JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
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 \(250\)
- D \(25\)
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}\) \(=\frac{\Delta \mathrm{H}^{\circ}+\mathrm{nFE}^{\circ}}{\mathrm{T}}\) \(=\frac{\left(-825.2 \times 10^{3}\right)+(2 \times 96487 \times 4.315)}{298}\)…
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