Chemistry - Electrochemistry Question with Solution | TestHub

$$\begin{gathered} \underline{\begin{array}{c}\mathrm{Anode} \Rightarrow \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)⟶{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\\ \mathrm{Cathode} \Rightarrow {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}⟶\mathrm{Ag}\left(\mathrm{s}\right)\end{array}} \\ \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\stackrel{\mathrm{n}=1}{⟶}{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+\mathrm{Ag}\left(\mathrm{s}\right) \end{gathered}$$

Relation between Gibbs Free Energy and EMF of a Cell is given by

$\mathrm{\Delta G}°=-{\mathrm{nFE}}_{\mathrm{cell}}^{°}=-1\times 96500\times 0.5332=51453.8 \mathrm{J}{ \mathrm{mol}}^{-1}=51.4538 \mathrm{kJ}{ \mathrm{mol}}^{-1}$