a
\(\Delta G \circ f H_{2} O(l)=-237.2 K J / m o l\)

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

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

In pentane-oxygen fuel cell following reaction takes place

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

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

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

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

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

\(=-1972-1423.2+8.2\)

\({=-3387 \mathrm{kJ} / \mathrm{mol}}\)

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

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

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

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