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

First, we calculate the standard Gibbs free energy change ($\Delta G^\circ_{\text{reaction}}$) for the reaction: $\Delta G^\circ_{\text{reaction}} = \sum \Delta G^\circ_f(\text{products}) - \sum \Delta G^\circ_f(\text{reactants})$ $\Delta G^\circ_{\text{reaction}} = [5 \times \Delta G^\circ_f(\text{CO}_2) + 6 \times \Delta G^\circ_f(\text{H}_2\text{O})] - [\Delta G^\circ_f(\text{C}_5\text{H}_{12}) + 8 \times \Delta G^\circ_f(\text{O}_2)]$ $\Delta G^\circ_{\text{reaction}} = [5 \times (-394.4) + 6 \times (-237.2)] - [-8.2 + 8 \times 0]$ $\Delta G^\circ_{\text{reaction}} = [-1972.0 - 1423.2] - [-8.2]$ $\Delta G^\circ_{\text{reaction}} = -3395.2 + 8.2 = -3387.0 \text{ kJ mol}^{-1} = -3387000 \text{ J mol}^{-1}$

Next, we determine the number of electrons transferred ($n$) in the balanced redox reaction. The oxidation state of Carbon in $\text{C}_5\text{H}_{12}$ is $-12/5 = -2.4$. The oxidation state of Carbon in $\text{CO}_2$ is $+4$. Change in oxidation state for one C atom = $+4 - (-2.4) = +6.4$ Total change for 5 C atoms = $5 \times 6.4 = 32$ electrons. Thus, $n = 32$.

Now, we use the relationship between $\Delta G^\circ$ and $E^\circ_{\text{cell}}$: $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ $E^\circ_{\text{cell}} = -\frac{\Delta G^\circ}{nF}$ $E^\circ_{\text{cell}} = -\frac{-3387000 \text{ J mol}^{-1}}{32 \times 96500 \text{ C mol}^{-1}}$ $E^\circ_{\text{cell}} = \frac{3387000}{3088000} \approx 1.0968 \text{ V}$