Consider the following cell reaction Cd_( s ) + Hg_2 SO_4( s ) +9/5 H_2 O_(l) ⇌ CdSO_4 · 9/5 H_2 O_( s ) +2 Hg_(l). The value of E_cell^0 is 4.315 V…

Consider the following cell reaction $$\mathrm{Cd}_{(\mathrm{s})}+\mathrm{Hg}_2 \mathrm{SO}_{4(\mathrm{~s})}+\frac{9}{5} \mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CdSO}_4 \cdot \frac{9}{5} \mathrm{H}_2 \mathrm{O}_{(\mathrm{s})}+2 \mathrm{Hg}_{(l)} . $$The value of $\displaystyle \mathrm{E}_{\text {cell }}^0$ is $\displaystyle 4.315$ V at $\displaystyle 25$°C. If $\displaystyle \Delta \mathrm{H}^{\circ}=-825.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the standard entropy change $\displaystyle \Delta \mathrm{S}^{\circ}$ in $\displaystyle \mathrm{J} \mathrm{K}^{-1}$ is \_\_\_\_. (Nearest integer) [Given : Faraday constant $\displaystyle =96487 \mathrm{C} \mathrm{mol}^{-1}$ ]