Question

Consider the following cell reaction:
$C{d}_{(s)}+H{g}_{2}S{O}_{4(s)}+\frac{9}{5}{H}_2{O}_{(l)}\rightleftharpoons CdS{O}_4\cdot \frac{9}{5}{H}_2{O}_{(s)}+2H{g}_{(l)}$
The value of $E_{\text{cell }}^0$ is $4.315V$ at $25^{\circ }C$. If $\Delta H^{\circ }=-825.2kJmo{l}^{-1}$, the standard entropy change $\Delta S^{\circ }$ in $J{K}^{-1}$ is ______ (Nearest integer) $[$ Given : Faraday constant $=96487Cmo{l}^{-1}]$

Answer 1

Answer: 25

$\Delta G^{\circ }=-nF{E}^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$
$=\frac{\Delta H^{\circ }+nF{E}^{\circ }}{T}$
$=\frac{\left(-825.2\times 10^3\right)+(2\times 96487\times 4.315)}{298}$
$=\frac{-825.2\times 10^3+832.682\times 10^3}{298}$
$=\frac{7.483\times 10^3}{298}=25.11J{K}^{-1}mo{l}^{-1}$
$=\frac{7.483\times 10^3}{298}=25.11J{K}^{-1}mo{l}^{-1}$
$\therefore$ Nearest integer answer is $25$