Question

For the reaction taking place in the cell: $Pt ( s )\left| H _{2}( g )\right| H ^{+}( aq ) \| {Ag}^{+}( aq ) \mid Ag ( s )$ $E _{\text {Cell }}^{ o }=+0.5332\, V$.
The value of $\Delta_{ f } G ^{0}$ is ___$kJ\, mol ^{-1}$. (in nearest integer)

Answer 1

$\frac{1}{2} H _{2}+ Ag ^{+} \rightarrow H ^{+}+ Ag$
$\Delta G ^{\circ}=- nE ^{\circ} F$
$=-1 \times 0.5332 \times 96500\, J$
$=-51.35\, kJ$
$\left( n =2\right.$ for $\left.H _{2}+2 Ag ^{+} \rightarrow 2 H ^{+}+2 Ag \right)$