$$C{u^ + }\left( {aq} \right)$$  is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction $$2C{u^ + }\left( {aq} \right) \rightleftharpoons C{u^{2 + }}\left( {aq} \right) + Cu\left( s \right)$$         Choose the correct $${E^ \circ }$$  for above reaction if $$E_{\frac{{C{u^{2 + }}}}{{Cu}}}^ \circ = 0.34\,V$$   and $$E_{\frac{{C{u^{2 + }}}}{{C{u^ + }}}}^ \circ = 0.15\,V$$

Solution :
$$\eqalign{ & \Delta {G^ \circ } = - nF{E^ \circ } \cr & {\text{From given data,}} \cr & \left( {\text{i}} \right)\,Cu\left( s \right) \to C{u^{2 + }}\left( {aq} \right) + 2{e^ - }, \cr & \Delta G_1^ \circ = - 2\left( { - 0.34} \right) \times F \cr & \left( {{\text{ii}}} \right)\,C{u^{2 + }}\left( {aq} \right) + {e^ - } \to C{u^ + }\left( {aq} \right), \cr & \Delta G_2^ \circ = - 1\left( {0.15} \right) \times F \cr & {\text{On addition,}} \cr & Cu\left( s \right) \to C{u^ + }\left( {aq} \right) + {e^ - }, \cr & \Delta G_3^ \circ = - 1 \times {E^ \circ } \times F \cr & {\text{and}}\,\,\Delta G_3^ \circ = \Delta G_1^ \circ + \Delta G_2^ \circ \cr & - {n_3}F{E^ \circ } = - {n_1}FE_1^ \circ - {n_2}F{E_2} \cr & - {E^ \circ } = - 2\left( { - 0.34} \right) - 1\left( {0.15} \right) \cr & = \left( { - 2 \times - 0.34} \right) + \left( { - 1 \times 0.15} \right) \cr & - {E^ \circ } = + 0.68 - 0.15 = 0.53 \cr & {\text{or}}\,\,\,{E^ \circ } = - 0.53\,V \cr & C{u^ + }\left( {aq} \right) \rightleftharpoons C{u^{2 + }}\left( {aq} \right) + {e^ - }; \cr & \Delta {G_1} = - 1 \times \left( { - 0.15} \right) \times F \cr & C{u^ + }\left( {aq} \right) + {e^ - } \rightleftharpoons Cu\left( s \right); \cr & \Delta {G_2} = - 1 \times \left( { - 0.53} \right) \times F \cr & {\text{On adding above equation we get,}} \cr & 2C{u^ + } \rightleftharpoons C{u^{2 + }} + Cu;\,\Delta G \cr & \Delta G = \Delta {G_1} + \Delta {G_2} \cr & - nF{E^ \circ } = 0.15F + \left( { - 0.53F} \right) \cr & - F{E^ \circ } = - 0.38F \cr & {E^ \circ } = 0.38\,V \cr} $$
Thus, for the result reaction $${E^ \circ }$$ value is $$0.38\,V{\text{.}}$$