1) Cu2+ + 2e- = Cu
deltaG'=-2Fx0.34
2) Cu+ + e- = Cu
deltaG" =-Fx0.52
1)-2)得3):Cu2+ + e- = Cu+
deltaG = -FE
又,deltaG = deltaG'-deltaG"= -0.68F + 0.52F = -0.16F
所以,E=0.16V
3)式两边加 Br-得:
Cu2+ + Br- + e- = CuBr E"'=0.64V
E"'=E - 0.059logKsp
所以,logKsp= (E-E"')/0.059 = (0.16-0.64)/0.059= -8.14
Ksp=10^-8.14