Consider the following cell reaction\ $\mathrm{Cd}_{(s)}+\mathrm{Hg}_{2} \mathrm{SO}_{4(s)}+\frac{9

Question

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

Professor PC Thomas Classes · Answer

(25): $\Delta G^{\circ}=-n F E_{	ext {cell }}^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$\ $\Delta S^{\circ}=\frac{\Delta H^{\circ}+n F E_{	ext {cell }}^{\circ}}{T}$\ $=\frac{\left[-825.2+2 	imes 96487 	imes 4.315 	imes 10^{-3}ight]}{298}$ $$ =\frac{7.482 \mathrm{~kJ} / \mathrm{mol}}{298 \mathrm{~K}} $$ or $\frac{7482 \mathrm{~J} / \mathrm{mol}}{298 \mathrm{~K}}=25.1 \mathrm{~J} / \mathrm{mol} \mathrm{K}$\