Standard free energies of formation (in $\mathrm{kJ} / \mathrm{mol}$ ) at $298 \mathrm{~K}$ are $-237.2,-394…
Standard free energies of formation (in $\mathrm{kJ} / \mathrm{mol}$ ) at $298 \mathrm{~K}$ are $-237.2,-394.4$ and -8.2 for $\mathrm{H}_2 \mathrm{O}(\mathrm{l}), \mathrm{CO}_2(\mathrm{~g})$ and pentane $(\mathrm{g})$ respectively. The value of $\mathrm{E}_{\text {cell }}^{\circ}$ for the pentaneoxygen fuel cell is
$0.0968 \mathrm{~V}$
$1.968 \mathrm{~V}$
$2.0968 \mathrm{~V}$
$1.0968 \mathrm{~V}$
Solution
After calculation $\Delta \mathrm{G}^{\circ}$, use the formula, $\Delta \mathrm{G}^{\circ}=-n F E^{\circ}$
Here, $n=32$ is taken because balanced equation is
$\mathrm{C}_5 \mathrm{H}_{12}+8 \mathrm{O}_2 \rightarrow 5 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O}$