Question

What will be $\Delta H$ (in cal) for the reaction $Ag(s)+\frac{1}{2}H{g}_{2}C{l}_2(s)⟶AgCl(s)+Hg(l)$ at $25^{\circ }C$, if this reaction can be conducted in a cell for which the emf $=0.0455$ volt at this temperature with temperature coefficient $3.389\times 10^{-4}$ volt deg ${}^{-1}$.

Answer 1

Answer: 1280

$\Delta S^{\circ }=-\left[\frac{\partial \left(\Delta G^{\circ }\right)}{\partial T}\right]$
$=+nF\frac{\partial }{\partial T}\left(E^{\circ }\right)$
$\frac{\partial E^{\circ }}{\partial T}=+\frac{\Delta S^{\circ }}{nF}$
$\Rightarrow 3.389\times 10^{-4}=+\frac{\Delta S^{\circ }}{96500}$
$\Delta S^{\circ }=+3.389\times 10^{-4}\times 96500$
$=327038.5\times 10^{-4}J/K$
$=32.70385J/K$
$\Delta G^{\circ }=-nF{E}^{\circ }{}_{\text{cell }}=-96500\times 0.0455J/mol$
$=-4390.75J/mol$
$\Delta H^{\circ }=\Delta G^{\circ }+T\Delta S^{\circ }$
$=-4390.75+298\times 32.70385$
$=1280cal$