$\Delta G o f C O_{2}(g)=-394.4 K J / m o l$

$\Delta G$ of pentane $(g)=-8.2 k J / m o l$

In pentane-oxygen fuel cell following reaction takes place

$C_{5} H_{12}+10 H_{2} O(l) \rightarrow 5 C O_{2}+32 H^{+}+32 e^{-}$

$\frac{8 O_{2}+32 H^{+}+32 e^{-} \rightarrow 16 H_{2} O(l)}{C_{5} H_{12}+8 O_{2} \rightarrow 5 C O^{2}+6 H_{2} O(l), E^{0}=?}$

$\Delta G_{\text {reaction}}=\boldsymbol{\Sigma} \Delta G_{\text {product}}-\boldsymbol{\Sigma} \Delta G_{\text {reactant}}$

$=5 \times \Delta G_{\left(C O_{2}\right)}+6 \Delta G_{\left(H_{2} O\right)}-\left[\Delta G_{\left(C_{3} H_{12}\right)}+8 \times \Delta G_{O_{2}}\right]$

$=5 \times(-394.4)+6 \times(-237.2)-(-8.2+0)$

$=-1972-1423.2+8.2$

${=-3387 \mathrm{kJ} / \mathrm{mol}}$

$= {-3387 \times 10^{3} \mathrm{J} / \mathrm{mol}}$

${\Delta G=-n F E_{c e l l}^{0}}$

${-3387 \times 10^{3}=-32 \times 96500 \times E_{\text {cell }}^{0}}$

${E_{\text {cell }}^{0}=\frac{-3387 \times 10^{3}}{-32 \times 96500}=1.0968 \mathrm{V}}$