$\Delta G o f C O_{2}(g)=-394.4 K J / m o l$
$\Delta G$ of pentane $(g)=-8.2 k J / m o l$
In pentane-oxygen fuel cell following reaction takes place
$C_{5} H_{12}+10 H_{2} O(l) \rightarrow 5 C O_{2}+32 H^{+}+32 e^{-}$
$\frac{8 O_{2}+32 H^{+}+32 e^{-} \rightarrow 16 H_{2} O(l)}{C_{5} H_{12}+8 O_{2} \rightarrow 5 C O^{2}+6 H_{2} O(l), E^{0}=?}$
$\Delta G_{\text {reaction}}=\boldsymbol{\Sigma} \Delta G_{\text {product}}-\boldsymbol{\Sigma} \Delta G_{\text {reactant}}$
$=5 \times \Delta G_{\left(C O_{2}\right)}+6 \Delta G_{\left(H_{2} O\right)}-\left[\Delta G_{\left(C_{3} H_{12}\right)}+8 \times \Delta G_{O_{2}}\right]$
$=5 \times(-394.4)+6 \times(-237.2)-(-8.2+0)$
$=-1972-1423.2+8.2$
${=-3387 \mathrm{kJ} / \mathrm{mol}}$
$= {-3387 \times 10^{3} \mathrm{J} / \mathrm{mol}}$
${\Delta G=-n F E_{c e l l}^{0}}$
${-3387 \times 10^{3}=-32 \times 96500 \times E_{\text {cell }}^{0}}$
${E_{\text {cell }}^{0}=\frac{-3387 \times 10^{3}}{-32 \times 96500}=1.0968 \mathrm{V}}$
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$n = 3, l = 1, m_l = 0$
$STATEMENT-2$: The higher oxidation states for the group $14$ elements are more stable for the heavier members of the group due to 'inert pair effect'.