E(θZn)=-0.7621V,E(θCu)=0.3394V
当[Zn^2+]=0.01mol/L时,
E(Zn)=E(θZn)-(0.0591/2)*lg(1/[Zn^2+])=-0.8212V
当[Cu^2+]=0.01mol/L时,
E(Cu)=E(θCu)-(0.0591/2)*lg(1/[Cu^2+])=0.2803V
[Cu(NH3)4](2+)的K(稳)=[Cu(NH3)2^2+]/([Cu^2+]*[NH3]^2)
E1=E(Cu^2+)-(0.0591/2)*lg([Cu(NH3)2^2+]/[Cu^2+])-E(Zn^2+)
把E1=0.714V,E(Cu)=0.2803V,E(Zn)=-0.8212代入,得
[Cu(NH3)2^2+]/[Cu^2+]=10^13.1
因为[NH3]=1molL,所以[Cu(NH3)4](2+)的K(稳)=10^13.1
所以[Cu(NH3)4](2+)的K(不稳)=1/[Cu(NH3)4](2+)的K(稳)=7.94*10^-14
K(sp)=[Zn^2+][S^2-]=1.6*10^-24
因为[S2-]=1mol/L,所以[Zn^2+]=1.6*10^-24
E(Zn')=E(θZn)-(0.0591/2)*lg(1/[Zn^2])=-1.465V
E(Cu')=E1+E(Zn)=-0.1072V
E2=E(Cu')-E(Zn')=1.36V