# Consider the following cell reaction :

Question:

Consider the following cell reaction :

$\mathrm{Cd}_{(s)}+\mathrm{Hg}_{2} \mathrm{SO}_{4(s)}+\frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CdSO}_{4} \cdot \frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(s)}+2 \mathrm{Hg}_{(l)}$

The value of $\mathrm{E}_{\text {cell }}^{0}$ is $4.315 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. If

$\Delta \mathrm{H}^{\circ}=-825.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the standard entropy

change $\Delta \mathrm{S}^{\circ}$ in $\mathrm{J} \mathrm{K}^{-1}$ is ___________(Nearest integer)

[Given : Faraday constant $=96487 \mathrm{C} \mathrm{mol}^{-1}$ ]

Solution:

$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$

$=\frac{\Delta \mathrm{H}^{\circ}+\mathrm{nFE}^{\circ}}{\mathrm{T}}$

$=\frac{\left(-825.2 \times 10^{3}\right)+(2 \times 96487 \times 4.315)}{298}$

$=\frac{-825.2 \times 10^{3}+832.682 \times 10^{3}}{298}$

$=\frac{7.483 \times 10^{3}}{298}=25.11 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

$\therefore$ Nearest integer answer is 25