Consider the following cell reaction :
$C d_{(s)} + H g_{_{2}} S O_{4 (s)} + \frac{9}{5} H_{_{2}} O_{(l)} \overset{}{⇄} C d S O_{4} . \frac{9}{5} H_{_{2}} O_{_{(s)}} + 2 H g (l)$
The value of $E_{c e l l}^{0}$ is 4.315V at $25^{0} C$ .If $Δ H^{0} = - 825.2 K J m o l^{- 1},$ the standard entropy change $Δ S^{0}$ in $J K^{- 1}$ is ……………..( Nearest integer) [Given: Faraday constant =96487 $C m o l^{- 1}$ ]

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answer is 25.

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Detailed Solution

$\begin{array}{l} Δ G^{0} = - n F E^{0} = Δ H^{0} - T Δ S^{0} = \frac{Δ H^{0} + n F E^{0}}{T} = \frac{(- 825.2 X 10^{3}) + (2 X 96487 X 4.315)}{298} \\ = \frac{- 825.2 X 10^{3} + 832.682 X 10^{3}}{298} = \frac{7.483 X 10^{3}}{298} = 25.1 J K^{- 1} m o l^{- 1} . \end{array}$