ch9可逆电池电动势及应用答案-1.pdf

第九章 可逆电池的电动势及应用答案 一、选择题 1. [答] (D) 2. [答] (C) 3. [答] (B) 4. [答] (D) 5. [答] (C) 6. [答] (C) 7. [答] (C) 8. [答] (B) 9. [答] (D) 10. [答] (B) 二、填空题 12. [答] E 2+ + 2+ + = 0.153 V ， φ (Cu /Cu ) = 2φ (Cu /Cu) - φ (Cu /Cu) 13. [答] E= 0.323 V K= 8.46 ×1010 14. [答] Pb(s)+PbO (s)+2H SO (aq) = 2PbSO (s)+2H O(l) 2 2 4 4 2 15. [答] φ(Pb2+/Pb(Hg)) = φ (Pb2+/Pb) - RT/2F ×ln[a(Pb(Hg))/a(Pb2+)] φ(Pb2+/Pb) = φ (Pb2+/Pb) - RT/2F ×ln[a(Pb(s))/a(Pb2+)] φ 不相同， φ 相同 16. [答] Pt,Cl - - (p )|Cl (a )||Cl (a )|Cl (p ),Pt 2 1 2 2 17. [答] φ = 1.25 V 3 18. [答] 左；右；右；左 19. [答] Ni(s) │Ni(OH) (s) │NaOH(aq) │H (g) │Pt 2 2 三、计算题 20. [答] (a) Pb(s) + 2AgCl(s) = 2Ag(s) + PbCl2(s) (b) E= E = 0.49004 V Δ G = - zFE= -94.56 kJ ·mol-1 r m Δ G (2) = -zFE = -40.74 kJ ·mol-1 r m 2 Δ G = Δ G (1) - Δ G (2) = -53.87 kJ ·mol-1 r m r m r m Δ H (1) = Δ G (1) + zF(∂E/∂T) (1) r m r m p -1 = -105.3 kJ ·mol Δ H (2) = Δ G (2) + zFT(∂E/∂T) (2) r m r m p -1 = -48.05 kJ ·mol Δ H = Δ H (1) - Δ H (2) = -57.2 kJ ·mol-1 r m r m r m 21.