Chapter 20
Practice Exercises
Mg(s) Mg2+(aq) + 2e–
Fe2+(aq) + 2e– Fe(s)
Mg(s)|Mg2+(aq)||Fe2+(aq)|Fe(s)
20.1
anode:
cathode:
cell notation:
20.2
anode:
cathode:
overall:
20.3
E°cell = E°substance reduced – E°substance oxidized
2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)
E ocell = E oAg+ − E oCu 2+
Al(s) Al3+(aq) + 3e–
Ni2+(aq) + 2e– Ni(s)
3Ni2+(aq) + 2Al(s) 2Al3+(aq) + 3Ni(s)
E°cell = 0.80 V – 0.34 V = 0.46 V
2Ag+(aq) + Zn(s) 2Ag(s) + Zn2+(aq)
E ocell = E oAg+ − E oZn 2+
E°cell = 0.80 V – (–0.76 V) = 1.56 V
Zinc will have the larger value for E°cell.
20.4
E°cell = E°substance reduced – E°substance oxidized
Fe2+(aq) + Mg(s) Mg2+(aq) + Fe(s)
1.93 V = E°Fe2+ – (–2.37 V)
E°Fe2+ = 1.93 V + (–2.37 V) = –0.44 V
This agrees exactly with Table 20.1.
20.5
The half–reaction with the more positive value of E° (listed higher in Table 20.1) will occur as a reduction.
The half–reaction having the less positive (more negative) value of E° (listed lower in Table 20.1) will be
reversed and occur as an oxidation.
(a)
I2(aq) + 2e– 2I–(aq)
oxidation
Fe3+(aq) + e– Fe2+(aq)
reduction
212
Chapter 20
(b)
(c)
20.6
2I–(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq)
Mg2+(aq) + 2e– Mg(s)
oxidation
Cr3+(aq) + 3e– Cr(s)
reduction
3Mg(s) + 2Cr3+(aq) 3Mg2+(aq) + 2Cr(s)
Co2+(aq) + 2e– Co(s)
oxidation
SO42–(aq) + 4H+(aq) + 2e– H2SO3(aq) + H2O
reduction
SO42–(aq)+ 4H+(aq) + Co(s) H2SO3(aq) + H2O + Co2+(aq)
The half–reaction with the more positive value of E° (listed higher in Table 20.1) will occur as a reduction.
The half–reaction having the less positive (more negative) value of E° (listed lower in Table 20.1) will be
reversed and occur as an oxidation.
Br2(aq) + 2e– 2Br–(aq)
SO42–(aq) + 4H+(aq) + 2e– H2SO3(aq) + H2O
Br2(aq) + H2SO3(aq) + H2O 2Br–(aq) + SO42–(aq) + 4H+(aq)
reduction
oxidation
20.7
Either nickel(II) or iron(III) will be reduced, depending on which way the reaction proceeds. Iron(III) is
listed higher than nickel(II) in Table 20.1 (it has a greater reduction potential), so we would expect that the
reaction would not be spontaneous in the direction shown.
The spontaneous reaction is:
Ni(s) + 2Fe3+(aq) Ni2+(aq) + 2Fe2+(aq)
20.8
The half–reaction having the more positive value for E° will occur as a reduction. The other half–reaction
should be reversed, so as to appear as an oxidation.
NiO2(s) + 2H2O + 2e– Ni(OH)2(s) + 2OH–(aq)
Fe(s) + 2OH–(aq) 2e– + Fe(OH)2(s)
NiO2(s) + Fe(s) + 2H2O Ni(OH)2(s) + Fe(OH)2(s)
reduction
oxidation
net reaction
E°cell = E°substance reduced – E°substance oxidized
E°cell = E°NiO2 – E°Fe
E°cell = 0.49 – (–0.88) = 1.37 V
20.9
5{Cr(s)
Cr3+(aq) + 3e-}
3{MnO4 (aq) + 8H+(aq) + 5eNet reaction
5Cr(s) + 3MnO4−(aq) + 24H+(aq)
Mn2+(aq) + 4H2O(l)}
oxidation
reduction
3Mn2+(aq) + 12H2O(l) + 5Cr3+(aq)
E°cell = E°substance reduced – E°substance oxidized
o
E ocell = E oMnO− − E Cr
3+
4
E°cell = 1.51 – (–0.74) = 2.25 V
20.10
The half–reaction having the more positive value for E° will occur as a reduction. The other half–reaction
should be reversed, so as to appear as an oxidation.
3Cu2+(aq) + 2Cr(s) 3Cu(s) + 2Cr3+(aq)
E°cell = E°substance reduced – E°substance oxidized
o
o
E ocell = E Cu
2+ − E
Cr 3+
E°cell = 0.34 – (–0.74) = 1.08 V
213
Chapter 20
20.11
A reaction will occur spontaneously in the forward direction if the value of E° is positive. We therefore
evaluate E° for each reaction using:
E°cell = E°substance reduced – E°substance oxidized
(a)
Br2(aq) + 2e– 2Br–(aq) reduction
I2(s) + 2e– 2I–(aq)
oxidation, when reversed
E°cell = E°Br2 – E°I2
E°cell = 1.07 V – (0.54 V) = 0.53 V, ∴ spontaneous
(b)
MnO4–(aq) + 8H+ + 5e– Mn2+(aq) + 4H2O
reduction
5Ag+(aq) + 5e– 5Ag(s)
oxidation, when reversed
E ocell = E oMnO− − E oAg+
4
E°cell = 1.51 V – (0.80V) = +0.71 V, ∴ spontaneous
20.12
A reaction will occur spontaneously in the forward direction if the value of E° is positive. We therefore
evaluate E° for each reaction using:
E°cell = E°substance reduced – E°substance oxidized
(a)
Br2(aq) + 2e– 2Br–(aq) reduction
Cl2(aq) + 2H2O 2HOCl(aq) + 2H+(aq) + 2e–
oxidation
E°cell = E°Br2 – E°HOCl
E°cell = 1.07 V – (1.63 V) = –0.56 V,
(b)
∴ non-spontaneous
2Cr3+(aq) + 6e– 2Cr(s) reduction
3Zn(s) 3Zn2+(aq) + 6e– oxidation
E°cell = E°Cr3+ – E°Zn2+
E°cell = –0.74 V – (–0.76 V) = +0.02 V,
∴ spontaneous
20.13
From the equation ∆G° = –nFE°cell
∆G° = –30.9 kJ or –30,900 J
F = 96,500 C mol–1
E°cell = 0.107 V
–30,900 J = –n(96,500 C mol–1)(0.107 V)
n = 2.99 which rounds to 3
Therefore, 3 moles of electrons are transferred in the reaction.
20.14
∆G° = –n F E°cell
From Practice Exercise 11 (a): n = 2 e–, E°cell = 0.53 V
∆G° = –n F E°cell = –(2 e–)(96,500 F)(0.53 V) = –102,000 J = –102 kJ
From Practice Exercise 11 (b): n = 5 e–, E°cell = 0.71 V
∆G° = –n F E°cell = –(5 e–)(96,500 F)(0.71 V) = –342,600 J = –343 kJ
From Practice Exercise 12 (a): n = 2 e–, E°cell = –0.56 V
∆G° = –n F E°cell = –(2 e–)(96,500 F)(–0.56 V) = 108080 J = 108 kJ
From Practice Exercise 12 (b): n = 6 e–, E°cell = 0.02 V
214
Chapter 20
∆G° = –n F E°cell = –(6 e–)(96,500 F)(0.02 V) = 11,600 J = –11.6 kJ
20.15
Using Equation 20.7,
RT
Eocell = n F ln K c
−0.46 V =
(8.314 J mol−1K −1 )(298 K)
2(96,500 C mol−1 )
ln K c = −35.83
Taking the antilog (ex) of both sides of the above equation gives
Kc = 2.7 × 10–16.
ln K c
This very small value for the equilibrium constant means that the products of the reaction are not formed
spontaneously. The equilibrium lies far to the left, favoring reactants, and we do not expect much product
to form.
The reverse reaction will be spontaneous, therefore, the value for Kc for the spontaneous reaction will be:
Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)
1
1
K c' =
=
= 3.7 × 1015
K c 2.7 × 10−16
20.16
Ag+(aq) + e– Ag(s)
E° = 0.80 V
E° = 0.07 V
AgBr(s) + e– Ag(s) + Br–(aq)
Equation for the spontaneous reaction:
Ag+(aq) + Br–(aq) AgBr(s)
E°cell = 0.73 V
1
K=
 Ag +   Br − 



( )(
)
−
−1
E ocell n F ( 0.73 V ) 1 e 96,500 C mol
=
= 28.43
ln Kc =
RT
8.314 J mol −1 K −1 ( 298 K )
(
)
Kc = 2.23 × 1012
Ksp for AgBr is 5.4 × 10–13
1
1
=
= 4.5 × 10–13
K c 2.23 × 1012
The Kc is the inverse of the Ksp. So the cell potential value agrees with Ksp table.
20.17
Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq)
 Mg 2 + 
RT

o
ln 
Ecell = E cell −
nF
Cu 2 + 


E°cell = 0.34 – (–2.37 V) = 2.71 V
(8.314 J mol K ) ( 298 K )
Ecell = 2.71 V –
( 2 ) ( 96,500 C mol )
−1
−1
−1
 2.2 × 10−6 
 = 2.82 V
ln 
[0.015]
215
Chapter 20
20.18
Ecell = Eocell
−
[1]
RT
ln
2
nF
H+ 
 
(8.314 J mol K ) ( 298 K )
0.00 = 0.14 V –
( 2 ) ( 96,500 C mol )
−1
−1
−1
ln
[1]
H+ 
 
2
0.14 V = -2.57 x 10-2 ln[H+]
5.45 = -ln[H+] = -2.303 log[H+]
2.37 = -log[H+] = pH
20.19
Zn(s) Zn2+(aq) + 2e–
Cu2+(aq) + 2e– 2Cu(s)
oxidation
reduction
E°cell = E°Cu2+ – E°Zn2+
E°cell = +0.34 V – (–0.76 V) = +1.10 V
The Nernst equation for this cell is:
o
Ecell = E cell
 Zn 2 + 
RT

ln 
−
2
+
nF
 Cu 


E cell = 1.10 V −
(8.314 J mol−1K −1 )(298 K)
−1
2(96,500 C mol )
=1.10 V – 0.01284(4.605) = 1.04 V
20.20
Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq)
 Mg 2 + 
RT

ln 
Ecell = Eocell −
nF
 Cu 2+ 


[1.0]
[ 0.010]
E°cell = 0.34 – (–2.37 V) = 2.71 V
(8.314 J mol K ) ( 298 K )
2.79 = 2.71 V –
( 2 ) ( 96,500 C mol )
−1
ln
−1
−1
 Mg 2 + 

ln 
[ 0.015]
 Mg 2+ 
 Mg 2+ 



0.08 V = –0.0128 ln
ln 
[0.015]
[0.015]
 Mg 2+ 

–6.25 = ln 
[0.015]
 Mg 2 + 

e–6.25 = 
[0.015]
[Mg2+] = 2.95 × 10–5 M
20.21
Cu(s) Cu2+(aq) + 2e–
Ag+(aq) + e– Ag(s)
oxidation
reduction
216
Chapter 20
E°cell = E°Ag+ – E°Cu2+
E°cell = +0.80 V – (+0.34 V) = +0.46 V
Ecell = Eocell −
Cu 2+ 
RT

ln 
2
nF
+
 Ag 


Cu 2+ 
 = (Eo − E )/  RT 
ln 
cell
cell 

2
+
nF
 Ag 


=
( 0.46 V
− 0.57 V )
0.01284
= − 8.5670
Cu 2+ 

 = e −8.5670 = 1.9 × 10−4
2
+
 Ag 


Since the [Ag+] = 0.225 M, [Cu2+] = 9.6 × 10–6 M
Substituting the second value of 0.82 V into the same expression gives
[Cu2+] = 3.4 × 10–14 M
20.22
We are told that, in this galvanic cell, the chromium electrode is the anode, meaning that oxidation occurs
at the chromium electrode.
Now in general, we have the equation:
E°cell = E°reduction – E°oxidation
which becomes, in particular for this case:
E°cell = E°Ni2+ – E°Cr3+
The net cell reaction is given by the sum of the reduction and the oxidation half–reactions, multiplied in
each case so as to eliminate electrons from the result:
3 × [Ni2+(aq) + 2e– Ni(s)]
reduction
2 × [Cr(s) Cr3+(aq) + 3e–]
oxidation
3Ni2+(aq) + 2Cr(s) 2Cr3+(aq) + 3Ni(s)
net reaction
In this reaction, n = 6, and the Nernst equation becomes:
2
Ecell = Eocell
Cr 3+ 
RT

−
ln 
3
nF
2
+
 Ni 


2
Cr 3+ 
 = (Eo − E )/  RT 
ln 
cell
cell 

3
nF
 Ni 2 + 


217
Chapter 20
2
Cr 3+ 
 = ( 0.487 V − 0.552 V )
ln 
3
0.004279
 Ni 2 + 


2
Cr 3+ 
 = − 15.190
ln 
3
+
2
 Ni 


Cr 3+ 


2
2+ 
3
 Ni


= e−15.190 = 2.5 × 10−7
Substituting [Ni2+] = 1.20 M, we solve for [Cr3+] and get: [Cr3+] = 6.6 × 10–4 M.
20.23
Fe3+(aq) + e– Fe2+(aq)
2I2(s) + e– 2I–(aq)
O2(g) + 4H+(aq) + 4e– 2H2O
Eo = 0.77 V
Eo = +0.54 V
Eo = +1.23 V
The reaction with the least positive reduction potential will be the easiest to oxidize, and its product will be
the product at the anode. I2 will be produced.
20.24
The cathode is always where reduction occurs. We must consider which species could be candidates for
reduction, then choose the species with the highest reduction potential from Table 20.1.
Cd2+(aq) + 2e– 2Cd(s)
Sn2+(aq) + 2e– 2Sn(s)
2H2O + 2e– H2(g) + 2OH–(aq)
E° = –0.40 V
E° = –0.14 V
E° = –0.83 V
Tin(II) has the highest reduction potential, so we would expect it to be reduced in this environment. We
expect Sn(s) at the cathode.
20.25
The number of Coulombs is: 4.00 A × 180 s = 720 C
The number of moles is:
mol OH − = 720 C ×
20.26
The number of moles of Au to be deposited is: 3.00 g Au ÷ 197 g/mol = 0.0152 mol Au. The number of
Coulombs (A × s) is:
3F
96,500 C
Coulombs = 0.0152 mol Au ×
×
= 4.40 × 103 C
1 mol Au
1F
The number of minutes is:
min =
20.27
1F
1 mol OH −
= 7.46 × 10–3 mol OH−
×
96,500 C
1F
4.40 × 103 A • s
1 min
×
= 7.33 min
10.0 A
60 s
As in Practice Exercise 26 above, the number of Coulombs is 4.40 × 103 C. This corresponds to a current
of:
A=
4.40 × 103 A • s
1 min
×
= 3.67 A
20.0 min
60 s
218
Chapter 20
20.28
The number of Coulombs is:
0.100 A × 1.25 hr(3600 s/hr) = 450 C
The number of moles of copper ions produced is:
mol Cu 2+ = 450 C ×
1 mol e −
1 mol Cu 2 +
= 0.00233 mol Cu2+
×
96, 500 C
2 mol e −
Therefore, the increase in concentration is:
M = mol/L = (0.00233 mol Cu2+)/(0.125 L) = + 0.0187 M
Review Questions
20.1
A galvanic cell is one in which a spontaneous redox reaction occurs, producing electricity. A half–cell is
one of either the cathode or the anode, together with the accompanying electrolyte.
20.2
The salt bridge connects two half–cells, and allows for electrical neutrality to be maintained by a flow of
appropriate ions.
20.3
These must be kept separate, because otherwise Ag+ ions would be reduced directly by Cu metal, and no
external current would be produced.
20.4
The anode is the electrode at which oxidation takes place, and the cathode is the electrode at which
reduction takes place. The charges of the electrodes in a galvanic cell are opposite to those in the
electrolysis cell; the cathode is positive and the anode is negative.
20.5
In both the galvanic and the electrolysis cells, the electrons move away from the anode and toward the
cathode.
20.6
The anions move away from the cathode toward the anode, and the cations move away from the anode
toward the cathode.
20.7
Aluminum metal constitutes the anode: Al(s) Al3+(aq) + 3e–
The cathode is tin metal: Sn2+(aq) + 2e– Sn(s)
20.8
A cell potential is a standard potential only if the temperature is 25 °C, the pressure is 1 atm, and all ions
have a concentration of 1 M.
20.9
The cell potential for the anode half–reaction is subtracted from the cell potential for the cathode half–cell:
o
o
E ocell = Esubstance
reduced – Esubstance oxidized
E ocell = Eoreduction – Eooxidation
20.10
The standard hydrogen electrode is diagramed in Figure 20.5 of the text. It consists of a platinum wire in
contact with a solution having [H+] equal to 1 M, and hydrogen gas at a pressure of 1 atm is placed over the
system. The half–cell potential is 0 V.
20.11
A positive reduction potential indicates that the substance is more easily reduced than the hydrogen ion.
Conversely, a negative reduction potential indicates that the substance comprising the half–cell is less
easily reduced than the hydrogen ion.
20.12
The difference between the reduction potentials for hydrogen and copper is a constant that is independent
of the choice for the reference potential. In other words, the reduction half–cell potential for copper is to be
0.34 units higher for copper than for hydrogen, regardless of the chosen point of reference. If E° for copper
is taken to be 0 V, then E° for hydrogen must be –0.34 V.
219
Chapter 20
20.13
The negative terminal of the voltmeter must be connected to the anode in order to obtain correct readings of
the voltage that is generated by the cell.
20.14
The metals are placed into the activity series based on their values of standard reduction potentials.
20.15
The silver will be reduced, according to the standard reduction potentials.
External circuit
electron flow
(+)
(–)
Salt Bridge
Ag
Ag+ (aq)
Fe
Fe3+ (aq)
Cathode
Anode
The net cell reaction is:
Fe(s) + 3Ag+(aq)
→
Fe3+(aq) + 3Ag(s)
20.16
External circuit
(–)
electron flow
(+)
Salt Bridge
Pt
Fe2+(aq)
Fe3+(aq)
Anode
Pt
Br2(aq)
Br–(aq)
Cathode
2Fe2+(aq) + Br2(aq) 2 Fe3+(aq) + 2Br–(aq)
20.17
∆G° = –nF Eocell
20.18
E ocell =
0.0592
log K c
n
220
Chapter 20
20.19
The Nernst equation:
RT
Ecell = E ocell −
ln Q
nF
–1
R = 8.314 J mol K–1
T = 25 °C = 298 K
F = 96,494 C mol–1
RT
= 0.02567 J C–1
F
RT
Ecell = E ocell −
ln Q
nF
If the system is at equilibrium, Q = Kc.
And Ecell = 0
RT
0 = E ocell −
ln K c
nF
RT
E ocell =
ln K c
nF
20.20
We begin by separating the reaction into its two half–reactions, in order to obtain the value of n.
Pb(s) + SO42–(aq) PbSO4(s) + 2e–
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– PbSO4(s) + 2H2O
Thus, n is equal to 2, and the equation that we are to use is:
0.0592
log Q
n
0.0592
1
= 2.05 V −
log
4
2
2
 H +  SO4 2 − 
  

E cell = E ocell −
E cell
20.21
A concentration cell consists of two almost identical half–cells, the two half–cells are composed of the
same substances, but have difference concentrations of the solute species.
Ecell = E ocell −
RT [ ion ]dilute
ln
nF [ion ]conc
E ocell = 0 because the standard cell potential is the reduction potential of the substance being reduce less the
reduction potential of the substance being oxidized, and the two are equal to each other because they are the
same substance.
20.22
Pb(s) + SO42–(aq) PbSO4(s) + 2e–
PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– PbSO4(s) + 2H2O
Connecting six cells in series produces 12 volts.
anode
cathode
20.23
PbSO4(s) + 2e– Pb(s) + SO42–(aq)
PbSO4(s) + 2H2O 2e– + PbO2(s) + 4H+(aq) + SO42–(aq)
cathode
anode
20.24
This is diagramed in Figure 20.11. The float inside the hydrometer sinks to a level that is inversely
proportional to the density of the liquid that is drawn into it. This works because the concentration of the
sulfuric acid (and hence the state of charge of the battery) is proportional to the concentration of sulfuric
acid in the battery.
20.25
The anodic reaction is: Zn(s) Zn2+(aq) + 2e–
221
Chapter 20
Several reactions take place at the cathode; one of the important ones is:
2MnO2(s) + 2NH4+(aq) + 2e– Mn2O3(s) + 2NH3(aq) + H2O
20.26
Zn(s) + 2OH–(aq) ZnO(s) + H2O + 2e–
2MnO2(s) + H2O + 2e– Mn2O3(s) + 2OH–(aq)
anode
cathode
20.27
Cd(s) + 2OH–(aq) Cd(OH)2(s) + 2e–
2e– + NiO2(s) + 2H2O Ni(OH)2(s) + 2OH–(aq)
anode
cathode
The overall cell reaction on discharge of the battery is:
Cd(s) + NiO2(s) + 2H2O Cd(OH)2(s) + Ni(OH)2(s)
On charging the battery, the reactions are reversed.
20.28
The hydrogen is held in a metal alloy, Mg2Ni, which has the ability to absorb and hold substantial amounts
of hydrogen. The electrolyte is KOH.
20.29
MH(s) + OH–(aq) M(s) +H2O + e–
NiO(OH)(s) + H2O + e– Ni(OH)2(s) + OH–(aq)
MH(s) + NiO(OH)(s) M(s) + Ni(OH)2(s)
anode
cathode
overall reaction
The reactions are reversed upon charging.
20.30
Lithium has the most negative reduction potential of any metal, so it is very easy to oxidize making it an
excellent material for an anode, and it is a very lightweight metal. The major problem with lithium in a cell
is that it reacts vigorously with water. Also, lithium batteries often have a large negative ∆H.
20.31
In a typical primary lithium cell, the electrodes are lithium as the anode and manganese(IV) oxide as the
cathode.
Li Li+ + e–
anode
MnO2 + Li+ + e– LiMnO2
cathode
Li + MnO2 LiMnO2
net cell reaction
20.32
The electrode materials in a typical lithium ion cell are graphite and cobalt oxide. When the cell is charged,
Li+ ions leave LiCoO2 and travel through the electrolyte to the graphite. When the cell discharges, the Li+
ions move back through the electrolyte to the cobalt oxide while electrons move through the external circuit
to keep the charge in balance.
20.33
O2(g) + 2H2O + 4e– 4OH–(aq)
H2(g) + 2OH–(aq) 2H2O + 2e–
2H2(g) + O2(g) 2H2O
20.34
Fuel cells are more efficient thermodynamically, and more of the energy of the reaction can be made
available for useful work provided that the supply of reactants is maintained. The only product formed by
the cell is water.
20.35
In an electrolytic cell, the cathode is negative, and the anode is positive. The opposite is true of a galvanic
cell. An inert electrode is an electrode which does not chemically react in the measurement of
electrochemical data.
20.36
The flow of electrons in the external circuit must be accompanied by the electrolysis reaction. Otherwise
the electrodes would accumulate charge, and the system would cease to function.
20.37
In solid NaCl, the ions are held in place and cannot move about. In molten NaCl, the crystal lattice of the
solid has been destroyed; the ions are free to move, and consequently to conduct current by migrating either
to the anode or to the cathode.
Anode:
2Cl–(l) Cl2(g) + 2e–
cathode
anode
net cell reaction
222
Chapter 20
Cathode:
Net:
Na+(l) + e– Na(l)
2Na+(l) + 2Cl–(l) 2Na(l) + Cl2(g)
20.38
oxidation:
reduction:
2H2O 4H+(aq) + 4e– + O2(g)
2H2O + 2e– H2(g) + 2OH–(aq)
20.39
It is reduction that occurs at the cathode, and near it, the pH increases due to the formation of OH–(aq). At
the anode, where the oxidation of water occurs, the pH decreases due to the production of H+(aq). See the
equations given in the answer to Review Question 20.38. The overall change in pH is 0 since the amount
of H+ formed and the amount of OH– formed are equal. K2SO4 serves as charge carriers to balance the
charge that occurs upon electrolysis of the K2SO4 solution.
20.40
One Faraday (F) is equivalent to one mole of electrons. Also, one Faraday is equal to 96,500 Coulombs,
and a Coulomb is equivalent to an Ampere⋅second:
1 F = 96,500 C and 1 C = 1 A⋅s
20.41
The deposition of 0.10 mol of Cr from a Cr3+ solution will take longer than the deposition of 0.10 mol Cu
from a Cu2+ solution because the Cr3+ requires 1.5 times as many electrons for deposition than Cu2+. This
is due to the difference in charges on the two ions.
20.42
The Ag+ solution will give more metal deposited since it is in the +1 state while the Cu2+ solution will give
half as much since the copper is in the +2 state.
20.43
Copper has a larger atomic mass than iron; therefore, the copper will deposit a greater mass of metal. Both
metals are in the same +2 state.
20.44
Electroplating is a procedure by which a metal is deposited on another conducting surface.
20.45
Al2O3(s) is dissolved in molten cryolite, Na3AlF6. The liquid mixture is electrolyzed to drive the following
reaction: 2Al2O3(l) 4Al(l) + 3O2(g)
The two half–reactions are:
anode:
cathode:
overall cell reaction
2O2– O2(g) + 4e–
Al3+(l) + 3e– Al(l)
6O2– + 4Al3+ 3O2(g) + 4Al(l)
223
Chapter 20
20.46
The oxygen that is formed slowly decomposes the anode.
20.47
Sodium is obtained from electrolysis of molten NaCl using the Downs cell. Some of the uses of sodium are
to make tetraethyl lead, for sodium vapor lamps, and as a coolant in nuclear reactors.
Na+ + e– Na(l)
Cl– 1/2Cl2(g) + e–
NaCl(l) Na(l) + 1/2Cl2(g)
20.48
cathode
anode
net reaction
This is shown in a photo and a diagram (Figure 20.24) of the text. Impure copper is the anode, which
dissolves during the process. Pure copper is deposited at the cathode. Anode sludge contains precious
metals, whose value makes the process cost effective.
Typical reactions occurring at the anode are:
Cu(s) Cu2+(aq) + 2e–
Zn(s) Zn2+(aq) + 2e–
Fe(s) Fe2+(aq) + 2e–
The reaction that occurs at the cathode is:
Cu2+(aq) + 2e– Cu(s)
20.49
One of the various methods arise diagramed in Figure 20.25 of the text. The physical apparatus influences
the products that are obtained. The cathode reaction is the same in stirred and unstirred cells:
2H2O + 2e– H2(g) + 2OH–(aq)
The unstirred anode reaction is:
2Cl–(aq) Cl2(g) + 2e–
The net reaction is:
2NaCl(aq) + 2H2O 2NaOH(aq) + Cl2(g) + H2(g)
In a stirred cell, the Cl2(g) that is produced reacts with the OH–(aq) present forming Cl–(aq), OCl–(aq) and
water. The anode reaction in the stirred cell is therefore,
Cl–(aq) + 2OH–(aq) OCl–(aq) + H2O + 2e–
The net reaction in a stirred cell is:
NaCl(aq) + H2O NaOCl(aq) + H2(g)
Review Problems
20.50
20.51
(a)
anode:
cathode:
cell:
Cd(s) Cd2+(aq) + 2e–
Au3+(aq) + 3e– Au(s)
3Cd(s) + 2Au3+(aq) 3Cd2+(aq) + 2Au(s)
(b)
anode:
cathode:
cell:
Fe(s) Fe2+(aq) + 2e–
Br2(aq) + 2e– 2Br–(aq)
Fe(s) + Br2(aq) Fe2+(aq) + 2Br–(aq)
(c)
anode:
cathode:
cell:
Cr(s) Cr3+(aq) + 3e–
Cu2+(aq) + 2e– Cu(s)
2Cr(s) + 3Cu2+(aq) 2Cr3+(aq) + 3Cu(s)
(a)
anode:
cathode:
Zn(s) Zn2+(aq) + 2e–
Cr3+(aq) + 3e– Cr(s)
224
Chapter 20
cell:
3Zn(s) + 2Cr3+(aq) 3Zn2+(aq) + 2Cr(s)
(b)
anode:
cathode:
cell:
Pb(s) + HSO4–(aq) PbSO4(s)+ H+(aq) + 2e–
PbO2(s) + HSO4–(aq) + 3H+(aq) + 2e– PbSO4(s) + 2H2O
Pb(s) + PbO2(s) + 2HSO42–(aq) + 2H+(aq) 2PbSO4(s) + 2H2O
(c)
anode:
cathode:
cell:
Mg(s) Mg2+(aq) + 2e–
Sn2+(aq) + 2e– Sn(s)
Mg(s) + Sn2+(aq) Mg2+(aq) + Sn(s)
20.52
(a)
(b)
(c)
Pt(s)|Fe2+(aq),Fe3+(aq)||NO3–(aq), H+(aq)|NO(g)|Pt(s)
Pt(s)|Br2(aq),Br–(aq)||Cl–(aq),Cl2(g)|Pt(s)
Ag(s)|Ag+(aq)||Au3+(aq)|Au(s)
20.53
(a)
(b)
(c)
Fe(s)|Fe2+(aq)||Cd2+(aq)|Cd(s)
Ag(s)|Ag+(aq)|| Ni2+(aq),H+(aq)| NiO2(s)|Pt(s)
Mg(s)|Mg2+(aq)||Cd2+(aq)|Cd(s)
20.54
(a)
Sn(s)
20.55
a)
MnO4–(aq)
20.56
(a)
(b)
(c)
Eocell = 0.96 V – (0.77) V = 0.19 V
Eocell = 1.07 V – (1.36 V) = –0.29 V
Eocell = 1.42 V – (0.80 V) = 0.62 V
20.57
(a)
(b)
E ocell = –0.40 V – (–0.44 V) = 0.04 V
The data for this question is not available in the textbook.
E ocell = 1.59 V – (0.80 V) = 0.79 V
(c)
E ocell = –0.40 V – (–2.37 V) = 1.97 V
20.58
20.59
Br–(aq) (c)
(b)
Zn(s)
Au3+(aq)
(d)
(c)
I–(aq)
PbO2(s)
(d)
HOCl(aq)
The reactions are spontaneous if the overall cell potential is positive.
Eocell = Esubstance reduced − Esubstance oxidized
(a)
Eocell = 1.42 V – (0.54 V) = 0.88 V
spontaneous
(b)
Eocell = 1.07 V – (0.17 V) = 0.90 V
spontaneous
(c)
Eocell = –0.74 V – (–2.76 V) = 2.02 V
spontaneous
A reaction is spontaneous if its net cell potential is positive:
o
o
E ocell = Esubstance
reduced – Esubstance oxidized
(a)
E ocell = 1.07 V – (1.36 V) = –0.29 V
not spontaneous
(b)
E ocell
E ocell
= –0.44 V – (0.96 V) = –1.40 V
not spontaneous
= –0.25 V – (–0.44 V) = +0.19 V
spontaneous
(c)
20.60
(b)
The half–cell with the more positive E ocell will appear as a reduction, and the other half–reaction is
reversed, to appear as an oxidation:
reduction
BrO3–(aq) + 6H+(aq) + 6e– Br–(aq) + 3H2O
225
Chapter 20
3 × (2I–(aq) I2(s)+ 2e–)
BrO3–(aq) + 6I–(aq) + 6H+(aq) 3I2(s) + Br–(aq) + 3H2O
oxidation
net reaction
Eocell = Esubstance reduced − Esubstance oxidized or
Eocell = Eoreduction − Eooxidation = 1.44 V – (0.54 V) = 0.90 V
20.61
The half–reaction having the more positive reduction potential is the reduction half–reaction, and the other
is reversed to become the oxidation half–reaction:
reduction
MnO2(s) + 4H+(aq) + 2e– Mn2+(aq) + 2H2O
Pb(s) + 2Cl–(aq) PbCl2(s) + 2e–
oxidation
MnO2(s) + 4H+(aq) + Pb(s) + 2Cl–(aq) Mn2+(aq) + 2H2O + PbCl2(aq)
net reaction
o
o
E ocell = Esubstance
reduced – Esubstance oxidized = 1.23 – (–0.27) = 1.50 V
20.62
The half–reaction having the more positive standard reduction potential is the one that occurs as a
reduction, and the other one is written as an oxidation:
2 × (2HOCl(aq) + 2H+(aq) + 2e– Cl2(g) + 2H2O) reduction
3H2O + S2O32–(aq) 2H2SO3(aq) + 2H+(aq) + 4e– oxidation
4HOCl(aq) + 4H+(aq) + 3H2O + S2O32–(aq) 2Cl2(g) + 4H2O + 2H2SO3(aq) + 2H+(aq)
which simplifies to give the following net reaction:
4HOCl(aq) + 2H+(aq) + S2O32–(aq) 2Cl2(g) + H2O+ 2H2SO3(aq)
20.63
Br2(aq) + 2e– 2Br–(aq)
Eocell = 1.07 V
I2(s) + 2e– 2I–(aq)
Eocell = 0.54 V
Since the first of these has the larger reduction half–cell potential, it occurs as a reduction, and the second is
reversed to become an oxidation:
Br2(aq) + 2I–(aq) I2(s) + 2Br–(aq)
20.64
The two half–reactions are:
SO42–(aq) + 2e–+ 4H+(aq) H2SO3(aq) + H2O(l)
2I–(aq) I2(s) + 2e–
reduction
oxidation
Eocell = Eoreduction − Eooxidation = 0.17 V – (0.54 V) = –0.37 V
Since the overall cell potential is negative, we conclude that the reaction is not spontaneous in the direction
written.
20.65
The two half–reactions are:
S2O82– + 2e– 2SO42–
Ni(OH)2 + 2OH– NiO2 + 2H2O + 2e–
reduction
oxidation
o
o
E ocell = Esubstance
reduced – Esubstance oxidized = 2.01 V – (0.49 V) = 1.52 V
Since the overall cell potential is positive, we conclude that the reaction is spontaneous in the direction
written.
20.66
First, separate the overall reaction into its two half–reactions:
2Br–(aq) Br2(aq) + 2e–
oxidation
226
Chapter 20
I2(s) + 2e– 2I–(aq)
reduction
Eocell = Eoreduction − Eooxidation = 0.54 V – (1.07 V) = –0.53 V
The value of n is 2: ∆G° = –n F Eocell = –(2)(96,500 C)(–0.53 J/C)
= 1.0 × 105 J = 1.0 × 102 kJ
20.67
Using the equation ∆G° = –n F E ocell , we have ∆G° = –n(96,500 C/mol e–)(1.69 V) for which we need n.
Upon writing the two half–reactions, i.e.,
MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O
reduction
HCHO2(l) CO2(g) + 2H+(aq) + 2e–
oxidation
we see that we need to multiply the reduction half–reaction by 2 and the oxidation reaction by 5 in order to
balance the eqaution:
2MnO4–(aq) + 16H+(aq) + 10e– 2Mn2+(aq) + 8H2O
reduction
5HCHO2(l) 5CO2(g) + 10H+(aq) + 10e–
oxidation
The net reaction has n = 10. So, ∆G° = –(10 mol e–)(96,500 C/mol e–)(1.69 V) = –1.63 × 103 kJ.
20.68
(a)
(b)
(c)
Eocell = Eoreduction − Eooxidation = 2.01 V – (1.47 V) = 0.54 V
Since n = 10, ∆G° = –n F Eocell = –(10)(96,500 C)(0.54 J/C) = –5.2 × 105 J
∆G° = –5.2 × 102 kJ
RT
Eocell = nF ln K c
0.54 V =
(8.314 J mol-1K -1 )(298 K)
ln K c
10(96,500 C mol-1 )
ln K c = 210.3
Taking the exponential of both sides of this equation:
Kc = 2.1 × 1091
20.69
Ni2+ is reduced by two electrons and Co is oxidized by two electrons.
o
Eocell = Eoreduction − Eoxidation
= –0.25 – (–0.28) = +0.03 V
E ocell =
0.0592
log K c
n
+0.03 = (0.0592/2) × log Kc
log Kc = 1 and Kc = 101 = 10
20.70
Sn is oxidized by two electrons and Ag is reduced by two electrons:
0.0592
log K c
n
–0.015 V = (0.0592 V/2) × log Kc
log Kc = –0.51
Kc = antilog(–0.51) = 0.31
Eocell =
20.71
First, separate the overall reaction into two half–reactions:
2H2O 4H+ + 4e– + O2 oxidation
2 × (Cl2 + 2e– 2Cl–)
reduction
o
Eocell = Eoreduction − Eoxidation
= 1.36 – (1.23) = 0.13 V
227
Chapter 20
0.0592
log K c
n
+0.13 V = (0.0592 V/4) log Kc
log Kc = 8.78
Kc = antilog(8.78) = 6.08 × 108.
E ocell =
20.72
This reaction involves the oxidation of Ag by two electrons and the reduction of Ni by two electrons. The
concentration of the hydrogen ion is derived from the pH of the solution: [H+] = antilog (–pH) = antilog (–
2) = 1 × 10–2 M
2
E cell
 Ag +   Ni 2 + 
0.0592 V
 

= 2.48 V −
log 
4
2
H+ 
 
2
3.0 × 10−2  3.0 × 10−2 
0.0592 V
 

= 2.48 V −
log 
4
2
−
2
1.0 × 10 


Ecell = 2.48 V – 0.101 V = 2.38 V
20.73
The following half–reactions indicate that the value of n is 30:
5Cr2O72– + 70H+ + 30e– 10Cr3+ + 35H2O
3I2 + 18H2O 6IO3– + 36H+ + 30e–
6
E cell
10
 IO3−   Cr 3+ 
0.0592 V

= 0.135 V −
log  34 
5
30
+
2
 H  Cr2 O7 − 
  

6
10
[ 0.00010] [0.0010]
0.0592 V
= 0.135 V −
log
30
[0.10]34 [ 0.010]5
= 0.135 V −
0.0592 V
log 1.0 × 10−10
30
= 0.155 V
20.74
2+
RT  Mg 
ln
Ecell = E cell −
n F Cd 2+ 


o
E cell = 1.97 −
(8.314 J mol-1K -1 )(298 K)
-1
2(96,500 C mol )
1.54 V = 1.97 V − 0.01284 ln
ln
[1.00]
Cd 2+ 


1
Cd 2+ 


ln(1/[Cd2+]) = 33.489
Taking (ex) of both sides:
1/[Cd2+] = 3.50 × 1014
[Cd2+] = 2.86 × 10–15 M
20.75
Since the copper half–cell is the cathode; this is the half–cell in which reduction takes place. The silver
half–cell is therefore the anode, where oxidation of silver occurs. The standard cell potential is:
228
Chapter 20
o
Eocell = Eoreduction − Eoxidation
= 0.3419 V – 0.2223 V = 0.1196 V. The overall cell reaction is:
Cu2+(aq) + 2Ag(s) + 2Cl–(aq) Cu(s) + 2AgCl(s), and the Nernst equation becomes:
E cell = 0.1196 V −
0.0592 V
1
log
2
2
Cu 2+  Cl− 



If we use the values given in the exercise, we arrive at:
0.0895 V = 0.1196 V – 0.0296 V × log(1/[Cl–]2), which rearranges to give: log(1/[Cl–]2) = 1.017, [Cl–] =
0.310 M
20.76
In the iron half–cell, we are initially given:
0.0500 L × 0.100 mol/L = 5.00 × 10–3 mol Fe2+(aq)
The precipitation of Fe(OH)2(s) consumes some of the added hydroxide ion, as well as some of the iron
ion: Fe2+(aq) + 2OH–(aq) Fe(OH)2(s). The number of moles of OH– that have been added to the iron
half–cell is:
0.500 mol/L × 0.0500 L = 2.50 × 10–2 mol OH–
The stoichiometry of the precipitation reaction requires that the following number of moles of OH– be
consumed on precipitation of 5.00 × 10–3 mol of Fe(OH)2(s):
5.00 × 10–3 mol Fe(OH)2 × (2 mol OH–/mol Fe(OH)2) = 1.00 × 10–2 mol OH–
The number of moles of OH– that are unprecipitated in the iron half–cell is:
2.50 × 10–2 mol – 1.00 × 10–2 mol = 1.50 × 10–2 mol OH–
Since the resulting volume is 50.0 mL + 50.0 mL, the concentration of hydroxide ion in the iron half–cell
becomes, upon precipitation of the Fe(OH)2:
[OH–] = 1.50 × 10–2 mol/0.100 L = 0.150 M OH–
The standard cell potential is:
Eocell = Eoreduction − Eooxidation = 0.3419 V – (–0.447 V) = 0.7889 V
The Nernst equation is:
o
Ecell = E cell
2+
RT  Fe 
ln
−
n F Cu 2 + 


1.175 = 0.7889 −
(8.314 J mol-1K -1 )(298 K)
2(96,500 C mol-1 )
 Fe 2+ 

ln 
[1.00]
1.175 = 0.7889 − 0.01284 ln  Fe 2+ 


ln[Fe2+] = –30.08
[Fe2+] = 8.66 × 10–14 M
20.77
The half–cell reactions and the overall cell reaction are:
Cu2+(aq) + 2e– Cu(s) E°red = +0.3419 V
H2(g) 2H+(aq) + 2e–
E°ox = 0.0000 V
Cu2+(aq) + H2(g) Cu(s) + 2H+(aq)
(a)
The standard cell potential is:
229
Chapter 20
o
Eocell = Eoreduction − Eoxidation
= 0.3419 V – 0 V = +0.3419 V
The Nernst equation for this system is:
2
E cell =
E ocell
H+ 
0.0592 V
−
log  
2
 Cu 2+ 


which becomes, under the circumstances defined in the problem:
E cell = E ocell −
2
0.0592 V
log  H + 
 
2
Rearranging the last equation gives:
o
( 2 ) ( E cell − E cell
)
0.0592 V
= − log  H + 
 
2
which becomes the desired relationship:
(E
cell
− E ocell
0.0592 V
(b)
)=
− log  H +  = pH
 
The equation derived in the answer to part (a) of this question is conveniently rearranged to give:
o
Ecell = (0.0592 V)(pH) + Ecell
= (0.0592 V)(5.15) + 0.3419 V = 0.647 V
(c)
The equation that was derived in the answer to part (a) of this question may be used directly:
pH =
20.78
o
E =
E ocell
(E
cell
o
− E cell
0.0592 V
E = 0−
( 0.645 V
− 0.3419 V )
0.0592 V
 +
RT  Ag  dilute
−
ln
n F  Ag + 

 conc
(8.314 J mol K ) ( 298 K ) ln [0.015]
(1 mol ) ( 9.65 ×10 C mol ) [0.50]
−1
o
)=
−1
4
−1
Eo = 0.090 V
o
E =
E ocell
 +
RT  Ag  dilute
−
ln
n F  Ag + 

 conc
(8.314 J mol K ) ( 348 K ) ln [0.015]
(1 mol ) ( 9.65 ×10 C mol ) [0.50]
−1
Eo = 0 −
−1
4
−1
Eo = 0.105 V
20.79
At 25 °C (298 K)
230
= 5.12
Chapter 20
o
Eo = E cell
−
 2+ 
RT Cu  dilute
ln
n F  Cu 2 + 

 conc
(8.314 J mol K ) ( 298 K ) ln [0.015]
= 0−
(1 mol ) ( 9.65 ×10 C mol ) [0.50]
−1
E
o
−1
4
−1
Eo = 0.090 V
At 75 °C (348 K)
(8.314 J mol K ) ( 348 K ) ln [0.015]
= 0−
(1 mol ) ( 9.65 ×10 C mol ) [0.50]
−1
E
o
−1
4
−1
Eo = 0.10 V
20.80
(a)
(b)
(c)
(d)
20.81
(a)
Fe2+(aq) + 2e– Fe(s)
0.20 mol Fe2+ × 2 mol e–/mol Fe2+ = 0.40 mol e–
Cl–(aq) 1/2Cl2(g) + e–
0.70 mol Cl– × 1 mol e–/mol Cl– = 0.70 mol e–
Cr3+(aq) + 3e– Cr(s)
1.50 mol Cr3+ × 3 mol e–/mol Cr3+ = 4.50 mol e–
Mn2+(aq) + 4H2O(l) MnO4–(aq) + 8H+(aq) + 5e–
1.0 × 10–2 mol Mn2+ × 5 mol e–/mol Mn2+ = 5.0 × 10–2 mol e–
Mg2+(aq) + 2e– Mg(s)
 1 mol Mg   2 mol e − 
−
mol e− = ( 5.00 g Mg ) 
 = 0.411 mol e
 
 24.31 g Mg   1 mol Mg 
(b)
Cu2+(aq) + 2e– Cu(s)
 1 mol Cu   2 mol e − 
−
mol e − = ( 41.0 g Cu ) 
 = 1.29 mol e
 
63.55
g
Cu
1
mol
Cu



20.82
Fe(s) + 2OH–(aq) Fe(OH)2(s) + 2e–
The number of Coulombs is: 12.0 min × 60 s/min × 8.00 C/s = 5.76 × 103 C. The number of grams of
Fe(OH)2 is:
 1 mol e −   1 mol Fe(OH)2   89.86 g Fe(OH)2 
g Fe(OH)2 = 5.76 × 103 C 
  1 mol Fe(OH) 
 96500 C   2 mol e −

2 


= 2.68 g Fe(OH)2
(
20.83
2Cl–(l) Cl2(g) + 2e–
The number of Coulombs is: 4.25 A × 35.0 min × 60 s/min = 8.92 × 103 C
The number of grams of Cl2 that will be produced is:
 1 mol e−   1 mol Cl2   70.91 g Cl2 
g Cl2 = 8.92 × 103 C 
 
 = 3.28 g Cl 2
− 

 96,500 C   2 mol e   1 mol Cl2 
Cr3+(aq) + 3e– Cr(s)
The number of Coulombs that will be required is:
(
20.84
)
)
231
Chapter 20
 1 mol Cr   3 mol e−   96,500 C 
Coulombs = ( 75.0 g Cr ) 
= 4.18 × 105 C
 
 

−
 52.00 g Cr   1 mol Cr   1 mol e 
The time that will be required is:
 1 s   1 hr 
hr = 4.18 × 105 C 

 = 51.5 hr
 2.25 C   3600 s 
(
20.85
)
The number of Coulombs that will be required is:
 1 mol Pb   2 mol e −   96,500 C 
Coulombs = ( 35.0 g Pb ) 
= 3.26 × 104 C
 
 
− 
 207.2 g Pb   1 mol Pb   1 mol e 
The time that will be required is:
 1 s   1 hr 
hr = 3.26 × 10 4 C 

 = 6.04 hr
 1.50 C   3600 s 
Mg2+ + 2e– Mg(l)
The number of Coulombs that will be required is:
 1 mol Mg   2 mol e −   96,500 C 
Coulombs = ( 60.0 g Mg ) 
= 4.76 × 105 C
 
 
− 
 24.31 g Mg   1 mol Mg   1 mol e 
The number of amperes is: 4.76 × 105 C ÷ 7200 s = 66.2 amp
(
20.86
20.87
)
Al3+ + 3e– Al(s)
The number of Coulombs that are required is:
 1 mol Al   3 mol e −   96,500 C 
Coulombs = 409 × 103 g Al 
= 4.39 × 109 C
 
  1 mol e− 
26.98
g
Al
1
mol
Al



(
)
The number of amperes is: 4.39 × 109 C ÷ 8.64 × 104 s = 5.08 × 104 A
(Note: There are 8.64 × 104 s in 24.0 hr.)
20.88
The electrolysis of NaCl solution results in the reduction of water, together with the formation of hydroxide
ion: 2H2O + 2e– H2(g) + 2OH–(aq). The number of Coulombs is: 2.00 A × 20.0 min × 60 s/min = 2.40
× 103 C. The number of moles of OH– is:
 1 mol e −  2 mol OH − 
mol OH − = 2.40 × 103 C 
= 0.0249 mol OH −
 96,500 C 
 2 mol e − 



[OH-] = 0.0249 mol/0.250 L = 0.0996 M
(
20.89
)
The electrolysis of NaCl solution results in the reduction of water, together with the formation of hydroxide
ion: 2H2O + 2e– H2(g) + 2OH–(aq). The number of seconds is: 25.0 min × 60 s/min = 1.50 × 103 s.
The number of moles of OH– is:
 1.45 C   1 mol e−  2 mol OH −
mol OH − = 1.50 x 103 s 


−
 s   96,500 C 
 2 mol e
(
)
2.25 × 10−3 mol OH −
−2
−

OH -  =  2.25 x 10 mol OH

 
0.100 L

2 Cl−
→
Cl2 + 2e−

 = 0.225 M

232

 =

Chapter 20
 1 mol Cl2 
mol OH − ) 
= 1.125 x 10 −2 mol Cl2
− 
 2 mol OH 
Assuming Cl2 and OH- react to make OCl−
( 2.25 x 10
OCl-  =
20.90
20.91
20.92
−2
1.125 x 10−2 mol
= 0.122 M
0.100 L
Possible cathode reactions:
Al3+ + 3e–
Al(s)
2H2O + 2e–
H2(g) + 2OH–(aq)
E° = –1.66 V
E° = –0.83 V
Possible anode reactions:
S2O82– + 2e–
2SO42–
O2 + 4H+ + 4e–
2H2O
E° = +2.05 V
E° = +1.23 V
Cathode reaction:
2H2O + 2e–
H2(g) + 2OH–(aq)
Anode reactions:
2H2O
O2 + 4H+ + 4e–
E° = –1.23 V
Net cell reaction:
2H2O
2H2(g) + O2(g)
E° = –2.06 V
Possible cathode reactions:
Cd2+ + 2e–
Cd(s)
–
2H2O + 2e
H2(g) + 2OH–(aq)
E° = –0.40 V
E° = –0.83 V
Possible anode reactions:
O2 + 4H+ + 4e–
2H2O
–
–
2I
I2(s) + 2e
E° = +1.23 V
E° = +0.54 V
Cathode reaction:
Cd2+ + 2e–
Anode reaction:
2I–
Net reaction:
Cd2+ +2I–
Cd(s)
I2(s) + 2e–
I2(s) + Cd(s)
Cu(s)
Possible anode reactions:
S2O82– + 2e–
2SO42–
2Br–
Br2 + 2e–
2H2O
O2(g) + 4H+(aq) + 4e–
Anode reaction: 2Br–
E° = –0.40 V
E° = –0.54 V
The answers to the previous Review Problems guide us here:
Possible cathode reactions:
K + + e–
K(s)
E° = –2.92 V
Cu2+ + 2e–
Cu(s)
E° = +0.34 V
2H2O + 2e–
H2(g) + 2OH–(aq)
E° = –0.83 V
Cathode reaction: Cu2+ + 2e–
E° = –0.83 V
E° = –2.01 V
E° = –1.07 V
E° = –1.23 V
Br2 + 2e–
233
E° = –0.94 V
Chapter 20
Overall reaction: Cu2+ + 2Br–
Br2 + Cu(s)
20.93
At the cathode, where reduction occurs, we expect Cu(s). At the anode, where oxidation occurs, we expect
I2(aq).
The net cell reaction would be Cu2+(aq) + 2I–(aq) Cu(s) + I2(aq)
20.94
In aqueous solution the following reduction of water can occur:
2H2O(l) + 2e−
→
H2(g) + 2OH-(aq)
Eo = − 0.83 V
Reactions that are less positive than this cannot occur at the cathode.
Therefore, Al3+, Mg2+, Na+, Ca2+, K+, and Li+ would not be reduced at the cathode.
20.95
In aqueous solution the following oxidation of water can occur:
2H2O(l)
→
O2(g) + 4H+(aq) + 4e−
E(oxidation) = −1.23 V
Reactions that are more negative than this cannot occur at the anode.
Therefore, Cl−, Au, Br−, Pb2+, Mn2+, Cl2, PbSO4, SO42−, and F− would not be oxidized at the anode.
Additional Exercises
20.96
∆G° = –n F Eocell , Eocell = 1.34 V = 1.34 J/C, and n = 2
∆G° = –(2)(96,500 C)(1.34 J/C) = –2.59 × 105 J per mol of HgO
The maximum amount of work that can be derived from this cell, on using 1.00 g of HgO, is thus:
 1 mol HgO   2.59 × 105 J 
3
J = (1.00 g HgO ) 
 = 1.20 × 10 J
 
216.6
g
HgO
1
mol
HgO



Now, since 1 watt = 1 J s–1, then 2 × 10–3 watt = 2 × 10–3 J s–1, and the time required for this process is:
1s

  1 min   1 hr 
hr = 1.20 × 103 J 
= 167 hr
−3   60 s   60 min 
2
×
10
J




(
20.97
)
The initial numbers of moles of Ag+ and Zn2+ are: 1.00 mol/L × 0.100 L = 0.100 mol. The number of
Coulombs (A × s) that have been employed is: 0.10 C/s × 15.00 hr × 3600 s/hr = 5.4 × 103 C. The number
of moles of electrons is: 5.4 × 103 C ÷ 96,500 C/mol = 5.6 × 10–2 mol electrons.
For Ag+, there is 1 mol per mole of electrons, and for Zn2+, there are two moles of electrons per mol of Zn.
This means that the number of moles of the two ions that have been consumed or formed is given by:
5.6 × 10–2 mol e– × 1 mol Ag+/1 mol e– = 5.6 × 10–2 mol Ag+ reacted.
5.6 × 10–2 mol e– × 1 mol Zn2+/2 mol e– = 2.8 × 10–2 mol Zn2+ formed
The number of moles of Ag+ that remain is: 0.100 – 0.056 = 0.044 mol of Ag+
The final concentration of silver ion is: [Ag+] = 0.044 mol/0.100 L = 0.44 M
The number of moles of Zn2+ that are present is: 0.100 + 0.028 = 0.128 mol Zn2+
234
Chapter 20
The final concentration of zinc ion is: [Zn2+] = 0.128 mol/0.100 L = 1.28 M
o
The standard cell potential should be: Eocell = Eoreduction − Eoxidation
= 0.80 – (–0.76) = 1.56 V
We now apply the Nernst equation:
E cell = E ocell −
0.0592 V
1.28
log
2
( 0.44 )2
Ecell = 1.56 V – 0.024 V = 1.54 V
20.98
The concentration of Br– in solution will be 0.10 M, since the Ksp of AgBr is so small, the amount of
dissociation of AgBr from the electrode will be negligible.
The reduction reaction will be
AgBr(s) + e–
Ag(s) + Br–(aq)
E° = 0.070 V
The oxidation reaction will occur at the standard hydrogen electrode.
The net cell reaction is:
2AgBr(s) + H2 2Ag(s) + 2Br–(aq) + 2H+(aq)
Eocell = 0.070 V – 0.000 V = 0.070 V
The potential for the constructed cell is calculated using the Nernst equation:
2
2
0.0592 V
log  Br −   H + 




n
0.0592
2
2
E cell = 0.070 V −
log [ 0.1] [1]
2
Ecell = 0.070 V – (–0.0592)
Ecell = 0.129 V
o
E cell = E cell
−
20.99
Our strategy will be thus:
Step A:
Pressure H2 (wet) partial pressure H2 mol H2 # e– used
Step B:
Find total charge used = (current)(time)
Step C:
The charge per electron can be arrived at by:
Charge per e– = total charge/total #e– used = Step A/Step B
Step A:
Pressure H2 (wet) partial pressure H2 mol H2 # e– used
The total pressure of wet hydrogen is 767 torr, but some of this is provided by water vapor. Consulting the
water vapor pressure table in the appendices, we find that at 27 ˚C, the vapor pressure of water is 26.7 torr.
Therefore pressure solely due to hydrogen gas (the partial pressure of hydrogen gas) is:
PH2 = 767 – 26.7 = 740 torr
740 torr(1 atm/760 torr) = 0.974 atm
Using the ideal gas law,
235
Chapter 20
PV = nRT
(0.974 atm)(0.288 L) = n(0.0821 L·atm mol–1·K–1)(27 + 273K)
n = 0.0114 mol H2
According to the electrolysis equation 2H–+ + 2e–
H2 gas formed. Therefore,
H2(g), 2 moles of electrons are required per mol of
electrons = 0.0114 mol H2(2 mol e–/1 mol H2)(6.022 × 1023 electrons/mol)
= 1.37 × 1022 electrons
Step B:
Total charge used = (1.22 A)(1800 s) = 2200 C
Step C:
Charge per e– = total charge/total # e– used = 2200 C/1.37 × 1022 electrons
= 1.61 × 10–19 C per electron
(This is a very good estimate; the accepted value is 1.60 × 10–19 C.)
20.100 The oxidation reaction is:
Fe2+ + 2e– Fe
The reduction reaction is:
2H+ + 2e– H2
The overall cell reaction is
2H+ + Fe H2 + Fe2+
E° = –0.44 V
E° = 0.00 V
Eocell = 0.44 V
 Fe 2 + 
0.0592 V

log 
2
n
+
H 
 
[Fe2+] = 0.10 M
[H+] is from the ionization of acetic acid
 H +   C 2 H 3O 2 − 
 = 1.8× 10–5
Ka =   
HC 2 H 3O 2 
[H+] = x
[C2H3O2–] = x
[ x ][ x ]
1.8 × 10–5 =
[0.10 − x ]
Ecell = Eocell –
[HC2H3O2] = 0.10 – x
Assume x is small compared to the concentration of HC2H3O2
x = 1.34 × 10–3 M = [H+] = [C2H3O2–]
[HC2H3O2] = 0.10 M
n=2
[ 0.10]
0.0592 V
log
2
2
1.34 × 10 −3 


Ecell = 0.44 V – 0.14 V
Ecell = 0.30 V
Ecell = 0.44 V –
20.101
The approach is as follows:
Area x thickness Cr volume Cr mass Cr moles Cr # e– needed current
VCr = (area)(thickness) = (1.00 m2)(5.0 × 10–5 m) = 5.0 × 10–5 m3
3
 100 cm   7.19 g Cr   1 mol Cr   6 F   96,500 C 
e − = 5.0 × 10 −5 m3 

 



 1 m   1 cm 3   52.0 g Cr   1 mol Cr   1 F 
(
)
236
Chapter 20
= 4.00 × 106 C
This is done in 4.50 hr (16,200 s). So the current must be:
Current = charge/time = 4.00 × 106 C/16,200 s = 247 A
20.102 (a)
First, we calculate the number of Coulombs:
1.50 A × 30.0 min × 60 s/min = 2.70 × 103 A·s = 2.70 × 103 C
Then we determine the number of moles of electrons:
 1 mol e − 
mole e– = (2.70 × 103 C) 
= 0.0280 mol e–
 96,500 C 


(b)
(c)
0.475 g ÷ 50.9 g/mol = 9.33 × 10–3 mol V
(2.80 × 10–2 mol e–)/(9.33 × 10–3 mol V) = 3.00 mol e–/mol V
The original oxidation state was V3+.
20.103
F2
Cl2
Br2
I2
+2.87
+1.36
+1.07
+0.54
Electron
Affinitiy
(kJ.mol)
-328
-348
-325
-295
Li+
Na+
K+
Rb+
Cs+
-3.05
-2.71
-2.92
-2.93
-2.92
-60
-53
-48
-47
-45
520
496
419
403
376
1.0
1.0
0.9
0.9
0.9
Mg2+
Ca2+
Sr2+
Ba2+
-2.32
-2.76
-2.89
-2.90
+230
+155
+176
+50
738
590
549
503
1.3
1.1
1.0
0.9
Eo (V)
Ionization
Energy1
(kJ/mol)
1682
1251
1140
1008
Electronegativity
4.1
2.9
2.8
2.2
(1) wiki.chemeddl.org
(a) Reduction potentials decrease in the same order as electronegativity and ionization energy. Electron
affinity follows no pattern.
(b) Reduction potentials and the properties decrease going down the family, though Li is out of order due to
its small size and larger effective nuclear charge compared to the other members of the family.
(c) Reduction potentials and the properties decrease going down the family though the electron affinity
value for Sr2+ is out of order.
20.104
237
Chapter 20
RT
ln Cl − 


nF
(8.314 J mol −1K −1 )(298 K)
0.0478 V = 0.2223 V −
ln  Cl− 


1(96,500 C mol−1 )
o
E cell = E cell
−
−0.1745 V =
(8.314 J mol−1K −1 )(298 K)
1(96,500 C mol −1 )
ln Cl − 


−6.797 = ln Cl− 


−3
−
1.117 × 10 = Cl 


20.105 The half–reactions diagrammed in this problem are:
Ag(s) Ag+(aq) + e–
anode
E° = –0.80 V
3+
–
2+
Fe (aq) + e Fe (aq)
cathode
E° = 0.77 V
Eocell = Eoreduction – Eooxidation = 0.77 V – 0.80 V = –0.03 V
 Fe2 +   Ag + 
0.0592 V


log 
1
Fe3+
3.0 × 10−4   4.0 × 10−2 
0.0592 V


= − 0.03 V −
log 
1
1.1 × 10−3 


= + 0.09 V
o
E cell =E cell
−
E cell
As stated, being a galvanic cell, and using conventions in cell notation, the left-side half–cell is the anode
and negatively charged, and the right–side half–cell is the cathode and positively charged. The equation for
the spontaneous cell reaction is Fe3+(1.1 × 10–3 M) + Ag(s) Ag+(3.0 × 10–4 M) + Fe2+(0.040 M). This is
an example of a concentration cell.
Multi-Concept Problems
20.106 The electrolysis of NaCl solution results in the reduction of water, together with the formation of hydroxide
ion: 2H2O + 2e– H2(g) + 2OH–(aq). The number of Coulombs is: 2.00 A × 20.0 min × 60 s/min = 2.40
× 103 C. The number of moles of OH– is:
 1 mol e −  2 mol OH − 
mol OH − = 2.40 × 103 C 
= 0.0249 mol OH −
 96,500 C 
 2 mol e − 



(
)
The volume of acid solution that will neutralize this much OH– is:
 1 mol HCl   1000 mL HCl 
mL HCl = 0.0249 mol OH − 

 = 40.1 mL HCl
 1 mol OH −   0.620 mol HCl 
(
)
20.107 The electrolysis of NaCl solution results in the reduction of water, together with the formation of chlorine
gas and hydroxide ion: 2Cl–(aq) Cl2(g) + 2e–. The number of Coulombs is: 2.50 A × 15.0 min × 60
s/min = 2.25 × 103 C. The number of moles of Cl2 gas collected is:
 1 mol e−   1 mol Cl2 
mol Cl2 = 2.25 × 103 C 
= 0.0117 mol Cl2
 96,500 C   2 mol e− 


The volume of Cl2 gas that is collected is:
(
)
238
Chapter 20
(
)
L atm
298 K )
( 0.0117 mol ) 0.0821 mol
nRT
K (
=
= 0.299 L = 299 mL
1 atm 
P
( 750 torr − 23.76 torr ) 

 760 torr 
+
–
20.108 2H (aq) + 2e H2(g)
The number of Coulombs is: 15.00 min × 60 s/min × 0.750 C/s = 675 C.
The number of moles of H2 is:
V=
 1 mol e −   1 mol H 2 
mol H 2 = ( 675 C ) 
= 3.50 × 10−3 mol H 2
 96,500 C   2 mol e− 


Finally, we calculate the volume of H2 gas:
(
)
L atm
293 K )
( 0.00350 mol ) 0.0821 mol
nRT
K (
V=
=
= 0.0870 L = 87.0 mL
1 atm 
P
( 735 torr ) 

 760 torr 
20.109 In the iron half–cell, we are initially given:
0.0500 L × 0.100 mol/L = 5.00 × 10–3 mol Fe2+(aq)
The precipitation of Fe(OH)2(s) consumes some of the added hydroxide ion, as well as some of the iron
ion: Fe2+(aq) + 2OH–(aq) Fe(OH)2(s). The number of moles of OH– that have been added to the iron
half–cell is:
0.500 mol/L × 0.0500 L = 2.50 × 10–2 mol OH–
The stoichiometry of the precipitation reaction requires that the following number of moles of OH– be
consumed on precipitation of 5.00 × 10–3 mol of Fe(OH)2(s):
5.00 × 10–3 mol Fe(OH)2 × (2 mol OH–/mol Fe(OH)2) = 1.00 × 10–2 mol OH–
The number of moles of OH– that are unprecipitated in the iron half–cell is:
2.50 × 10–2 mol – 1.00 × 10–2 mol = 1.50 × 10–2 mol OH–
Since the resulting volume is 50.0 mL + 50.0 mL, the concentration of hydroxide ion in the iron half–cell
becomes, upon precipitation of the Fe(OH)2:
[OH–] = 1.50 × 10–2 mol/0.100 L = 0.150 M OH–
We have assumed that the iron hydroxide that forms in the above precipitation reaction is completely
insoluble. This is not accurate, though, because some small amount does dissolve in water according to the
following equilibrium:
Fe2+(aq) + 2OH–(aq)
Fe(OH)2(s)
This means that the true [OH–] is slightly higher than 0.150 M as calculated above. Thus we must set up
the usual equilibrium table, in order to analyze the extent to which Fe(OH)2(s) dissolves in 0.150 M OH–
solution:
I
C
E
[Fe2+]
–
+x
+x
[OH–]
0.150
+2x
0.150+2x
The quantity x in the above table is the molar solubility of Fe(OH)2 in the solution that is formed in the iron
half–cell.
239
Chapter 20
Ksp = [Fe2+][OH–]2 = (x)(0.150 + 2x)2
The standard cell potential is:
o
Eocell = Eoreduction − Eoxidation
= 0.3419 V – (–0.447 V) = 0.7889 V
The Nernst equation is:
E cell =
o
E cell
 Fe2+ 
RT

−
ln 
nF
Cu 2+ 


1.175 = 0.7889 −
(8.314 J mol-1K -1 )(298 K)
2(96,500 C mol-1 )
 Fe 2+ 

ln 
1.00
[ ]
1.175 = 0.7889 − 0.01284 ln  Fe 2+ 


ln[Fe2+] = –30.07
[Fe2+] = 8.72 × 10–14 M
This is the concentration of Fe2+ in the saturated solution, and it is the value to be used for x in the above
expression for Ksp.
Ksp = (x)(0.150 + 2x)2 = (8.72 × 10–14)[0.150 + (2)(8.72 × 10–14)]2
Ksp = 1.96 × 10–15
20.110 The desired reaction is:
Ag+(aq) + Br−(aq)
AgBr(s)
While this does not look like an oxidation-reduction reaction, we can use Hess’s law and the half-reaction
provided in the problem to construct the desired reaction. Reverse the direction of the first reaction and
keep the second reaction in the given direction. Be sure to change the sign of the cell potential when the
reaction direction is changed.
Ag
→
Ag+ + e−
Eo = − 0.80 V
AgBr
→
Ag + Br− + e−
Eo = + 0.07 V
Sum the two reactions and their cell potentials:
AgBr → Ag+ + Br− Eocell = − 0.73 V
Now use the Nernst equation to determine the equilibrium constant for the reaction.
At equilibrium, Ecell = 0
0.0592 V
log  Ag +   Br − 
1
- 0.73 V
log  Ag +   Br −  = log K sp ( AgBr ) =
= −12.3
0.0592V
K sp = 4.7 x 10−13
o
E cell =E cell
−
20.111 The range of mmoles of Cl− is:
240
Chapter 20
0.096 M x 3.00 mL = 0.288 mmol Cl−
0.106 M x 3.00 mL = 0.318 mmol Cl−
−
 1 mol   1 mol e
Coulombs = 0.288 mmol x 


 1000 mmol   1 mol Cl→
The time that will be required is:
 1s 
seconds = ( 27.8 C ) 
 = 55.6 s
 0.500 C 
  96,500 C 
= 27.8 C
 
− 
  1 mol e 
−
 1 mol   1 mol e
Coulombs = 0.318 mmol x 


 1000 mmol   1 mol Cl→
  96,500 C 
= 30.7 C
 
− 
  1 mol e 
 1s 
seconds = ( 30.7 C ) 
 = 61.4 s
 0.500 C 
The range of time is thus, 55.6 to 61.4 seconds to precipitate the chloride ions as AgCl.
The electrolysis of NaCl solution results in the reduction of water, together with the formation of hydroxide
20.112
ion: 2H2O + 2e– H2(g) + 2OH–(aq). The number of seconds is: 25.0 min × 60 s/min = 1.50 × 103 s.
The number of moles of OH– is:
 0.250 mol H +  1 mol OH − 
mol OH − = 15.5 mL H + 
= 3.87 × 10−3 mol OH −
 1000 mL H + 
 1 mol H + 



The number of Coulombs that will form this much OH– is:
(
)
 2 mol e−
Coulombs = 3.87 × 10−3 mol OH − 
 2 mol OH −

The average current in amperes is:
(
Current =
3.74 × 102 C
1.50 × 103 s
)
  96,500 C 
= 3.74 × 102 C
 
− 
  1 mol e 
= 0.250 A
20.113
1
H 2 ( g ) + O 2 ( g ) → H 2 O(g)
2
o
o
∆GT ≈ ∆H 289 - T∆So298
−1
 1000 J 
−1
−1
 - 383 K{(1 mol H 2 O)(188.7 J mol K )
 kJ 
= (1 mol H 2 O )(-241.8 kJ mol ) 
− [(1 mol H 2 )(130.6 J mol −1 K −1 ) + (
1
2
mol O 2 )(205.0 J mol−1 K −1 )]}
= − 2.248 x 105 J or 224.8 kJ
The is the maximum amount of work obtained by the reaction of one mole of hydrogen gas.
241
Chapter 20

( 2.248 x 10 J )  1Jwatt
 = 2.248 x 10
s 
5
−1
5
watt s
Therefore, one mole of hydrogen gas will produce 224.8 kW s. We want enough hydrogen gas to
produce 1 kW s.
The mass of H2, at 100% efficiency would be:
1 mol H 2   2.02 g 
−3

 = 8.98 x 10 g H 2

 224.8 kW s   mol H 2 
(1 kW s ) 
Since the efficiency is only 70% we will need:
8.98 x 10−3 g
= 1.28 x 10−2 g H 2
0.70
242