CHAPTER 21 ELECTROCHEMISTRY:
CHEMICAL CHANGE AND ELECTRICAL
WORK
FOLLOW–UP PROBLEMS
21.1A
Plan: Follow the steps for balancing a redox reaction in acidic solution:
1. Divide into half-reactions
2. For each half-reaction balance
a) Atoms other than O and H,
b) O atoms with H2O,
c) H atoms with H+ and
d) Charge with e–.
3. Multiply each half-reaction by an integer that will make the number of electrons lost
equal to the number of electrons gained.
4. Add the half-reactions and cancel substances appearing as both reactants and products.
Then, add another step for basic solution.
5. Add hydroxide ions to neutralize H+. Cancel water.
Solution:
1. Divide into half-reactions: group the reactants and products with similar atoms.
MnO4–(aq)  MnO42–(aq)
I–(aq)  IO3–(aq)
2. For each half-reaction balance
a) Atoms other than O and H
Mn and I are balanced so no changes needed.
b) O atoms with H2O
MnO4–(aq)  MnO42–(aq) O already balanced
I–(aq) + 3H2O(l)  IO3–(aq) Add 3 H2O to balance oxygen.
c) H atoms with H+
MnO4–(aq)  MnO42–(aq) H already balanced
I–(aq) + 3H2O(l)  IO3–(aq) + 6H+(aq) Add 6 H+ to balance hydrogen.
d) Charge with e–; Total charge of reactants is –1 and of products is –2, so add 1 e– to
reactants to balance charge:
MnO4–(aq) + e–  MnO42–(aq)
Total charge is –1 for reactants and +5 for products, so add 6 e– as product:
I–(aq) + 3H2O(l)  IO3–(aq) + 6 H+(aq) + 6e–
3. Multiply each half-reaction by an integer that will make the number of electrons lost equal to the
number of electrons gained. One electron is gained and 6 are lost so reduction must be multiplied by 6
for the number of electrons to be equal.
6 {MnO4–(aq) + e–  MnO42–(aq)}
I–(aq) + 3H2O(l)  IO3–(aq) + 6H+(aq) + 6 e–
4. Add half-reactions and cancel substances appearing as both reactants and products.
6MnO4–(aq) + 6e–  6MnO42–(aq)
I–(aq) + 3H2O(l)  IO3–(aq) + 6H+(aq) + 6 e–
Overall: 6MnO4–(aq) + I–(aq) + 3H2O(l)  6MnO42–(aq) + IO3–(aq) + 6H+(aq)
5. Add hydroxide ions to neutralize H+. Cancel water. The 6 H+ are neutralized by adding 6 OH–. The
same number of hydroxide ions must be added to the reactants to keep the balance of O and H atoms on
both sides of the reaction.
6MnO4–(aq) + I–(aq) + 3H2O(l) + 6OH–(aq)  6MnO42–(aq) + IO3–(aq) + 6H+(aq) +6 OH–(aq)
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21-1
The neutralization reaction produces water: 6 {H+ + OH–  H2O}.
6MnO4–(aq) + I–(aq) + 3H2O(l) + 6OH–(aq)  6MnO42–(aq) + IO3–(aq) + 6H2O(l)
Cancel water:
6MnO4–(aq) + I–(aq) + 3H2O(l) + 6OH–(aq)  6MnO42–(aq) + IO3–(aq) + 6H2O(l)
Balanced reaction is
6MnO4–(aq) + I–(aq) + 6OH–(aq)  6MnO42–(aq) + IO3–(aq) + 3H2O(l)
Balanced reaction including spectator ions is
6KMnO4(aq) + KI(aq) + 6KOH(aq)  6K2MnO4(aq) + KIO3(aq) + 3H2O(l)
21.1B
Plan: Follow the steps for balancing a redox reaction in acidic solution:
1. Divide into half-reactions
2. For each half-reaction balance
a) Atoms other than O and H,
b) O atoms with H2O,
c) H atoms with H+ and
d) Charge with e–.
3. Multiply each half-reaction by an integer that will make the number of electrons lost
equal to the number of electrons gained.
4. Add the half-reactions and cancel substances appearing as both reactants and products.
Then, add another step for basic solution.
5. Add hydroxide ions to neutralize H+. Cancel water.
Solution:
1. Divide into half-reactions: group the reactants and products with similar atoms.
Cr(OH)3(aq)  CrO42–(aq)
IO3–(aq)  I–(aq)
2. For each half-reaction balance
a) Atoms other than O and H
Cr and I are balanced so no changes needed.
b) O atoms with H2O
H2O(l) + Cr(OH)3(aq)  CrO42–(aq) Add 1 H2O to balance oxygen.
IO3–(aq)  I–(aq) + 3H2O(l) Add 3 H2O to balance oxygen.
c) H atoms with H+
H2O(l) + Cr(OH)3(aq)  CrO42–(aq) + 5H+(aq) Add 5 H+ to balance hydrogen.
6H+(aq) + IO3–(aq)  I–(aq) + 3H2O(l) Add 6 H+ to balance hydrogen.
d) Charge with e–; total charge of reactants is 0 and of products is +3, so add 3 e– to products
to balance charge:
H2O(l) + Cr(OH)3(aq)  CrO42–(aq) + 5H+(aq) + 3e–
Total charge is –1 for products and +5 for reactants, so add 6 e– as reactant:
6e– + 6H+(aq) + IO3–(aq)  I–(aq) + 3H2O(l)
3. Multiply each half-reaction by an integer that will make the number of electrons lost equal to the
number of electrons gained. Six electrons are gained and 3 are lost so oxidation must be multiplied by 2
for the number of electrons to be equal.
2 {H2O(l) + Cr(OH)3(aq)  CrO42–(aq) + 5H+(aq) + 3e–}
6e– + 6H+(aq) + IO3–(aq)  I–(aq) + 3H2O(l)
4. Add half-reactions and cancel substances appearing as both reactants and products.
2H2O(l) + 2Cr(OH)3(aq)  2CrO42–(aq) + 104H+(aq) + 6e–
6e– + 6H+(aq) + IO3–(aq)  I–(aq) + 31H2O(l)
Overall: 2Cr(OH)3(aq) + IO3–(aq)  2CrO42–(aq) + I–(aq) + H2O(l) + 4H+(aq)
5. Add hydroxide ions to neutralize H+. Cancel water. The 4 H+ are neutralized by adding 4 OH–. The
same number of hydroxide ions must be added to the reactants to keep the balance of O and H atoms on
both sides of the reaction.
2Cr(OH)3(aq) + IO3–(aq) + 4OH–(aq) 2CrO42–(aq) + I–(aq) + H2O(l) + 4H+(aq) + 4OH–(aq)
The neutralization reaction produces water: 4 {H+ + OH–  H2O}.
2Cr(OH)3(aq) + IO3–(aq) + 4OH–(aq) 2CrO42–(aq) + I–(aq) + 5H2O(l)
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21-2
Balanced reaction is
2Cr(OH)3(aq) + IO3–(aq) + 4OH–(aq) 2CrO42–(aq) + I–(aq) + 5H2O(l)
Balanced reaction including spectator ions is
2Cr(OH)3(aq) + NaIO3(aq) + 4NaOH(aq) 2Na2CrO4(aq) + NaI(aq) + 5H2O(l)
21.2A
Plan: Given the solution and electrode compositions, the two half cells involve the transfer of electrons
1) between chromium in Cr2O72– and Cr3+ and 2) between Sn and Sn2+. The negative electrode is the anode so the
tin half-cell is where oxidation occurs. The graphite electrode with the chromium ion/chromate solution is where
reduction occurs. In the cell diagram, show the electrodes and the solutes involved in the half-reactions. Include
the salt bridge and wire connection between electrodes. Set up the two half-reactions and balance. (Note that the
Cr3+/Cr2O72– half-cell is in acidic solution.) Write the cell notation placing the anode half-cell first, then the salt
bridge, then the cathode half-cell.
Solution: Cell diagram:
Voltmeter


e
e
Salt bridge
Sn
()
Sn2+
C
(+)
Cr3+, H+
Cr2O72
Balanced equations:
Anode is Sn/Sn2+ half-cell. Oxidation of Sn produces Sn2+:
Sn(s)  Sn2+(aq)
All that needs to be balanced is charge:
Sn(s)  Sn2+(aq) + 2e–
Cathode is the Cr3+/Cr2O72– half-cell. Check the oxidation number of chromium in each substance to determine
which is reduced. Cr3+ oxidation number is +3 and chromium in Cr2O72– has oxidation number +6. Going from +6
to +3 involves gain of electrons so Cr2O72– is reduced.
Cr2O72–(aq)  Cr3+(aq)
Balance Cr:
Cr2O72–(aq)  2Cr3+(aq)
Balance O:
Cr2O72–(aq)  2Cr3+(aq) + 7H2O(l)
Balance H:
Cr2O72–(aq) + 14H+(aq)  2Cr3+(aq) + 7H2O(l)
Balance charge:
Cr2O72–(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l)
Add two half-reactions, multiplying the tin half-reaction by 3 to equalize the number of electrons transferred.
3{Sn(s)  Sn2+(aq) + 2e–}
Cr2O72–(aq) + 14H+(aq) + 6e–  2 Cr3+(aq) + 7H2O(l)
3Sn(s) + Cr2O72–(aq) + 14H+(aq)  3Sn2+(aq) + 2Cr3+(aq) + 7H2O(l)
Cell notation:
Sn(s)  Sn2+(aq)  H+(aq), Cr2O72–(aq), Cr3+(aq)  C(graphite)
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21-3
21.2B
Plan: Given the solution and electrode compositions, the two half cells involve the transfer of electrons
1) between Cl– and ClO3– and 2) between Ni and Ni2+. The negative electrode is the anode so the nickel half-cell
is where oxidation occurs. The graphite electrode with the chloride ion/chlorate solution is where reduction
occurs. In the cell diagram, show the electrodes and the solutes involved in the half-reactions. Include the salt
bridge and wire connection between electrodes. Set up the two half-reactions and balance. (Note that the
Cl–/ClO3– half-cell is in acidic solution.) Write the cell notation placing the anode half-cell first, then the salt
bridge, then the cathode half-cell.
Solution: Cell diagram:
Voltmeter


e
e
Salt bridge
Ni
()
Ni2+
C
(+)
ClO3,
H+, Cl
Balanced equations:
Anode is Ni/Ni2+ half-cell. Oxidation of Ni produces Ni2+:
Ni(s)  Ni2+(aq)
All that needs to be balanced is charge:
Ni(s)  Ni2+(aq) + 2e–
Cathode is the Cl–/ClO3– half-cell. Check the oxidation number of chlorine in each substance to determine which
is reduced. Cl– oxidation number is –1 and chlorine in ClO3– has oxidation number +5. Going from +5 to –1
involves gain of electrons so ClO3– is reduced.
ClO3–(aq)  Cl– (aq)
Cl is balanced. Balance O:
ClO3–(aq)  Cl– (aq) + 3H2O(l)
Balance H:
6H+(aq) + ClO3–(aq)  Cl– (aq) + 3H2O(l)
Balance charge:
6H+(aq) + ClO3–(aq) + 6e–  Cl– (aq) + 3H2O(l)
Add two half-reactions, multiplying the nickel half-reaction by 3 to equalize the number of electrons transferred.
3{Ni(s)  Ni2+(aq) + 2e–}
6H+(aq) + ClO3–(aq) + 6e–  Cl– (aq) + 3H2O(l)
3Ni(s) + ClO3–(aq) + 6H+(aq)  3Ni2+(aq) + Cl– (aq) + 3H2O(l)
Cell notation:
Ni(s)  Ni2+(aq)  H+(aq), ClO3–(aq), Cl– (aq)  C(graphite)
21.3A



Plan: Use the relationship Ecell
= Ecathode
– Eanode
. E  values are found in Appendix D. Spontaneous reactions

have Ecell
> 0.
Solution:
Oxidation:
Reduction:
Overall reaction:
2{Ag(s)  Ag+(aq) + e–}
Cu2+(aq) + 2e–  Cu(s)
2+
Cu (s) + 2Ag(s)  Cu(s) + 2Ag+(aq)
E°(anode) = 0.80 V
E°(cathode) = 0.34 V

Ecell
= 0.34 V  (0.80 V) = 0.46 V

Reaction is not spontaneous under standard-state conditions because Ecell
is negative.
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21-4
21.3B



Plan: Use the relationship Ecell
= Ecathode
– Eanode
. E  values are found in Appendix D. Spontaneous reactions

have Ecell
> 0.
Solution:
Oxidation:
Reduction:
Overall reaction:
2{Fe2+(aq)  Fe3+(aq) + e–}
Cl2(g) + 2e–  2Cl–(aq)
Cl2(g) + 2Fe2+(aq)  2Cl–+(aq) + 2Fe3+(aq)
E°(anode) = 0.77 V
E°(cathode) = 1.36 V

Ecell
= 1.36 V  (0.77 V) = +0.59 V

Reaction is spontaneous under standard-state conditions because Ecell
is positive.
21.4A
Plan: Divide the reaction into half-reactions showing that Br2 is reduced and V3+ is oxidized. Use the equation




Ecell
= Ecathode
– Eanode
to solve for Eanode
.
Solution:
Half-reactions:

Reduction (cathode): Br2(aq) + 2e–  2 Br–(aq)
Ecathode
= 1.07 V from Appendix D
3+
2+
+
Oxidation (anode): 2V (aq) + 2H2O(l)  2VO (aq) + 4H (aq) + 2e–
Overall:
Br2(aq) + 2V3+(aq) + 2H2O(l)  2VO2+(aq) + 4H+(aq) + 2Br–(aq)

Ecell
= 0.73 V (given)



Ecell
= Ecathode
– Eanode



Eanode
= Ecathode
– Ecell
= 1.07 V – 0.73 V = 0.34 V
21.4B
Plan: Divide the reaction into half-reactions showing that nitrate is reduced and V3+ is oxidized. Use the equation




Ecell
= Ecathode
– Eanode
to solve for Eanode
.
Solution:
Half-reactions:
Reduction (cathode): 4H+(aq) + NO3–(aq) + 3e–  NO(g) + 2H2O(l)
Oxidation (anode): 3{V3+(aq) + H2O(l)  VO2+(aq) + 2H+(aq) + e–}
E°(anode) = 0.34 V
from Follow Up Problem 21.4A
Overall:
NO3–(aq) + 3V3+(aq) + H2O(l)  3VO2+(aq) + 2H+(aq) + NO(g)

Ecell
= 0.62 V (given)



Ecell
= Ecathode
– Eanode



Ecathode
= Ecell
+ Eanode
= 0.62 V + 0.34 V = 0.96 V
21.5A
Plan: To write a spontaneous reaction, examine two reduction half-reactions. The half-reaction with the smaller Eo
is reversed (to become an oxidation). Reducing strength increases with decreasing E°.
Solution:
a) Combining reactions (1) and (2), reaction (1) is reversed because it has the smaller Eo.
Oxidation:
3{2Ag(s) + 2OH–(aq)  Ag2O(s) + H2O(l) + 2e–}
E°(anode) = 0.34 V
E°(cathode) = 0.61 V
Reduction:
BrO3–(g) + 3H2O(l) + 6e–  Br–(aq) + 6OH–(aq)
Overall reaction:
6Ag(s) + BrO3–(g)  3Ag2O(s) + Br–(aq)

Ecell
= 0.61 V  (0.34 V) = +0.27 V
Combining reactions (1) and (3), reaction (3) is reversed because it has the smaller Eo.
Oxidation:
Zn(s) + 2OH–(aq)  Zn(OH)2(s) + 2e–
E°(anode) = –1.25 V
E°(cathode) = 0.34 V
Reduction:
Ag2O(s) + H2O(l) + 2e–  2Ag(s) + 2OH–(aq)
Overall reaction:
Zn(s) + Ag2O(s) + H2O(l)  Zn(OH)2(s) + 2Ag(s)

Ecell
= 0.34 V  (–1.25 V) = +1.59 V
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21-5
Combining reactions (2) and (3), reaction (3) is reversed because it has the smaller Eo.
Oxidation:
3{Zn(s) + 2OH–(aq)  Zn(OH)2(s) + 2e–}
Reduction:
BrO3–(g) + 3H2O(l) + 6e–  Br–(aq) + 6OH–(aq)
Overall reaction:
3Zn(s) + BrO3–(g) + 3H2O(l)  3Zn(OH)2(s) + Br–(aq)
E°(anode) = –1.25 V
E°(cathode) = 0.61 V

Ecell
= 0.61 V  (–1.25 V) = +1.86 V
b) Oxidizing agents are substances that cause another substance to be oxidized. In other words, oxidizing agents
are, themselves, reduced. The more likely a substance is to be reduced (the stronger the oxidizing agent), the
greater the value of its reducing potential, Eo. Examining, the reduction half-reactions given in the problem
statement, we can see that Zn(OH)2 has the lowest reducing potential, Ag2O has the next highest reducing
potential, and BrO3 has the greatest reducing potential. Therefore, we can rank these substances by their
strengths as oxidizing agents: BrO3 > Ag2O > Zn(OH)2.
Reducing agents are substances that cause another substance to be reduced. In other words, reducing agents are,
themselves, oxidized. The more likely a substance is to be oxidized (the stronger the reducing agent), the greater
the value of its oxidizing potential. The oxidizing potentials of substances in this problem can be obtained by
reversing the provided reduction half-reactions and changing the sign of the half-reaction potential. When we do
this, we find that Ag has an oxidizing potential of 0.34 V, Br has an oxidizing potential of 0.61 V, and Zn has
an oxidizing potential of 1.25 V. The more positive the oxidizing potential, the stronger the reducing agent.
Therefore, we can rank these substances by their strengths as reducing agents: Zn > Ag > Br.
21.5B


Plan: To determine if the reaction is spontaneous, divide into half-reactions and calculate Ecell
. If Ecell
is
negative, the reaction is not spontaneous, so reverse the reaction to obtain the spontaneous reaction. Reducing
strength increases with decreasing E°.
Solution: Divide into half-reactions and balance:
Reduction: Fe2+(aq) + 2e–  Fe(s)

Ecathode
= –0.44 V

Oxidation: 2 {Fe2+(aq)  2Fe3+(aq) + e–}
Eanode
= 0.77 V
The first half-reaction is reduction, so it is the cathode half-cell. The second half-reaction is oxidation, so it is the
anode half-cell. Find the half-reactions in Appendix D.



Ecell
= Ecathode
– Eanode
= –0.44 V – 0.77 V = –1.21 V
The reaction is not spontaneous as written, so reverse the reaction:
Fe(s) + 2Fe3+(aq)  3Fe2+(aq)

Ecell
is now +1.21 V, so the reversed reaction is spontaneous under standard state conditions.
When a substance acts as a reducing agent, it is oxidized. Both Fe and Fe2+ can be oxidized, so they can act as
reducing agents. Since Fe3+ cannot lose more electrons, it cannot act as a reducing agent. The stronger reducing
agent between Fe and Fe2+ is the one with the smaller standard reduction potential. E° for Fe is –0.44, which is
less than E° for Fe2+, +0.77. Therefore, Fe is a stronger reducing agent than Fe2+. Ranking all three in order of
decreasing reducing strength gives Fe > Fe2+ > Fe3+.
21.6A

Plan: Find the number of moles of electrons transferred in the balanced equation. Given Ecell
, both K and G°
can be calculated using the relationships log K =

nEcell

and G° = nFEcell
.
0.0592 V
Solution:
The balanced equation is:
2MnO4–(aq) + 4H2O(l) + 3Cu(s)  2MnO2(s) + 8OH–(aq) + 3Cu2+(aq)
In this reaction, 3 moles of copper atoms are oxidized to become 3 moles of copper(II) ions. This process requires
the loss of 3 x 2 = 6 electrons. Therefore, 6 electrons are transferred in this reaction, and n = 6.
log K =

nEcell
=
0.0592 V
.
.
= 25.3378
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21-6
K = 2.1769x1025 = 2.2x1025

G° = nFEcell
= – (6 mol e–)(96,485 C/mol e–)(0.25 J/C) = –1.4473x105 = –1.4x105 J = –1.4x102 kJ
21.6B

can be calculated using the
Plan: Reaction is Cd(s) + Cu2+(aq)  Cd2+(aq) + Cu(s). Given G°, both K and Ecell

relationships G° = –RT ln K and G° = nFEcell
.
Solution:

  103 J 
G
143 kJ

ln K = 
= 
 = 57.71779791
  8.314 J / mol • K    273  25  K    1 kJ 
RT


K = 1.1655238x1025 = 1.2x1025
n = 2 mol e for this reaction

G° = nFEcell

 103 J   C 
143 kJ
G 
= 

 
 = 0.7410478 = 0.741 V


nF
  2 mol  96485 C / mol    1 kJ   J / V 
0.0592 V


log K
is to use Ecell
=
Note that an alternative way to calculate Ecell
n

Ecell
= 
21.7A
Plan: Write a balanced equation for the spontaneous reaction between Cr and Sn (we know the reaction is
spontaneous because we are told the reaction occurs in a voltaic cell). Determine the number of moles of electrons
0.0592


. Use the Nernst equation, Ecell = Ecell
–
log Q to find Ecell.
transferred, and calculate Ecell
n
Solution:

Determining the cell reaction and Ecell
:
2+
The Cr/Cr reduction half-reaction has a smaller Eo. Therefore, it is reversed (and becomes the oxidation halfreaction).
Oxidation:
Cr(s)  Cr2+(aq) + 2e–
E° (anode) = –0.91 V
E° (cathode) = –0.14 V
Reduction:
Sn2+(aq) + 2e–  Sn(s)
Overall
Cr(s) + Sn2+(aq)  Cr2+(aq) + Sn(s)
(2 moles of electrons transferred)



Ecell
= Ecathode
– Eanode
= –0.14 V – (–0.91 V) = 0.77 V
2+
2+
For the reaction Q = [Cr ]/[Sn ], so the Nernst equation is:

Ecell = Ecell

.
log
Substituting in values from the problem:
Ecell = 0.77 V 
.
log
.
.
Ecell = 0.74 V
21.7B

Plan: The problem is asking for the concentration of iron ions when Ecell = Ecell
+ 0.25 V.

Use the Nernst equation, Ecell = Ecell
–
0.0592
log Q to find [Fe2+].
n
Solution:

Determining the cell reaction and Ecell
:
2+
Oxidation:
Fe(s)  Fe (aq) + 2e–
Reduction:
Cu2+(aq) + 2e–  Cu(s)
Overall Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
E° = –0.44 V
E° = 0.34 V



Ecell
= Ecathode
– Eanode
= 0.34 V – (–0.44 V) = 0.78 V
2+
2+
For the reaction Q = [Fe ]/[Cu ], so the Nernst equation is:
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21-7
0.0592
[Fe 2+ ]
log
n
[Cu 2+ ]
Substituting in values from the problem:


Ecell = Ecell

+ 0.25 V = 0.78 V + 0.25 V = 1.03 V
Ecell = Ecell
1.03 V = 0.78 V 
0.25 V = 
0.0592
[Fe 2+ ]
log
n
[Cu 2+ ]
0.0592
[Fe2+ ]
log
2
[0.30]
–8.44595 = log
[Fe2+ ]
[0.30]
[Fe2+ ]
[0.30]
2+
[Fe ] = 1.074423x10–9 = 1.1x10–9 M
3.58141x10–9 =
21.8A
Plan: Half-cell B contains a higher concentration of nickel ions, so the ions will be reduced to decrease the
concentration while in half-cell A, with a lower [Ni2+], nickel metal will be oxidized to increase the concentration
of nickel ions. In the overall cell reaction, the lower [Ni2+] appears as a product and the higher [Ni2+] appears as a

reactant. This means that Q = [Ni2+]lower/[Ni2+]higher. Use the Nernst equation with Ecell
= 0 to find Ecell. Oxidation
of nickel metal occurs at the anode, half-cell A, which is negative.
Solution:
0.0592

Ecell = Ecell
–
log Q
n


Ecell = Ecell
Ecell = 0 
21.8B
.
.
n = 2 mol e– for the reduction of Ni2+ to Ni
log
log
.
.
= 0.04221 = 0.042 V
Plan: Half-cell B contains a higher concentration of gold ions, so the ions will be reduced to decrease the
concentration while in half-cell A, with a lower [Au3+], gold metal will be oxidized to increase the concentration
of gold ions. In the overall cell reaction, the lower [Au3+] appears as a product and the higher [Au3+] appears as a

reactant. This means that Q = [Au3+]lower/[Au3+]higher. Use the Nernst equation with Ecell
= 0 to find Ecell.
Solution:
0.0592

Ecell = Ecell
–
log Q
n


Ecell = Ecell
[Au 3+lower ]
0.0592
log
n
[Au 3+ higher ]
n = 3 mol e– for the reduction of Au3+ to Au
0.0592
[7.0x104 ]
= 0.0306427 = 0.031 V
log
3
[2.5x10 2 ]
Oxidation of gold metal occurs at the anode, half-cell A, which is negative.
Ecell = 0 
21.9A
Plan: In the electrolysis, the more easily reduced metal ion will be reduced at the cathode. Compare Cs+ and Li+ as
to their location on the periodic table to find which has the higher ionization energy (IE increases up and across
the table). Higher ionization energy of the metal means more easily reduced. At the anode, the more easily
oxidized nonmetal ion will be oxidized. For oxidation, compare the electronegativity of the two nonmetals. The
ion from the less electronegative element will be more easily oxidized. Note that the standard reduction potentials
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21-8
cannot be used in the case of molten salts because the E° values are based on aqueous solutions of ions, not liquid
salts.
Solution:
Li is above Cs, so Li+ is more easily reduced than Cs+. The reaction at the cathode is
Li+(l) + e–  Li(s)
F is the most electronegative element, so I– must be more easily oxidized than F–. The reaction at the anode is
2 I– (l)  I2(g) + 2e–
To add the two half-reaction, the number of electrons must be equal:
2 {Li+(l) + e–  Li(s)}
2I– (l)  I2(g) + 2e–
Overall: 2Li+(l) + 2I–(l)  2Li(s) + I2(g)
This is the overall reaction. Li(s) forms at the cathode, and I2(g) forms at the anode.
21.9B
Plan: In the electrolysis, the more easily reduced metal ion will be reduced at the cathode. Compare K+ and Al3+
as to their location on the periodic table to find which has the higher ionization energy (IE increases up and across
the table). Higher ionization energy of the metal means more easily reduced. At the anode, the more easily
oxidized nonmetal ion will be oxidized. For oxidation, compare the electronegativity of the two nonmetals. The
ion from the less electronegative element will be more easily oxidized. Note that the standard reduction potentials
cannot be used in the case of molten salts because the E° values are based on aqueous solutions of ions, not liquid
salts.
Solution:
Al is above and to the right of K, so Al3+ is more easily reduced than K+. The reaction at the cathode is
Al3+(l) + 3e–  Al(s)
F is the most electronegative element, so Br– must be more easily oxidized than F–. The reaction at the anode is
2 Br– (l)  Br2(g) + 2e–
To add the two half-reaction, the number of electrons must be equal:
2 {Al3+(l) + 3e–  Al(s)}
3 {2Br– (l)  Br2(g) + 2e–}
Overall: 2Al3+(l) + 6Br– (l)  2Al(s) + 3Br2(g)
This is the overall reaction. Al(s) forms at the cathode, and Br2(g) forms at the anode.
21.10A Plan: In aqueous Pb(NO3)2, the species present are Pb2+(aq), NO3–(aq), H2O, and very small amounts of H+ and
OH–. The possible half-reactions are reduction of either Pb2+ or H2O and oxidation of either NO3– or H2O.
Whichever reduction and oxidation half-reactions are more spontaneous will take place, with consideration of the
overvoltage.
Solution: The two possible reductions are
Pb2+(aq) + 2e–  Pb(s)
E° = –0.13 V
E = –0.42 V
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
The reduction of lead ions occurs because it has a higher reduction potential (more spontaneous) than reduction of
water.
The two possible oxidations are
NO3–(aq)  no reaction (nitrate compounds are always soluble)
E = 0.82 V
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Water will be oxidized at the anode since no reaction is possible for the nitrate ion.
The two half-reactions that are predicted are
Cathode:
Pb2+(aq) + 2e–  Pb(s)
E° = –0.13 V
E = 0.82 V
Anode:
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Thus, Pb2+ is reduced at the cathode, and H2O is oxidized at the anode. Pb forms at the cathode, and O2 forms
at the anode.
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21-9
21.10B Plan: In aqueous AuBr3, the species present are Au3+(aq), Br–(aq), H2O, and very small amounts of H+ and OH–.
The possible half-reactions are reduction of either Au3+ or H2O and oxidation of either Br– or H2O. Whichever
reduction and oxidation half-reactions are more spontaneous will take place, with consideration of the
overvoltage.
Solution: The two possible reductions are
Au3+(aq) + 3e–  Au(s)
E° = 1.50 V
E = –1 V (with overvoltage)
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
The reduction of gold ions occurs because it has a higher reduction potential (more spontaneous) than reduction of
water.
The two possible oxidations are
2Br–(aq)  Br2(l) + 2e–
E° = 1.07 V
E = 1.4 V (with overvoltage)
2H2O(l)  O2(g) + 4H+(aq) + 4e–
The oxidation with the less positive reduction potential is the more spontaneous oxidation so bromide ions are
oxidized at the anode.
The two half-reactions that are predicted are
Cathode:
Au3+(aq) + 3e–  Au(s)
E° = 1.50 V
E° = 1.07 V
Anode:
2Br–(aq)  Br2(l) + 2e–
Thus, Au3+ is reduced at the cathode, and Br– is oxidized at the anode. Au forms at the cathode, and Br2 forms
at the anode.
21.11A Plan: To find the charge transferred, first write the balanced half-reaction and then calculate the charge from the
grams of copper and moles of electrons transferred per molar mass of copper. Current is charge per time, so to
find the time, divide the charge by the current.
Solution:
Cu2+(aq) + 2e–  Cu(s), so 2 mol e– per mole of Cu
 1 mol Cu   2 mol e  96485 C   A   1  1 min 
Time (min) = 1.50 g Cu  


 



 

 63.55 g Cu   1 mol Cu   1 mol e   C/s   4.75 A  60 s 
= 15.9816 = 16.0 min
21.11B Plan: Convert mass of Zn to moles and use the mole ratio in the balanced reaction to find the number of moles of
electrons required for every mole of Zn produced. The Faraday constant is used to find the charge of the electrons
in coulombs. To find the current (in units of amperes, A, where one ampere equals 1 coulomb per second), the
charge is divided by the time in seconds.
Solution:
 1 mol Zn   2 mol e   96485 C 
3
Charge (C) = 1.25g Zn  

 = 3689.39278 = 3.69x10 C
 
 

 65.38 g Zn   1 mol Zn   1 mol e 
 3.69x103 C   1 min 
Current (A) = 
 
 = 7.0286 = 7.03 A
 8.75 min   60 s 
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B21.1
Plan: Reduction is the gain of electrons while oxidation is the loss of electrons.
Solution:
a) Reduction: Fe3+ + e– → Fe2+
Oxidation: Cu+ → Cu2+ + e–
b) Overall: Fe3+ + Cu+ → Fe2+ + Cu2+
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21-10
B21.2

to calculate the free energy change.
Plan: Use the relationship G° = nFEcell
Solution:

G° (J/mol) = nFEcell
= – (2 mol e–)(96485 C/mol e–)(0.224 J/C) = –4.322528x105 J


 1 kJ 
G° (kJ/mol) = 4.322528x105 J  3  = –43.22528 = –43.2 kJ/mol
 10 J 
END–OF–CHAPTER PROBLEMS
21.1
Oxidation is the loss of electrons (resulting in a higher oxidation number), while reduction is the gain of electrons
(resulting in a lower oxidation number). In an oxidation-reduction reaction, electrons transfer from the oxidized
substance to the reduced substance. The oxidation number of the reactant being oxidized increases while the
oxidation number of the reactant being reduced decreases.
21.2
An electrochemical process involves electron flow. At least one substance must lose electron(s) and one
substance must gain electron(s) to produce the flow. This electron transfer is a redox process.
21.3
No, one half-reaction cannot take place independently of the other because there is always a transfer of electrons
from one substance to another. If one substance loses electrons (oxidation half-reaction), another substance must
gain those electrons (reduction half-reaction).
21.4
O2 is too strong a base to exist in H 2O. The reaction O2 + H 2O  2OH  occurs. Only species actually existing
in solution can be present when balancing an equation.
21.5
Multiply each half-reaction by the appropriate integer to make e lost equal to e gained.
21.6
To remove protons from an equation, add an equal number of hydroxide ions to both sides to neutralize the H+
and produce water: H+(aq) + OH(aq)  H2O(l).
21.7
No, spectator ions are not used to balance the half-reactions. Add spectator ions to the balanced ionic equation to
obtain the balanced molecular equation.
21.8
Spontaneous reactions, Gsys < 0, take place in voltaic cells, which are also called galvanic cells.
Nonspontaneous reactions take place in electrolytic cells and result in an increase in the free energy of the cell
(Gsys > 0).
21.9
a) True
b) True
c) True
d) False, in a voltaic cell, the system does work on the surroundings.
e) True
f) False, the electrolyte in a cell provides a solution of mobile ions to maintain charge neutrality.
21.10
Plan: Assign oxidation numbers; the species with an atom whose oxidation number has increased is being
oxidized and is the reducing agent. The species with an atom whose oxidation number has decreased is being
reduced and is the oxidizing agent. Electrons flow from the reducing agent to the oxidizing agent. To write the
molecular equation, pair K+ ions with anions and SO42– with cations in the equation to write neutral molecules.
Solution:
–8
+2
+1
+7 –2
–1
+2
0
+1 –2
16H+(aq) + 2MnO4(aq) + 10Cl(aq) → 2Mn2+(aq) + 5Cl2(g) + 8H2O(l)
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21-11
a) To decide which reactant is oxidized, look at oxidation numbers. Cl is oxidized because its oxidation number
increases from –1 in Cl to 0 in Cl2.
b) MnO4 is reduced because the oxidation number of Mn decreases from +7 in MnO4 to +2 in Mn2+.
c) The oxidizing agent is the substance that causes the oxidation by accepting electrons. The oxidizing agent is the
substance reduced in the reaction, so MnO4 is the oxidizing agent.
d) Cl is the reducing agent because it loses the electrons that are gained in the reduction.
e) From Cl, which is losing electrons, to MnO4, which is gaining electrons.
f) 8H2SO4(aq) + 2KMnO4(aq) + 10KCl(aq)  2MnSO4(aq) + 5Cl2(g) + 8H2O(l) + 6K2SO4(aq)
21.11
2CrO2(aq) + 2H 2O(l) + 6ClO(aq)  2CrO42(aq) + 3Cl2(g) + 4OH (aq)
a) The CrO2 is the oxidized species because Cr increases in oxidation state from +3 to +6.
b) The ClO is the reduced species because Cl decreases in oxidation state from +1 to 0.
c) The oxidizing agent is ClO; the oxidizing agent is the substance reduced.
d) The reducing agent is CrO2; the reducing agent is the substance oxidized.
e) Electrons transfer from CrO2 to ClO.
f) 2NaCrO2(aq) + 6NaClO(aq) + 2H 2O(l) 2Na2CrO4(aq) + 3Cl2(g) + 4NaOH (aq)
21.12
Plan: Divide the reaction into the two half-reactions, balance elements other than oxygen and hydrogen, and
then balance oxygen by adding H2O and hydrogen by adding H+. Balance the charge by adding electrons and
multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons
gained. Add the half-reactions together, canceling substances that appear on both sides. For basic solutions, add
one OH– ion to each side of the equation for every H+ ion present to form H2O and cancel excess H2O
molecules. The substance that gains electrons is the oxidizing agent while the substance that loses electrons is the
reducing agent.
Solution:
a) Divide into half-reactions:
ClO3(aq)  Cl(aq)
I(aq)  I2(s)
Balance elements other than O and H:
chlorine is balanced
ClO3(aq)  Cl(aq)
2I(aq)  I2(s)
iodine now balanced
Balance O by adding H2O:
add three waters to add three O atoms to product
ClO3(aq)  Cl(aq) + 3H2O(l)
2I(aq)  I2(s)
no change
Balance H by adding H+:
add six H+ to reactants
ClO3(aq) + 6H+(aq)  Cl(aq) + 3H2O(l)

2I (aq)  I2(s)
no change
Balance charge by adding e–:
add 6e– to reactants for a –1 charge on each side
ClO3(aq) + 6H+(aq) + 6e–  Cl(aq) + 3 H2O(l)

–
2I (aq)  I2(s) + 2e
add 2e– to products for a –2 charge on each side
Multiply each half-reaction by an integer to equalize the number of electrons:
multiply by one to give 6e–
ClO3(aq) + 6H+(aq) + 6e–  Cl(aq) + 3H2O(l)

–
3{2I (aq)  I2(s) + 2e } or
multiply by three to give 6e–

–
6I (aq) 3I2(s) + 6e
Add half-reactions to give balanced equation in acidic solution:
ClO3(aq) + 6H+(aq) + 6I(aq)  Cl(aq) + 3H2O(l) + 3I2(s)
Check balancing:
Reactants:
1 Cl
Products:
1 Cl
3O
3O
6H
6H
6I
6I
–1 charge
–1 charge
Oxidizing agent is ClO3 and reducing agent is I.
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21-12
b) Divide into half-reactions:
MnO4(aq)  MnO2(s)
SO32(aq)  SO42(aq)
Balance elements other than O and H:
MnO4(aq)  MnO2(s)
Mn is balanced
S is balanced
SO32(aq)  SO42(aq)
Balance O by adding H2O:
MnO4(aq)  MnO2(s) + 2H2O(l)
add two H2O to products
add one H2O to reactants
SO32(aq) + H2O(l)  SO42(aq)
Balance H by adding H+:
MnO4(aq) + 4H+(aq)  MnO2(s) + 2H2O(l)
add four H+ to reactants
+
2
2
add two H+ to products
SO3 (aq) + H2O(l)  SO4 (aq) + 2H (aq)
–
Balance charge by adding e :
MnO4(aq) + 4H+(aq) + 3e–  MnO2(s) + 2H2O(l) add 3e– to reactants for a 0 charge on each side
add 2e– to products for a –2 charge on each side
SO32(aq) + H2O(l)  SO42(aq) + 2H+(aq) + 2e–
Multiply each half-reaction by an integer to equalize the number of electrons:
multiply by two to give 6e–
2{MnO4(aq) + 4H+(aq) + 3e– MnO2(s) + 2H2O(l)} or
+
–

2MnO4 (aq) + 8H (aq) + 6e  2MnO2(s) + 4H2O(l)
multiply by three to give 6e–
3{SO32(aq) + H2O(l)  SO42(aq) + 2H+(aq) + 2e–} or
+
–
2
2
3SO3 (aq) + 3H2O(l)  3SO4 (aq) + 6H (aq) + 6e
Add half-reactions and cancel substances that appear as both reactants and products:
2 MnO4(aq) + 8H+(aq) + 3SO32(aq) + 3H2O(l)  2MnO2(s) + 4H2O(l) + 3SO42(aq) + 6H+(aq)
The balanced equation in acidic solution is:
2MnO4(aq) + 2H+(aq) + 3SO32(aq)  2MnO2(s) + H2O(l) + 3SO42(aq)
To change to basic solution, add OH to both sides of equation to neutralize H+
2MnO4(aq) + 2H+(aq) + 2OH(aq) + 3SO32(aq)  2MnO2(s) + H2O(l) + 3SO42(aq) + 2OH(aq)
2MnO4(aq) + 2H2O( l) + 3SO32(aq)  2MnO2(s) + H2O(l) + 3SO42(aq) + 2OH(aq)
Balanced equation in basic solution:
2MnO4(aq) + H2O(l) + 3SO32(aq)  2MnO2(s) + 3SO42(aq) + 2OH(aq)
Check balancing:
Reactants:
2 Mn
Products:
2 Mn
18 O
18 O
2H
2H
3S
3S
–8 charge
–8 charge
Oxidizing agent is MnO4 and reducing agent is SO32.
c) Divide into half-reactions:
MnO4(aq)  Mn2+(aq)
H2O2(aq)  O2(g)
Balance elements other than O and H:
Mn is balanced
MnO4(aq)  Mn2+(aq)
No other elements to balance
H2O2(aq)  O2(g)
Balance O by adding H2O:
add four H2O to products
MnO4(aq)  Mn2+(aq) + 4H2O(l)
O is balanced
H2O2(aq)  O2(g)
Balance H by adding H+:
add eight H+ to reactants
MnO4(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l)
+
add two H+ to products
H2O2(aq)  O2(g) + 2H (aq)
–
Balance charge by adding e :
add 5e– to reactants for +2 on each side
MnO4(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(l)
+
–
add 2e– to products for 0 charge on each side
H2O2(aq)  O2(g) + 2H (aq) + 2e
Multiply each half-reaction by an integer to equalize the number of electrons:
multiply by two to give 10e–
2{MnO4(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(l)} or
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21-13
21.13
21.14
2MnO4(aq) + 16H+(aq) + 10e– 2Mn2+(aq) + 8H2O(l)
or
multiply by five to give 10e–
5{H2O2(aq)  O2(g) + 2H+(aq) + 2e–}
+
–
5H2O2(aq)  5O2(g) + 10H (aq) + 10e
Add half-reactions and cancel substances that appear as both reactants and products:
2MnO4(aq) + 16H+(aq) + 5H2O2(aq)  2Mn2+(aq) + 8H2O(l) + 5O2(g) + 10H+(aq)
The balanced equation in acidic solution:
2MnO4(aq) + 6H+(aq) + 5H2O2(aq)  2Mn2+(aq) + 8H2O(l) + 5O2(g)
Check balancing:
Reactants:
2 Mn
Products:
2 Mn
18 O
18 O
16 H
16 H
+4 charge
+4 charge
Oxidizing agent is MnO4– and reducing agent is H2O2.
a) 3O2(g) + 4NO(g) + 2H 2O(l)  4NO3–(aq) + 4H +(aq)
Oxidizing agent is O2 and reducing agent is NO.
b) 2CrO42–(aq) + 8H 2O(l) + 3Cu(s)  2Cr(OH )3(s) + 3Cu(OH)2(s) + 4OH –(aq)
Oxidizing agent is CrO42– and reducing agent is Cu.
3–
c) AsO4 (aq) + NO2–(aq) + H 2O(l)  AsO2–(aq) + NO3–(aq) + 2OH –(aq)
Oxidizing agent is AsO43– and reducing agent is NO2–.
Plan: Divide the reaction into the two half-reactions, balance elements other than oxygen and hydrogen, and
then balance oxygen by adding H2O and hydrogen by adding H+. Balance the charge by adding electrons and
multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons
gained. Add the half-reactions together, canceling substances that appear on both sides. For basic solutions, add
one OH– ion to each side of the equation for every H+ ion present to form H2O and cancel excess H2O
molecules. The substance that gains electrons is the oxidizing agent while the substance that loses electrons is the
reducing agent.
Solution:
a) Balance the reduction half-reaction:
Cr2O72(aq)  2Cr3+(aq)
balance Cr
balance O by adding H2O
Cr2O72(aq)  2Cr3+(aq) + 7H2O(l)
balance H by adding H+
Cr2O72(aq) + 14H+(aq)  2Cr3+(aq) + 7H2O(l)
+
–
3+
2
Cr2O7 (aq) + 14H (aq) + 6e  2Cr (aq) + 7H2O(l)
balance charge by adding 6e–
Balance the oxidation half-reaction:
Zn(s)  Zn2+(aq) + 2e–
balance charge by adding 2e–
Add the two half-reactions multiplying the oxidation half-reaction by three to equalize the electrons:
Cr2O72(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l)
3Zn(s)  3Zn2+(aq) + 6e–
Add half-reactions and cancel substances that appear as both reactants and products:
Cr2O72(aq) + 14H+(aq) + 3Zn(s)  2Cr3+(aq) + 7H2O(l) + 3Zn2+(aq)
Oxidizing agent is Cr2O72 and reducing agent is Zn.
b) Balance the reduction half-reaction:
balance O by adding H2O
MnO4(aq)  MnO2(s) + 2H2O(l)
MnO4(aq) + 4H+(aq)  MnO2(s) + 2H2O(l)
balance H by adding H+
+
–

MnO4 (aq) + 4H (aq) + 3e  MnO2(s) + 2H2O(l)
balance charge by adding 3 e–
Balance the oxidation half-reaction:
Fe(OH)2(s) + H2O(l)  Fe(OH)3(s)
balance O by adding H2O
balance H by adding H+
Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H+(aq)
+

balance charge by adding 1e–
Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H (aq) + e
Add half-reactions after multiplying oxidation half-reaction by 3:
MnO4(aq) + 4H+(aq) + 3e–  MnO2(s) + 2H2O(l)
3Fe(OH)2(s) + 3H2O(l)  3Fe(OH)3(s) + 3H+(aq) + 3e
Add half-reactions and cancel substances that appear as both reactants and products:
MnO4(aq) + 4H+(aq) + 3Fe(OH)2(s) + 3H2O(l)  MnO2(s) + 2H2O(l) + 3Fe(OH)3(s) + 3H+(aq)
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21-14
MnO4(aq) + H+(aq) + 3Fe(OH)2(s) + H2O(l)  MnO2(s) + 3Fe(OH)3(s)
Add OH to both sides to neutralize the H+ and convert H+ + OH–  H2O:
MnO4(aq) + H+(aq) + OH–(aq) + 3Fe(OH)2(s) + H2O(l)  MnO2(s) + 3Fe(OH)3(s) + OH–(aq)
MnO4(aq) + 3Fe(OH)2(s) + 2H2O(l)  MnO2(s) + 3Fe(OH)3(s) + OH(aq)
Oxidizing agent is MnO4 and reducing agent is Fe(OH)2.
c) Balance the reduction half-reaction:
balance N
2NO3(aq)  N2(g)
balance O by adding H2O
2NO3(aq)  N2(g) + 6H2O(l)
balance H by adding H+
2NO3(aq) + 12H+(aq)  N2(g) + 6H2O(l)
+


balance charge by adding 10e
2NO3 (aq) + 12H (aq) + 10e  N2(g) + 6H2O(l)
Balance the oxidation half-reaction:
Zn(s)  Zn2+(aq) + 2e
balance charge by adding 2e
–
Add the half-reactions after multiplying the reduction half-reaction by one and the oxidation half-reaction by five:
2NO3(aq) + 12H+(aq) + 10e  N2(g) + 6H2O(l)
5Zn(s)  5Zn2+(aq) + 10e
Add half-reactions and cancel substances that appear as both reactants and products:
2NO3(aq) + 12H+(aq) + 5Zn(s)  N2(g) + 6H2O(l) + 5Zn2+(aq)
Oxidizing agent is NO3 and reducing agent is Zn.
21.15
a) 3BH 4–(aq) + 4ClO3–(aq)  3H 2BO3–(aq) + 4Cl–(aq) + 3H 2O(l)
Oxidizing agent is ClO3– and reducing agent is BH 4–.
b) 2CrO42–(aq) + 3N2O(g) + 10H +(aq)  2Cr3+(aq) + 6NO(g) + 5H 2O(l)
Oxidizing agent is CrO42– and reducing agent is N2O.
c) 3Br2(l) + 6OH –(aq)  BrO3–(aq) + 5Br–(aq) + 3H 2O(l)
Oxidizing agent is Br2 and reducing agent is Br2.
21.16
Plan: Divide the reaction into the two half-reactions, balance elements other than oxygen and hydrogen, and
then balance oxygen by adding H2O and hydrogen by adding H+. Balance the charge by adding electrons and
multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons
gained. Add the half-reactions together, canceling substances that appear on both sides. For basic solutions, add
one OH– ion to each side of the equation for every H+ ion present to form H2O and cancel excess H2O
molecules. The substance that gains electrons is the oxidizing agent while the substance that loses electrons is the
reducing agent.
Solution:
a) Balance the reduction half-reaction:
balance O by adding H2O
NO3(aq)  NO(g) + 2H2O(l)
NO3(aq) + 4H+(aq)  NO(g) + 2H2O(l)
balance H by adding H+
NO3(aq) + 4H+(aq) + 3e–  NO(g) + 2H2O(l)
balance charge by adding 3e–
Balance oxidation half-reaction:
4Sb(s)  Sb4O6(s)
balance Sb
balance O by adding H2O
4Sb(s) + 6H2O(l)  Sb4O6(s)
balance H by adding H+
4Sb(s) + 6H2O(l)  Sb4O6(s) + 12H+(aq)
+
–
4Sb(s) + 6H2O(l)  Sb4O6(s) + 12H (aq) + 12e
balance charge by adding 12e–
Multiply each half-reaction by an integer to equalize the number of electrons:
4{NO3(aq) + 4H+(aq) + 3e–  NO(g) + 2H2O(l)}
multiply by four to give 12e–
+
–
multiply by one to give 12e–
1{4Sb(s) + 6H2O(l)  Sb4O6(s) + 12H (aq) + 12e }
This gives:
4NO3(aq) + 16H+(aq) + 12e–  4NO(g) + 8H2O(l)
4Sb(s) + 6H2O(l)  Sb4O6(s) + 12H+(aq) + 12e–
Add half-reactions. Cancel common reactants and products:
4 NO3(aq) + 16H+(aq) + 4Sb(s) + 6H2O(l)  4NO(g) + 8H2O(l) + Sb4O6(s) + 12H+(aq)
Balanced equation in acidic solution:
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21-15
4NO3(aq) + 4H+(aq) + 4Sb(s)  4NO(g) + 2H2O(l) + Sb4O6(s)
Oxidizing agent is NO3– and reducing agent is Sb.
b) Balance reduction half-reaction:
balance O by adding H2O
BiO3(aq)  Bi3+(aq) + 3H2O(l)
BiO3(aq) + 6H+(aq)  Bi3+(aq) + 3H2O(l)
balance H by adding H+
+
–
3+

balance charge to give +3 on each side
BiO3 (aq) + 6H (aq) + 2e  Bi (aq) + 3H2O(l)
Balance oxidation half-reaction:
Mn2+(aq) + 4H2O(l)  MnO4(aq)
balance O by adding H2O
balance H by adding H+
Mn2+(aq) + 4H2O(l)  MnO4(aq) + 8H+(aq)
2+
+
–

balance charge to give +2 on each side
Mn (aq) + 4H2O(l)  MnO4 (aq) + 8H (aq) + 5e
Multiply each half-reaction by an integer to equalize the number of electrons:
5{BiO3(aq) + 6H+(aq) + 2e–  Bi3+(aq) + 3H2O(l)}
multiply by five to give 10e–
2+
+
–

2{Mn (aq) + 4H2O(l)  MnO4 (aq) + 8H (aq) + 5e }
multiply by two to give 10e–
This gives:
5BiO3(aq) + 30H+(aq) + 10e–  5Bi3+(aq) + 15H2O(l)
2Mn2+(aq) + 8H2O(l)  2MnO4(aq) + 16H+(aq) + 10e–
Add half-reactions. Cancel H2O and H+ in reactants and products:
5BiO3(aq) + 30H+(aq) + 2Mn2+(aq) + 8H2O(l)  5Bi3+(aq) + 15H2O(l) + 2MnO4(aq) + 16H+(aq)
Balanced reaction in acidic solution:
5BiO3(aq) + 14H+(aq) + 2Mn2+(aq)  5Bi3+(aq) + 7H2O(l) + 2MnO4(aq)
BiO3– is the oxidizing agent and Mn2+ is the reducing agent.
c) Balance the reduction half-reaction:
Pb(OH)3(aq)  Pb(s) + 3H2O(l)
balance O by adding H2O
balance H by adding H+
Pb(OH)3(aq) + 3H+(aq)  Pb(s) + 3H2O(l)
balance charge to give 0 on each side
Pb(OH)3(aq) + 3H+(aq) + 2e–  Pb(s) + 3H2O(l)
Balance the oxidation half-reaction:
Fe(OH)2(s) + H2O(l)  Fe(OH)3(s)
balance O by adding H2O
balance H by adding H+
Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H+(aq)
+
–
Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H (aq) + e
balance charge to give 0 on each side
Multiply each half-reaction by an integer to equalize the number of electrons:
multiply by 1 to give 2e–
1{Pb(OH)3(aq) + 3H+(aq) + 2e–  Pb(s) + 3H2O(l)}
+
–
2{Fe(OH)2(s) + H2O(l)  Fe(OH)3(s) + H (aq) + e }
multiply by 2 to give 2e–
This gives:
Pb(OH)3(aq) + 3H+(aq) + 2e–  Pb(s) + 3H2O(l)
2Fe(OH)2(s) + 2H2O(l)  2Fe(OH)3(s) + 2H+(aq) + 2e–
Add the two half-reactions. Cancel H2O and H+:
Pb(OH)3(aq) + 3H+(aq) + 2Fe(OH)2(s) + 2H2O(l)  Pb(s) + 3H2O(l) + 2Fe(OH)3(s) + 2H+(aq)
Pb(OH)3(aq) + H+(aq) + 2Fe(OH)2(s)  Pb(s) + H2O(l) + 2Fe(OH)3(s)
Add one OH– to both sides to neutralize H+:
Pb(OH)3(aq) + H+(aq) + OH(aq) + 2Fe(OH)2(s)  Pb(s) + H2O(l) + 2 Fe(OH)3(s) + OH(aq)
Pb(OH)3(aq) + H2O(l) + 2Fe(OH)2(s)  Pb(s) + H2O(l) + 2Fe(OH)3(s) + OH(aq)
Balanced reaction in basic solution:
Pb(OH)3(aq) + 2Fe(OH)2(s)  Pb(s) + 2Fe(OH)3(s) + OH(aq)
Pb(OH)3 is the oxidizing agent and Fe(OH)2 is the reducing agent.
21.17
a) 2NO2(g) + 2OH (aq)  NO3(aq) + NO2(aq) + H 2O(l)
Oxidizing agent is NO2 and reducing agent is NO2.
b) 4Zn(s) + 7OH (aq) + NO3(aq) + 6H 2O(l)  4Zn(OH )42(aq) + NH 3(aq)
Oxidizing agent is NO3 and reducing agent is Zn.
c) 24H 2S(g) + 16NO3(aq) + 16H +(aq)  3S8(s) + 16NO(g) + 32H 2O(l)
Oxidizing agent is NO3 and reducing agent is H 2S.
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21-16
21.18
Plan: Divide the reaction into the two half-reactions, balance elements other than oxygen and hydrogen, and
then balance oxygen by adding H2O and hydrogen by adding H+. Balance the charge by adding electrons and
multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons
gained. Add the half-reactions together, canceling substances that appear on both sides. For basic solutions, add
one OH– ion to each side of the equation for every H+ ion present to form H2O and cancel excess H2O
molecules. The substance that gains electrons is the oxidizing agent while the substance that loses electrons is the
reducing agent.
Solution:
a) Balance reduction half-reaction:
MnO4(aq)  Mn2+(aq) + 4H2O(l)
balance O by adding H2O
balance H by adding H+
MnO4(aq) + 8H+(aq)  Mn2+(aq) + 4H2O(l)
+

2+

balance charge by adding 5e
MnO4 (aq) + 8H (aq) + 5e  Mn (aq) + 4H2O(l)
Balance oxidation half-reaction:
As4O6(s)  4AsO43(aq)
balance As
balance O by adding H2O
As4O6(s) + 10H2O(l)  4AsO43(aq)
balance H by adding H+
As4O6(s) + 10H2O(l)  4AsO43(aq) + 20H+(aq)
+

3
balance charge by adding 8e
As4O6(s) + 10H2O(l)  4AsO4 (aq) + 20H (aq) + 8e
Multiply reduction half-reaction by 8 and oxidation half-reaction by 5 to transfer 40 e in overall reaction.
8MnO4(aq) + 64H+(aq) + 40e  8Mn2+(aq) + 32H2O(l)
5As4O6(s) + 50H2O(l)  20AsO43(aq) + 100H+(aq) + 40e
Add the half-reactions and cancel H2O and H+:
5As4O6(s) + 8MnO4(aq) + 64H+(aq) + 50H2O(l)  20AsO43(aq) + 8Mn2+(aq) + 32H2O(l) + 100H+(aq)
Balanced reaction in acidic solution:
5As4O6(s) + 8MnO4(aq) + 18H2O(l)  20AsO43(aq) + 8Mn2+(aq) + 36H+(aq)
Oxidizing agent is MnO4 and reducing agent is As4O6.
b) The reaction gives only one reactant, P4. Since both products contain phosphorus, divide the half-reactions so
each includes P4 as the reactant.
Balance reduction half-reaction:
P4(s)  4PH3(g)
balance P
balance H by adding H+
P4(s) + 12H+(aq)  4PH3(g)
P4(s) + 12H+(aq) + 12e  4PH3(g)
balance charge by adding 12 e
Balance oxidation half-reaction:
balance P
P4(s)  4HPO32(aq)
P4(s) + 12H2O(l)  4HPO32(aq)
balance O by adding H2O
balance H by adding H+
P4(s) + 12H2O(l)  4HPO32(aq) + 20H+(aq)
balance charge by adding 12 e
P4(s) + 12H2O(l)  4HPO32(aq) + 20H+(aq) + 12e
+
Add two half-reactions and cancel H :
2P4(s) + 12H+(aq) + 12H2O(l)  4HPO32(aq) + 4PH3(g) + 20H+(aq)
Balanced reaction in acidic solution:
2P4(s) + 12H2O(l)  4HPO32(aq) + 4PH3(g) + 8H+(aq) or
P4(s) + 6H2O(l)  2HPO32(aq) + 2PH3(g) + 4H+(aq)
P4 is both the oxidizing agent and reducing agent.
c) Balance the reduction half-reaction:
balance O by adding H2O
MnO4(aq)  MnO2(s) + 2H2O(l)
balance H by adding H+
MnO4(aq) + 4H+(aq)  MnO2(s) + 2H2O(l)
+


MnO4 (aq) + 4H (aq) + 3e  MnO2(s) + 2H2O(l)
balance charge by adding 3e
Balance oxidation half-reaction:
balance O by adding H2O
CN(aq) + H2O(l)  CNO(aq)
CN(aq) + H2O(l)  CNO(aq)+ 2H+(aq)
balance H by adding H+
balance charge by adding 2e
CN(aq) + H2O(l)  CNO(aq)+ 2H+(aq) + 2e
Multiply the oxidation half-reaction by three and reduction half-reaction by two to transfer 6e in overall reaction.
2MnO4(aq) + 8H+(aq) + 6e  2MnO2(s) + 4H2O(l)
3CN(aq) + 3H2O(l)  3CNO(aq)+ 6H+(aq) + 6e
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21-17
Add the two half-reactions. Cancel the H2O and H+:
2MnO4(aq) + 3CN(aq) + 8H+(aq) + 3H2O(l)  2MnO2(s) + 3CNO (aq) + 6H+(aq) + 4H2O(l)
2MnO4(aq) + 3CN(aq) + 2H+(aq)  2MnO2(s) + 3CNO (aq) + H2O(l)
Add 2 OH to both sides to neutralize H+ and form H2O:
2MnO4(aq) + 3CN(aq) + 2H+(aq) + 2OH(aq) 2MnO2(s) + 3CNO(aq) + H2O(l) + 2OH(aq)
2MnO4(aq) + 3CN(aq) + 2H2O(l)  2MnO2(s) + 3CNO(aq) + H2O(l) + 2OH(aq)
Balanced reaction in basic solution:
2MnO4(aq) + 3CN(aq) + H2O(l)  2MnO2(s) + 3CNO(aq) + 2OH(aq)
Oxidizing agent is MnO4 and reducing agent is CN-.
21.19
a) SO32(aq) + 2OH (aq) + Cl2(g)  SO42(aq) + 2Cl(aq) + H 2O(l)
Oxidizing agent is Cl2 and reducing agent is SO32.
3
b) 7Fe(CN)6 (aq) + Re(s) + 8OH (aq) + 7e  7Fe(CN)64(aq) + ReO4(aq) + 4H 2O(l) + 7e
Oxidizing agent is Fe(CN)63 and reducing agent is Re.
c) 2MnO4(aq) + 5H COOH(aq) + 6H +(aq)  2Mn2+(aq) + 5CO2(g) + 8H 2O(l)
Oxidizing agent is MnO4 and reducing agent is H COOH.
21.20
5Fe2+(aq) + MnO4–(aq) + 8H +(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H 2O(l)
21.21
a) Balance reduction half-reaction:
NO3–(aq)  NO2(g) + H2O(l)
balance O by adding H2O
balance H by adding H+
NO3–(aq) + 2H+(aq)  NO2(g) + H2O(l)
balance charge to give 0 on each side
NO3–(aq) + 2H+(aq) + e–  NO2(g) + H2O(l)
Balance oxidation half-reaction:
balance Cl
Au(s) + 4Cl–(aq)  AuCl4–(aq)
balance charge to –4 on each side
Au(s) + 4Cl–(aq)  AuCl4–(aq) + 3e–
Multiply each half-reaction by an integer to equalize the number of electrons:
multiply by three to give 3e–
3{NO3–(aq) + 2H+(aq) + e–  NO2(g) + H2O(l)}
–
–
–
1{ Au(s) + 4Cl (aq)  AuCl4 (aq) + 3e }
multiply by one to give 3e–
This gives:
3NO3–(aq) + 6H+(aq) + 3e–  3NO2(g) + 3H2O(l)
Au(s) + 4Cl–(aq)  AuCl4–(aq) + 3e–
Add half-reactions:
Au(s) + 3NO3–(aq) + 4Cl–(aq) + 6H+(aq)  AuCl4–(aq) + 3NO2(g) + 3H2O(l)
b) Oxidizing agent is NO3– and reducing agent is Au.
c) The HCl provides chloride ions that combine with the unstable gold ion to form the stable ion, AuCl4–.
21.22
Plan: The oxidation half-cell (anode) is shown on the left while the reduction half-cell (cathode) is shown on the
right. Remember that oxidation is the loss of electrons and electrons leave the oxidation half-cell and move
towards the positively charged cathode. If a metal is reduced, it will plate out on the cathode.
Solution:
a) A is the anode because by convention the anode is shown on the left.
b) E is the cathode because by convention the cathode is shown on the right.
c) C is the salt bridge providing electrical connection between the two solutions.
d) A is the anode, so oxidation takes place there. Oxidation is the loss of electrons, meaning that electrons are
leaving the anode.
e) E is assigned a positive charge because it is the cathode.
f) E gains mass because the reduction of the metal ion produces the solid metal which plates out on E.
21.23
Unless the oxidizing and reducing agents are physically separated, the redox reaction will not generate electrical
energy. This electrical energy is produced by forcing the electrons to travel through an external circuit.
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21-18
21.24
The purpose of the salt bridge is to maintain charge neutrality by allowing anions to flow into the anode
compartment and cations to flow into the cathode compartment.
21.25
An active electrode is a reactant or product in the cell reaction, whereas an inactive electrode is neither a reactant
nor a product. An inactive electrode is present only to conduct electricity when the half-cell reaction does not
include a metal. Platinum and graphite are commonly used as inactive electrodes.
21.26
a) The metal A is being oxidized to form the metal cation. To form positive ions, an atom must always lose
electrons, so this half-reaction is always an oxidation.
b) The metal ion B is gaining electrons to form the metal B, so it is displaced.
c) The anode is the electrode at which oxidation takes place, so metal A is used as the anode.
d) Acid oxidizes metal B and metal B oxidizes metal A, so acid will oxidize metal A and bubbles will form when
metal A is placed in acid. The same answer results if strength of reducing agents is considered. The fact that metal
A is a better reducing agent than metal B indicates that if metal B reduces acid, then metal A will also reduce acid.
21.27
Plan: The oxidation half-cell (anode) is shown on the left while the reduction half-cell (cathode) is shown on the
right. Remember that oxidation is the loss of electrons and electrons leave the oxidation half-cell and move
towards the positively charged cathode. Anions from the salt bridge flow into the oxidation half-cell, while
cations from the salt bridge flow into the reduction half-cell.
Solution:
a) Electrons flow from the anode to the cathode, so from the iron half-cell to the nickel half-cell, left to right in
the figure. By convention, the anode appears on the left and the cathode on the right.
b) Oxidation occurs at the anode, which is the electrode in the iron half-cell.
c) Electrons enter the reduction half-cell, the nickel half-cell in this example.
d) Electrons are consumed in the reduction half-reaction. Reduction takes place at the cathode, nickel electrode.
e) The anode is assigned a negative charge, so the iron electrode is negatively charged.
f) Metal is oxidized in the oxidation half-cell, so the iron electrode will decrease in mass.
g) The solution must contain nickel ions, so any nickel salt can be added. 1 M NiSO4 is one choice.
h) KNO3 is commonly used in salt bridges, the ions being K+ and NO3–. Other salts are also acceptable answers.
i) Neither, because an inactive electrode could not replace either electrode since both the oxidation and the
reduction half-reactions include the metal as either a reactant or a product.
j) Anions will move towards the half-cell in which positive ions are being produced. The oxidation half-cell
produces Fe2+, so salt bridge anions move from right (nickel half-cell) to left (iron half-cell).
k)
Oxidation half-reaction:
Fe(s)  Fe2+(aq) + 2e–
Reduction half-reaction:
Ni2+(aq) + 2e–  Ni(s)
Overall cell reaction:
Fe(s) + Ni2+(aq)  Fe2+(aq) + Ni(s)
21.28
a) The electrons flow left to right.
b) Reduction occurs at the electrode on the right.
c) Electrons leave the cell from the left side.
d) The zinc electrode generates the electrons.
e) The cobalt electron has the positive charge.
f) The cobalt electrode increases in mass.
g) The anode electrolyte could be 1 M Zn(NO3)2.
h) One possible pair would be K+ and NO3–.
i) Neither electrode could be replaced because both electrodes are part of the cell reaction.
j) The cations move from left to right to maintain charge neutrality.
k)
Reduction: Co2+(aq) + 2e–  Co(s)
Oxidation: Zn(s)  Zn2+(aq) + 2e–
Overall: Zn(s) + Co2+(aq)  Co(s) + Zn2+(aq)
21.29
Plan: The anode, at which the oxidation takes place, is the negative electrode. Electrons flow from the anode to
the cathode. Anions from the salt bridge flow into the oxidation half-cell, while cations from the salt bridge flow
into the reduction half-cell.
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21-19
Solution:
a) If the zinc electrode is negative, it is the anode and oxidation takes place at the zinc electrode:
Zn(s)  Zn2+(aq) + 2e–
Reduction half-reaction: Sn2+(aq) + 2e–  Sn(s)
Overall reaction: Zn(s) + Sn2+(aq)  Zn2+(aq) + Sn(s)
b)
Voltmeter
e
e
Salt bridge
Zn
Sn
()
(+)
1M
Zn2+
21.30
a)
(red half-rxn)
(ox half-rxn)
(overall rxn)
Anion
flow
Cation
flow
1M
Sn2+
Ag+(aq) + e–  Ag(s)
Pb(s)  Pb2+(aq) + 2e–
2Ag+(aq) + Pb(s)  2Ag(s) + Pb2+(aq)
b)
e
Voltmeter
e
Salt bridge
Pb
Ag
()
1M
Pb2+
21.31
(+)
Anion
flow
Cation
flow
1M
Ag+
Plan: The cathode, at which the reduction takes place, is the positive electrode. Electrons flow from the anode to
the cathode. Anions from the salt bridge flow into the oxidation half-cell, while cations from the salt bridge flow
into the reduction half-cell.
Solution:
a) The cathode is assigned a positive charge, so the iron electrode is the cathode.
Reduction half-reaction: Fe2+(aq) + 2e–  Fe(s)
Oxidation half-reaction: Mn(s)  Mn2+(aq) + 2e–
Overall cell reaction:
Fe2+(aq) + Mn(s)  Fe(s) + Mn2+(aq)
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21-20
b)
e–
Voltmeter
e–
Salt bridge
Mn
Fe
()
1M
+
Mn2
+
21.32
a)
(+)
Anion
flow
(red half-rxn)
(ox half-rxn)
(overall rxn)
b)
Cation
flow
1M
Fe2+
Cu2+(aq) + 2e–  Cu(s)
Ni(s)  Ni2+(aq) + 2e–
Cu2+(aq) + Ni(s)  Cu(s) + Ni2+(aq)
Voltmeter
e–
e–
bridge
Ni
Cu
()
1M
Ni2+
(+)
1M
Cu2+
Anion
flow
Cation
flow
21.33
Plan: In cell notation, the oxidation components of the anode compartment are written on the left of the salt bridge
and the reduction components of the cathode compartment are written to the right of the salt bridge. A double
vertical line separates the anode from the cathode and represents the salt bridge. A single vertical line separates
species of different phases. Anode || Cathode
Solution:
a) Al is oxidized, so it is the anode and appears first in the cell notation. There is a single vertical line
separating the solid metals from their solutions.
Al(s) | Al3+(aq) || Cr3+(aq) | Cr(s)
2+
b) Cu is reduced, so Cu is the cathode and appears last in the cell notation. The oxidation of SO2 does not
include a metal, so an inactive electrode must be present. Hydrogen ion must be included in the oxidation
half-cell.
Pt | SO2(g) | SO42–(aq), H+(aq) || Cu2+(aq) | Cu(s)
21.34
a) Mn(s) + Cd2+(aq)  Mn2+(aq) + Cd(s)
b) 3Fe(s) + 2NO3–(aq) + 8H +(aq)  3Fe2+(aq) + 2NO(g) + 4H 2O(l)
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21-21
21.35
An isolated reduction or oxidation potential cannot be directly measured. However, by assigning a standard halfcell potential to a particular half-reaction, the standard potentials of other half-reactions can be determined relative
to this reference value. The standard reference half-cell is a standard hydrogen electrode defined to have an E°
value of 0.000 V.
21.36

A negative Ecell
indicates that the cell reaction is not spontaneous, G° > 0. The reverse reaction is spontaneous

with Ecell
> 0.
21.37
Similar to other state functions, the sign of E° changes when a reaction is reversed. Unlike G°, H° and S°, E° is
an intensive property, the ratio of energy to charge. When the coefficients in a reaction are multiplied by a factor,
the values of G°, H° and S° are multiplied by the same factor. However, E° does not change because both the
energy and charge are multiplied by the factor and their ratio remains unchanged.
21.38
Plan: Divide the balanced equation into reduction and oxidation half-reactions and add electrons. Add water and



hydroxide ion to the half-reaction that includes oxygen. Use the relationship Ecell
= Ecathode
– Eanode
to find the unknown E° value.
Solution:
a) Oxidation:
Se2–(aq)  Se(s) + 2e–
Reduction:
2SO32–(aq) + 3H2O(l) + 4e–  S2O32–(aq) + 6OH–(aq)



= Ecathode
– Eanode
b) Ecell



Eanode
= Ecathode
– Ecell
= –0.57 V – 0.35 V = –0.92 V
21.39
a) Reduction:
Oxidation:
O3(g) + 2H +(aq) + 2e–  O2(g) + H 2O(l)
Mn2+(aq) + 2H 2O(l)  MnO2(s) + 4H +(aq) + 2e–



b) Ecell
= Ecathode
– Eanode



Eanode
= Ecathode
– Ecell
= 2.07 V – 0.84 V= 1.23 V
21.40
Plan: The greater (more positive) the reduction potential, the greater the strength as an oxidizing agent.
Solution:
a) From Appendix D:
Fe3+(aq) + e–  Fe2+(aq)
E° = 0.77 V
E° = 1.07 V
Br2(l) + 2e–  2Br–(aq)
E° = 0.34 V
Cu2+(aq) + e–  Cu(s)
When placed in order of decreasing strength as oxidizing agents: Br2 > Fe3+ > Cu2+.
b) From Appendix D:
Ca2+(aq) + 2e–  Ca(s)
E° = –2.87 V
E° = 1.33 V
Cr2O72–(aq) + 14H+(aq) 6e–  2Cr3+(aq) + 7H2O(l)
E° = 0.80 V
Ag+(aq) + e–  Ag(s)
When placed in order of increasing strength as oxidizing agents: Ca2+ < Ag+ < Cr2O72–.
21.41
a) When placed in order of decreasing strength as reducing agents: SO2 > MnO2 > PbSO4
b) When placed in order of increasing strength as reducing agents: Hg < Sn < Fe
21.42



Plan: Use the relationship Ecell
= Ecathode
– Eanode
. E  values are found in Appendix D. Spontaneous reactions

have Ecell
> 0.
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21-22
Solution:
a)
Oxidation:
Reduction:
Overall reaction:
Co(s)  Co2+(aq) + 2e–
2H+(aq) + 2e–  H2(g)
Co(s) + 2H+(aq)  Co2+(aq) + H2(g)
E° = –0.28 V
E° = 0.00 V

Ecell
= 0.00 V (–0.28 V) = 0.28 V
b)

Reaction is spontaneous under standard-state conditions because Ecell
is positive.
2+
–
2+
Oxidation:
Hg2 (aq)  2Hg (aq) + 2e
E° = +0.92 V
E° = +0.85 V
Reduction:
Hg22+(aq) + 2e–  2Hg(l)
Overall:
2Hg22+(aq)  2Hg2+(aq) + 2Hg(l)

Ecell
= 0.85 V 0.92 V = –0.07 V
or Hg22+(aq)  Hg2+(aq) + Hg(l)

Negative Ecell
indicates reaction is not spontaneous under standard-state conditions.
21.43
a) Mn2+(aq) + 2H 2O(l) + 2Co3+(aq)  MnO2(s) + 4H +(aq) + 2Co2+(aq)



Ecell
= Ecathode
– Eanode
= 1.82 V – (1.23 V) = 0.59 V
The reaction is spontaneous.
b) 3AgCl(s) + NO(g) + 2H 2O(l)  3Ag(s) + 3Cl–(aq) + NO3–(aq) + 4H +(aq)



Ecell
= Ecathode
– Eanode
= 0.22 V – (0.96 V) = –0.74 V
The reaction is nonspontaneous.
21.44



Plan: Use the relationship Ecell
= Ecathode
– Eanode
. E  values are found in Appendix D. Spontaneous reactions

have Ecell
> 0.
Solution:
a)
Oxidation:
3{Cd(s)  Cd2+(aq) + 2e–}
E° = –0.40 V
Reduction: Cr2O72–(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l)
E° = +1.33 V
Overall: Cr2O72–(aq) + 3Cd(s) + 14H+(aq)  2Cr3+(aq) + 3Cd2+(aq) + 7H2O(l)

Ecell
= +1.33 V– (–0.40 V) = +1.73 V
b)
The reaction is spontaneous.
Oxidation:
Pb(s)  Pb2+(aq) + 2e–
Reduction:
Ni2+(aq) + 2e–  Ni(s)
Overall:
Pb(s) + Ni2+(aq)  Pb2+(aq) + Ni(s)
E° = –0.13 V
E° = –0.25 V

Ecell
= –0.25 V– (–0.13 V) = –0.12 V
The reaction is not spontaneous.
21.45
a) 2Cu+(aq) + PbO2(s) + 4H +(aq) + SO42–(aq)  2Cu2+(aq) + PbSO4(s) + 2H 2O(l)



Ecell
= Ecathode
– Eanode
= 1.70 V – (0.15 V) = 1.55 V
The reaction is spontaneous.
b) H 2O2(aq) + Ni2+(aq)  2H +(aq) + O2(g) + Ni(s)



Ecell
= Ecathode
– Eanode
= –0.25 V – (0.68 V) = –0.93 V
The reaction is nonspontaneous.
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21-23
21.46

> 0. All three reactions are written as reductions. When two halfPlan: Spontaneous reactions have Ecell
reactions are paired, one half-reaction must be reversed and written as an oxidation. Reverse the half-reaction that




will result in a positive value of Ecell
using the relationship Ecell
= Ecathode
– Eanode
. To balance each reaction,
multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons
gained and then add the half-reactions. The greater (more positive) the reduction potential, the greater the
strength as an oxidizing agent.
Solution:
Adding (1) and (2) to give a spontaneous reaction involves converting (1) to oxidation:
Oxidation: 2{Al(s)  Al3+(aq) + 3e–}
E° = –1.66 V
E° = +0.867 V
Reduction: 3{N2O4(g) + 2e–  2NO2–(aq)}
3N2O4(g) + 2Al(s)  6NO2–(aq) + 2Al3+(aq)

Ecell
= 0.867 V – (–1.66 V) = 2.53 V
3+
Oxidizing agents: N2O4 > Al ; reducing agents: Al > NO2–
Adding (1) and (3) to give a spontaneous reaction involves converting (1) to oxidation:
Oxidation: 2{Al(s)  Al3+(aq) + 3e–}
E° = –1.66 V
E° = +0.93 V
Reduction: 3{SO42–(aq) + H2O(l) + 2e– SO32–(aq) + 2OH–(aq)}
2Al(s) + 3SO42–(aq) + 3H2O(l)  2Al3+(aq) + 3SO32–(aq) + 6OH–(aq)

Ecell
= 0.93 V – (–1.66 V) = 2.59 V
Oxidizing agents: SO4 > Al ; reducing agents: Al > SO32–
Adding (2) and (3) to give a spontaneous reaction involves converting (2) to oxidation:
Oxidation: 2NO2–(aq)  N2O4(g) + 2e–
E° = 0.867 V
E° = 0.93 V
Reduction: SO42–(aq) + H2O(l) + 2e–  SO32–(aq) + 2OH–(aq)
SO42–(aq) + 2NO2–(aq) + H2O(l)  SO32–(aq) + N2O4(g) + 2OH–(aq)
2–
3+

Ecell
= 0.93 V – 0.867 V = 0.06 V
Oxidizing agents: SO4 > N2O4; reducing agents: NO2– > SO32–
Rank oxidizing agents (substance being reduced) in order of increasing strength:
Al3+ < N2O4 < SO42–
Rank reducing agents (substance being oxidized) in order of increasing strength:
SO32– < NO2– < Al
2–
21.47
3N2O(g) + 6H +(aq) + 2Cr(s)  3N2(g) + 3H 2O(l) + 2Cr3+(aq)

Ecell
= 1.77 V – (–0.74 V) = 2.51 V
Oxidizing agents: N2O > Cr3+; reducing agents: Cr > N2
3Au+(aq) + Cr(s)  3Au(s) + Cr3+(aq)

Ecell
= 1.69 V – (–0.74 V) = 2.43 V
Oxidizing agents: Au+ > Cr3+; reducing agents: Cr > Au
N2O(g) + 2H +(aq) + 2Au(s)  N2(g) + H 2O(l) + 2Au+(aq)

Ecell
= 1.77 V – (1.69 V) = 0.08 V
Oxidizing agents: N2O > Au+; reducing agents: Au > N2
Oxidizing agents: N2O > Au+ > Cr3+; reducing agents: Cr > Au > N2
21.48

Plan: Spontaneous reactions have Ecell
> 0. All three reactions are written as reductions. When two halfreactions are paired, one half-reaction must be reversed and written as an oxidation. Reverse the half-reaction that




will result in a positive value of Ecell
using the relationship Ecell
= Ecathode
– Eanode
. To balance each reaction,
multiply each half-reaction by an integer so that the number of electrons lost equals the number of electrons
gained and then add the half-reactions. The greater (more positive) the reduction potential, the greater the
strength as an oxidizing agent.
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
21-24
Solution:
Adding (1) and (2) to give a spontaneous reaction involves converting (2) to oxidation:
E° = +1.20 V
Oxidation: Pt(s)  Pt2+(aq) + 2e–
E° = +1.63 V
Reduction: 2HClO(aq) + 2H+(aq) + 2e– Cl2(g) + 2H2O(l)
2HClO(aq) + Pt(s) + 2H+(aq)  Cl2(g) + Pt2+(aq) + 2H2O(l)

Ecell
= 1.63 V – 1.20 V = 0.43 V
Oxidizing agents: HClO > Pt2+; reducing agents: Pt > Cl2
Adding (1) and (3) to give a spontaneous reaction involves converting (3) to oxidation:
E° = –0.31 V
Oxidation: Pb(s) + SO42–(aq)  PbSO4(s) + 2e–
E° = +1.63 V
Reduction: 2HClO(aq) + 2H+(aq) + 2e–  Cl2(g) + 2H2O(l)
2HClO(aq) + Pb(s) + SO42–(aq) + 2H+(aq)  Cl2(g) + PbSO4(s) + 2H2O(l)

Ecell
= 1.63 V – (–0.31 V) = 1.94 V
Oxidizing agents: HClO > PbSO4; reducing agents: Pb > Cl2
Adding (2) and (3) to give a spontaneous reaction involves converting (3) to oxidation:
Oxidation: Pb(s) + SO42–(aq)  PbSO4(s) + 2e–
E° = –0.31 V
E° = 1.20 V
Reduction: Pt2+(aq) + 2e– Pt(s)
Pt2+(aq) + Pb(s) + SO42–(aq)  Pt(s) + PbSO4(s)

Ecell
= 1.20 V – (–0.31 V) = 1.51 V
Oxidizing agents: Pt2+ > PbSO4; reducing agents: Pb > Pt
Order of increasing strength as oxidizing agent: PbSO4 < Pt2+ < HClO
Order of increasing strength as reducing agent: Cl2 < Pt < (Pb + SO42–)
21.49
S2O82–(aq) + 2I –(aq)  2SO42–(aq) + I 2(s)

Ecell
= 2.01 V – (0.53 V) = 1.48 V
Oxidizing agents: S2O82– > I 2; reducing agents: I – > SO42–
Cr2O72–(aq) + 14H +(aq) + 6I –(aq)  2Cr3+(aq) + 7H 2O(l) + 3I 2(s)

Ecell
= 1.33 V – (0.53 V) = 0.80 V
Oxidizing agents: Cr2O72– > I 2; reducing agents: I – > Cr3+
3S2O82–(aq) + 2Cr3+(aq) + 7H 2O(l)  6SO42–(aq) + Cr2O72–(aq) + 14H +(aq)

Ecell
= 2.01 V – (1.33 V) = 0.68 V
Oxidizing agents: S2O82– > Cr2O72–; reducing agents: Cr3+ > SO42–
Oxidizing agents: S2O82– > Cr2O72– > I 2; reducing agents: I – > Cr3+ > SO42–
21.50
Metal A + Metal B salt  solid colored product on metal A
Conclusion: Product is solid metal B. B is undergoing reduction and plating out on A. A is a better
reducing agent than B.
Metal B + acid  gas bubbles
Conclusion: Product is H2 gas produced as result of reduction of H+. B is a better reducing agent than
acid.
Metal A + Metal C salt  no reaction
Conclusion: C is not undergoing reduction. C must be a better reducing agent than A.
Since C is a better reducing agent than A, which is a better reducing agent than B and B reduces acid, then C
would also reduce acid to form H2 bubbles.
The order of strength of reducing agents is: C > A > B.
21.51
a) Copper metal is coating the iron.
b) The oxidizing agent is Cu2+ and the reducing agent is Fe.
c) Yes, this reaction, being spontaneous, may be made into a voltaic cell.
d) Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq)
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
21-25



= Ecathode
– Eanode
e) Ecell
= 0.34 V – (–0.44 V)
= 0.78 V
21.52
21.53
0.0592 V
Q
log   and G = –nFEcell
n
K
a) When Q/K < 1, the reaction is preceding to the right; Ecell > 0 and G < 0 and the reaction is spontaneous.
When Q/K = 1, the reaction is at equilibrium; Ecell = 0 and G = 0.
When Q/K > 1, the reaction is preceding to the left; Ecell < 0 and G > 0 and the reaction is not spontaneous.
b) Only when Q/K < 1 will the reaction proceed spontaneously and be able to do work.
Ecell = –
At the negative (anode) electrode, oxidation occurs so the overall cell reaction is
A(s) + B+(aq)  A+(aq) + B(s) with Q = [A+]/[B+].
a) The reaction proceeds to the right because with Ecell > 0 (voltaic cell), the spontaneous reaction occurs. As the
cell operates, [A+] increases and [B+] decreases.
b) Ecell decreases because the cell reaction takes place to approach equilibrium, Ecell = 0.
 
RT  A 


c) Ecell and Ecell are related by the Nernst equation: Ecell = Ecell –
.
ln
nF
 B 
 
 
A 
RT  A 

Ecell = Ecell when
= 0. This occurs when ln   = 0. Recall that e0 = 1, so [A+] must equal [B+]
ln
nF
 B 
 B 
 
 
for Ecell to equal

Ecell
.

d) Yes, it is possible for Ecell to be less than Ecell
when [A+] > [B+].
21.54

a) Examine the Nernst equation: Ecell = Ecell
–
Ecell =
RT
ln Q.
nF
RT
RT
ln K –
ln Q
nF
nF
RT  K 
RT  Q 
ln
 ln  = –
nF  Q 
nF  K 
If Q/K < 1, Ecell will decrease with a decrease in cell temperature. If Q/K > 1, Ecell will increase
(become less negative) with a decrease in cell temperature.
 Active ion at anode
RT

–
ln
b) Ecell = Ecell
nF
 Active ion at cathode
Ecell =
Ecell will decrease as the concentration of an active ion at the anode increases.
c) Ecell will increase as the concentration of an active ion at the cathode increases.
d) Ecell will increase as the pressure of a gaseous reactant in the cathode compartment increases.
21.55
In a concentration cell, the overall reaction takes place to decrease the concentration of the more concentrated
electrolyte. The more concentrated electrolyte is reduced, so it is in the cathode compartment.
21.56
Plan: The equilibrium constant can be found by using ln K =


nFEcell
nEcell
or log K =
. Use E° values from
0.0592
RT




the Appendix to calculate Ecell
( Ecell
= Ecathode
– Eanode
) and then calculate K. The substances given in the
problem must be the reactants in the equation.
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21-26
Solution:
a) Oxidation: Ni(s)  Ni2+(aq) + 2e–
Reduction: 2{Ag+(aq) + 1e–  Ag(s)}
Ni(s) + 2Ag+(aq)  Ni2+(aq) + 2Ag(s)
E° = –0.25 V
E° = +0.80 V



Ecell
= Ecathode
– Eanode
= 0.80 V – (–0.25 V) = 1.05 V; two electrons are transferred.

2 1.05 V 
nEcell
=
= 35.47297
0.0592
0.0592 V
K = 1035.47297
K = 2.97146x1035 = 3x1035
b) Oxidation: 3{Fe(s)  Fe2+(aq) + 2e–}
Reduction: 2{Cr3+(aq) + 3e–  Cr(s)}
3Fe(s) + 2Cr3+(aq)  3Fe2+(aq) + 2Cr(s)
log K =
E° = –0.44 V
E° = –0.74 V



Ecell
= Ecathode
– Eanode
= –0.74 V – (–0.44 V) = –0.30 V; six electrons are transferred.
log K =

6  0.30 V 
nEcell
=
= –30.4054
0.0592
0.0592 V
K = 10–30.4054
K = 3.936x10–31 = 4x10–31
21.57
a) 2Al(s) + 3Cd2+(aq)  2Al3+(aq) + 3Cd(s)

Ecell
=
log K =

Ecathode

nEcell

– Eanode
=
= –0.40 V – (–1.66 V) = 1.26 V
6 1.26 V 
0.0592
0.0592 V
K = 5.04316x10127 = 5x10127
b) I2(s) + 2Br–(aq)  Br2(l) + 2I –(aq)

Ecell
=

Ecathode

nEcell
n=6

– Eanode
=
= 127.7027
n=2
0.53 V – (1.07 V) = –0.54 V
2  0.54 V 
=
= –18.24324
0.0592
0.0592 V
K = 5.7116x10–19 = 6x10–19
log K =
21.58
Plan: The equilibrium constant can be found by using ln K =


nFEcell
nEcell
or log K =
. Use E° values from
0.0592
RT




the Appendix to calculate Ecell
( Ecell
= Ecathode
– Eanode
) and then calculate K. The substances given in the
problem must be the reactants in the equation.
Solution:
a) Oxidation: 2{Ag(s)  Ag+(aq) + 1e–}
E° = +0.80 V
Reduction: Mn2+(aq) + 2e–  Mn(s)
E° = –1.18 V
2Ag(s) + Mn2+(aq)  2Ag+(aq) + Mn(s)



Ecell
= Ecathode
– Eanode
= –1.18 V – (0.80 V) = –1.98V; two electrons are transferred.
log K =

2  1.98 V 
nEcell
=
= –66.89189
0.0592
0.0592 V
K = 10–66.89189
K = 1.2826554x10–67 = 1x10–67
b) Oxidation: 2Br–(aq)  Br2(l) + 2e–
Reduction: Cl2(g) + 2e–  2Cl–(aq)
2Br–(aq) + Cl2(g)  Br2(l) + 2Cl–(aq)
E° = 1.07 V
E° = 1.36 V
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21-27



Ecell
= Ecathode
– Eanode
= 1.36 V – 1.07 V = 0.29 V; two electrons transferred.

2  0.29 V 
nEcell
=
= 9.797297
0.0592
0.0592 V
log K =
K = 109.797297
K = 6.2704253x109 = 6x109
21.59
a) 2Cr(s) + 3Cu2+(aq)  2Cr3+(aq) + 3Cu(s)

Ecell
=
log K =

Ecathode

nEcell

– Eanode
=
n=6
= 0.34 V – (–0.74 V) = 1.08 V
6 1.08 V 
= 109.4594595
0.0592
0.0592 V
K = 2.880444x10109 = 3x10109
n=2
b) Sn(s) + Pb2+(aq)  Sn2+(aq) + Pb(s)



Ecell
= Ecathode
– Eanode
= –0.13 V – (–0.14 V) = 0.01V

2  0.01 V 
nEcell
=
= 0.33784
0.0592
0.0592 V
K = 2.1769 = 2
log K =
21.60


Plan: Use G° = nFEcell
to calculate G°. Substitute J/C for V in the unit for Ecell
.
Solution:

a) G° = nFEcell
= – (2 mol e–)(96,485 C/mol e–)(1.05 J/C) = –2.026185x105 = –2.03x105 J

b) G° = nFEcell
= – (6 mol e–)(96,485 C/mol e–)(–0.30 J/C) = 1.73673x105 = 1.7x105 J
21.61

a) G° = nFEcell
= – (6 mol e–)(96,485 C/mol e–)(1.26 J/C) = –7.294266x105 = –7.29x105 J

b) G° = nFEcell
= – (2 mol e–)(96,485 C/mol e–)(–0.54 J/C) = 1.042038x105 = 1.0x105 J
21.62


Plan: Use G° = nFEcell
to calculate G°. Substitute J/C for V in the unit for Ecell
.
Solution:

a) G° = nFEcell
= – (2 mol e–)(96,485 C/mol e–)(–1.98 J/C) = 3.820806x105 = 3.82x105 J

b) G° = nFEcell
= – (2 mol e–)(96,485 C/mol e–)(0.29 J/C) = –5.59613x104 = –5.6x104 J
21.63

a) G° = nFEcell
= – (6 mol e–)(96,485 C/mol e–)(1.08 J/C) = –6.252228x105 = –6.25x105 J

b) G° = nFEcell
= – (2 mol e–)(96,485 C/mol e–)(0.01 J/C) = –1.9297x103 = –2x103 J
21.64

Plan: Use Ecell
=
0.0592 V

log K to find Ecell
and then G° = –RT ln K to find G°.
n
Solution:
T = (273 + 25)K = 298 K
0.0592 V
0.0592 V

log K =
log 5.0x104 = 0.278179 = 0.28 V
Ecell
=
n
1
G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (5.0x104) = –2.68067797x104 = –2.7x104 J


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21-28
21.65
0.0592 V

log K to find Ecell
and then G° = –RT ln K to find G°.
n
T = (273 + 25)K = 298 K
0.0592 V
0.0592 V

log K =
log 5.0x106 = –0.31382 = –0.31 V
Ecell
=
n
1
G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (5.0x10–6) = 3.0241424x104 = 3.0x104 J

=
Use Ecell

21.66

Plan: Use Ecell
=

0.0592 V

log K to find Ecell
and then G° = –RT ln K to find G°.
n
Solution:
T = (273 + 25)K = 298 K
0.0592
0.0592 V

log K =
log 65 = 0.0536622 = 0.054 V
Ecell
=
n
2
G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (65) = –1.03423x104 = –1.0x104 J
0.0592 V

log K to find Ecell
and then G° = –RT ln K to find G°.
n
T = (273 + 25)K = 298 K
0.0592
0.0592 V

log K =
log 0.065 = –0.03513776 = –0.035 V
Ecell
=
n
2
G° = –RT ln K = – (8.314 J/mol•K)(298 K) ln (0.065) = 6.772116x103 = 6.8x103 J
21.67

Use Ecell
=
21.68
Plan: The standard reference half-cell is the H2/H+ cell. Since this is a voltaic cell, a spontaneous reaction is
occurring. For a spontaneous reaction between H2/H+ and Cu/Cu2+, Cu2+ must be reduced and H2 must be
0.0592


oxidized. Write the balanced reaction and calculate Ecell
. Use the Nernst equation, Ecell = Ecell
–
log Q,
n
to find [Cu2+] when Ecell = 0.22 V.
Solution:
Oxidation: H2(g)  2H+(aq) + 2e–
E° = 0.00 V
E° = 0.34 V
Reduction: Cu2+(aq) + 2e–  Cu(s)
Cu2+(aq) + H2(g)  Cu(s) + 2H+(aq)



Ecell
= Ecathode
– Eanode
= 0.34 V – 0.00 V = 0.34 V

Ecell = Ecell
–
0.0592
log Q
n
2
H  
0.0592
 
Ecell =
–
log
n
 Cu 2    H 2 


For a standard hydrogen electrode [H+] = 1.0 M and [H2] = 1.0 atm.
0.0592
1.0
log
0.22 V = 0.34 V –
2
 Cu 2   1.0


0.0592
1.0
log
0.22 V – 0.34 V = –
2
 Cu 2   1.0


0.0592
1.0
log
–0.12 V = –
2
 Cu 2   1.0



Ecell
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21-29
4.054054 = log
1.132541x104 =
1.0
 Cu 2   1.0


1
Raise each side to 10x .
[Cu 2+ ]
[Cu2+] = 8.82970x10–5 = 8.8x10–5 M
21.69
The cell reaction is: Pb2+(aq) + Mn(s)  Pb(s) + Mn2+(aq)



Ecell
= Ecathode
– Eanode
= –0.13 V – (–1.18 V) = 1.05 V
0.0592
log Q
n
 Mn 2  
0.0592


Ecell = Ecell –
log 
n
 Pb 2  


1.4
0.0592
0.44 V = 1.05 V –
log
2
 Pb2  



Ecell = Ecell
–
(0.44 V – 1.05 V)(–2/0.0592) = log
1.4
1.4
 Pb2  


= 20.608108
= 4.056094x1020
 Pb2  


[Pb2+] = 3.45160x10–21 = 3.5x10–21 M
21.70 Plan: Since this is a voltaic cell, a spontaneous reaction is occurring. For a spontaneous reaction between Ni/Ni2+

and Co/Co2+, Ni2+ must be reduced and Co must be oxidized. Write the balanced reaction and calculate Ecell
.
0.0592
log Q, to find Ecell at the given ion concentrations. Then
n
the Nernst equation can be used to calculate [Ni2+] at the given Ecell. To calculate equilibrium concentrations,
recall that at equilibrium Ecell = 0.00.
Solution:
a) Oxidation: Co(s)  Co2+(aq) + 2e–
E° = –0.28 V
E° = –0.25 V
Reduction: Ni2+(aq) + 2e–  Ni(s)
Ni2+(aq) + Co(s)  Ni(s) + Co2+(aq)

Use the Nernst equation, Ecell = Ecell
–



Ecell
= Ecathode
– Eanode
= –0.25 V – (–0.28 V) = 0.03 V
0.0592
log Q
n
 Co 2  
0.0592


Ecell = Ecell –
log 
n
 Ni 2  


0.20
0.0592
log
Ecell = 0.03 V –
2
0.80

Ecell = Ecell
–
n = 2e–
Ecell = 0.047820976 V = 0.05 V
b) From part a), notice that an increase in [Co2+] leads to a decrease in cell potential. Therefore, the
concentration of cobalt ion must increase further to bring the potential down to 0.03 V. Thus, the new
concentrations will be [Co2+] = 0.20 M + x and [Ni2+] = 0.80 M – x (there is a 1:1 mole ratio).
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21-30
 Co 2  


 Ni 2  


0.20 + x 
0.0592
0.03 V = 0.03 V –
log
2
0.80  x 

–
Ecell = Ecell
0=–
0.0592
log
n
0.20 + x 
0.0592
log
2
0.80  x 
0.20 + x 
0.80  x 
0.20 + x 
0.80  x 
0 = log
1=
Raise each side to 10x .
0.20 + x = 0.80 – x
x = 0.30 M
[Ni2+] = 0.80 – x = 0.80 – 0.30 = 0.50 M
c) At equilibrium Ecell = 0.00; to decrease the cell potential to 0.00, [Co2+] increases and [Ni2+] decreases.
0.20 + x 
0.0592
log
0.00 V = 0.03 V –
2
0.80  x 
–0.03 V = –
0.20 + x 
0.0592
log
2
0.80  x 
0.20 + x 
0.80  x 
0.20 + x 
10.316051 =
0.80  x 
1.0135135 = log
Raise each side to 10x .
x = 0.71163
[Co2+] = 0.20 + 0.71163 = 0.91163 = 0.91 M
[Ni2+] = 0.80 – 0.71163 = 0.08837 = 0.09 M
21.71
The spontaneous reaction (voltaic cell) is Cd2+(aq) + Mn(s)  Cd(s) + Mn2+(aq) with



Ecell
= Ecathode
– Eanode
= –0.40 V – (–1.18 V) = 0.78 V.
0.0592

a) Ecell = Ecell
–
log Q
n
 Mn 2  
0.0592


Ecell = Ecell
–
log 
n = 2e–

2
n
Cd 


0.090
0.0592
Ecell = 0.78 –
log
2
0.060
Ecell = 0.774787699 V = 0.77 V
b) For the [Cd2+] to decrease from 0.060 M to 0.050 M, a change of 0.010 M, the [Mn2+] must increase by the
same amount, from 0.090 M to 0.100 M.
0.10
0.0592
Ecell = 0.78 –
log
2
0.050
Ecell = 0.771089512 V = 0.77 V
c) Increase the manganese and decrease the cadmium by equal amounts. Total = 0.150 M
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21-31
 Mn 2  


Cd 2  


0.0592
log
2
0.055 = 0.78 –
 Mn 2  

 = 3.1134597x1024
Cd 2  


[Mn2+] = (3.1134597x1024)[Cd2+]
[Mn2+] + [Cd2+] = 0.150 M
(3.1134597x1024)[Cd2+] + [Cd2+] = 0.150 M
[Cd2+] = 4.81779x10–26 M
[Mn2+] = 0.150 M – 4.81779x10–26 M = 0.150 M
d) At equilibrium Ecell = 0.00.
 Mn 2  
0.0592

log 
0.00 = 0.78 –
2
Cd 2  


2

 Mn 

 = 2.2456980x1026
Cd 2  


2+
[Mn ] = (2.2456980x1026)[Cd2+]
[Mn2+] + [Cd2+] = 0.150 M
(2.2456980x1026)[Cd2+] + [Cd2+] = 0.150 M
[Cd2+] = 6.6794378x10–28 = 7x10–28 M
[Mn2+] = 0.150 M – [Cd2+] = 0.150 M
21.72
Plan: The overall cell reaction proceeds to increase the 0.10 M H+ concentration and decrease the 2.0 M H+

concentration. Use the Nernst equation to calculate Ecell. Ecell
= 0 V for a concentration cell since the halfreactions are the same.
Solution:
Half-cell A is the anode because it has the lower concentration.
Oxidation: H2(g;0.95 atm)  2H+(aq; 0.10 M) + 2e–
E° = 0.00 V
E° = 0.00 V
Reduction: 2H+(aq; 2.0 M) + 2 e–  H2(g; 0.60 atm)
2H+(aq; 2.0 M) + H2(g; 0.95 atm)  2H+(aq; 0.10 M) + H2(g; 0.60 atm)

Ecell
= 0.00 V

–
Ecell = Ecell
n = 2e–
0.0592
log Q
n
2
Q for the cell equals
Ecell = 0.00 V –
21.73
H 
P
  anode H2 (cathode)
2
H 
P
  cathode H2 (anode)
Ecell =
 0.10 2  0.60 
 2.0 2  0.95
= 0.00157895
0.0592
log (0.00157895) = 0.0829283 = 0.083 V
2
Sn2+(0.87 M)  Sn2+(0.13 M)
Half-cell B is the cathode.
0.0592

Ecell = Ecell
–
log Q
n

Ecell
=

Ecell
= 0.00 V
n = 2e–
Sn 2 
0.0592

 anode
–
log
2

2
Sn 

 cathode
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
21-32
Ecell = 0.00 V –
 0.13
0.0592
log
2
 0.87 
Ecell = 0.024437 = 0.024 V
21.74
Electrons flow from the anode, where oxidation occurs, to the cathode, where reduction occurs. The electrons
always flow from the anode to the cathode, no matter what type of cell.
21.75
The electrodes are separated by an electrolyte paste, which for the ordinary dry cell battery contains ZnCl2,
NH4Cl, MnO2, starch, graphite, and water.
21.76
A D-sized battery is much larger than an AAA-sized battery, so the D-sized battery contains a greater amount of
the cell components. The potential, however, is an intensive property and does not depend on the amount of the
cell components. (Note that amount is different from concentration.) The total amount of charge a battery can
produce does depend on the amount of cell components, so the D-sized battery produces more charge than the
AAA-sized battery.
21.77
a) Alkaline batteries = (6.0 V)(1 alkaline battery/1.5 V) = 4 alkaline batteries.
b) Voltage = (6 Ag batteries)(1.6 V/Ag battery) = 9.6 V
c) The usual 12-V car battery consists of six 2 V cells. If two cells are shorted only four cells remain.
Voltage = (4 cells)(2 V/cell) = 8 V
21.78
The Teflon spacers keep the two metals separated so the copper cannot conduct electrons that would promote the
corrosion of the iron skeleton. Oxidation of the iron by oxygen causes rust to form and the metal to corrode.
21.79
Bridge supports rust more rapidly at the water line due to the presence of large concentrations of both O2 and
H2O.
21.80
Chromium, like iron, will corrode through the formation of a metal oxide. Unlike the iron oxide, which is
relatively porous and easily cracks, the chromium oxide forms a protective coating, preventing further corrosion.
21.81
Plan: Sacrificial anodes are metals with E° values that are more negative than that for iron, –0.44 V, so they are
more easily oxidized than iron.
Solution:
a) E°(aluminum) = –1.66 V. Yes, except aluminum resists corrosion because once a coating of its oxide covers it,
no more aluminum corrodes. Therefore, it would not be a good choice.
b) E°(magnesium) = –2.37 V. Yes, magnesium is appropriate to act as a sacrificial anode.
c) E°(sodium) = –2.71 V. Yes, except sodium reacts with water, so it would not be a good choice.
d) E°(lead) = –0.13 V. No, lead is not appropriate to act as a sacrificial anode because its E° value is too high.
e) E°(nickel) = –0.25 V. No, nickel is inappropriate as a sacrificial anode because its E° value is too high.
f) E°(zinc) = –0.76 V. Yes, zinc is appropriate to act as a sacrificial anode.
g) E°(chromium) = –0.74 V. Yes, chromium is appropriate to act as a sacrificial anode.
21.82
a) Oxidation occurs at the left electrode (anode).
b) Elemental M forms at the right electrode (cathode).
c) Electrons are being released by ions at the left electrode.
d) Electrons are entering the cell at the right electrode.
21.83
3Cd2+(aq) + 2Cr(s)  3Cd(s) + 2Cr3+(aq)

Ecell
= –0.40 V – (–0.74 V) = 0.34 V
To reverse the reaction requires 0.34 V with the cell in its standard state. A 1.5 V supplies more than enough
potential, so the cadmium metal oxidizes to Cd2+ and chromium plates out.
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
21-33
21.84

values because in pure water, the [H +] and [OH –] are
The Ehalf–cell values are different than the Ehalf-cell
–7
1.0 x 10 M rather than the standard-state value of 1 M.
21.85
The oxidation number of nitrogen in the nitrate ion, NO3–, is +5 and cannot be oxidized further since nitrogen
has only five electrons in its outer level. In the nitrite ion, NO2–, on the other hand, the oxidation number of
nitrogen is +3, so it can be oxidized at the anode to the +5 state.
21.86
Due to the phenomenon of overvoltage, the products predicted from a comparison of electrode potentials are not
always the actual products. When gases (such as H2(g) and O2(g)) are produced at metal electrodes, there is an
overvoltage of about 0.4 to 0.6 V more than the electrode potential indicates. Due to this, if H2 or O2 is the
expected product, another species may be the true product.
21.87
Plan: Oxidation occurs at the anode, while reduction occurs at the cathode.
Solution:
a) At the anode, bromide ions are oxidized to form bromine (Br2). 2Br–(l) → Br2(l) + 2e–
b) At the cathode, sodium ions are reduced to form sodium metal (Na). Na+(l) + e– → Na(s)
21.88
a) At the negative electrode (cathode) barium ions are reduced to form barium metal (Ba).
b) At the positive electrode (anode), iodide ions are oxidized to form iodine (I2).
21.89
Plan: Oxidation occurs at the anode, while reduction occurs at the cathode. Decide which anion is more likely to
be oxidized and which cation is more likely to be reduced. The less electronegative anion holds its electrons less
tightly and is more likely to be oxidized; the cation with the higher ionization energy has the greater attraction for
electrons and is more likely to be reduced.
Solution:
Either iodide ions or fluoride ions can be oxidized at the anode. The ion that more easily loses an electron will
form. Since I is less electronegative than F, I– will more easily lose its electron and be oxidized at the anode. The
product at the anode is I2 gas. The iodine is a gas because the temperature is high to melt the salts.
Either potassium or magnesium ions can be reduced at the cathode. Magnesium has greater ionization energy than
potassium because magnesium is located up and to the right of potassium on the periodic table. The greater
ionization energy means that magnesium ions will more readily add an electron (be reduced) than potassium ions.
The product at the cathode is magnesium (liquid).
21.90
Liquid strontium (Sr) forms at the negative electrode, and gaseous bromine (Br2) forms at the positive electrode.
21.91
Plan: Oxidation occurs at the anode, while reduction occurs at the cathode. Decide which anion is more likely to
be oxidized and which cation is more likely to be reduced. The less electronegative anion holds its electrons less
tightly and is more likely to be oxidized; the cation with the higher ionization energy has the greater attraction for
electrons and is more likely to be reduced.
Solution:
Bromine gas forms at the anode because the electronegativity of bromine is less than that of chlorine. Calcium
metal forms at the cathode because its ionization energy is greater than that of sodium.
21.92
Liquid calcium forms at the negative electrode and chlorine gas forms at the positive electrode.
21.93
Plan: Compare the electrode potentials of the species with those of water. The reduction half-reaction with the
more positive electrode potential occurs at the cathode, and the oxidation half-reaction with the more negative
electrode potential occurs at the anode.
Solution:
Possible reductions:
E° = +0.34 V
Cu2+(aq) + 2e–  Cu(s)
E° = –2.90 V
Ba2+(aq) + 2e–  Ba(s)
Al3+(aq) + 3e–  Al(s)
E° = –1.66 V
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21-34
E = –1 V with overvoltage
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Copper can be prepared by electrolysis of its aqueous salt since its reduction half-cell potential is more positive
than the potential for the reduction of water. The reduction of copper is more spontaneous than the reduction of
water. Since the reduction potentials of Ba2+ and Al3+ are more negative and therefore less spontaneous than the
reduction of water, these ions cannot be reduced in the presence of water since the water is reduced instead.
Possible oxidations:
2Br–(aq)  Br2(l) + 2e–
E° = +1.07 V
E = 1.4 V with overvoltage
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Bromine can be prepared by electrolysis of its aqueous salt because its reduction half-cell potential is more
negative than the potential for the oxidation of water with overvoltage. The more negative reduction potential for
Br– indicates that its oxidation is more spontaneous than the oxidation of water.
21.94
Strontium is too electropositive to form from the electrolysis of an aqueous solution. The elements that
electrolysis will separate from an aqueous solution are gold, tin, and chlorine.
21.95
Plan: Compare the electrode potentials of the species with those of water. The reduction half-reaction with the
more positive electrode potential occurs at the cathode, and the oxidation half-reaction with the more negative
electrode potential occurs at the anode.
Solution:
Possible reductions:
E° = –3.05 V
Li+(aq) + e–  Li(s)
E° = –0.76 V
Zn2+(aq) + 2e–  Zn(s)
Ag+(aq) + e–  Ag(s)
E° = 0.80 V
E = –1 V with overvoltage
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Zinc and silver can be prepared by electrolysis of their aqueous salt solutions since their reduction half-cell
potentials are more positive than the potential for the reduction of water. The reduction of zinc and silver is more
spontaneous than the reduction of water. Since the reduction potential of Li+ is more negative and therefore less
spontaneous than the reduction of water, this ion cannot be reduced in the presence of water since the water is
reduced instead.
Possible oxidations:
2I–(aq)  I2(l) + 2e–
E° = +0.53 V
E = 1.4 V with overvoltage
2H2O(l)  O2(g) + 4H+(aq) + 4e–
Iodine can be prepared by electrolysis of its aqueous salt because its reduction half-cell potential is more negative
than the potential for the oxidation of water with overvoltage. The more negative reduction potential for I–
indicates that its oxidation is more spontaneous than the oxidation of water.
21.96
Electrolysis will separate both iron and cadmium from an aqueous solution.
21.97
Plan: Compare the electrode potentials of the species with those of water. The reduction half-reaction with the
more positive electrode potential occurs at the cathode, and the oxidation half-reaction with the more negative
electrode potential occurs at the anode.
Solution:
a) Possible oxidations:
E = 1.4 V with overvoltage
2H2O(l)  O2(g) + 4H+(aq) + 4e–
2F–  F2(g) + 2e–
E° = 2.87 V
Since the reduction potential of water is more negative than the reduction potential for F–, the oxidation of water
is more spontaneous than that of F–. The oxidation of water produces oxygen gas (O2), and hydronium ions
(H3O+) at the anode.
Possible reductions:
E = –1 V with overvoltage
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Li+(aq) + e–  Li(s)
E° = –3.05 V
Since the reduction potential of water is more positive than that of Li+, the reduction of water is more spontaneous
than the reduction of Li+. The reduction of water produces H2 gas and OH– at the cathode.
b) Possible oxidations:
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21-35
E = 1.4 V with overvoltage
2H2O(l)  O2(g) + 4H+(aq) + 4e–
The oxidation of water produces oxygen gas (O2), and hydronium ions (H3O+) at the anode.
The SO42– ion cannot oxidize as S is already in its highest oxidation state in SO42–.
Possible reductions:
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
E = –1 V with overvoltage
E° = –0.14 V
Sn2+(aq) + 2e–  Sn(s)
E = –0.63 V (approximate)
SO42–(aq) + 4H+(aq) + 2e–  SO2(g) + 2H2O(l)
The potential for sulfate reduction is estimated from the Nernst equation using standard-state concentrations and
pressures for all reactants and products except H+, which in pure water is 1x10–7 M.
E = 0.20 V – (0.0592/2) log [1/(1x10–7)4] = –0.6288 = –0.63 V
The most easily reduced ion is Sn2+ with the most positive reduction potential, so tin metal forms at the cathode.
21.98
a) Solid zinc (Zn) forms at the cathode and liquid bromine (Br2) forms at the anode.
b) Solid copper (Cu) forms at the cathode and both oxygen gas (O2) and aqueous hydrogen ions (H +) form at the
anode.
21.99
Plan: Compare the electrode potentials of the species with those of water. The reduction half-reaction with the
more positive electrode potential occurs at the cathode, and the oxidation half-reaction with the more negative
electrode potential occurs at the anode.
Solution:
a) Possible oxidations:
2H2O(l)  O2(g) + 4H+(aq) + 4e–
E = 1.4 V with overvoltage
The oxidation of water produces oxygen gas (O2), and hydronium ions (H3O+) at the anode.
NO3– cannot oxidize since N is in its highest oxidation state in NO3–.
Possible reductions:
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
E = –1 V with overvoltage
E° = –0.74 V
Cr3+(aq) + 3e–  Cr(s)
E = +0.13 V (approximate)
NO3–(aq) + 4H+(aq) + 3e–  NO(g) + 2H2O(l)
The potential for nitrate reduction is estimated from the Nernst equation using standard-state concentrations and
pressures for all reactants and products except H+, which in pure water is 1x10–7 M.
E = 0.96 V – (0.0592/2) log [1/(1x10–7)4] = 0.1312 = 0.13 V
The most easily reduced ion is NO3–, with the most positive reduction potential so NO gas is formed at the
cathode.
b) Possible oxidations:
E = 1.4 V with overvoltage
2H2O(l)  O2(g) + 4H+(aq) + 4e–
2Cl–(aq)  Cl2(g) + 2e–
E° = 1.36 V
The oxidation of chloride ions to produce chlorine gas occurs at the anode. Cl– has a more negative reduction
potential showing that it is more easily oxidized than water.
Possible reductions:
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
E = –1 V with overvoltage
E° = –1.18 V
Mn2+(aq) + 2e–  Mn(s)
It is easier to reduce water than to reduce manganese ions, so hydrogen gas and hydroxide ions form at the
cathode. The reduction potential of Mn2+ is more negative than that of water showing that its reduction is less
spontaneous than that of water.
21.100 a) Solid iron (Fe) forms at the cathode, and solid iodine (I2) forms at the anode.
b) Gaseous hydrogen (H2) and aqueous hydroxide ion (OH –) form at the cathode, while gaseous oxygen (O2) and
aqueous hydrogen ions (H +) form at the anode.
21.101 Plan: Write the half-reaction for the reduction of Mg2+. Convert mass of Mg to moles and use the mole ratio in
the balanced reaction to find the number of moles of electrons required for every mole of Mg produced. The
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21-36
Faraday constant is used to find the charge of the electrons in coulombs. To find the current, the charge is divided
by the time in seconds.
Solution:
Mg2+ + 2e–  Mg
 1 mol Mg   2 mol e  
–
a) Moles of electrons =  45.6 g Mg  
 = 3.751542575 = 3.75 mol e
 
24.31
g
Mg
1
mol
Mg



96,
485
C


b) Charge = 3.751542575 mol e  
= 3.619676x105 = 3.62x105 C
 
 mol e 
 3.619676x105 C   1 h   A 
 = 28.727587 = 28.7 A
c) Current = 
 


3.50 h

  3600 s   C s 


21.102 Na+ + 1e–  Na
 1 mol Na   1 mol e  
–
a) Moles of electrons =  215 g Na  
 = 9.351892127 = 9.35 mol e
 
 22.99 g Na   1 mol Na 


 96, 485 C 
b) Charge = 9.351892127 mol e  
= 9.0231731x105 = 9.02x105 C
 
 mol e 
 9.0231731x105 C   1 h   A 
 = 26.383547 = 26.4 A
c) Current = 
 


9.50 h

  3600 s   C s 
21.103 Plan: Write the half-reaction for the reduction of Ra2+. Use the Faraday constant to convert charge in coulombs
to moles of electrons provided; the balanced reduction reaction converts moles of electrons to moles of radium
produced. Multiply moles of radium by its molar mass to obtain grams.
Solution:
Ra2+ + 2e–  Ra
In the reduction of radium ions, Ra2+, to radium metal, the transfer of two electrons occurs.
 1 mol e    1 mol Ra   226 g Ra 
Mass (g) of Ra =  235 C  
 
 = 0.275224 = 0.275 g Ra
 

 96, 485 C   2 mol e   1 mol Ra 
21.104 Al3+ + 3e–  Al
In the reduction of aluminum ions, Al3+, to aluminum metal, the transfer of three electrons occurs.
 1 mol e    1 mol Al   26.98 g Al 
Mass (g) of Al =  305 C  
= 0.028428944 = 0.0284 g Al
 96, 485 C   3 mol e    1 mol Al 




21.105 Plan: Write the half-reaction for the reduction of Zn2+. Convert mass of Zn to moles and use the mole ratio in
the balanced reaction to find the number of moles of electrons required for every mole of Zn produced. The
Faraday constant is used to find the charge of the electrons in coulombs. To find the time, the charge is divided
by the current.
Solution:
Zn2+ + 2e–  Zn
 1 mol Zn   2 mol e    96, 485 C   1   1 A 
3
3
Time (s) =  65.5 g Zn  
 
 
  C  = 9.20169x10 = 9.20x10 s
 
65.41
g
Zn
1
mol
Zn
21.0
A


1
mol
e


 s 


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21-37
21.106 Ni2+ + 2e–  Ni
 1 mol Ni   2 mol e    96, 485 C   1   1 A 
 = 391.1944859 = 391 s
Time (s) = 1.63 g Ni  
 
 

 
 58.69 g Ni   1 mol Ni   1 mol e   13.7 A   C s 
21.107 a) The sodium sulfate ionizes to produce Na+ and SO42– ions which make the water conductive; therefore the
current will flow through the water to complete the circuit, increasing the rate of electrolysis. Pure water, which
contains very low (10–7 M) concentrations of H+ and OH–, conducts electricity very poorly.
b) The reduction of H2O has a more positive half-potential (–1 V) than the reduction of Na+ (–2.71 V); the
more spontaneous reduction of water will occur instead of the less spontaneous reduction of sodium ion. The
oxidation of H2O is the only oxidation possible because SO42– cannot be oxidized under these conditions. In other
words, it is easier to reduce H2O than Na+ and easier to oxidize H2O than SO42–.
21.108 Iodine is formed first since the oxidation potential of I– is more positive than the oxidation potential of Br–.
21.109 Plan: Write the half-reaction for the reduction of Zn2+. Find the charge in coulombs by multiplying the current
by the time in units of seconds. Use the Faraday constant to convert charge in coulombs to moles of electrons
provided; the balanced reduction reaction converts moles of electrons to moles of zinc produced. Multiply moles
of zinc by its molar mass to obtain grams.
Solution:
Zn2+ + 2e–  Zn
C 
 24 h   3600 s   1 mol e    1 mol Zn   65.41 g Zn 
Mass (g) of Zn =  0.855 A   s   2.50 day  

 

 
 
 A 
 1 day   1 h   96, 485 C   2 mol e   1 mol Zn 


= 62.59998 = 62.6 g Zn
21.110 Mn2+(aq) + 2H2O(l)  MnO2(s) + 4H+(aq) + 2e–
 103 g   1 mol MnO 2
Time (h) = 1.00 kg MnO 2  
 1 kg   86.94 g MnO
2


  2 mol e 
 
  1 mol MnO 2
  96485 C   1   A
 

 

  1 mol e   25.0 A   C s

  1 h 
  3600 s 

= 24.66196 = 24.7 h
The MnO2 product forms at the anode, since the half-reaction is an oxidation.
21.111 Plan: Write the reduction half-reaction for H +. Use the ideal gas law to find the moles of H2 produced. The mole
ratio in the balanced reduction reaction gives the moles of electrons required for that amount of H2 and the
Faraday constant converts moles of electrons to charge in coulombs. To convert coulombs to energy in joules,
remember that 1 V equals 1 J/C; multiply the charge in coulombs by volts to obtain joules. Convert this energy to
units of kilojoules and use the given conversion factor between mass of oil and energy to find the mass of oil
combusted to provide the needed amount of energy.
Solution:
The half-reaction is: 2H +(aq) + 2e–  H2(g)
a) First, find the moles of hydrogen gas.


12.0 atm  3.5x106 L
PV
=
= 1.716682x106 mol H2
L•atm 
RT

 0.0821 mol•K    273  25 K 


Then, find the coulombs knowing that there are two electrons transferred per mol of H2.
 2 mol e    96485 C 
Coulombs = 1.716682x106 mole H 2 
= 3.3126813x1011 = 3.3x1011 C
 1 mol H   1 mol e  
2 


n=


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21-38


 1.44 J 
11
11
11
b) Energy (J) = 
 3.31268x10 C = 4.77026x10 = 4.8x10 J
 C 
1 kg
 1 kJ  

4
4
c) Mass (kg) = 4.77026x1011 J  3  
 = 1.19257x10 = 1.2x10 kg
4
 10 J   4.0x10 kJ 


21.112 a) The half-reactions are:
H2O(l) + Zn(s)  ZnO(s) + 2H+(aq) + 2e–
2H+(aq) + MnO2(s) + 2e–  Mn(OH)2(s)
Two moles of electrons flow per mole of reaction.
 1 mol Zn  1 mol MnO2   86.94 g MnO2 
b) Mass (g) of MnO2 =  4.50 g Zn  
 = 5.981196 = 5.98 g MnO2


 65.41 g Zn  1 mol Zn  1 mol MnO2 
 1 mol Zn  1 mol H 2 O   18.02 g H2 O 
Mass (g) of H2O =  4.50 g Zn  
 = 1.23972 = 1.24 g H2O


 65.41 g Zn  1 mol Zn  1 mol H2 O 
c) Total mass of reactants = 4.50 g Zn + 5.981196 g MnO2 + 1.23972 g H2O = 11.720916 = 11.72 g
 1 mol Zn   2 mol e    96, 485 coulombs 
4
4
d) Charge =  4.50 g Zn  
 = 1.32757x10 = 1.33x10 C
 

 
65.41
g
Zn
1
mol
Zn
1
mol
e




e) An alkaline battery consists of more than just reactants. The case, electrolyte paste, cathode, absorbent, and
unreacted reactants (less than 100% efficient) also contribute to the mass of an alkaline battery.
21.113 The reaction is: Ag+(aq) + e–  Ag(s)
Mass (g) of Ag = 1.8016 g – 1.7854 g = 0.0162 g Ag produced
 1 mol Ag   1 mol e    96, 485 coulombs 
Coulombs =  0.0162 g Ag  
 
 = 14.48616 = 14.5 C
 
1 mol e 
 107.9 g Ag   1 mol Ag  

21.114 Plan: Write balanced half-reactions for the reduction of each metal ion. From the current, 65.0% of the moles of
product will be copper and 35.0% zinc. Assume a current of exactly 100 C. The amount of current used to
generate copper would be (65.0%/100%)(100 C) = 65.0 C, and the amount of current used to generate zinc would
be (35.0%/100%)(100 C) = 35.0 C. Convert each coulomb amount to moles of electrons using the Faraday
constant and use the balanced reduction reactions to convert moles of electrons to moles and then mass of each
metal. Divide the mass of copper produced by the total mass of both metals produced and multiply by 100 to
obtain mass percent.
Solution:
The half-reactions are: Cu2+(aq) + 2e–  Cu(s) and Zn2+(aq) + 2e–  Zn(s)
 1 mol e    1 mol Cu   63.55 g Cu 
Mass (g) of copper =  65.0 C  
= 0.021406177 g Cu
 96, 485 C   2 mol e    1 mol Cu 




 1 mol e    1 mol Zn   65.41 g Zn 
Mass (g) of zinc =  35.0 C  
= 0.01186376 g Zn
 96, 485 C   2 mol e    1 mol Zn 




0.021406177 g Cu
Mass % of copper =
100  = 64.340900 = 64.3% Cu
0.021406177 g Cu + 0.01186376 g Zn
21.115 Plan: Write the reduction half-reaction for Au3+. Use the equation for the volume of a cylinder to find the volume
of gold required; use the density to convert volume of gold to mass and then moles of gold. The mole ratio in the
balanced reduction reaction is used to convert moles of gold to moles of electrons required and the Faraday
constant is used to convert moles of electrons to coulombs. Divide the coulombs by the current to obtain time in
seconds, which is converted to time in days. To obtain the cost, start by multiplying the moles of gold from part
a) by four to get the moles of gold needed for the earrings. Convert this moles to grams, then to troy ounces, and
finally to dollars.
Solution:
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21-39
The reaction is: Au3+(aq) + 3e–  Au(s)
a) V = r2h
2
 103 m   1 cm 
 4.00 cm 
3
V (cm3) =  
 0.25 mm  
  2  = 0.314159265 cm

2
1
mm



  10 m 
 19.3 g Au   1 mol Au 
Moles of Au = 0.314159265 cm3 
 = 0.03077804 mol Au

 1 cm3   197.0 g Au 


 3 mol e    96, 485 C   A

Time (days) =  0.03077804 mol Au  
 1 mol Au   1 mol e    C
 s



  1 h   1 day 
1




  0.013 A   3600 s   24 h 

= 7.931675 = 8 days
b) The time required doubles once for the second earring of the pair and doubles again for the second side, thus it
will take four times as long as one side of one earring.
Time = (4)(7.931675 days) = 31.7267 = 32 days
 197.0 g Au   1 troy oz   $1210 
c) Cost =  4  0.03077804 mol Au  


 = 943.60918 = $940
 1 mol Au   31.10 g   troy oz 
21.116 a) The half-reaction is: 2H2O(l)  O2(g) + 4H +(aq) + 4e–
Determine the moles of oxygen from the ideal gas equation. Use the half-reaction and the current to
convert the moles of oxygen to time.




  103 Pa 
PV
 99.8 kPa 10.0 L 
1 atm

n=
= 
 = 0.398569697 mol O2

5
L•atm 
RT

  1.01325 x10 Pa   1 kPa 
  0.0821 mol•K    273  28  K  



  96, 485 C   A   1   1 min 


 


  C

1
mol
e
  s   1.3 A   60 s 

= 1.97210x103 = 2.0x103 min
b) The balanced chemical equation is 2H2O(l)  2H2(g) + O2(g).
The moles of oxygen determined previously and this chemical equation leads to the mass of hydrogen.
 2 mol H 2  2.016 g H 2 
Mass (g) of H2 =  0.398569697 mol O2  

 = 1.60703 = 1.61 g H2
 1 mol O2  1 mol H 2 
 4 mol e 
Time (min) =  0.398569697 mol O 2  
 1 mol O
2

21.117 Corrosion is an oxidation process. This would be favored by the metal being in contact with the moist ground. To
counteract this, electrons should flow into the rails and away from the overhead wire, so the overhead wire should
be connected to the positive terminal.
21.118 Plan: Write the half-reactions and cell reaction for the silver battery. Convert mass of zinc to moles of zinc,
keeping in mind that only 80% of the zinc will react; from the moles of zinc, the moles of electrons required is
obtained. The Faraday constant is used to convert moles of electrons to charge in coulombs which is divided by
the current to obtain the time in seconds. The moles of zinc is also used to find the moles of Ag2O consumed and
the amount of Ag needed for that amount of Ag2O. Convert this mass of silver to troy ounces and then to dollars.
Solution:
The half-reactions and the cell reaction are:
Zn(s) + 2OH– (aq)  ZnO(s) + H2O(l) + 2e–
Ag2O(s) + H2O(l) + 2e–  2Ag(s) + 2OH–(aq)
Zn(s) + Ag2O(s)  ZnO(s) + 2Ag(s)
 80%   1 mol Zn 
Moles of Zn =  0.75 g Zn  
 = 0.00917291 mol Zn

 100%   65.41 g Zn 
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21-40
 2 mol e    96, 485C   A

a) Time (days) =  0.00917291 mol Zn  
 1 mol Zn   1 mol e    C
 s




  1 h   1 day 
1
  1 A  



6
  10 A   0.85  A   3600 s   24 h 

= 2.410262x104 = 2.4x104 days
 1 mol Ag 2 O   100%   2 mol Ag   107.9 g Ag 
b) Mass (g) of Ag =  0.00917291 mol Zn  




 1 mol Zn   95%   1 mol Ag 2 O   1 mol Ag 
= 2.0836989 = 2.1 g Ag

 95%   1 troy oz   $23.00  
c) Cost =  2.0836989 g Ag  





4
 100%   31.10 g Ag   troy oz   2.410262x10 days 
= 6.073818x10–5 = $6.1x10–5/day
21.119 Cu2+(aq) + 2e–  Cu(s)
 C   3600 s 
 1 mol e    1 mol Cu   63.55 g Cu 
Theoretical amount of copper =  5.8 A   s  
 
 10 h  

 
 A  1 h 
 96, 485 C   2 mol e   1 mol Cu 


= 68.7632 g Cu
53.4 g Cu
actual yield
Efficiency =
100  = 77.6578 = 78%
100  =
68.7632 g Cu
theoretical yield
The final assumes that 10 h has two significant figures.
21.120
Anode product
Cathode product
Species reduced
Species oxidized
a) Molten electrolysis
Cl2(g)
Na(l)
Na+(l)
Cl–(l)
b) Aqueous electrolysis
Cl2(g)
H2(g) and OH –(aq)
H2O(l)
Cl–(aq)

21.121 Plan: Since the cells are voltaic cells, the reactions occurring are spontaneous and will have a positive Ecell
.
Write the two half-reactions. When two half-reactions are paired, one half-reaction must be reversed and written

as an oxidation. Reverse the half-reaction that will result in a positive value of Ecell
using the relationship



Ecell
= Ecathode
– Eanode
. E  values are found in Appendix D. The oxidation occurs at the negative electrode
(the anode). Use the Nernst equation to find cell potential at concentrations other than 1 M.
Solution:
a) Cell with SHE and Pb/Pb2+:
Oxidation: Pb(s) → Pb2+(aq) + 2e–
E  = – 0.13 V
Reduction: 2H+(aq) + 2e– → H2(g)
E  = 0.0 V



Ecell
= Ecathode
– Eanode
= 0.0 V – (– 0.13 V) = 0.13 V
Cell with SHE and Cu/Cu2+:
Oxidation: H2(g) → 2 H+(aq) + 2e–
E  = 0.0 V
Reduction: Cu2+(aq) + 2e–→ Cu(s)
E  = 0.34 V



Ecell
= Ecathode
– Eanode
= 0.34 V– 0.00 V = 0.34 V
b) The anode (negative electrode) in cell with SHE and Pb/Pb2+ is Pb.
The anode in cell with SHE and Cu/Cu2+ is platinum in the SHE.
c) The precipitation of PbS decreases [Pb2+]. Use Nernst equation to see how this affects potential. Cell reaction
is:
Pb(s) + 2H+(aq)  Pb2+(aq) + H2(g)
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21-41

–
Ecell = Ecell
0.0592
log Q
n

Ecell = Ecell
–
[Pb2+ ]PH2
0.0592
log
n
[H + ]2
n = 2e–
[Pb2+ ]PH2
0.0592
Ecell =
–
log
2
[H + ]2
Decreasing the concentration of lead ions gives a negative value for the term:
[Pb2+ ]PH2
0.0592
log
2
[H + ]2

Ecell

When this negative value is subtracted from Ecell
, cell potential increases.
+
d) The [H ] = 1.0 M and the H2 = 1 atm in the SHE.
Cell reaction: Cu2+(aq) + H2(g)  Cu(s) + 2H+(aq)
0.0592

Ecell = Ecell
–
log Q
n
0.0592
[H + ]2

Ecell = Ecell
–
log
n = 2e–
2+
n
[Cu ]PH
2
0.0592
1
Ecell = 0.34 V –
log

16
2
1x10
1 atm 


Ecell = –0.1336 = –0.13 V
 1 cm 
21.122 a) Volume (cm3) = 2.0 cm2 7.5x106 m  2  = 0.0015 cm3
 10 m 
3
3
b) Mass = (0.0015 cm )(10.5 g Ag/1 cm ) = 0.01575 = 0.016 g
 1 mol Ag   1 mol e    96, 485 C   A   1 mA  
  1 min 
1


c) Time =  0.01575 g Ag  
 

 


  C
3
107.9
g
Ag
1
mol
Ag
12.0
mA


1
mol
e
10
A


  60 s 
 s 


= 19.56079 = 20. min
 1 troy oz   $23.00   100 cents 
d) Cost =  0.01575 g Ag  


 = 1.164791 = 1.2 ¢
$1

 31.10 g Ag   1 troy oz  



21.123 The reduction of H2O to H2 and OH – is easier than the reduction of Al3+ to Al.
21.124 The three steps equivalent to the overall reaction M+(aq) + e–  M(s) are:
Energy is –Hhydration
1) M+(aq)  M+(g)
2) M+(g) + e–  M(g)
Energy is –IE or –Hionization
3) M(g)  M(s)
Energy is –Hdeposition
The energy for step 3 is similar for all three elements, so the difference in the energy for the overall reaction
depends on the values for –Hhydration and –IE. The lithium ion has a more negative hydration energy than Na+ and
K+ because it is a smaller ion with large charge density that holds the water molecules more tightly. The amount
of energy required to remove the waters surrounding the lithium ion offsets the lower ionization energy to make
the overall energy for the reduction of lithium larger than expected.
21.125 The key factor is that the table deals with electrode potentials in aqueous solution. The very high and low standard
electrode potentials involve extremely reactive substances, such as F2 (a powerful oxidant), and Li (a powerful
reductant). These substances react directly with water, rather than according to the desired half-reactions. An
alternative (essentially equivalent) explanation is that any aqueous cell with a voltage of more than 1.23 V has the
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21-42
ability to electrolyze water into hydrogen and oxygen. When two electrodes with 6 V across them are placed in
water, electrolysis of water will occur.
21.126 Wherever the tin surface is scratched to expose the iron, the iron corrosion occurs more rapidly because iron is a
better reducing agent (compare E° values). An electrochemical cell is set up where iron becomes the anode and tin
the cathode. A coating on the inside of the can separates the tin from the normally acidic contents, which would
react with the tin and add an unwanted “metallic” taste.
21.127 Plan: Write the half-reaction for the reduction of Al3+. Convert mass of Al to moles and use the mole ratio in
the balanced reaction to find the number of moles of electrons required for every mole of Al produced. The
Faraday constant is used to find the charge of the electrons in coulombs. To find the time, the charge is divided
by the current. To calculate the electrical power, multiply the time by the current and voltage, remembering that
1 A = 1 C/s (thus, 100,000 A is 100,000 C/s) and 1 V = 1 J/C (thus, 5.0 V = 5.0 J/C). Change units of J to kW • h.
To find the cost of the electricity, convert the mass of aluminum from lb to kg and use the kW • h per 1000 kg of
aluminum calculated in part b) to find the kW • h for that mass of aluminum, keeping in mind the 90.% efficiency.
Solution:
a) Aluminum half-reaction: Al3+(aq) + 3 e–  Al(s), so n = 3. Remember that 1 A = 1 C/s.
 103 g   1 mol Al   3 mol e    96, 485 C   A  

1


Time (s) = 1000 kg Al  
 1 kg   26.98 g Al   1 mol Al   1 mol e    C   100, 000 A 


 s 



5
5
= 1.0728503x10 = 1.073x10 s
The molar mass of aluminum limits the significant figures.
b)
 100,000 C   5.0 J   1 kJ   1 kW•h 
4
4
Power = 1.0728503x105 s 
 = 1.4900699x10 = 1.5x10 kW•h
 C  3 
3
s
10
J
3.6x10
kJ





4
 1 kg   1.4900699x10 kW•h   0.123 cents   100% 
c) Cost = 1 lb Al  
 
 

 = 0.923551 = 0.92 ¢/lb Al
1000 kg Al
 2.205 lb  
  1 kW•h   90.% 


21.128 a) Electrons flow from magnesium bar to the iron pipe since magnesium is more easily oxidized than iron.
b) The magnesium half-reaction is: Mg(s)  Mg2+(aq) + 2e–
Current is charge per time. The mass of magnesium can give the total charge. Convert the mass of magnesium to
moles of magnesium and multiply by two moles of electrons produced for each mole of magnesium and by the
Faraday constant to convert the moles of electrons to coulombs of charge. Time must be in seconds, so convert the
8.5 yr to s.
 103 g   1 mol Mg   2 mol e    96, 485 C 
12 kg Mg  




1 kg   24.31 g Mg   1 mol Mg   mol e    A 
ch arg e


 = 0.35511 = 0.36 A
Current =
=
C 
time
 365.25 days   24 h   3600 s 
 s
8.5 yr  



1 yr

  1 days   1 h 
21.129 Plan: When considering two substances, the stronger reducing agent will reduce the other substance.
Solution:
Statement: Metal D + hot water  reaction Conclusion: D reduces water to produce H2(g). D is a stronger
reducing agent than H+.
Statement: D + E salt  no reaction
Conclusion: D does not reduce E salt, so E reduces D salt. E is better
reducing agent than D.
Statement: D + F salt  reaction
Conclusion: D reduces F salt. D is better reducing agent than F.
If E metal and F salt are mixed, the salt F would be reduced producing F metal because E has the greatest
reducing strength of the three metals (E is stronger than D and D is stronger than F). The ranking of increasing
reducing strength is F < D < E.
21.130 3Pt(s) + 16H +(aq) + 4NO3–(aq) + 18Cl–(aq)  3PtCl62–(aq) + 4NO(g) + 8H2O(l)
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21-43
21.131 Plan: Examine the change in oxidation numbers in the equations to find n, the moles of electrons transferred. Use
G =  nFE to calculate G. Substitute J/C for V in the unit for E . Convert G to units of kJ and divide by the
total mass of reactants to obtain the ratio.
Solution:
a) Cell I: Oxidation number (O.N.) of H changes from 0 to +1, so one electron is lost from each of four hydrogen
atoms for a total of four electrons. O.N. of oxygen changes from 0 to –2, indicating that two electrons are gained
by each of the two oxygen atoms for a total of four electrons. There is a transfer of four electrons in the reaction.
G = –nFE = – (4 mol e–)(96,485 C/mol e–)(1.23 J/C) = –4.747062x105 = –4.75x105 J
Cell II: In Pb(s)  PbSO4, O.N. of Pb changes from 0 to +2 and in PbO2  PbSO4, O.N. of Pb changes from +4
to +2. There is a transfer of two electrons in the reaction.
G = –nFE = – (2 mol e–)(96,485 C/mol e–)(2.04 J/C) = –3.936588x105 = –3.94x105 J
Cell III: O.N. of each of two Na atoms changes from 0 to +1 and O.N. of Fe changes from +2 to 0. There is a
transfer of two electrons in the reaction.
G = –nFE = – (2 mol e–)(96,485 C/mol e–)(2.35 J/C) = –4.534795x105 = –4.53x105 J
 2.016 g H 2 
 32.00 g O2 
b) Cell I: Mass of reactants =  2 mol H2  
  1 mol O2  
 = 36.032 g
 1 mol H2 
 1 mol O2 
 4.747062x105 J   1 kJ 
wmax
= 
  3  = –13.17457 = –13.2 kJ/g

reactan t mass
36.032 g

  10 J 
Cell II: Mass of reactants =
 239.2 g PbO2 
 98.09 g H 2SO4 
 207.2 g Pb 
1 mol Pb  
   2 mol H 2SO4  

  1 mol PbO2  
 1 mol Pb 
 1 mol PbO2 
 1 mol H 2SO4 
= 642.58 g
 3.936588x105 J   1 kJ 
wmax
= 
  3  = –0.612622 = –0.613 kJ/g
reactan t mass 
642.58 g
  10 J 
 126.75 g FeCl2 
 22.99 g Na 
Cell III: Mass of reactants =  2 mol Na  
 = 172.73 g
  1 mol FeCl2  
 1 mol Na 
 1 mol FeCl2 
 4.534795x105 J   1 kJ 
wmax
= 
  3  = –2.625366 = –2.63 kJ/g
reactan t mass 
172.73 g
  10 J 
Cell I has the highest ratio (most energy released per gram) because the reactants have very low mass while Cell
II has the lowest ratio because the reactants are very massive.
21.132 The current traveling through both cells is the same, so the amount of silver is proportional to the amount of zinc
based on their reduction half-reactions:
Zn(s)  Zn2+(aq) + 2e– and Ag+(aq) + e–  Ag(s)
 1 mol Zn   2 mol e    1 mol Ag   107.9 g Ag 
Mass (g) of Ag = 1.2 g Zn  
 
 
 = 3.9590 = 4.0 g Ag
 

 65.41 g Zn   1 mol Zn   1 mol e   1 mol Ag 
21.133 A standard thermodynamic table would allow the calculation of the equilibrium constant by the equation
G° = –RT ln K. G° could be calculated from Gf values, or H f and S° values. A table of standard electrode
potentials would allow the calculation of the equilibrium constant by the following equation:
E° = (0.0592 V/n) log K
21.134 2Fe(s) + O2(g) + 4H+(aq)  2H2O(l) + 2Fe2+(aq)
An increase in temperature increases the cell potential for the above reaction. An increase in the temperature will
cause the reaction rate to increase. Finally, Kw for water increases with temperature, thus at a higher temperature
there will be a higher hydrogen ion concentration.
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21-44
21.135 Plan: Write the balanced equation. Multiply the current and time to calculate total charge in coulombs. Remember
that the unit 1 A is 1 C/s, so the time must be converted to seconds. From the total charge, the number of
electrons transferred to form copper is calculated by dividing total charge by the Faraday constant. Each mole of
copper deposited requires two moles of electrons, so divide moles of electrons by two to get moles of copper.
Then convert to grams of copper. The initial concentration of Cu2+ is 1.00 M (standard condition) and initial
volume is 345 mL. Use this to calculate the initial moles of copper ions, then subtract the number of moles of
copper ions converted to copper metal and divide by the cell volume to find the remaining [Cu2+].
Solution:
a) Since the cell is a voltaic cell, write a spontaneous reaction. The reduction of Cu2+ is more spontaneous
than the reduction of Sn2+: Cu2+(aq) + Sn(s)  Cu(s) + Sn2+(aq)
 C   3600 s 
 1 mol e    1 mol Cu   63.55 g Cu 
Mass (g) of Cu =  0.17 A   s  
 
  48.0 h  

 
 A  1 h 
 96, 485 C   2 mol e   1 mol Cu 


= 9.674275 = 9.7 g Cu

mol Cu 2 
b) Initial moles of Cu2+ =  1.00

L


 103 L 
2+
  345 mL  
 = 0.345 mol Cu
1
mL



 1 mol Cu 
2+
Moles of Cu2+ reduced =  9.674275 g Cu  
 = 0.1522309205 mol Cu
 63.55 g Cu 
Remaining moles of Cu2+ = initial moles – moles reduced = 0.345 mol – 0.1522309205 = 0.1927691 mol Cu2+
0.1927691 mol Cu 2 
M Cu2+ =
= 0.558751 = 0.56 M Cu2+
 103 L 
 345 mL  

 1 mL 
21.136 2H2O(l) + 2e–  H2(g) + 2OH–(aq)
0.0592

Ecell = Ecell
–
log Q
n
2
0.0592 V

Ecell = Ecell
–
log  PH 2  OH   


2


0.0592 V
Ecell = –0.83 V –
log [1(1.0x10–7)2]
2
= –0.83 V + 0.4144 = –0.4156 = –0.42 V
21.137 Plan: Examine each reaction to determine which reactant is the oxidizing agent; the oxidizing agent is the reactant
that gains electrons in the reaction, resulting in a decrease in its oxidation number.
Solution:
From reaction between U3+ + Cr3+  Cr2+ + U4+, find that Cr3+ oxidizes U3+.
From reaction between Fe + Sn2+  Sn + Fe2+, find that Sn2+ oxidizes Fe.
From the fact that there is no reaction that occurs between Fe and U4+, find that Fe2+ oxidizes U3+.
From reaction between Cr3+ + Fe  Cr2+ + Fe2+, find that Cr3+ oxidizes Fe.
From reaction between Cr2+ + Sn2+  Sn + Cr3+, find that Sn2+ oxidizes Cr2+.
Notice that nothing oxidizes Sn, so Sn2+ must be the strongest oxidizing agent. Both Cr3+ and Fe2+ oxidize U3+, so
U4+ must be the weakest oxidizing agent. Cr3+ oxidizes iron so Cr3+ is a stronger oxidizing agent than Fe2+.
The half–reactions in order from strongest to weakest oxidizing agent:
Sn2+(aq) + 2e–  Sn(s)
Cr3+(aq) + e–  Cr2+(aq)
Fe2+(aq) + 2e–  Fe(s)
U4+(aq) + e–  U3+(aq)
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21-45
21.138 The given half-reactions are:
(1) Fe3+(aq) + e–  Fe2+(aq)
(2) Fe2+(aq) + 2e–  Fe(s)
(3) Fe3+(aq) + 3e–  Fe(s)
E°2 = –0.44 V
(a) E°1 = +0.77 V
Adding half-reaction (1) and (2) gives half-reaction (3), thus
E°3 = E°1 + E°2 = +0.77 V + (–0.44 V) = 0.33 V
(b) G° = –nFE°
Reaction 1:
G° = – (1 mol e–)(96,485 C/mol e–)(0.77 J/C) = –7.429345x104 = –7.4x104 J
Reaction 2:
G° = – (2 mol e–)(96,485 C/mol e–)(–0.44 J/C) = 8.49068x104 = 8.5x104 J
c) G3° = G1° + G2° = –7.429345x104 J + 8.49068x104 J = 1.061335x104 = 1.1x104 J
d) E° = –G°/nF = – (1.061335x104 J)/(3 mol e–)(96,485 C/mol e–)[V/(J/C)] = –0.036667 = –0.037 V
e) The half-cells (1) and (2) add to (3) so their voltages should add to E°3.
(red half-reaction)
21.139 6e– + 14H + + Cr2O72–  2Cr3+ + 7H2O
(ox half-reaction)
CH3CH2OH + H2O  CH3COOH + 4H + + 4e–
2{6e– + 14H + + Cr2O72–  2Cr3+ + 7H2O}
3{CH3CH2OH + H2O  CH3COOH + 4H + + 4e– }
12e– + 28H + + 2Cr2O72–  4Cr3+ + 14H2O
3CH3CH2OH + 3H2O  3CH3COOH + 12H + + 12e–
Overall: 3CH3CH2OH + 2Cr2O72– + 16H +  3CH3COOH + 4Cr3+ + 11H2O
21.140 At STP hydrogen gas occupies 22.4 L/mol. The reduction of zinc and of hydrogen both required two moles of
electrons per mole of product, thus, the current percentages are equal to the mole percentages produced.
 103 g   1 mol Zn   8.50 mol H 2   22.4 L H 2 
= 31.81278578 = 31.8 L H2
Volume (L) of H2 = 1 kg Zn  
 1 kg   65.41 g Zn   91.50 mol Zn   1 mol H 


2 




21.141 Plan: Write a balanced equation that gives a positive Ecell
for a spontaneous reaction. Calculate the Ecell
and
use the Nernst equation to find the silver ion concentration that results in the given Ecell.
Solution:
a) The calomel half-cell is the anode and the silver half-cell is the cathode. The overall reaction is:
2Ag+(aq) + 2Hg(l) + 2Cl–(aq)  2Ag(s) + Hg2Cl2(s)



Ecell
= Ecathode
– Eanode
= 0.80 V – 0.24 V = 0.56 V with n = 2.
Use the Nernst equation to find [Ag+] when Ecell = 0.060 V.
0.0592

Ecell = Ecell
–
log Q
n
0.0592 V
1

–
log
Ecell = Ecell
2
2
2
 Ag   Cl 

 

0.0592V
1
log
0.060 V = 0.56 V –
2
2
2
 Ag   Cl 

 

The problem suggests assuming that [Cl–] is constant. Assume it is 1.00 M.
0.0592V
1
– 0.50 V = –
log
2
2

 Ag  1.00 2


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21-46
16.89189 = log
1016.89189 =
1
2
 Ag   1.00 2


1
 Ag  


7.7963232x1016 =
2
1
2
 Ag  


7.7963232x1016[Ag+]2 = 1
[Ag+]2 = 1.282656x10–17
[Ag+] = 3.581419x10–9 = 3.6x10–9 M
b) Again use the Nernst equation and assume [Cl–] = 1.00 M.
0.0592 V
1

Ecell = Ecell
–
log
2
2
2
 Ag   Cl 

 

0.0592 V
1
log
0.53 V = 0.56 V –
2
2
2

 Ag  Cl 

 

0.0592 V
1
log
– 0.03 V = –
2
2
2
 Ag   Cl 

 

1
1.0135135 = log
2

 Ag  1.00 2


1
101.0135135 =
2
 Ag  


1
10.31605146 =
2
 Ag  


10.31605146[Ag+]2 = 1
[Ag+]2 = 0.0969363
[Ag+] = 0.311346 = 0.3 M
21.142 Oxidation: Ag(s)  Ag+(aq) + e–
Reduction: AgCl(s) + e–  Ag(s) + Cl–(aq)
Overall:

Eanode
= +0.80 V

Ecathode
= + 0.22 V
AgCl(s)  Ag+(aq) + Cl–(aq)



Ecell
= Ecathode
– Eanode
= 0.22 V – 0.80 V = –0.58V

1 0.58 V 
nEcell
=
= –9.797297297
0.0592
0.0592 V
K = 1.594788x10–10 = 1.6x10–10
log Ksp =

and the cell potential for both the waste
21.143 Plan: Use the Nernst equation to write the relationship between Ecell
stream and for the silver standard.
Solution:
a) The reaction is Ag+(aq) → Ag(s) + 1e–
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21-47

Ecell = Ecell
–
0.0592
log Q
n
Nonstandard cell:
 0.0592 

Ewaste = Ecell
–
log [Ag+]waste
 
1
e


Standard cell:
 0.0592 

+
Estandard = Ecell
– 
 log [Ag ]standard
 1 e 
b) To find [Ag+]waste:
 0.0592 
 0.0592 

+
Ecell
= Estandard + 
log [Ag+]standard = Ewaste + 
 log [Ag ]waste
 
 1e 
 1 e 
[Ag + ]waste
 0.0592 
Ewaste – Estandard = – 
log

 1 e 
[Ag + ]standard
Estandard – Ewaste = (0.0592 V)(log [Ag+]waste – log [Ag+]standard)
Es tan dard  Ewaste
= (log [Ag+]waste – log [Ag+]standard)
0.0592 V
log [Ag+]waste =
Es tan dard  Ewaste
+ log [Ag+]standard
0.0592 V


E
 E waste     
[Ag+]waste =  antilog  standard
   Ag standard
0.0592 V



+
c) Convert M to ng/L for both [Ag ]waste and [Ag+]standard:
Ewaste – Estandard = (–0.0592 V) log

[Ag + ]waste
[Ag + ]standard
If both silver ion concentrations are in the same units, in this case ng/L, the “conversions”
cancel and the equation derived in part b) applies if the standard concentration is in ng/L.

E
 Ewaste   


Conc. (Ag+)waste =  antilog  s tan dard
   Conc. Ag s tan dard 
0.0592
V


  



d) Plug the values into the answer for part c).

 0.003  
[Ag+]waste =  antilog 
  1000.ng/L  = 889.8654 = 900 ng/L

 0.0592 V  
e) Temperature is included in the RT/nF term, which equals 0.0592 V/n at 25°C. To account for different
temperatures, insert the RT/nF term in place of 0.0592 V/n.
 2.303RT 
 2.303RT 
+
Estandard + 
log [Ag+]standard = Ewaste + 

 log [Ag ]waste
 nF 
 nF 
 2.303R 
+
+
Estandard – Ewaste = 
 (Twaste log [Ag ]waste – Tstandard log [Ag ]standard)
 nF 
 nF 
+
+
(Estandard – Ewaste) 
 = Twaste log [Ag ]waste – Tstandard log [Ag ]standard
 2.303R 
 nF 
+
+
(Estandard – Ewaste) 
 + Tstandard log [Ag ]standard = Twaste log [Ag ]waste
 2.303R 
 E
 
standard  Ewaste   nF / 2.303 R   Tstandard log  Ag 

standard
log [Ag ]waste =

T

waste

+





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21-48
 E
 
standard  Ewaste   nF / 2.303 R   Tstandard log  Ag 

standard
[Ag ]waste = anti log

Twaste


+
21.144 Reduction: Ag+(aq) + e–  Ag(s)
Oxidation: Ag(s) + 2NH3(aq)  Ag(NH3)2+(aq) + e–
Overall: Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)






Ecathode
= 0.80 V

Eanode
= 0.37 V



Ecell
= Ecathode
– Eanode
= 0.80 V  0.37 V = 0.43 V

1 0.43 V 
nEcell
=
= 7.26351
0.0592
0.0592 V
K = 1.834467x107 = 1.8x107
log Ksp =
21.145 Plan: Multiply the current in amperes by the time in seconds to obtain coulombs. Convert coulombs to
moles of electrons with the Faraday constant and use the mole ratio in the balanced half-reactions to convert
moles of electrons to moles and then mass of reactants. Divide the total mass of reactants by the mass of the
battery to find the mass percentage that consists of reactants.
Solution:
a) Determine the total charge the cell can produce.
 103 A   3600 s   1 C 
= 1.08x103 C
Capacity (C) =  300. mA•h  

 1 mA   1 h   1 A•s 


b) The half-reactions are:
Cd0  Cd2+ + 2e– and NiO(OH) + H2O(l) + e–  Ni(OH)2 + OH–
Assume 100% conversion of reactants.
 1 mol e    1 mol Cd   112.4 g Cd 
= 0.62907 = 0.629 g Cd
Mass (g) of Cd = 1080 C  
 96,485 C   2 mol e    1 mol Cd 


 1 mol e    1 mol NiO(OH)   91.70 g NiO(OH) 
Mass (g) of NiO(OH) = 1080 C  
  1 mol NiO(OH) 
 96,485 C  
1 mol e 




= 1.026439 = 1.03 g NiO(OH)
 1 mol e    1 mol H 2 O   18.02 g H 2 O 
= 0.20170596 = 0.202 g H2O
Mass (g) of H2O = 1080 C  
 96,485 C   1 mol e    1 mol H O 
2




Total mass of reactants = 0.62907 g Cd + 1.026439 g NiO(OH) + 0.20170596 g H2O
= 1.857215 = 1.86 g reactants
1.85721 g
c) Mass % reactants =
100  = 10.14872 = 10.1%
18.3 g
21.146 a) 2Zn  2Zn2+ + 4e– and
4e– + O2  2O2–
Four mol e– flow per mole of reaction.
Mass (g) of Zn = (1/10)(0.275 g battery) = 0.0275 g Zn (assuming the 1/10 is exact.)
Coulombs = (0.0275 g Zn)(1 mol Zn/65.41 g Zn)(4 mol e–/2 mol Zn)(96,485 C/1 mol e–) = 81.1294 = 81.1 C
b) Free energy (J) = (volts)(coulombs) = (1.3 V)(81.1294 C) = 105.46822 = 1.1x102 J
21.147 Plan: For a list of decreasing reducing strength, place the elements in order of increasing (more positive) E°.
Metals with potentials lower than that of water (–0.83 V) can displace hydrogen from water by reducing the
hydrogen in water. Metals with potentials lower than that of hydrogen (0.00 V) can displace hydrogen from acids
by reducing the H+ in acid. Metals with potentials above that of hydrogen (0.00 V) cannot displace (reduce)
hydrogen.
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21-49
Solution:
Reducing agent strength: Li > Ba > Na > Al > Mn > Zn > Cr > Fe > Ni > Sn > Pb > Cu > Ag > Hg > Au
These can displace H2 from water: Li, Ba, Na, Al, and Mn.
These can displace H2 from acid: Li, Ba, Na, Al, Mn, Zn, Cr, Fe, Ni, Sn, and Pb.
These cannot displace H2: Cu, Ag, Hg, and Au.

for the reaction given from the free energy:
21.148 Determine the Ecell

G° = –nF Ecell

Ecell
= –G°/nF = – [(–298 kJ/mol)(103 J/1 kJ)]/(2 mol e–)(96485 C/mol e–)[V/(J/C)] = 1.54428 V

Using E° for Cu2+ + 2 e–  Cu (cathode) and the Ecell
just found:



Ecell
= Ecathode
– Eanode
= 1.54428 V

1.54428 V = 0.34 V – Eanode

Eanode
= 0.34 V – 1.54428 V = –1.20428 = –1.20 V
The standard reduction potential of V2+ is –1.20 V.
The difference in the cell potentials of V2+ and Ti2+ is 0.43 V.
The cell potential of Ti2+ is either 0.43 V more positive than that of V2+: –1.20 V + 0.43 V = -0.77 V
Or the cell potential of Ti2+ is 0.43 V more negative than that of V2+: –1.20 V – 0.43 V = –1.63 V
Since the problem states that both Ti and V are reactive enough to displace H2 from water, the reduction
cell potentials of both Ti2+ and V2+ must be more negative than that of H2O, which is –0.83 V. Therefore, the
reduction potential of Ti2+ must be –1.63 V.
21.149 a) The iodine goes from –1 to 0, so it was oxidized.
Iodine was oxidized, so S4O62– is the oxidizing agent.
Iodine was oxidized, so I– is the reducing agent.
b) G° = –nFE° n = 2

Ecell
= –G°/nF = – [(87.8 kJ/mol)(103 J/1 kJ)]/(2 mol e–)(96,485 C/mol e–)[V/(J/C)] = – 0.454993 = – 0.455 V
c) S4O62–(aq) + 2e–  2S2O32– (aq)
Oxygen remains –2 throughout.
Sulfur is +2.5 in S4O62– (2.5 is an average).
Sulfur is +2 in S2O32– .
The potential for the iodine half-reaction comes from the Appendix. Since the iodine was oxidized, it is the
anode.



Ecell
= Ecathode
– Eanode
= – 0.454993 V

Ecathode
– (0.53 V) = –0.454993 V

Ecathode
= 0.53 V – 0.454993 V = 0.075007 = 0.08 V
21.150 a) The reference half-reaction is: Cu2+(aq) + 2e–  Cu(s)
E° = 0.34 V
Before the addition of the ammonia, Ecell = 0. The addition of ammonia lowers the concentration of copper ions
through the formation of the complex Cu(NH3)42+. The original copper ion concentration is [Cu2+]original, and the
copper ion concentration in the solution containing ammonia is [Cu2+]ammonia.
The Nernst equation is used to determine the copper ion concentration in the cell containing ammonia.
The reaction is Cu2+initial(aq) + Cu(s) → Cu(s) + Cu2+ammonia(aq).
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21-50
The half-cell with the larger concentration of copper ion (no ammonia added) is the reduction and the half-cell
with the lower concentration of copper ion due to the addition of ammonia and formation of the complex is the
oxidation.
0.0592

Ecell = Ecell
–
log Q
n
 Cu 2  
0.0592 V

 ammonia
0.129 V = 0.00 V –
log
2
 Cu 2  

 original
 Cu 2  
0.0592 V

 ammonia
log
0.129 V = –
2
0.0100original
–4.358108108 = log
 Cu 2  

 ammonia
0.0100

original
 Cu 2  

 ammonia
4.3842155x10 =
0.0100

original
–5
[Cu2+]ammonia = 4.3842155x10–7 M
This is the concentration of the copper ion that is not in the complex. The concentration of the complex and of the
uncomplexed ammonia must be determined before Kf may be calculated.
The original number of moles of copper and the original number of moles of ammonia are found from the original
volumes and molarities:
 0.0100 mol Cu(NO3 ) 2   1 mol Cu 2    103 L 
Original moles of copper = 

  90.0 mL 
 
L

  1 mol Cu(NO3 ) 2   1 mL 
= 9.00x10–4 mol Cu2+
 0.500 mol NH 3   10 3 L 
–3
Original moles of ammonia = 
 10.0 mL  = 5.00x10 mol NH3
 
L
1
mL



Determine the moles of copper still remaining uncomplexed.
 4.3842155x10 7 mol Cu 2    10 3 L 
–8
Remaining moles of copper = 
 
 100.0 mL  = 4.3842155x10 mol Cu

L
1
mL



The difference between the original moles of copper and the copper ion remaining in solution is the copper in the
complex (= moles of complex). The molarity of the complex may now be found.
Moles copper in complex = (9.00x10–4  4.3842155x10–8) mol Cu2+ = 8.9995616x10–4 mol Cu2+
 8.9995616x10 4 mol Cu 2    1 mol Cu(NH 3 ) 24    1 mL 
Molarity of complex = 
  3 
 

100.0 mL
1 mol Cu 2 


  10 L 
–3
2+
= 8.9995616x10 M Cu(NH3)4
The concentration of the remaining ammonia is found as follows:


3
4
2   4 mol NH 3 
 5.00x10 mol NH 3  8.9995616x10 mol Cu 
2  

 1 mol Cu    1 mL 
Molarity of ammonia = 
  103 L 
100.0 mL






= 0.014001754 M ammonia
The Kf equilibrium is:
Cu2+(aq) + 4NH3(aq)  Cu(NH3)42+(aq)

 

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21-51
 Cu(NH 3 ) 4 2  
8.9995616x10 3 
 =


Kf = 
= 5.34072x1011 = 5.3x10–11
 Cu 2    NH 3  4
 4.3842155x10 7  0.014001754 4




b) The Kf will be used to determine the new concentration of free copper ions.
Moles uncomplexed ammonia before the addition of new ammonia =
(0.014001754 mol NH3/L)(10–3 L/1 mL)(100.0 mL) = 0.0014001754 mol NH3
Moles ammonia added = 5.00x10–3 mol NH3 (same as original moles of ammonia)
From the stoichiometry:
Cu2+(aq)
+
4NH3(aq)

Cu(NH3)42+(aq)
–8
Initial moles
4.3842155x10 mol
0.001400175 mol
8.9995616x10–4 mol
–3
Added moles
5.00x10 mol
+(4.3842155x10–8 mol)
Cu2+ is limiting –(4.3842155x10–8 mol) –4(4.3842155x10–8 mol)
After the reaction
0
0.006400 mol
9.00000x10–4 mol
Determine concentrations before equilibrium:
[Cu2+] = 0
[NH3] = (0.006400 mol NH3/110.0 mL)(1 mL/10–3 L) = 0.0581818 M NH3
[Cu(NH3)42+] = (9.00000x10–4 mol Cu(NH3)42+/110.0 mL)(1 mL/10–3 L)
= 0.008181818 M Cu(NH3)42+
Now allow the system to come to equilibrium:
Cu2+(aq)
+
4NH3(aq)

Cu(NH3)42+(aq)
Initial molarity 0
0.0581818
0.008181818
Change
+x
+4x
–x
Equilibrium
x
0.0581818 + 4 x
0.008181818 – x
 Cu(NH 3 ) 4 2  
 =  0.008181818  x  = 5.34072x1011
Kf = 

2
 Cu   NH 3  4
 x 0.0581818  4x 4


Assume – x and + 4x are negligible when compared to their associated numbers:
0.008181818
Kf = 5.34072x1011 =
 x 0.05818184
x = [Cu2+] = 1.3369x10–9 M Cu2+
Use the Nernst equation to determine the new cell potential:
 Cu 2  
0.0592 V

 ammonia
E = 0.00 V –
log
2
 Cu 2  

 original
1.3369x109 
0.0592 V

E=–
log 
2
0.0100
E = 0.203467 = 0.20 V
c) The first step will be to do a stoichiometry calculation of the reaction between copper ions and hydroxide ions.

3
 0.500 mol NaOH   1 mol OH   10 L 
–3
–
Moles of OH– = 



  1 mol NaOH   1 mL  10.0 mL  = 5.00x10 mol OH
L




The initial moles of copper ions were determined earlier: 9.00x10–4 mol Cu2+
The reaction:
Cu2+(aq) +
2OH–(aq)

Cu(OH)2(s)
–4
Initial moles
9.00x10 mol
5.00x10–3 mol
Cu2+ is limiting –(9.00x10–4 mol)
–2(9.00x10–4 mol)
After the reaction
0
0.0032 mol
Determine concentrations before equilibrium:
[Cu2+] = 0
[NH3] = (0.0032 mol OH–/100.0 mL)(1 mL/10–3 L) = 0.032 M OH–
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21-52
Now allow the system to come to equilibrium:
Cu(OH)2(s)  Cu2+(aq)
+
2OH–(aq)
Initial molarity
0.0
0.032
Change
+x
+2x
Equilibrium
x
0.032 + 2 x
Ksp = 2.2x10–20 = [Cu2+][OH–]2
Ksp = 2.2x10–20 = [x][0.032 + 2x]2
Assume 2x is negligible compared to 0.032 M.
Ksp = 2.2x10–20 = [x][0.032]2
x = [Cu2+] = 2.1484375x10–17 = 2.1x10–17 M
Use the Nernst equation to determine the new cell potential:
 Cu 2  
0.0592 V

 hydroxide
E = 0.00 V –
log
2
 Cu 2  

 original
 2.1484375x1017 
0.0592 V

E=–
log 
2
0.0100
E = 0.434169 = 0.43 V
d) Use the Nernst equation to determine the copper ion concentration in the half-cell containing the hydroxide ion.
 Cu 2  
0.0592 V

 hydroxide
E = 0.00 V –
log
2
 Cu 2  

 original
 Cu 2  
0.0592 V

 hydroxide
log
0.340 = –
2
0.0100


 Cu 2  

 hydroxide
–11.486486 = log
0.0100
3.2622257x10
–12
 Cu 2  

 hydroxide
=
0.0100
[Cu2+]hydroxide = 3.2622257x10–14 M
Now use the Ksp relationship:
Ksp = [Cu2+][OH–]2 = 2.2x10–20
Ksp = 2.2x10–20 = [3.2622257x10–14][OH–]2
[OH–]2 = 6.743862x10–7
[OH–] = 8.2121x10–4 = 8.2x10–4 M OH– = 8.2x10–4 M NaOH
21.151 a) The half-reactions found in the Appendix are:
Au3+(aq) + 3e–  Au(s) E° = 1.50 V
Cr3+(aq) + 3e–  Cr(s)
E° = –0.74 V
Co2+(aq) + 2e–  Co(s) E° = –0.28 V
Zn2+(aq) + 2e–  Zn(s) E° = –0.76 V



Calculate Ecell
= Ecathode
– Eanode
Au/Cr

Ecell
= 1.50 V – (–0.74 V) = 2.24 V
Co/Zn

Ecell
= –0.28 V – (–0.76 V) = 0.48 V
b) Connecting the cells as two voltaic cells in series will add the voltages.



Eseries
= EAu/Cr
+ ECo/Zn
= 2.24 V + 0.48 V = 2.72 V
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21-53
c) Connecting the cells as one voltaic cell (Au/Cr) and one electrolytic (C /Zn) in series will result in the
difference in the voltages.



Eseries
= EAu/Cr
– ECo/Zn
= 2.24 V – 0.48 V = 1.76 V
d) In parts a-c), Au3+ is reduced in the Au/Cr cell.
In parts a-b), Co2+ is reduced in the Co/Zn cell.
The connection in part c), forces the Co/Zn cell to operate in reverse, thus, Zn2+ is reduced.
 1 mol Au   3 mol e    1 mol Zn   65.41 g Zn 
e) Mass (g) of Zn =  2.00 g Au  
 
 
 = 0.99609 = 0.996 g Zn
 

 197.0 g Au   1 mol Au   2 mol e   1 mol Zn 
21.152 a) The half-reactions found in the Appendix are:
E° = – 0.40 V
Oxidation:
Cd(s)  Cd2+(aq) + 2e–
Reduction: Cu2+(aq) + 2e–  Cu(s)
E° = 0.34 V
Overall: Cd(s) + Cu2+(aq)  Cd2+(aq) + Cu(s)



Ecell
= Ecathode
– Eanode
= 0.34 V – (–0.40 V) = 0.74 V
Note: Cd is a better reducing agent than Cu so Cu2+ reduces while Cd oxidizes.

G° = nFEcell
G° = – (2 mol e–)(96,485 C/mol e–)(0.74 J/C) = –1.427978x105 = –1.4x105 J
log K =
log K =

nEcell
0.0592
2  0.74 
= 25
0.0592
25
K = 1x10
b) The cell reaction is: Cu2+(aq) + Cd(s) Cu(s) + Cd2+(aq)
Use the Nernst equation:
0.0592

Ecell = Ecell
–
log Q
n
 Cd 2  
0.0592 V

E = 0.74 V –
log 
2
 Cu 2  


An increase in the cadmium concentration by 0.95 M requires an equal decrease in the copper
concentration since the mole ratios are 1:1. Thus, when [Cd2+] = 1.95 M,
[Cu2+] = (1.00 – 0.95) M = 0.05 M.
1.95
0.0592 V
E = 0.74 V –
log
2
0.05
E = 0.69290 = 0.69 V
c) At equilibrium, Ecell = 0, and G = 0
The Nernst equation is necessary to determine the [Cu2+].
Let the copper ion completely react to give [Cu2+] = 0.00 M and [Cd2+] = 2.00 M. The system can now go to
equilibrium giving [Cu2+] = +x M and [Cd2+] = (2.00 – x) M.
0.0592

Ecell = Ecell
–
log Q
n
 2.00  x 
0.0592 V
log
0.00 V = 0.74 V –
2
x
Assume x is negligible compared to 2.00.
 2.00 
25.0 = log
x
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21-54
1x1025 =
 2.00 
x
x = 2.0x10–25 M Cu2+
21.153 a) The chemical equation for the combustion of octane is:
2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g)
The heat of reaction may be determined from heats of formation.

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(16 mol CO2)( H f of CO2) + (18 mol H2O)( H f of H2O)]
– [(2 mol C8H18)( H f of C8H18) + (25 mol O2)( H f of O2)]

H rxn
= [(16 mol)(–393.5 kJ/mol) + (18 mol)(–241.826 kJ/mol)]
– [(2 mol)(–250.1 kJ/mol) + (25 mol)(0 kJ/mol)]

H rxn
= –10148.868 = –10148.9 kJ
The energy from 1.00 gal of gasoline is:
 4 qt   1 L   1 mL   0.7028 g   1 mol C8 H18   10148.9 kJ 
Energy (kJ) = 1.00 gal  



  3  

 1 gal   1.057 q t   10 L   mL   114.22 g C8 H18   2 mol C8 H18 
= –1.18158x105 = –1.18x105 kJ
b) The energy from the combustion of hydrogen must be found using the balanced chemical equation and the
heats of formation.
H2(g) + 1/2O2(g)  H2O(g)
With the reaction written this way, the heat of reaction is simply the heat of formation of water vapor,
and no additional calculations are necessary.
H° = –241.826 kJ
The moles of hydrogen needed to produce the energy from part a) are:
 1 mol H 2 
Moles of H2 = 1.18158x105 kJ 
 = 488.6075 mol
 241.826 kJ 


Finally, use the ideal gas equation to determine the volume.
L•atm 
 488.6075 mol H 2   0.0821
   273  25  K 
nRT
mol•K


V=
=
= 1.195417x104 = 1.20x104 L
P
1.00 atm 
c) This part of the problem requires the half-reaction for the electrolysis of water to produce hydrogen gas.
2H2O(l) + 2e–  H2(g) + 2OH–(aq)
Use 1 A = 1 C/s
 2 mol e    96, 485 C  
s

Time (s) =  488.6075 mol H 2  
= 9.42866x104 = 9.43x104 seconds
 1 mol H   1 mol e    1.00x103 C 
2 


d) Find the coulombs involved in the electrolysis of 488.6075 moles of H2.
 2 mol e    96485 C 
Coulombs =  488.6075 mol H 2  
= 94,286,589 C
 1 mol H   1 mol e  
2 


Joules = C x V = 94,286,589 C x 6.00 V = 565,719,534 J
 1 kW•h 
Power (kW • h) =  565,719,534 J  
 = 157.144 = 157 kW•h
 3.6x106 J 
e) The process is only 88.0% efficient, additional electricity is necessary to produce sufficient hydrogen. This is the
purpose of the (100%/88.0%) factor.
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21-55
 100%   0.123 cents 
Cost = 157.144 kW•h  

 = 21.964445 = 22.0 ¢
 88%   1 kW•h 




21.154 Plan: Write the half-reactions and the overall reaction. Calculate Ecell
by using Ecell
= Ecathode
– Eanode
and
+
+
then use the Nernst equation to find [H3O ] at a cell potential of 0.915 V. pH is obtained from [H3O ].
Solution:
The half-reactions are (from the Appendix):
E° = 0.00 V
Oxidation:
H2(g)  2H+(aq) + 2e–
Reduction:
2(Ag+(aq) + 1e–  Ag(s))
E° = 0.80 V

Overall:
2Ag+(aq) + H2(g)  2Ag(s) + 2H+(aq)
Ecell
= 0.80 V – 0.0 V = 0.80 V
The hydrogen ion concentration can now be found from the Nernst equation.
0.0592

Ecell = Ecell
–
log Q
n
2
H 
0.0592 V
 
log
0.915 V = 0.80 V –
2
2
 Ag   PH
2



2
H 
0.0592 V
 
log
0.915 V – 0.80 V = –
2
0.1002 1.00
2
H 
0.0592 V
 
0.115 V = –
log
2
0.1002 1.00
2
H 
 
–3.885135 = log
0.0100
2
H 
 
1.30276x10 =
0.0100
–4
[H+] = 1.1413851x10–3 M
pH = –log [H+] = –log (1.1413851x10–3) = 2.94256779 = 2.94
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21-56