无机化学电极电势的计算····按E cu2/+cu+计算为0.867

E0(Cu2+/Cu)=0.337v
显然E0(Cu2+/Cu+)=0.159v

Cu2+ +e- +I-=CuI
1mol/L 1mol/L
它的本质：
Cu2+ +e- =Cu+
1mol/L Ksp/1mol/L
E0（Cu2+/CuI）=E（Cu2+/Cu+）
=E0（Cu2+/Cu+）+0.0591v*lg[Cu2+]/[Cu+]
=E0（Cu2+/Cu+）+0.0591v*lg1/KspCuI
=0.159v +0.0591V*lg[1/(1.1*10^-12)]
=0.866v
因此:
正极:Cu2+ +e- +I-=CuI E0（Cu2+/CuI）=0.866v
负极:I2 = 2e-=2I- E0(I2/I-)=0.535v
电动势=0.866V-0.535V =0.331v>0
瓜完全进行
即 2Cu2+ + 4I-=2CuI +I2