STD 12 - 2. Electrochemistry — JEE STD 11 Science — Question

$E_1^o\, = \,0.15\,\,V;$       $\Delta G_1^o\, = \, - \,{n_1}E_1^oF$

$C{u^{2 + }}\, + \,2e\, \to \,Cu$

$E_2^o\, = \,0.34\,V;$         $\Delta G_2^o\, = \, - \,{n_2}E_2^oF$

On subracting eq.$(i)$ from eq. $(ii)$ we get

$C{u^ + }\, + \,{e^ - }\, \to \,Cu;$      $\Delta {G^o}\, = \,\Delta G_2^o\, - \,\Delta G_1^o\,$

$ - n{E^o}F\, = \, - \,({n_2}{E^o}F\, - \,{n_1}E_1^oF)$

${E^o}\, = \,\frac{{\,{n_2}{E^o}F\, - \,{n_1}E_1^oF}}{{nF}}$

$ = \,\frac{{\,2 \times 0.34 - 0.15}}{1}$

$=\,0.53\,V$