解：⑴ 阳极反应：Zn=Zn2+(0.555mol/kg)+2e?
阴极反应：2AgCl(s)+2e?=2Ag(s)+2Cl?(b)
电池反应：Zn+2AgCl(s)=2Ag(s)+ZnCl2(0.555mol/kg)
⑵E?E??E??E(Cl?AgClAg)?E(Zn2+Zn)= 0.2222V – (–0.762V) = 0.9842V ???G??zFE??2?96485.309?0.9842?33K?exp?rm??exp???exp???1.883?10
8.314?298.15???RT??RT?⑶
??E??4?1?1Qr,m?T?rSm?TzF???298.15?2?96485.309?(?4.02?10)J?mol??23.13kJ?mol
??T?pRTa(ZnCl2)?a2(Ag)RTln?E?lna(ZnCl2) ⑷ 由能斯特方程：E?E?zFa(Zn)?a2(AgCl)zF1.015?0.9842?8.314?298.15lna(ZnCl2)
2?96485.309解得：a(ZnCl2)?0.0909
3由a(ZnCl2)?a?得a??0.4496
b??41/3b?41/3?0.555mol/kg?0.881mol/kg
Qa?????????b?b
a??b0.4496?1??0.510 b?0.881
