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  • Consider the following cell reaction <img alt="\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)

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 \mathrm{\Delta\mathrm{S}^{\circ}\ in\ J K^{-1}} is_____________. (Nearest integer)

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

Answers (1)

best_answer

For the given cell reaction ,

\mathrm{E^{0}_{cell}= 4. 315\, V}

\mathrm{\therefore \Delta G^{0}= -2\times 96487\times 4. 315}

                \mathrm{= -832.68\: kJ}

Now,

\mathrm{\Delta G^{0}= \Delta H^{0}-T\Delta S^{0}}

\mathrm{\Rightarrow \Delta S^{0}= \frac{\Delta H^{0}-\Delta G^{0}}{T}= \frac{\left ( -825. 2+832. 68 \right )}{298}\frac{kJ}{mol\, K}}
                                                   \mathrm{= 0. 0251\times 10^{3}\: J\: K^{-1}\: mol^{-1}}
                                                   \mathrm{= 25.1\ J\: K^{-1}\: mol^{-1}}

Hence,the correct answer is 25

Posted by

sudhir.kumar

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