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1、第五章习题解答第 4 页 共 4 页第五章习题解答1(1) k=QLQ=KR=0.1413112.3=15.87 m1(2) Sm1(3) Sm2mol12 Sm1 Sm2mol1=(349.82+40.9)104=390.72104 Sm2mol1=0.042313=(1.351+4.295-4.211)1022=2.87102 Sm2mol1molm3若以为单位，c=0.0214 molm34a=0.032HA溶解浓度= ca= 0.010.032=3.2104 molkg1 molkg1HAKClNa+SO425=0.015 molkg1g+=0.5631g+=0.8662或 g=0.7
2、5046 (1) 2Ag+ (a1) +H2(p) = 2Ag + 2H+ (a2)负：H2(p) 2e = 2H+ (a2)正：2Ag+ (a1) +2e = 2Ag电池：Pt, H2(p) H+ (a2) Ag+ (a1)Ag(2) Sn + Pb2+(a1) = Sn2+ (a2) + Pb负：Sn2e = Sn2+ (a2)正：Pb2+(a1) +2e = Pb电池：SnSn2+ (a2) Pb2+ (a1)Pb(3) 负：正：AgCl +e = Ag + Cl(a)电池：Pt, H2(p) HCl (a) AgCl, Ag(4) Fe2+(a1) + Ag+(a3) = Fe3+(a
3、2) + Ag负：Fe2+(a1)e = Fe3+(a2)正：Ag+(a3) +e = Ag 电池：Pt, Fe2+(a1) , Fe3+(a2) Ag+(a3), Ag7 (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
4、0.4837=93.34 kJmol1Cu2+Cu为正极。9 DG=nFE=2F1.015 = 195.86 kJmol1=2F(4.92104)=94.94 JK1DHm=DGm+TDSm=224.27 kJmol1Qr=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=412981.4104=1.711 V11DSm=(96.2+77.4)(42.7+0.5195.6) = 33.1 JK1 VK1DG=DHTDS=nFEV12络合平衡反应：Ag+ + 2NH3 = Ag(NH3)2+设
5、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=1.5910713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=2F0.9265=178.8 kJDG2=DHTDS =258.81298.15(70.80.52058.1130.67)103=237.38 kJDG3=DG1DG2=98.58 kJ分解平衡常数=5.
6、391011=2.941016 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(j0.05915pH) = A + 0.05915pH(A=j甘j)ES = A + 0.05915pHS0.228=A+0.059153.98Ex = A + 0.05915pHx0.3459=A+0.5915pHxpHx=5.9616 PtH+ (a=1) 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，甘汞电极电位较大，Pb阴极电位教低，组成原电池应是：PbH2SO4(0.10 molkg1, g=0.265) 甘汞h 求算E= j甘j不可逆 = j甘(j可逆h)h= 0.6950 V
