复旦 物理化学 第五章 习题答案

1、第五章习题解答1(1) k=Q×LQ=K×R=0.1413´112.3=15.87 m1(2) S×m1(3) S×m2×mol12 S×m1 S×m2×mol1=(349.82+40.9)´104=390.72´104 S×m2×mol1=0.042313=(1.351+4.295-4.211)´102´2=2.87´102 S×m2×mol1mol×m3若以为单位，c=0.0214 mol×m

2、34a=0.032HA溶解浓度= c×a= 0.01´0.032=3.2´104 mol×kg1 mol×kg1HAKClNa+SO425=0.015 mol×kg1g+=0.5631g+=0.8662或 g±=0.75046 (1) 2Ag+ (a1) +H2(p) = 2Ag + 2H+ (a2)负：H2(p) 2e = 2H+ (a2)正：2Ag+ (a1) +2e = 2Ag电池：Pt, H2(p) ½H+ (a2)½½ Ag+ (a1)½Ag(2) Sn + Pb2+(a1)

3、 = Sn2+ (a2) + Pb负：Sn2e = Sn2+ (a2)正：Pb2+(a1) +2e = Pb电池：Sn½Sn2+ (a2)½½ Pb2+ (a1)½Pb(3) 负：正：AgCl +e = Ag + Cl(a)电池：Pt, H2(p) ½HCl (a)½ AgCl, Ag(4) Fe2+(a1) + Ag+(a3) = Fe3+(a2) + Ag负：Fe2+(a1)e = Fe3+(a2)正：Ag+(a3) +e = Ag 电池：Pt, Fe2+(a1) , Fe3+(a2) ½½Ag+(a3), A

4、g7 (1)负：Cd2e = Cd2+(a=0.01)正：Cl2(p°) + 2e = 2Cl(a=0.5)电池反应：Cd + Cl2(p°) = Cd2+(a=0.01) + 2Cl(a=0.5)(2) VVE=j+j=1.8378 V8负：Pb2e = Pb2+(a=0.10)正：Cu2+(a=0.50) + 2e = Cu电池反应：Pb + Cu2+(a=0.50) = Pb2+(a=0.10) + Cu=VDG=nFE=2F´0.4837=93.34 kJ×mol1Cu2+½Cu为正极。9 DG=nFE=2F´1.015 =

5、195.86 kJ×mol1=2F(4.92´104)=94.94 J×K1DHm=DGm+TDSm=224.27 kJ×mol1Qr=TDSm=28.31 kJ10 短路放电是热效应相当于DH。（T，p不变，W=0，Qp=DH）DH=40Qr即DG+TDS=40TDSnFE=41TDS=41´298´1.4´104=1.711 V11DSm=(96.2+77.4)(42.7+0.5´195.6) = 33.1 J×K1 V×K1DG=DHTDS=nFEV12络合平衡反应：Ag+ + 2NH3

6、= Ag(NH3)2+设计电池反应：负 Ag + 2NH3e = Ag(NH3)2+j°=0.373 V正 Ag+ + e = Agj°+=0.799 VE°=0.426 V=1.59´10713电池反应负H2(p°)2e = 2H+(aq)正HgO + H2O + 2e = Hg +2OHHgO + H2(p°) = Hg + H2O(1)DG1水生成反应(2)DG2HgO分解反应=(1)-(2)：(3)DG3DG1=nFE=2F´0.9265=178.8 kJDG2=DHTDS =258.81298.15´(7

7、0.80.5´2058.1130.67)´103=237.38 kJDG3=DG1DG2=98.58 kJ分解平衡常数=5.39´1011=2.94´1016 Pa14负：H2(p°)2e = 2H+ (m=1.0)正：电池反应g±=0.15715负极电位：Sb2O3 + 6H+ + 6e = 2Sb + 3H2O电动势E=j甘j= j甘(j°0.05915pH) = A + 0.05915pH(A=j甘j°)ES = A + 0.05915pHS0.228=A+0.05915´3.98Ex = A + 0.05915pHx0.3459=A+0.5915pHxpHx=5.9616 Pt½H+ (a=1)½ Pt阴：2H+ + 2e = H2(p°)阳：E分解=E可逆+h阴+h阳=1.229 + 0 + 0.487 = 1.716 V17电解时Pb阴极发生H+还原反应电极反应（H+浓度由一级电离决定H2SO4 ® H+ + HSO4）可逆电位不可逆电位j不可逆 = j可逆h测量时，甘汞电