Question
Given,
$$\left( {\text{i}} \right)C{u^{2 + }} + 2{e^ - } \to Cu,$$ $${E^ \circ } = 0.337\,V$$
$$\left( {{\text{ii}}} \right)C{u^{2 + }} + {e^ - } \to C{u^ + },$$ $${E^ \circ } = 0.153\,V$$
Electrode potential, $${E^ \circ }$$ for the reaction,
$$C{u^ + } + {e^ - } \to Cu,$$ will be
A.
0.52$$\;V$$
B.
0.90$$\,V$$
C.
0.30$$\,V$$
D.
0.38$$\,V$$
Answer :
0.52$$\;V$$
Solution :
$$\eqalign{
& \Delta {G^ \circ } = - nF{E^ \circ } \cr
& {\text{For reaction,}}\,C{u^{2 + }} + 2{e^ - } \to Cu,\,\,\,...\left( {\text{i}} \right) \cr
& \Delta {G^ \circ } = - 2 \times F \times 0.337 \cr
& {\text{For reaction,}}\,C{u^ + } \to C{u^{2 + }} + {e^ - },\,\,\,...\left( {{\text{ii}}} \right) \cr
& \Delta {G^ \circ } = - 1 \times F \times \left( { - 0.153} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = + 0.153\,F \cr
& {\text{Adding Eqs}}{\text{. (i) and (ii), we get}} \cr
& C{u^ + } + {e^ - } \to Cu,\,\Delta {G^ \circ } = - 0.521\,F \cr
& \Delta {G^ \circ } = - nF{E^ \circ } \cr
& \therefore \,\,\, - 0.521\,F = - nF{E^ \circ } \cr
& \therefore \,\,{E^ \circ } = 0.52\,V \cr} $$