MCQ
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 $=96487\, \mathrm{C}\, \mathrm{mol}^{-1}$ ]

  • A
    $0.25$
  • B
    $2.5$
  • C
    $250$
  • $25$

Answer

Correct option: D.
$25$
d
$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$

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

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

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

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

$\therefore$ Nearest integer answer is $25$

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