The standard free energies of formation (in $\text{kJ/mol}$) at $298 \text{ K}$ are $-237.2$, $-394.4$, and $-8.2$ for $\text{H}_2\text{O(l)}$, $\text{CO}_2\text{(g)}$, and pentane (g), respectively. The value of $E^\circ_{\text{cell}}$ for the pentane-oxygen fuel cell is:

The combustion reaction for the pentane-oxygen fuel cell is: $\text{C}_5\text{H}_{12}\text{(g)} + 8\text{O}_2\text{(g)} \rightarrow 5\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)}$

First, calculate the standard Gibbs free energy change ($\Delta_r G^\circ$) for the reaction: $\Delta_r G^\circ = \sum \Delta_f G^\circ(\text{products}) - \sum \Delta_f G^\circ(\text{reactants})$ $\Delta_r G^\circ = [5 \times \Delta_f G^\circ(\text{CO}_2) + 6 \times \Delta_f G^\circ(\text{H}_2\text{O})] - [\Delta_f G^\circ(\text{C}_5\text{H}_{12}) + 8 \times \Delta_f G^\circ(\text{O}_2)]$ $\Delta_r G^\circ = [5(-394.4) + 6(-237.2)] - [-8.2 + 8(0)]$ $\Delta_r G^\circ = [-1972.0 - 1423.2] + 8.2$ $\Delta_r G^\circ = -3395.2 + 8.2 = -3387.0 \text{ kJ/mol} = -3387000 \text{ J/mol}$

Next, determine the number of moles of electrons transferred ($n$) in the balanced reaction. In $8\text{O}_2$, there are $16$ oxygen atoms going from an oxidation state of $0$ to $-2$. Thus, $n = 16 \times 2 = 32$ electrons. Alternatively, for carbon in $\text{C}_5\text{H}_{12}$, the average oxidation state is $-12/5$, and it changes to $+4$ in $\text{CO}_2$. Total change for $5$ carbon atoms $= 5 \times [4 - (-12/5)] = 5 \times (32/5) = 32$.

Using the relation between $\Delta G^\circ$ and $E^\circ_{\text{cell}}$: $\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$ $-3387000 \text{ J/mol} = -32 \times 96500 \text{ C/mol} \times E^\circ_{\text{cell}}$ $E^\circ_{\text{cell}} = \frac{3387000}{32 \times 96500} = \frac{3387000}{3088000} \approx 1.0968 \text{ V}$.