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Calculate $\Delta \mathrm{G}^{\circ}$ for the reaction $\mathrm{Mg}_{(s)}+\mathrm{Sn}_{(\mathrm{aq})}^{++} \longrightarrow \mathrm{Mg}_{(\mathrm{mq})}^{++}+\mathrm{Sn}_{(s)}$ if $\mathrm{E}_{\text {cell }}^0$ is $2.23 \mathrm{~V}$.
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The correct answer is:
$-430.4 \mathrm{~kJ}$
$\begin{aligned} \Delta \mathrm{G}^{\circ} & =-\mathrm{nFE}_{\text {cell }}^{\circ} \\ & =-2 \times 96500 \times 2.23 \\ & =-430390 \mathrm{~J} \\ & =-430.4 \mathrm{~kJ}\end{aligned}$
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