For the cell reaction, $2 \mathrm{Al}_{(\mathrm{s})}+3 \mathrm{Cu}^{+2}{ }_{(\mathrm{aq})} \rightarrow 2 \mathrm{Al}^{+3}{ }_{(\mathrm{aq})}+3 \mathrm{Cu}_{(\mathrm{s})}$ If $\Delta \mathrm{G}^{\circ}=-1158 \mathrm{~kJ}$, What is $\mathrm{E}_{\text {cell }}^{\circ}$ ?

The standard cell potential $E^{\circ}_{\text{cell}}$ relates to the standard Gibbs free energy change $\Delta G^{\circ}$ through the fundamental equation $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$ where $n$ represents the number of electrons transferred and $F = 96485\text{ C/mol}$ is Faraday's constant.

Given $\Delta G^{\circ} = -1158 \times 10^3\text{ J}$, we first determine $n$ from the balanced redox reaction $2\mathrm{Al}_{(s)} + 3\mathrm{Cu}^{2+}_{(aq)} \rightarrow 2\mathrm{Al}^{3+}_{(aq)} + 3\mathrm{Cu}_{(s)}$.

The oxidation half-reaction $2\mathrm{Al} \rightarrow 2\mathrm{Al}^{3+} + 6\mathrm{e}^-$ shows $n = 6$ moles of electrons transferred.

Solving for $E^{\circ}_{\text{cell}}$ gives $E^{\circ}_{\text{cell}} = \frac{-\Delta G^{\circ}}{nF} = \frac{1158000}{6 \times 96485} = 2.0002\text{ V}$.

This value rounds to $2.0\text{ V}$, which corresponds to option C.