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1、普通化学第四章习题解答第四章 电化学与金属腐蚀-习题解答 4-2. 选择题: (1)b; (2) b; (3) b; (4) a,b 4-4. 答 (1) 正极: Fe2+ (aq) + 2e Fe (s) 负极: Zn (s) - 2e Zn (aq); 3+2+(2) 正极: Fe (aq) + e Fe (aq) 负极: 2I- (aq) - 2e I2 (s);. (3) 正极: Sn (aq) + e Sn (aq) 负极: Ni (s) - 2e Ni (aq); (4) 正极: MnO4 (aq) + 8H + 5e Mn (aq) + 4H2O 负极: Fe2+ (aq) -
2、e Fe3+ (aq); 4-7. 解 (1) (-) Sn Sn2+ (aq, 0.0100 moldm-3) Pb2+ (aq, 1.00 moldm-3) Pb (+) E = jPb2+/Pb - jSn2+/Sn + (RT/nF)ln(CPb2+/C)/(CSn2+/C) = (-0.126 (-0.138) + (0.0592/2)log(1/0.0100) = 0.0712 V 2+正极: Pb (aq) + 2e Pb (s) 负极: Sn (s) - 2e Sn2+ (aq); 总反应式: Pb (aq) + Sn (s) Pb (s) + Sn (aq) (2) (-)
3、Pb Pb2+ (aq, 1.00 moldm-3) Sn2+ (aq, 0.0100 moldm-3) Sn (+) E = j Sn2+/Sn - j Pb2+/Pb + (RT/nF)ln(CSn2+/C)/ (CPb2+/C) = (-0.138 (-0.126) + (0.0592/2)log(1/0.100) = 0.018 V 正极: Sn (s)+ 2e Sn2+ (aq) 负极: Pb2+ (aq) - 2e Pb (s); 总反应式: Pb (s) + Sn2+ (aq) Pb2+ (aq) + Sn (s) 4-9. 解 2Fe3+ (aq) + 2I- (aq) 2Fe
4、2+ (aq) + I2 (s) (1) E = j Fe/ Fe - j I2/ I = 0.775 0.535 = 0.236 (2) (-) Pt I2 (s) I- (aq, 1 moldm-3) Fe3+ (aq, 1 moldm-3), Fe2+ (aq, 1 moldm-3) Pt (+) (2) G = - nFE = -2964850.236 / 1000 (KJ) = -45.5 KJ (4) E = E + (RT)/nF ln(CFe3+/C)(CI-/C)/ (CFe2+/C) = 0.0236 + (0.0592/2)log (10-2)2/(10)2 = 0.05
5、84 V 4-10 -+-2+解 已知: 2MnO4(aq) + 16H(aq) + 10Cl(aq) = 2Mn(aq) + 5Cl2(g) + 8H2O(l) 正极为: MnO4-(aq) + 8H+(aq) + 5e- = Mn2+(aq) + 4H2O 负极为 2Cl(aq) = Cl2(g) + 2e pH = 5.0 时: j MnO4-/Mn2+ = j MnO4-/Mn2+ + (RT/nF)ln(C MnO4-/C) (CH+/C)8/ (Mn2+/C) = 1.033 V jCl2/Cl- = jCl2/Cl- = 1.358 V E = j MnO4-/Mn2+ - jC
6、l2/Cl- 0 -3+2+_ 2+2+-2+2+4+2+2+因此这个反应正向不能进行或逆向自发进行 4-11. 解 2H+ (aq) + Ni (s) = H2 (g) + Ni2+(aq) +2+22+根据题意: E = j H/H2 - jNi/Ni + (RT/nF)ln(CH/C)/ (CNi/C) (PH2/P) = 0.315 V 0 - jNi2+/Ni + (0.0592/2)log1/0.011 = 0.315 V jNi2+/Ni = -0.256 V 2-+ -3+4-14. 解 (1) Cr2O7 (aq) + 14H(aq) + 6Br (aq) = 2Cr (aq
7、) + 3Br2 (l) + 7H2O(l) Cr2O72- + 14H+ (aq) + 6e = 2Cr3+ (aq) Br2 (l) + 2e = 2Br-(aq) E = j Cr2O7/Cr - jBr2/Br + (RT/nF)ln(CCr2O7/C) (CH/C) (CBr/C)/ (CCr/C) = 1.232 1.066 + (0.0592/6) log (10-4)14 = -0.387 V 0 2-3+2-3+2-+143+2或 j Cr2O7/Cr = j Cr2O7/Cr + (RT/nF)ln(CCr2O7/C) (CH/C)/ (CCr/C) = 0.680 V j
8、Br2/Br- = jBr2/Br- = 1.066 V E = j Cr2O72-/Cr3+ - jBr2/Br- 0, 因此反应正向自发进行 (3) 从电池电动势计算公式看出: E = j Cl2/Cl- - jCo2+/Co + (RT/nF)ln (PCl2/P)/ (CCo2+/C) (CCl-/C)2 电池电动势与Cl2压力成正比, 氯气压力增大, 电池电动势增大: 氯气压力减小, 电池电动势减小. (4) 从电池电动势计算公式看出: E = j Cl2/Cl- - jCo2+/Co + (RT/nF)ln(PCl2/P)/(CCo2+/C) (CCl-/C) 2+2+ 电池电动势
9、与Co浓度成反比, Co浓度减小, 电池电动势增大. 当Co2+浓度从1 moldm-3减小到0.0100 moldm-3时, 电池电动势变化值为 E = 0.0592/2log (1/ 0.0100) = 0.0592 V E = 1.64 + 0.0592 = 1.70 4-18. 解 Cu(s) +2Fe3+ (aq) = Cu2+ (aq) + 2Fe2+ (aq) (-) Cu Cu2+ (aq, C1) Fe3+ (aq, C2), Fe2+ (aq, C3) Pt (+) log K = nE/0.059 = 2(0.771 0.342)0.0592 = 14.5 或者: ln K = nFE/RT = 296485(0.771 0.342)(8.314298.15) = 33.4 K = 3.110 4-22 答: 在微酸性水膜中,铁发生吸氧腐蚀,电极反应为: 阳极: Fe = Fe2+ (aq) + 2e- 14-2+2-3+-2-+14-63+2阴极: 1/2O2(g) + H2O(l) +2e = 2OH (aq) 在稀硫酸介质中,铁发生析氢腐蚀,电极反应为: 阳极: Fe = Fe2+ (aq) + 2e- +-阴极: 2H (aq) + 2e = H2 (g) -
