Unit 9 HW Answers

20.20 a. Mo+3 + 3e- → Mo reduction b. H2O + H2SO3 → SO42- + 4H+ + 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 → CrO4-2 + 4H2O + 3e- oxidation g. 4e- + 2H2O + O2 → 4OH- reduction

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+ + 2Cr2O7-2 + 3CH3OH → 4Cr+3 + 3HCO2H + 11H2O

20.22 d. 2(5e- + 8H+ + MnO4- + → Mn+2 + 4H2O) 5(2Cl- → Cl2 + 2e-) 16H+ + 2MnO4- + 10Cl- → 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

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

20.34 a. Cd  Cd+2 + 2e- PdCl42- + 2e-  Pd + 4Cl- b. 1.03 = X – (-0.403) X = 0.627V

20.36 a. 2.87 – 0 = 2.87 V b. 0.337 – (-2.87) = 3.207 V c. -0.440 – 0.771 = -1.211 V d. 0.789 – 0.153 = 0.636 V

20.38 a. 3IO- + 3H2O + 2Au + 8Br-  3I- + 6OH- + 2AuBr4- 1.348 V b. Sn+2 + 2Eu+2  Sn + 2Eu+3 0.29 V

20.50 a. 2I- + Hg2+2  I2 + 2Hg 0.789 – 0.536 = 0.253 V ΔG = -(2)(96485)(0.253) = -48821.4 J ΔG = -RTlnK -48821.4 = -(8.314)(298)lnK K = 3.6 x 108 b. 3Cu+ + NO3- + 4H+  3Cu+2 + NO + 2H2O 0.96 – 0.153 = 0.807 V ΔG = -(3)(96485)(0.807) = -233600 J ΔG = -RTlnK -233600 = -(8.314)(298)lnK K = 8.83 x 1040 c. 2Cr(OH)3 + 4OH- + 3ClO-  2CrO4-2 + 2H2O + 3Cl- 0.89 – (-0.13) = 1.02 V ΔG = -(6)(96485)(1.02) = -590500 J ΔG = -RTlnK -48821.4 = -(8.314)(298)lnK K = 3.2 x 10103

20.52 K = 8.7 x 104 n = 1 ΔG = -RTlnK = -(8.314)(298)ln(8.7 x 104) = -28179.1 J ΔG = -nFE -28179.1 = -(1)(96485)E E = 0.292 V

20.54 a. 0.799 – 0.337 = 0.462 V ΔG = -(2)(96485)(0.462) = -89152.14 J ΔG = -RTlnK -89152.14 = -(8.314)(298)lnK K = 4.24 x 1015 b. 1.61 – 0.32 = 1.29 V ΔG = -(3)(96485)(1.29) = -373396.95 J ΔG = -RTlnK -373396.95 = -(8.314)(298)lnK K = 2.84x 1065 c. 0.36 – (-0.23) = 0.59 V ΔG = -(4)(96485)(0.59) = -227704.6 J ΔG = -RTlnK -227704.6 = -(8.314)(298)lnK K = 8.2 x 1039

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