For the cell reaction,2 Al_{(s)} + 3 Cu^{2+}_ | vedclass

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(C) The relationship between standard Gibbs free energy change and standard cell potential is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{

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(C) The relationship between standard Gibbs free energy change and standard cell potential is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$.Here,the reaction is $2 Al + 3 Cu^{2+} ightarrow 2 Al^{3+} + 3 Cu$.The number of electrons transferred $(n)$ is $6$ (since $Al ightarrow Al^{3+} + 3e^-$ and $Cu^{2+} + 2e^- ightarrow Cu$,multiplied by $2$ and $3$ respectively).Given $\Delta G^{\circ} = -1158 \ kJ = -1158000 \ J$.$F = 96500 \ C \ mol^{-1}$.Substituting the values: $-1158000 = -(6) 	imes (96500) 	imes E^{\circ}_{cell}$.$E^{\circ}_{cell} = \frac{1158000}{6 	imes 96500} = \frac{1158000}{579000} = 2 \ V$.

For the cell reaction,$2 Al_{(s)} + 3 Cu^{2+}_{(aq)} \rightarrow 2 Al^{3+}_{(aq)} + 3 Cu_{(s)}$. If $\Delta G^{\circ} = -1158 \ kJ$,what is $E^{\circ}_{cell}$ (in $V$)?

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