1. Unit 9 HW
Answers

2. 20.20
a. Mo+3 + 3e- → Mo reduction
b. H2O + H2SO3 → SO H+ + 2e- oxidation
c. 3e- + 4H+ + NO3- → NO + 2H2O reduction
d. 4e- + 4H+ + O2 → 2H2O reduction
e. 4OH- + Mn+2 → MnO2 + 2H2O + 2e- oxidation
f. 5OH- + Cr(OH)3 → CrO H2O + 3e- oxidation
g. 4e- + 2H2O + O2 → 4OH- reduction

3. 20.22
a. 3(NO2- + H2O → NO3- + 2H+ +2e-)
6e- + 14H+ + Cr2O7-2 → 2Cr+3 + 7H2O
3NO2- + 8H+ + Cr2O7-2 → 3NO3- + 2Cr+3 + 4H2O
b. 2(3H2O + S → H2SO3 + 4H+ + 4e-)
8e- + 8H+ + 2HNO3 → N2O + 5H2O
H2O + 2S + 2HNO3 → 2H2SO3 + N2O
c. 2(6e- + 14H+ + Cr2O7-2 → 2Cr+3 + 7H2O)
3(H2O + CH3OH → HCO2H + 4H+ + 4e-)
16H+ + 2Cr2O CH3OH → 4Cr+3 + 3HCO2H + 11H2O

4. 20.22 d. 2(5e- + 8H+ + MnO4- + → Mn+2 + 4H2O) 5(2Cl- → Cl2 + 2e-)
16H+ + 2MnO Cl- → 2Mn+2 + 8H2O + 5Cl2
e. 6e- + 6H2O + NO2- → NH4+ + 8OH-
2(4OH- + Al → AlO2- + 2H2O + 3e-)
2H2O + 2Al + NO2- → 2AlO2- + NH4+
f. 2OH- + H2O2 → O2 + 2H2O + 2e-
2(e- + ClO2 → ClO2-)
2OH- + H2O2 + 3ClO2 → 2ClO2- + O2 + 2H2O

5. 20.6 a. B is being oxidized, A is being reduced
b. A half reaction would be at the cathode, B half reaction would be at the anode
c. reduction half reaction is at a higher potential, A half reaction
d. Since E° is a positive number, then ΔG has to be a negative number

6. 20.34 a. Cd  Cd+2 + 2e- PdCl42- + 2e-  Pd + 4Cl-
b = X – (-0.403)
X = 0.627V

7. 20.36
a – 0 = 2.87 V
b – (-2.87) = V
c – = V
d – = V

8. 20.38 a. 3IO- + 3H2O + 2Au + 8Br-  3I- + 6OH- + 2AuBr4- 1.348 V
b. Sn+2 + 2Eu+2  Sn + 2Eu V

9. 20.50
a. 2I- + Hg2+2  I2 + 2Hg
0.789 – = V ΔG = -(2)(96485)(0.253) = J
ΔG = -RTlnK = -(8.314)(298)lnK K = 3.6 x 108
b. 3Cu+ + NO3- + 4H+  3Cu+2 + NO + 2H2O
0.96 – = V ΔG = -(3)(96485)(0.807) = J
ΔG = -RTlnK = -(8.314)(298)lnK K = 8.83 x 1040
c. 2Cr(OH)3 + 4OH- + 3ClO-  2CrO H2O + 3Cl-
0.89 – (-0.13) = 1.02 V ΔG = -(6)(96485)(1.02) = J
ΔG = -RTlnK = -(8.314)(298)lnK K = 3.2 x 10103

10. 20.52
K = 8.7 x 104
n = 1
ΔG = -RTlnK = -(8.314)(298)ln(8.7 x 104) = J
ΔG = -nFE = -(1)(96485)E E = V

11. 20.54
a – = V ΔG = -(2)(96485)(0.462) = J ΔG = -RTlnK = -(8.314)(298)lnK K = 4.24 x 1015 b – 0.32 = 1.29 V ΔG = -(3)(96485)(1.29) = J ΔG = -RTlnK = -(8.314)(298)lnK K = 2.84x 1065 c – (-0.23) = 0.59 V ΔG = -(4)(96485)(0.59) = J ΔG = -RTlnK = -(8.314)(298)lnK K = 8.2 x 1039

12. 20.56
K = 5.5 x 105
E = 0.17 V
-RTlnK = -nFE
-(8.314)(298)ln(5.5 x 105) = -n(96485)(0.17)
n = 2