Homework #3
Chapter 11
Electrochemistry
Chapter 4
73.
a)
b)
c)
Oxidation ½ Reaction
Fe + HClHFeCl4
Fe + 4HClHFeCl4
Fe + 4HClHFeCl4 + 3H+
Fe + 4HClHFeCl4 + 3H+ + 3eReduction ½ Reaction
 H2
2H+  H2
2H+ + 2e-  H2
Balanced Reaction
2(Fe + 4HClHFeCl4 + 3H+ + 3e-)
3(2H+ + 2e-  H2)
2Fe(s) + 8HCl(aq)  2HFeCl4(aq) + 3H2(g)
Oxidization ½ Reaction
I-  I3 3I-  I33I-  I3- + 2eReduction ½ Reaction
IO3-  I33IO3- I33IO3- I3- + 9H2O
3IO3- + 18H+I3- + 9H2O
3IO3- + 18H++ 16e-I3- + 9H2O
Balanced Reaction
3IO3- + 18H++ 16e-I3- + 9H2O
8(3I-  I3- + 2e-)
3IO3-(aq) + 18H+(aq) + 24I-(aq)  9I3-(aq) + 9H2O(l)
Divide trough by 3
IO3-(aq) + 6H+(aq) + 8I-(aq)  3I3-(aq) + 3H2O(l)
Oxidation ½ Reaction
Cr(NCS)64-  Cr3+ + NO3- + CO2 + SO42Cr(NCS)64-  Cr3+ + 6NO3- + 6CO2 + 6SO42Cr(NCS)64- + 54H2O  Cr3+ + 6NO3- + 6CO2 + 6SO42Cr(NCS)64- + 54H2O  Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+
Cr(NCS)64- + 54H2O  Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ + 97e-
1
d)
e)
Reduction ½ Reaction
Ce4+  Ce3+
Ce4+ + e-  Ce3+
Balanced Reaction
Cr(NCS)64- + 54H2O  Cr3+ + 6NO3- + 6CO2 + 6SO42- + 108H+ + 97e97(Ce4+ + e-  Ce3+)
Cr(NCS)64-(aq)+ 54H2O(l) + 97Ce4+(aq)
 Cr3+(aq) + 6NO3-(aq) + 6CO2(g) + 6SO42(aq) + 108H+(aq) + 97Ce3+(aq)
Oxidation ½ Reaction
CrI3  CrO42- + IO4CrI3  CrO42- + 3IO4CrI3 + 16H2O  CrO42- + 3IO4CrI3 + 16H2O  CrO42- + 3IO4- + 32H+
CrI3 + 16H2O  CrO42- + 3IO4- + 32H+ + 27eReduction ½ Reaction
Cl2  ClCl2  2ClCl2 + 2e-  2ClBalanced Equation
2(CrI3 + 16H2O  CrO42- + 3IO4- + 32H+ + 27e-)
27(Cl2 + 2e-  2Cl-)
2CrI3(s) + 32H2O(l) + 27Cl2(g)  2CrO42-(aq) + 6IO4-(aq) + 64H+(aq) + 54Cl-(aq)
The solution is basic not acidic, add 64 OH- to both sides
2CrI3(s) + 27Cl2(g) + 64OH-(aq)  2CrO42-(aq) + 6IO4-(aq) + 32H2O(l) + 54Cl-(aq)
Oxidation ½ Reaction
Fe(CN)64-  Fe(OH)3 + CO32- + NO3Fe(CN)64-  Fe(OH)3 + 6CO32- + 6NO3Fe(CN)64- + 39H2O  Fe(OH)3 + 6CO32- + 6NO3Fe(CN)64- + 39H2O  Fe(OH)3 + 6CO32- + 6NO3- + 75H+
Fe(CN)64- + 39H2O  Fe(OH)3 + 6CO32- + 6NO3- + 75H+ + 61eReduction ½ Reaction
Ce4+  Ce(OH)3
Ce4+ + 3H2O  Ce(OH)3
Ce4+ + 3H2O  Ce(OH)3 + 3H+
Ce4+ + 3H2O + e- Ce(OH)3 + 3H+
Balance Reaction
Fe(CN)64- + 39H2O  Fe(OH)3 + 6CO32- + 6NO3- + 75H+ + 61e61(Ce4+ + 3H2O + e- Ce(OH)3 + 3H+)
Fe(CN)64-(aq) + 222H2O(l) + 61Ce4+(aq)
 Fe(OH)3(s) + 6CO32-(aq) + 6NO3-(aq)+ 258H+(aq) + 61Ce(OH)3(s)
2
The solution is basic not acidic, add 258OH- to both sides
Fe(CN)64-(aq) + 258OH-(aq) + 61Ce4+(aq)
 Fe(OH)3(s) + 6CO32-(aq) + 6NO3-(aq)+ 61Ce(OH)3(s) + 36H2O(l)
Chapter 11
3.
Reactions of Interest
Ni2+ + 2e-  Ni
E°=-0.23 V
Cu+ + e-  Cu
E°=0.52 V
Cu2+ + 2e-  Cu
E°=0.34 V
2+
-
Zn + 2e  Zn
E°=-0.76 V
In order to plate out Ni you need the Ni reaction to be the reduction ½ reaction (cathode). In
addition, you also need the cell to be galvanic (E°>0). The E°cell of the nickel/copper cell is
-0.75 V or -0.57 V depending on the ion of copper that is used. Therefore, neither of these cells
would be a galvanic cell, resulting in copper not being an appropriate material. The E°cell of the
nickel/zinc cell is 0.53 V. Therefore, the nickel/zink cell would plate out Ni.
10.
E° is the reaction potential at the standard state. The standard state is 1 M concentration of
aqueous solutions or 1 atm pressure of gasses. By convention E° is set to 0 for hydrogen’s
reduction ½ reaction (2H+ + 2e-  H2). E is the reaction potential not at the standard state. E is
0 when the cell is at equilibrium.
19.
a)
Reactions of interest
Cr2O72- + 14H+ + 6e-  2Cr3+ + 7H2O
Cl2 + 2e- 2Cl3(Cl2 + 2e- 2Cl-)
3+
E° = 1.33 V
E° = 1.36 V
E° = 1.36 V
2-
+
-
2Cr + 7H2O Cr2O7 + 14H + 6e
E° =- 1.33 V
3Cl2(g) + 2Cr3+(aq) + 7H2O(l)  6Cl-(aq) + Cr2O72-(aq) + 14H+(aq) E°=1.36 V-1.33 V=0.03V
3
b)
Reactions of Interest
2+
Cu2+ + 2e-  Cu
E°= 0.34 V
Mg2+ + 2e-  Mg
E°=-2.37 V
-
Cu + 2e  Cu
2+
E°= 0.34 V
-
Mg Mg + 2e
E° =2.37 V
2+
2+
Mg(s) + Cu (aq)  Cu(s) + Mg (aq)
c)
E° = 0.34 V + 2.37 V = 2.71 V
Reactions of Interest
IO3- + 6H+ + 5e-  ½I2 +3H2O
3+
-
2+
Fe + e  Fe
E°=1.20 V
E°= 0.77 V
IO3- + 6H+ + 5e-  ½I2 +3H2O
E° = 1.20 V
5(Fe2+  Fe3+ + e-)
-
+
E°=-0.77V
2+
3+
IO3 (aq) + 6H (aq) + 5Fe (aq)  ½I2(s) +3H2O(l) + 5Fe (aq)
d)
E°=1.20V-0.77V= 0.43 V
Note: I2 does not conduct electricity; therefore, the Pt electrode is used
Reactions of Interest
Ag+ + e-  Ag
E°= 0.80 V
Zn2+ + 2e-  Zn
E°= -0.76 V
2(Ag+ + e-Ag)
2+
E° = 0.80 V
-
Zn  Zn + 2e
+
E° = 0.76 V
2+
2Ag (aq) + Zn(s)  2Ag(s) + Zn (aq)
E° = 0.80 V + 0.76 V = 1.56 V
4
21.
a)
Cl2 + 2e-  2Cl-
E°= 1.36 V
2Br- Br2 + 2e-
E°=-1.09 V
Cl2(g) + 2Br-(aq)  2Cl-(aq) + Br2(aq)
E°=1.36 V – 1.09 V= 0.27 V
Note: Br2 is a liquid at room temperature
b)
5(IO3- + 2H+ + 2e-  IO3- +H2O)
E°= 1.60 V
2(Mn2+ + 4H2O  MnO4- + 8H+ + 5e-)
-
2+
-
E°= -1.51 V
-
+
5IO3 (aq)+2Mn (aq)+3H2O(l)5IO3 (aq)+2MnO4 (aq)+6H (aq)
E°= 1.60V–1.51V=0.09V
5
c)
d)
22.
a)
H2O2 +2H+ + 2e-  2H2O
E°= 1.78 V
H2O2  O2 + 2H+ + 2e-
E°= -0.68 V
2H2O2(aq)  O2(g) + 2H2O(l)
E°= 1.78 V -0.68 V = 1.10 V
2(Fe3+ + 3e-  Fe)
3(Mn  Mn2+ + 2e-)
E°= -0.036V
E° = 1.18 V
2Fe3+(aq) + 3Mn(s)  2Fe(s) + Mn2+(aq)
E° = -0.036 V + 1.18 V = 1.14 V
For a galvanic cell E° must be positive therefore Cu+ must be oxidized
given:
Au3+ + 3e-  Au
2+
-
+
Cu +e  Cu
3+
Au
b)
E° = 1.50V
E° = 0.16 V
-
+ 3e  Au
E° = 1.50V
3(Cu+  Cu2+ +e- )
E° = -0.16 V
Au3+(aq) + 3Cu+(aq)  Au(s) + 3Cu2+(aq)
E° = 1.34 V
For a galvanic cell E° must be positive therefore Cd must be oxidized
given:
VO2+ + 2H+ + e-  VO2+ + H2O
E° = 1.00V
Cd2+ +2e-  Cd
E° = -0.40 V
6
2(VO2+ + 2H+ + e-  VO2+ + H2O)
E° = 1.00V
Cd  Cd2+ +2e+
E° = 0.40 V
+
2+
2+
2VO2 (aq) +4H (aq) + Cd(s)  2VO (aq) +2H2O(l) + Cd (aq)
24.
a)
E° = 1.40 V
Cu2+ + 2e-  Cu
E° = 0.34 V
In order for the cell to be a galvanic cell Cu2+ must be reduced, therefore, SCE is
oxidized and at the anode.
E°cell = 0.34 V + -0.242 V = 0.10 V
b)
Fe3+ + e-  Fe2+
E° = 0.77 V
In order for the cell to be a galvanic cell, Fe3+ must be reduced, therefore, SCE is oxidized
and at the anode.
E°cell = 0.77 V + -0.242 V = 0.53 V
c)
AgCl + e-  Ag + ClE° = 0.22 V
In order for the cell to be a galvanic cell, Ag must be oxidized, therefore, SCE is reduced
and at the cathode.
E°cell = 0.242 V – 0.22 V = 0.02 V
d)
Al3+ + 3e-  Al E° = -1.66 V
In order for the cell to be a galvanic cell, Al must be oxidized, therefore, SCE is reduced
and at the cathode.
E°cell = 0.242 V + 1.66 V = 1.90 V
e)
Ni2+ + 2e-  Ni E° = -0.23 V
In order for the cell to be a galvanic, cell Ni must be oxidized, therefore, SCE is reduced
and at the cathode.
E°cell = 0.242 V + 0.23 V = 0.47 V
25.
a)
Cu  Cu2+ +2e-
E°= -0.34 V
2H+ + 2e-  H2
E°= 0.0 V
+
No H cannot oxidize Cu
b)
2I- I2 +2e3+
-
E°=-0.54 V
2+
Fe + e  Fe
E°= 0.77 V
3+
Yes Fe is capable of oxidizing I- if it is going to Fe2+
2I- I2 +2e3+
E°=-0.54 V
-
Fe + 3e  Fe
E°= -0.04 V
3+
No Fe cannot oxidize I if it is going to Fe
c)
Ag+ + e-  Ag
E°=0.80 V
H2  2H+ +2eE°= 0.0 V
Yes H2 is capable of reducing Ag+
7
d)
Cr3+ + e-  Cr2+
E° = -0.50 V
Fe2+  Fe3+ + eE°=-0.77 V
2+
No Fe is not capable of reducing Cr3 + to Cr2+
26.
The oxidizing agent is the species that is reduced. Therefore, the best oxidizing reagent is the
species that has the largest E° value for the reduction ½ reaction.
K+ < H2O < Cd2+ < I2 < AuCl4 - < IO3-
27.
The reducing agent is the species that is oxidized. Therefore, the best reducing agent is the
species that has the smallest E° value for the reduction ½ reaction.
F- < H2O < I2 < Cu+ < H- < K
28.
Choices
Br2 + 2e-  2Br+
a)
b)
c)
d)
-
E°=1.09 V
2H + 2e  H2
E°=0.0 V
Cd+ + 2e-  Cd
E°=-0.40 V
La3+ + 3e-  La
E°=-2.37 V
Ca2+ + 2e-  Ca
E°=-2.76 V
Underlined are possible answers
The oxidizing agent is the species that is being reduced or species in the reduction
reaction. Therefore Br-, H2, Cd, and Ca can be eliminated because if they were on the
reactants side of the reaction, therefore, they can only be in an oxidization reaction.
Out of the remaining species the best oxidizing agent is the one with the largest E° value
in the reduction ½ reactions.
Br2
The reducing agent is a species that is being oxidized or the species in the oxidizing
reaction. Therefore Br2, H+, La3+ can be eliminated because they were on the reactant
side of the equation, therefore, they can only be in a reduction reaction. Out of the
remaining species the best oxidizing agent is the one with the smallest E° value in the
reduction ½ reactions.
Ca
MnO42- + 8H+ + 5e-  Mn2+ + 4H2O
E°= 1.51 V
2In order for MnO4 to oxidize a species the species has to be in an oxidation reaction;
therefore, Br2, H+, and La3+ can be eliminated because they were on the reactant side of
the equation, therefore, they can only be in a reduction reaction. Of the remaining
species the E°value of the reduction ½ reaction must be smaller than 1.51 V.
Br-, H2, Cd, and Ca
Zn2+ + 2e-  Zn
Reaction of interest
E°=-0.76 V
Zn  Zn2+ + 2eE°= 0.76 V
In order for Zn to reduce a species it must be in a reduction reaction. Therefore Br-, H2,
8
Cd, and Ca can be eliminated because if they were on the reactants side of the reaction ,
therefore, they can only be in an oxidization reaction. Of the remaining species the E°
value of the reduction ½ reaction must be greater than -0.76 V.
Br2 or Ag+
29.
a)
Br2 + 2e-  2Br-
E° = 1.09 V
Cl2 + 2e- 2ClReactions of interest
E°=1.36 V
2Br-  Br2 + 2e-
E° = -1.09 V
2Cl- Cl2 + 2e-
E°=-1.36 V
In order to oxidize Br- and not Cl- the E° value of the reduction ½ reaction must be
greater than 1.09 V but smaller than 1.36 V
Cr2O72- + 14H+ + 6e-  2Cr3+ + 7H2O
+
-
O2 + 4H + 4e  2H2O
+
-
E° = 1.23 V
2+
MnO2 + 4H + 4e  Mn + 2H2O
-
+
E° = 1.33 V
E° = 1.21 V
-
IO3 + 6H + 5 e  ½I2 + 3H2O
E° = 1.20 V
2Therefore, Cr2O7 , O2, MnO2, and IO3 could oxidize Br- to Br2 but not oxidize Cl- to Cl2.
b)
Mn2+ + 2e-  Mn
E° = -1.18 V
Ni2+ + 2e-  Ni
Reactions of interest
E° = -0.23 V
Mn  Mn2+ + 2e-
E° = 1.18 V
Ni  Ni2+ + 2e-
E° = 0.23 V
In order to oxidize Mn and not Ni the E° value of the reduction ½ reaction must be
greater than -1.18 V but smaller than -0.23 V
PbSO4 + 2e-  Pb + SO42-
E° = -0.35 V
Cd2+ +2e-  Cd
E °= -0.40 V
Fe2+ + 2e-  Fe
E° = -0.44 V
3+
-
3+
-
Cr + e  Cr
2+
E° = -0.50 V
Cr + e  Cr
E° = -0.73 V
Zn2+ + 2e-  Zn
E°=-0.76 V
2H2O + 2e-  H2 + 2OHE° = -0.83 V
2+
2+
3+
2+
Therefore, PbSO4, Cd , Fe , Cr , Zn , and H2O are capable of oxidizing Mn to Mn2+ but
not oxidizing Ni to Ni2+.
30.
a)
Cu2+ + 2e-  Cu
2+
-
+
Cu + e  Cu
E° = 0.34 V
E° = 0.16 V
9
In order to reduce Cu2+ to Cu but not to Cu+ the E˚ values of the reduction ½ reaction
must be greater than 0.16 V and less than 0.34 V. The species also must be on the
product side of the reduction reaction.
HgCl2 + 2e-  2Hg + 2Cl-
E° = 0.27 V
AgCl + e-  Ag + 4ClE° = 0.22 V
2+
SO4 + 4H + 2e  H2SO4 + H2O
E°= 0.20 V
Therefore, Hg/Cl-, Ag/Cl-, and H2SO3 are capable of reducing Cu2+ to Cu but not to Cu+.
b)
E °= 1.09 V
Br2 + 2e- 2Br-
-
I2 + 2e  2I
E°=0.54 V
In order to reduce Br2 to Br- but not I2 to I- the E˚ values of the reduction ½ reaction
must be great than 0.54 V and less than 1.09 V. The species also must be on the product
side of the reduction reaction.
VO2+ + 2H+ + e-  VO2+ + H2O
E° = 1.00V
AuCl4- + 3e-  Au + 4Cl-
E° = 0.99V
NO3- + 4H+ + 3e- NO + 2H2O
E° = 0.96V
-
ClO2 + e  ClO2
2+
-
-
2Hg + 2e  Hg2
E°=0.954 V
2+
E°=0.91 V
Ag+ + e-  Ag
E° = 0.80 V
2Hg2+ + 2e-  2Hg
E°=0.80 V
Fe3+ + 2e-  Fe2+
E° = 0.77 V
+
-
O2 + 2H + 2e  H2O2
-
-
E°=0.68 V
2-
MnO4 + e  MnO4
E°=0.56 V
2+
2+
Therefore, VO ,Au/Cl , NO, ClO2 , Hg2 , Ag, Hg, Fe , H2O2, and MnO4- are capable of
reducing Br2 to Br- but not I2 to I-.
2+
40.
a)
2(ClO2-  ClO2 +e-)
E°=-0.95 V
Cl2 + 2e-  2Cl-
E°= 1.36 V
2ClO2-(aq) + Cl2(g)  2ClO2(g) + 2Cl-(aq)
E°=1.36 V - 0.95 V = 0.41 V
+
Note: The Na in the equation is just there as a spectator ion

G  nFE    2mol e  96, 485 molC e
  0.41V   79000J  79kJ
RT
ln  K 
nF
8.3145 molJ K   298 K 

0.41V 
ln  K 
 2mol e  96, 485 molC e
E 


K  7.4 1013
b)
Assume acidic conditions
Reduction ½ Reaction
10
ClO2  ClClO2  Cl- + 2H2O
ClO2 + 4H+ Cl- +2H2O
ClO2 + 4H+ + 5e-  Cl- +2H2O
Oxidation ½ Reaction
ClO2  ClO3ClO2 + H2O ClO3ClO2 + H2O ClO3- + 2H+
ClO2 + H2O ClO3- + 2H+ + eBalanced Reaction
ClO2 + 4H+ + 5e-  Cl- +2H2O
5(ClO2 + H2O ClO3- + 2H+ + e-)
6ClO2(g) + 3H2O(l)  Cl-(aq) + 5ClO3-(aq) + 6H+(aq)
41.
a)
Reaction 1:
Unbalanced
Mn(s) + NO3-(aq)  NO(g) + Mn2+(aq)
Oxidation ½ reaction
Mn  Mn2+
Mn  Mn2+ + 2eReduction ½ reaction (The problem told you that you have nitric acid as a
reactant. To determine what species NO3- forms, use the standard reduction
potentials)
NO3-  NO
NO3-  NO + 2H2O
NO3- + 4H+  NO + 2H2O
NO3- + 4H+ + 3e- NO + 2H2O
Balanced Reaction
3(Mn  Mn2+ + 2e-)
2(NO3- + 4H+ + 3e- NO + 2H2O)
3Mn(s) + 2NO3-(aq) +8H+(aq)  3Mn2+(aq) + 2NO(g) + 4H2O(l)
Reaction 2:
Unbalanced
Mn2+(aq)+ IO4-(aq) MnO4-(aq) + IO3-(aq)
Oxidation ½ reaction
Mn2+  MnO4Mn2+ + 4H2O MnO4Mn2+ + 4H2O MnO4- + 8H+
Mn2+ + 4H2O MnO4- + 8H+ + 5eReduction ½ Reaction
IO4-  IO3IO4-  IO3- + H2O
IO4- + 2H+ + 2e- IO3- + H2O
IO4- + 2H+ + 2e- IO3- + H2O
Balanced Reaction
11
2(Mn2+ + 4H2O MnO4- + 8H+ + 5e-)
5(IO4- + 2H+ + 2e-  IO3- + H2O)
2Mn2+(aq) + 5IO4-(aq) +3H2O(l) 2MnO4-(aq) + 5IO3-(aq) + 6H+(aq)
b)
Reaction 1
3(Mn Mn2+ + 2e-)
E° = 1.18 V
2(NO3- + 4H+ + 3e- NO + 2H2O)
E° = 0.96 V
+
2+
3Mn(s) + 2NO3 (aq) +8H (aq)  3Mn (aq) + 2NO(g) + 4H2O(l)
E˚cell = 1.18 V + 0.96 V = 2.14 V

G  nFE    6mol e  96, 485 molC e
E  
  2.14   1.24 10 J  1240kJ
6
J
C
RT
ln  K 
nF
C
 6  96,485 mol
 2.14 CJ 
8.3145 molJK  298 K 
nF E 
RT
K e
 10
 e500.
The K number is most likely too large for your calculator
Reaction 2
2(Mn2+ + 4H2O MnO4- + 8H+ + 5e- )
E° = -1.51 V
5(IO4- + 2H+ + 2e-  IO3- + H2O)
E° = 1.60 V
2+
2Mn (aq) + 5IO4 (aq) +3H2O(l) 2MnO4-(aq) + 5IO3-(aq) + 6H+(aq)

G  nFE   10mol e  96, 485 molC e
E  
K e
42.
a)
E°cell= 1.60 V + - 1.51 V = 0.09 V
  0.09   9 10 J  90kJ
J
C
4
RT
ln  K 
nF
nF E 
RT
 10
C
10  96,485 mol
 0.09 CJ 
J
8.3145 molK  298 K 
 2 1015
Given
2H+ + 2e-  H2
+
E°=0.00 V
-
O2 + 4H + 4e  2H2O
E° = 1.23 V
2(H22H+ + 2e-)
E°=-0.00 V
O2 + 4H+ + 4e-  2H2O
E° = 1.23 V
2H2 + O2  2H2O
E°cell= 1.23 V + 0.00 V = 1.23 V


G  nFE    4mol e  96, 485 molC e 1.23 CJ   4.75 105 J  475kJ
b)
Two gases go to a liquid, therefore, the positional probability decreases causing ΔS˚ to
be negative.
G  H   T S 
c)
If ΔS˚ is negative the only way for ΔG˚ to be negative is if ΔH˚ is negative.
ΔG˚ is the maximum possible work that can be done. Increasing the temperature
increases the TΔS˚ term, therefore, since this term is +, less possible work can be done.
12
43.
G  nFE
G  H   T S 
nFE   H   T S 
H  S 
E  

T
nF
nF
S 
H 
E 
T
nF
nF
H
If E° vs. T was plotted on a graph, the intercept would equal  nF
and the slope of the line
would equal nFS  The smaller the  nFS  term is the smaller the temperature dependence of E.
Therefore, cells that have small ΔS° terms are relatively temperature independent.
45.
a)
Cu+  Cu2+ + e-
E°= -0.16 V
Cu+ + e-  Cu
E°= 0.52 V
2Cu+  Cu2+ + Cu
E°= -0.16 V + 0.52 V = 0.36 V
This reaction is spontaneous
C
kJ
G  nFE   1  96, 485 mol
  0.36V   35000 molJ  35 mol
E 
RT
ln  K 
nF
K e
b)
nFE 
RT
e
C
1 96,485 mol
 0.36V 
J
8.3145 molK  298 K 
 1.2 106
2(Fe2+  Fe3+ +e- )
E°= -0.44 V
Fe2+ +2e-  Fe
E°= -0.77 V
3Fe2+  2Fe3+ + Fe
E°=-0.44V+ - 0.77V=-1.21 V
This reaction is not spontaneous
c)
HClO2 + H2O  ClO3- + 3H+ + 2e-
E°=-1.21 V
HClO2 + 2H+ + 2e-  HClO + H2O
E°=1.65 V
2HClO2(aq)  ClO3-(aq) + H+(aq) + HClO(aq)
E°= -1.21 V + 1.65 V =0.44 V
C
kJ
0.44V   85,000 molJ  85 mol
G  nFE   296,485 mol
E 
RT
ln  K 
nF
K e
nFE 
RT
e
C
 2 96,485 mol
 0.44V 
8.3145 molJK  298 K 
 7.6 1014
13
53.
a)
For galvanic cell E° must be positive
Au3+ +3e-  Au
E° = 1.50 V
3(Tl  Tl+ + e-)
E° = 0.34 V
Au3+(aq) + 3Tl(s)  Au(s) + 3Tl+(aq)
C
kJ
G  nFE    3  96485 mol
 1.84V   5.32 105 molJ  532 mol
b)
E 
RT
ln  K 
nF
K e
nFE 
RT
e
C
 3 96,485 mol
1.84V 
J
8.3145
298 K 

molK 
 2.26 1093
8.3145 molJK   298.14K  ln  1.0104    2.04V
RT
RT  Tl   
ln  Q   E 
ln   Au3    1.84V 
 1.0102 
nF
nF    
 3  96485 molC 


3
3
c)
54.
E° = 1.50 V + 0.34 V = 1.84 V
E  E 
2(Cr2+ Cr3+ + e-)
Co2+ + 2e-  Co
2Cr2+ + Co2+  2Cr3+
8.3145 molJK   298K  ln 2.79 107  0.220V
RT
E 
ln  K  


nF
 2   96, 485 molC 
E  E 
2
8.3145 molJK   298K  ln
RT
RT  Cr 3  
ln  Q   E  
ln  2 2 2   0.220V 
nF
nF  Cr  Co  
 2   96, 485 molC 

 2.02
 0.302  0.20
  0.151V
C
G  nFE    2  96, 485 mol
  0.151V   29100J   29.1kJ
55.
Given
Al3+ + 3e- Al
E° = -1.66 V
Pb2+ + 2e-  Pb
E° =- 0.13 V
2(Al Al3+ + 3e-)
2+
E° = 1.66 V
-
3(Pb + 2e  Pb)
E° = -0.13 V
2+
3+
2Al(s) + 3Pb (aq)  2Al (aq) + 2Pb(s)
Calculate the final Pb2+ concentration:
E° = 1.66V + -0.13V= 1.53V
Pb2+
Al3+
Initial
1.00
1.00
Change
-3x
+2x=0.60
Final
1.00-3x
1.00+0.60=1.60
x must equal 0.30 M, therefore, the Pb2+ concentration will change by 0.90 M and the
final Pb2+ concentration equals 1.00-0.90=0.10 M.
14
Calculate E of cell
RT
RT   Al 3  
E  E 
ln  Q   E  
ln
3
nF
nF   Pb2  
8.3145 molJK   298K  ln 1.602  1.50V
E  1.53V 
 0.103
 6   96, 485 molC 
2
 
61.
Since the same material is on both sides of the cell E°= 0
In order to be a galvanic cell the concentration on the anode side of Ni2+ must be smaller than
the concentration of Ni2+ on the cathode side
Ni2+ + 2e-  Ni
RT
RT   Ni2 ( anode ) 
ln  K   
ln  2

nF
nF   Ni ( cathode ) 
8.3145 molJK   298K  ln 1.0 M  0.0V
E
 1.0M 
 2  96, 485 molC 
E  E 
a)
No electron flow.
b)
8.3145 molJK   298K 

M
E
ln  1.0
2.0 M   0.0089V
 2 96, 485 molC 
Anode is on the left (1.0 M) the cathode is on the right (2.0 M) and electrons flow from
left to right on the diagram.
c)
8.3145 molJ K   298K 

0.1M
E
ln  1.0
M   0.030V
 2   96, 485 molC 
Anode is on the right (0.1 M) the cathode is on the left (1.0 M) and electrons flow from
right to left on the diagram.
d)
E
8.3145 molJK   298K  ln 4.010 M  0.13V
 1.0M 
 2  96, 485 molC 
5
Anode is on the right (4.0×10-5 M) the cathode is on the left (1.0 M) and electrons flow
from right to left on the diagram.
e)
E
8.3145 molJK   298K  ln 2.5M  0.0V
 2.5M 
 2  96, 485 molC 
No electron flow.
68.
a)
Al3+ + 3e-  Al
1.0kg Al

1000 g
1kg

1mol Al
26.98 g Al

3mol e
1mol Al
  111mol e 
It
F
C
  1.07 105 s  29.7h
nF 111mol   96, 485 mol
t

I
100.0 A
n
b)
Ni2+ + 2e-  Ni
15
1.0 g Ni
t
c)

1mol Ni
58.69 g Ni

5.0mol Ag

1mol e
1mol Ag
  5.0mol e
69.
a)
Co2+ + 2e-  Co
2I-  I2 +2e-
0.56mol e
d)


+
1mol Co
2 mol e
1mol Hf
4 mol e
1mol I 2
2 mol e
-
60s

58.93 g Co
1mol Co
  17 g Co

178.49 g Hf
1mol Hf
  25g Hf

253.80 g I 2
1mol I 2
CrO3 + 6H + 6e  Cr + 3H2O
0.56mol e
70.

Hf4+ + 4e-  Hf
0.56mol e
c)

It 15 A 1.0h  1h  1min  

 0.56mol e
C
F
96, 485 mol
0.56mol e
b)

C
  4800s  1.3h
nF  5.0mol   96, 485 mol

I
100.0 A
60min
n
  0.034mol e
C
  33s
nF  0.034mol   96, 485 mol

I
100.0 A
Ag+ + e-  Ag
t
2 mol e
1mol Ni

1mol Cr
6 mol e

52.00 g Cr
1mol Cr
  71g I
2
  4.9g Cr
The question wants you to calculate the molarity of the initial solution
c
n
V
We know the volume so we need to calculate the mole of Ag +.
The ½ reaction that we are interested in is
Ag+ +e-  Ag
If we know the moles of electrons we can get the moles of Ag +. Calculate the moles of
electrons.
60s
It  2.00 A  2.30 min  1min  
n 
 0.00286mol e
C
F
96, 485 mol
Calculate the moles of Ag+
0.00286mol e

1mol Ag 
1mol e
  0.00286 Ag

Calculate molarity
c
n 0.00286mol

 0.0114M Ag 
V
0.250 L
16
71.
The question asked you to calculate time to plate out 10 g Bi.
To calculate the time use
n
It
F
Therefore, need to determine the number of e-. To do this you must find a relationship between
e- and Bi. Balance the equation.
BiO+  Bi
BiO+  Bi + H2O
BiO+ + 2H+  Bi + H2O
BIO+ + 2H+ + 3e-  Bi + H2O
Note: If you balanced it in basic conditions you would also come out with 3 electrons
Determine the number of e-
n  10.0 g Bi

1mol Bi
209.0 g

3mol e
1mol Bi
  0.144mol
Determine time
C
  556s
nF  0.144mol   96485 mol
t

I
 25.0 A
73.
For electrolysis reaction E°cell is a)
b)
Cathode reaction
Ni2+ + 2e-  Ni
E°= -0.23 V
Anode reaction
2Br-  Br2 + 2e-
E°= -1.09 V
Cathode reaction
Anode reaction
c)
Cathode reaction
3+
-
Al + 3e  Al
-
E°= -1.66 V
-
2F  F2 + 2e
2+
E°= -2.87 V
-
Mn + 2e  Mn
E°= -1.18 V
Anode reaction
2I-  I2 + 2eE°= -0.54 V
For the aqueous solution we must also consider the reactions with H2O
d)
Cathode reaction
Ni2+ + 2e-  Ni
E°= -0.23 V
2H2O + 2e-  H2 + 2OH-
E°= -0.83 V
2+
Since it is easier to reduce Ni than H2O the nickel reaction will occur at the
cathode.
Anode reaction
e)
2Br-  Br2 + 2e-
E°= -1.09 V
H2O  O2 + 4H+ + 4e-
E°= -1.23 V
Since it is easier to oxidize B- than H2O the bromine reaction will occur at the
anode.
Cathode reaction
Al3+ + 3e-  Al
E°=-1.66 V
17
2H2O + 2e-  H2 + 2OH-
E°= -0.83 V
3+
Since it is easier to reduce H2O than Al the water reaction will occur at the cathode.
Anode reaction
f)
2F-  F2 + 2e-
E°= -2.87 V
H2O  O2 + 4H+ + 4e-
E°= -1.23 V
Since it is easier to oxidize H2O than F- the water reaction will occur at the anode.
Cathode reaction
Mn2+ + 2e-  Mn
E°= -1.18 V
2H2O + 2e-  H2 + 2OH-
E°= -0.83 V
2+
Since it is easier to reduce H2O than Mn the water reaction will occur at the cathode.
Anode reaction
2I-  I2 + 2e-
E°=-0.54 V
H2O  O2 + 4H+ + 4e-
E°=-1.23 V
-
Since it is easier to oxidize I than H2O the iodine reaction will occur at the anode.
76.
The question wants you to determine the charge on the ruthenium. The reaction that we are
interested in is: Run+ + ne-  Ru. Therefore, we should be able to identify n by comparing the
moles of Ru to the moles of e-.
Calculate the moles of Ru
2.618g Ru

1mol Ru
101.1g Ru
  0.0259mol Ru
Calculate the moles of e-
It  2.50 A  50.0 min  1min  

 0.0777mol e
C
F
96, 485 mol
60s
n
Because the ratio between the moles of e- and moles of Ru is 3:1 the charge on the ruthenium
must be 3+.
84.
The question wants you to calculate the volume of O 2 and H2 gas produced in the electrolysis of
water. The question gives you the time and the current. With this information you can
calculate the mole of electrons.
60s S
It  2.50 A 15.0 min  1min  
n 
 0.0233mol e
C
F
96, 485 mol
You need to find a relationship between the moles of electrons and the moles of O 2 and H2, to
do this look at the equations for the electrolysis of water.
2(2H+ + 2e-  H2)
2H2O O2 + 4H+ + 4e2H2O  2H2 + O2
Calculate the moles of H2
0.0233mol e

2 mol H 2
4 mol e
  0.0117mol H
2
18
Calculate the volume of H2
V
Latm
nRT  0.0117mol   0.08206 mol
 K   273.15 K 

 0.262 L H 2
P
1.00atm 
Calculate moles of O2
0.0233mol e

1mol O2
4 mol e
  0.00583mol O
2
Calculate the volume of O2
Latm
nRT  0.00583mol   0.08206 mol
 K   273.15 K 
V

 0.131L O2
P
1.00atm 
88.
A battery is a galvanic cell. When a battery is new there are many more reactants than
products in the battery and therefore, Q is small. As the battery is used it turns reactants into
product and Q gets larger. Since
E  E 
RT
ln Q 
nF
as the battery is used up E gets smaller and smaller until it is 0. A battery is not a system at
equilibrium otherwise E = 0.
Fuel cells and batteries are both galvanic cells. The difference between them is that batteries
are enclosed and therefore, eventually reach equilibrium. For fuel cells the reactants are
continually supplied and therefore, never reach equilibrium.
90.
The reaction that goes on in a hydrogen oxygen fuel cell is
2H2(g) + O2(g) 2H2O(l)
O2 + 4H+ + 4e-  2H2O
E°= 1.23 V
2(H2 2H+ + 2e-)
E°= -0.0 V
O2(g) + 2H2(g) 2H2O(l)
E°= 1.23 V + 0.0 V = 1.23 V
wmax  G  nFE
Because standard conditions
wmax  nFE 
C
kJ
1.23V   4.75  105 molJ  475 mol
wmax  496,485 mol
This is the wmax for 2 moles of H2O. Need to find the work mass for 1.00 kg of H2O
1.00kg

1000 g
1kg

1mol H 2O
18.02 g H 2O

475 kJ
2 mol H 2O
  13, 200kJ
The work done can be no larger than -13,200 kJ. Usually the max work is not done and some of
the energy is lost to heat. Fuel cells are more efficient in converting chemical energy into
electrical energy than combustion reaction. The main disadvantage of fuel cells is their cost.
19