Question

​Given the standard potentials E^o(Cu^2-/Cu ) and E^o(Cu^-/Cu) as 0.340 V and 0.522V respectively,the value of E^o(Cu^2-/Cu^-) is

Answer 1

Dear Student

As the electrode potential for the half cell reaction is :
$$\begin{gathered} C{u}^{2-}+\hspace{0.17em}2e^{-}\to Cu{E}^o=0.34V..............................\left(1\right) \\ Asthe{E}^{o}isintensiveproperty,hence∆G^{o}iscalculated \\ ∆G^o=-nF{E}^o \\ ∆G^{o}forthisreactionis=-2F\times 0.34=-0.68F \\ C{u}^{-}+e^{-}\to Cu{E}^o=0.522V.............................\left(2\right) \\ ∆G^{o}forthisreactionis=-1F\times 0.522=-0.522F\hspace{0.17em} \end{gathered}$$
Now the desired reaction is
$$\begin{gathered} C{u}^{2-}+e^{-}\to C{u}^{-}∆G_3=-1F{E_3}^o \\ Togetthisreactioneq\left(1\right)-eq\left(2\right)\hspace{0.17em} \\ Hence, \\ ∆{G^o}_3=∆{G^o}_1-∆{G^o}_2 \\ -F{E_3}^o=-0.68F+0.522F \\ {E_3}^o=0.68-0.522=0.158V\hspace{0.17em} \end{gathered}$$

So the electrode potential for the desired reaction is E(Cu^2-/Cu^-) = 0.158 V

Regards.