$\mathrm{Cd}_{(s)}+\mathrm{Hg}_{2} \mathrm{SO}_{4(s)}+\frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CdSO}_{4} \cdot \frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(s)}+2 \mathrm{Hg}_{(l)}$
$25^{\circ} {C}$ પર ${E}_{\text {cell }}^{0}$નું મૂલ્ય $4.315\, {~V}$ છે.
જો $\Delta {H}^{\circ}=-825.2\, {~kJ} \,{~mol}^{-1}$, પ્રમાણિત એન્ટ્રોપી ફેરફાર $\Delta {S}^{\circ}$ ${J} \,{K}^{-1}$માં $........$ છે. (નજીકના પૂર્ણાંકમાં) [આપેલ: ફેરાડે અચળાંક $ = 96487 \, {C} \, {mol}^{-1} $]
\(=\frac{\Delta {H}^{\circ}+{nFE}^{\circ}}{{T}}\)
\(=\frac{\left(-825.2 \times 10^{3}\right)+(2 \times 96487 \times 4.315)}{298}\)
\(=\frac{-825.2 \times 10^{3}+832.682 \times 10^{3}}{298}\)
\(=\frac{7.483 \times 10^{3}}{298}=25.11\, {JK}^{-1}\, {~mol}^{-1}\)
\(\therefore\) Nearest integer answer is \(25\)