O) but requires a more expensive and explosive fuel (hydrogen); moreover, current battery technologies incorporate materials that are still very expensive and bulky.

Think about It The determining factor whether alternate fuels and power systems get used will be the cost of petroleum used to produce gasoline, which traditionally has been much less expensive than alternative energy sources.


We are to describe three limitations of fuel cells for cars.

Solve The safe handling of explosive hydrogen fuel and the need for hydrogen fuel stations, the limited range of fuel cell vehicles, the bulkiness of the fuel cells themselves, and their high cost of manufacture all limit the use of fuel cells in cars. Think about It Perhaps as fuel cell technology develops, the bulkiness of the fuel cells will be reduced and better hydrogen storage systems will be developed.


Given the information that methane may be used in fuel cells, we are to consider why these fuel cells are likely to produce less CO2 emissions per mile than an internal combustion engine fueled by methane.

Analyze Both the methane fuel cell and the combustion of methane in a combustion reaction have the balanced equation

CH4(g) + 2 O2(g) → 2 H2O(g) + CO2(g) Solve Fuel cells burn methane fuel more efficiently. Electric engines are more efficient by converting more of the energy into motion instead of losing it as heat. Therefore, less CO2 is produced per mile with fuel cells.

Think about It The bulkiness and short range of fuel cells currently limit the use of fuel cells in transportation.


We are to describe some of the benefits of systems that convert propane or methanol to H2 to use in a fuel cell.

328 | Chapter 17

Solve Fuel cells are inherently more efficient than internal combustion engines and, therefore, less fuel is used to power the vehicle. Also, fuel cells eliminate NOx and SOx emissions.

Think about It The biggest disadvantage to fuel cells is their space requirements (bulkiness) in vehicles.


For the reactions of CH4 and CO with water, we are to assign oxidation numbers to the C and H atoms in all the species in the reactions and calculate ∆Grxn

° for each and for the overall reaction

CH4(g) + 2 H2O(g) → 4 H2(g) + CO2(g)

Analyze We can use the usual rules of assigning oxidation states from Chapter 4. To calculate ∆Grxn

° we use ∆Gf ° values

from Appendix 4 in the following equation:

ΔGrxn

∗ = n∑ ΔGf,products∗ − n∑ ΔGf,reactants

∗ Solve

(a) 4 +1 +1 +2 0

4 2 2

2 +1 0 4

2 2 2

CH ( ) H O( ) CO( ) 3H ( )

CO( ) H O( ) H ( ) CO ( )

−

+ +

+ → +

+ → +

g g g g

g g g g

(b) For the reaction of CH4 with H2O:

ΔG = (1 mol CO × −137.2 kJ/mol)+ (3 mol H2 × 0.0 kJ/mol⎡⎣ ⎤⎦ −

(1 mol CH4 × −50.8 kJ/mol)+ (1 mol H2O × −228.6 kJ/mol)⎡⎣ ⎤⎦ = 142.2 kJ

For the reaction of CO with H2O:

ΔG = (1 mol CO2 × −394.4 kJ/mol)+ (1 mol H2 × 0.0 kJ/mol⎡⎣ ⎤⎦ −

(1 mol CO × −137.2 kJ/mol)+ (1 mol H2O × −228.6 kJ/mol)⎡⎣ ⎤⎦ = −28.6 kJ

For the overall reaction:

ΔGoverall

= ΔGrxn1

+ ΔGrxn2

= 113.6 kJ

Think about It The spontaneity of the second reaction is not enough to overcome the positive free energy of the first reaction, so the overall reaction is nonspontaneous.

17.86. Collect and Organize To calculate the amount of electrical energy released when 1 mol of H2 is converted into water in a fuel cell, we can use the formula for electrical work:

welec = C × E

Analyze To determine the value for welec we need the value of Ecell° .

2 × [H2(g) → 2 H+(aq) + 2 e–] Eanode °

= 0.000 V 4 e– + 4 H+(aq) + O2(g) → 2 H2O(ℓ) Ecathode

° = 1.229 V

2 H2(g) + O2(g) → 2 H2O(ℓ) Ecell ° = Ecathode

° – Eanode ° = 1.229 V – 0.000 V = 1.229 V

We also need to calculate the coulombs required to consume 1 mol of H2. From the balanced equation, we see that we need 4 mol of e– for every 2 mol of H2 used, so the coulombs required are

– 45

2 –2

4 mol e 9.65 10 C1 mol H 1.93 10 C2 mol H mol e

×× × = ×


Solve welec = 1.93 × 105 C × 1.229 V = 2.372 × 105 J, or 237.2 kJ

Because the fuel cell is 60% efficient, the electrical energy obtained is 237.2 kJ × 0.60 = 140 kJ

Think about It We could also calculate the energy released by using free-energy values from Appendix 4.

17.87. Collect and Organize / Analyze The values of Ered° in Appendix 6 use the standard hydrogen electrode (SHE):

2 H+(aq) + 2 e– → H2(g) as Ered° = 0.000 V. We are asked how the values in Appendix 6 would change if

2 H2O(ℓ) + 2 e– → H2(g) + 2 OH–(aq) were defined as Ered° = 0.000 V.

Solve The Ered° of the reduction of water to give H2 versus the SHE is – 0.8278 V. If this reduction reaction were redefined to have Ered° = 0.000 V, all the values in Appendix 6 would increase by + 0.828 V.

Think about It Any reduction half-reaction could have been chosen as the standard half-reaction and the voltage of its reduction defined as 0.000 V.


We are to explain how sacrificial anodes work and describe what properties of zinc make it useful as a sacrificial anode.

Analyze We are given the example that zinc is often used to protect other metals from corrosion. Corrosion is the oxidation of the metal. Solve A metal such as zinc that is more easily oxidized than the protected metal such as iron in steel preferentially corrodes (that is, has a more negative reduction potential and is therefore more easily oxidized) and is thus “sacrificed” to protect the desired metal. For zinc the reduction reaction and potential are Zn2+(aq) + 2 e– → Zn(s) Ered° = –0.762 V This is more negative than Ered° for Fe (– 0.447 V). Also the corrosion of the zinc provides electrons to the metal it protects, forcing the protected metal to act as a cathode and keeping the metal in a reducing environment, thus preventing its oxidation.

Think about It Once the sacrificial anode is used up, the protected metal begins to corrode.

17.89. Collect and Organize We are asked to write the overall cell reaction for the reaction of Zn with O2 in the zinc–air battery to form Zn(OH)4

2–.

330 | Chapter 17

Analyze For each half-reaction, we need to have OH– or H2O as a reactant. We balance each for both atoms and charge before adding the two half-reactions together.

Solve

Cathode O2(g) + 2 H2O(ℓ) + 4 e– → 4 OH–(aq) Anode [Zn(s) + 4 OH–(aq) → Zn(OH)4

2–(aq) + 2 e–] × 2 O2(g) + 2 H2O(ℓ) + 2 Zn(s) + 4 OH–(aq) → 2 Zn(OH)4

2–(aq)

Think about It This differs from the other overall reaction for the zinc–air battery shown in Figure 19.6 in that water and OH– are reactants to give soluble Zn(OH)4

2– instead of solid ZnO.

17.90. Collect and Organize We are asked to write the half-reactions and overall cell reaction for the reaction of Cd with NiO(OH) in the nicad battery to form Cd2+ and Ni(OH)2.

Analyze For each half-reaction, we need to have OH– or H2O as a reactant. We balance each for both atoms and charge before adding the two half-reactions together. Solve

Cathode [e– + NiO(OH)(s) + H2O(ℓ) → Ni(OH)2(s) + OH–(aq)] × 2 Anode Cd(s) + 2 OH–(aq) → Cd(OH)2(s) + 2 e–

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

Think about It Because the NiO(OH) is reduced by gaining one electron and Cd is oxidized by losing two electrons, we need to multiply the nickel half-reaction in the battery by a factor of 2 to obtain the overall balanced redox equation.

17.91. Collect and Organize We can write the overall cell reaction for the battery by adding the equations together. The overall cell potential is Ecell

° = Ecathode ° – Eanode° .

Analyze Because the reduction reaction and the oxidation reaction each involve the exchange of one electron, we do not need to multiply either reaction to write the overall reaction.

Solve (a) NiO(OH)(s) + TiZr2H(s) → TiZr2(s) + Ni(OH)2(s) (b) Ecell

° =1.32 V – 0.00 V = 1.32 V

Think about It This reaction is spontaneous with a ∆G value of

ΔG = –1 mol× 9.65×104 C

mol× 1.32 J

C× 1 kJ

1000 J= –127 kJ


For a lithium–ion battery that has a LiFePO4 cathode when fully discharged, we are asked to deduce the formula of the cathode material when the battery is charged, determine the redox property of iron as the battery discharges, and discuss whether we expect the cell potential of this battery to be different from one based on LiCoO2.


Analyze (a) A cathode fully discharged has all its species in the reduced state. (b) By examining the oxidation numbers of Fe in the cathode material in the charged and discharged states, we can tell whether Fe is oxidized or reduced during the process of discharging the battery.

Solve (a) When fully charged, Li is not in the formula of the cathode material. The cathode is FePO4 when the battery is fully charged. (b) In discharging, the Fe in the cathode material is reduced from Fe3+ in FePO4 to Fe2+ in LiFePO4. (c) Because the reduction of Co4+ to Co3+ is likely to have a different potential as CoO2 is reduced to LiCoO2 from the potential for the reduction of FePO4 to LiFePO4, we expect the cell potentials of these two batteries to differ.

Think about It New battery materials continue to be developed that are more powerful and longer lasting while providing stability under extreme conditions.


We are asked to describe the meaning of the cell diagram given. Solve The cell consists of a tin anode that is immersed in a solution of 0.025 M Sn2+. This half-cell is connected via an outside circuit and salt bridge to a silver cathode immersed in a 0.010 M solution of Ag+.

Think about It A convention of the cell diagram is also that the reduction reaction is on the right-hand side. So, in this cell, Sn is oxidized and Ag+ is reduced.


For the components of a NiMH battery we are to provide the oxidation states. Solve (a) In NiO(OH) the nickel is +3. (b) In MH the hydrogen is –1. (c) In MH the metal is +1. (d) In H2O the hydrogen is +1.

Think about It Oxidation states, remember, are a formalism that helps us to keep track of electrons. In salts and compounds, the actual charges on atoms may be different.


For a Mg–Mo3S4 battery for which we are given the half-reaction potential of the anode reaction (2.37 V) and the overall cell potential (1.50 V), we are to calculate Ered° of Mo3S4. We are also to consider why Mg2+ is added to the battery’s electrolyte and determine the oxidation states and electron configurations of Mo in Mo3S4 and MgMo3S4.

Analyze Because Ecell


° the reduction potential for Mo3S4 will be Ecell

° + Eanode ° .

Solve

(a) Ecathode ° = 1.50 V + (–2.37 V) = –0.87 V

(b) Sulfur usually carries a 2– charge, so each Mo atom in Mo3S4 has a calculated charge of 2.67+. This, therefore, is a mixed oxidation state compound where it is likely that two of the Mo atoms have a 3+ charge and one Mo atom has a 2+ charge. In MgMo3S4, the Mg atom has a 2+ charge, so the Mo atoms have a 2+ charge.

332 | Chapter 17

The electron configurations for the two oxidation states of Mo are Mo in +2 oxidation state [Kr]4d4

Mo in +3 oxidation state [Kr]4d

3 (c) Mg2+ is added to the electrolyte to better carry the charge in the cell. This cation is produced at the anode and consumed at the cathode.

Think about It This battery resembles the lithium–ion battery in that the migration of a cation in the cell generates the electrical current.


The concentration difference in Na+ in red blood cells compared to that in plasma sets up both an electrochemical potential and an osmotic pressure between the cells and the surrounding plasma. To calculate the electrochemical potential, we can use the Nernst equation. Here, we have to use the form that includes temperature as a variable because in the body T is not 298 K, but 310 K (37°C).

Ecell = Ecell

− RTnF

logQ

Analyze The electrochemical cell potential established between a red blood cell and the plasma can be described by the two half-reactions

Na+(140 mM, aq) + e– → Na(s) Na(s) → Na+(11 mM, aq) + e– Na+(140 mM, aq) → Na+(11 mM, aq)

The Ecell ° = 0.00 V when the cell is at standard conditions (298 K, 1 atm, 1 M). In the Nernst equation

Q = 11 mM/140 mM and n = 1. Solve

cell 4

8.314 J/mol K 310 K 11 m0.00 V log 0.030 V140 m1 (9.65 10 C)

MEM

⋅ ×= − =

× ×

Think about It Because [Na+] is greater in plasma than in red blood cells, some transport mechanism must maintain this concentration difference in the cells.

17.97. Collect and Organize We consider the thermodynamic and electrochemical properties of the synthesis of F2 both chemically and in an electrolysis reaction.

Analyze (a) We can use the usual rules described in Chapter 4 to assign oxidation numbers to all the elements in the reactants and products for the chemical synthesis of F2. From the change in oxidation numbers we can deduce the number of electrons involved in the process. (b) To calculate ∆Hrxn

° we use

ΔHrxn = n∑ ΔHf,products

− n∑ ΔHf,reactants

(c) When we assume ∆S ≈ 0 ∆G = ∆H – T∆S ≈ ∆H

so we can use ∆H for ∆G in the equation to calculate Ecell° . Solve (a) In K2MnF6: K = +1, Mn = +4, F = –1 In SbF5: Sb = +5, F = –1 In KSbF6: K = +1, Sb = +5, F = –1


In MnF3: Mn = +3, F = –1 In F2: F = 0 This is a one-electron process.

(b) ΔHrxn = 2 mol KSbF6 × −2080 kJ/mol( ) + 1 mol MnF3 × −1579 kJ/mol( ) + 1

2 mol F2 × 0.0 kJ/mol( )⎡⎣ ⎤⎦ − 1 mol K2MnF6 × −2435 kJ/mol( ) + 2 mol SbF5 × −1324 kJ/mol( )⎡⎣ ⎤⎦ = −656 kJ

(c) Ecell ≈ 6.56×105 J

1 mol e– × 9.65×104 C/mol= 6.80 V

(d) If ∆S is positive, then the ∆G estimate is too positive; ∆G would be more negative, giving a more positive Ecell° . Therefore, our calculated Ecell

° using only the ∆Hrxn° value would be too low.

(e) In H2: H = 0 In F2: F = 0 In KF: K = +1, F = –1 In KHF2: K = +1, H = +1, F = –1 This is a two-electron process.

Think about It The chemical synthesis of F2 relies on the oxidation of F– by the strong oxidant MnF6

2–. This process, however, is not very practical, and fluorine is prepared industrially by the electrolysis reaction described in part e.

17.98. Collect and Organize For the reaction of Cu with heavily chlorinated water, we are to write balanced equations for the half-reactions and calculate Ecell° and ∆Grxn

° .

Analyze (a) In the reaction Cu is oxidized to Cu+ and Cl2 is reduced to Cl–. (b) From the Ecathode

° and ∆Eanode° , for the half-reactions, we can calculate Ecell° . To determine ∆Grxn

° we use

ΔGrxn = –nFEcell

Solve (a) 2 × [Cu(s) → Cu+(aq) + e–] 𝐸𝐸anode° = 0.521 V

2 e– + Cl2(aq) → 2 Cl–(aq) 𝐸𝐸cathode ° = 1.3583 V 2 Cu(s) + Cl2(g) → 2 CuCl(s)

(b) Ecell


° = 1.3583 – 0.521 = 0.837 V

∆Grxn° = –2 × (9.65 × 104 C/mol) × 0.837 J/C = –1.616 × 105

J, or –162 kJ Think about It The negative value of free energy for this reaction shows that the corrosion of copper pipes in highly chlorinated water is a spontaneous process.


Given the balanced reactions of UO2 with HF and UF4 with Mg to ultimately produce U(s), we are to identify the reducing agent and the element that is reduced. We will also determine the highest possible value for the reduction of UF4 and determine whether 1.00 g of Mg could produce 1.00 g of solid uranium

Analyze (a) and (b) A species that is reduced is one which decreases (makes less positive) its oxidation state. A reducing agent is that species that causes some other species to be reduced; it is itself oxidized. (b) The reduction of UF4 must be positive when combined with the oxidation of magnesium. (c) We can determine whether 1.00 g of Mg is sufficient to give 1.00 g of U through the stoichiometry of the given balanced equation for the process. Solve

334 | Chapter 17

(a) Mg is the reducing agent; it is oxidized to Mg2+. (b) Uranium in UF4 is reduced from a +4 to a 0 oxidation state. (c) The half-cell reduction potential for Mg2+ is –2.37. This means that the reduction of UF4 must have a reduction potential greater than –2.37 V for the reaction to have a positive cell potential and be spontaneous. (d) Yes, 1.00 g of Mg will be sufficient to produce 1.00 g of U:

1.00 g Mg × 1 mol Mg24.305 g

× 1 mol U2 mol Mg

× 238.03 g U1 mol

= 4.90 g U could be produced

Think about It The first reaction in this process where UF4 is produced is not a redox reaction, but an acid–base reaction.

17.100. Collect and Organize For the reaction described in Equation 17.13 that describes a lithium–ion battery, we are to identify the oxidized and reduced elements.

Analyze The cell reaction is

Li1–xCoO2(s) + LixC6(s) → 6 C(s) + LiCoO2(s) The element that is reduced will have a less positive oxidation number at the end of the reaction, and the element that is oxidized will have a more positive oxidation number at the end of the reaction. Solve In LixC6, Li is less electronegative than C, so it carries a positive charge; carbon in this compound carries a negative charge. In the product C, however, the oxidation number of carbon is 0. Therefore, carbon is oxidized in this cell reaction. In Li1–xCoO2, Co shares positive charge (balanced by two oxygen atoms in the compound) with Li. As more lithium is added, the oxidation state of the cobalt atom must decrease (become less positive) to maintain charge balance. Therefore, Co is oxidized in this cell reaction. Think about It This battery discharges by migrating the Li+ from the carbon anode to the cobalt oxide cathode. A fully discharged battery has no Li+ at the carbon electrode.


We consider electrolysis of a molten Mg2+ salt from evaporated seawater (so it may contain NaCl).

Analyze The possible reactions are (with E° values when listed in Appendix 6)

Mg2+(ℓ) + 2 e– → Mg(s) Na+(ℓ) + e– → Na(s) 2 H2O(ℓ) + 2 e– → H2(g) + 2 OH–(aq) Ered° = –0.8277 V 2 H2O(ℓ) → O2(g) + 4 H+(aq) + 4 e– Ered° = 1.229 V Mg2+(aq) + 2 e– → Mg(s) Ered° = –2.37 V Na+(aq) + e– → Na(s) Ered° = –2.71 V

Solve (a) Mg2+ undergoes a reduction reaction that occurs at the cathode. Mg forms at the cathode. (b) No. Mg2+, with a higher positive charge, has a lower (less negative) reduction potential than Na+, so the Mg2+ would not need to be separated from the NaCl in seawater first. (c) No. The electrolysis of MgCl2(aq) would not produce Mg(s) because water, with a less negative reduction potential, would be electrolyzed. (d) H2 and O2 gases would be produced.


Think about It Because different components are reduced at different potentials, electrolysis of molten salts is one way to separate components (such as metals) from each other.

17.102. Collect and Organize We consider the removal of silver tarnish (Ag2S) with silver polish [Al in OH–(aq)].

Analyze When we polish silver, the Ag+ in Ag2S is reduced to Ag and the Al is oxidized to Al(OH)3.

Solve (a) [Ag2S(s) + 2 e– → 2 Ag(s) + S2–(aq)] × 3

[Al(s) + 3 OH–(aq) → Al(OH)3(s) + 3 e–] × 2

3 Ag2S(s) + 2 Al(s) + 6 OH–(aq) → 6 Ag(s) + 3 S2–(aq) + 2 Al(OH)3(s)

(b) For the silver tarnish half-reaction:

2 Ag+(aq) + 2 e– → 2 Ag(s) Ered° = 0.7996 V Ag2S(s) 2 Ag+(aq) + S2–(aq) Ksp = 1.6 × 10–49

Using these values, we can calculate ∆Grxn° for each of these reactions:

ΔG = –2 mol e– × 9.65×104C/mol× 0.7996 J/C

= –1.543×105 J

ΔG = –RT lnKsp = –8.314 J/mol ⋅K × 298 K × ln(1.6×10– 49 )

= 2.784×105 J

Overall, ∆Grxn° for the half-reaction is

Ag2S(s) + 2 e– → 2 Ag(s)+ S2–(aq) ΔGrxn = 1.241 × 105 J

For the silver polish half-reaction:

Al(s)→ Al3+ (aq)+ 3 e– Eox = 1.662 V

Al3+ (aq)+ 3 OH– (aq) Al(OH)3(s) 1Ksp

= 11.9×10–33

= 5.263 × 1032

Using these values, we can calculate ∆Grxn° for each of these reactions:

ΔG = –3 mol e– × 9.65×104 C/mol×1.662 J/C

= –4.811×105 J

ΔG = –RT ln K(1/ Ksp ) = –8.314 J/mol ⋅K × 298 K × ln(5.263×1032 )

= –1.867 ×105 J

Overall, ∆Grxn° for the half-reaction is

Al(s) + 3 OH–(aq) → Al(OH)3(s) + 3 e– ∆Grxn° = –6.678 × 105 J

For the overall reaction of Ag2S with Al in OH– we have [Ag2S(s) + 2 e– → 2 Ag(s) + S2–(aq)] × 3 ∆Grxn

° = 3 × 124.1 kJ = 372.3 kJ [Al(s) + 3 OH–(aq) → Al(OH)3(s) + 3 e–] × 2 ∆Grxn

° = 2 × –667.8 kJ = –1335.6 kJ ∆Grxn

° = –963.3 kJ

The potential of the reaction is

Ecell =

−ΔGrxn

nF= 9.633×105 J

6 mol e– × 9.65×104 C/mol= 1.66 V

Think about It Under standard conditions, this reaction is spontaneous.

chapter 17 | electrochemistry: the quest for …postonp/ch223/pdf/chemat_ism_ch17.pdf292 chapter 17...

Documents