Given at 298 K: E^ominus_{Fe^{2+}/Fe} = X V | vedclass

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(B) $1$. For the reaction $Fe^{2+} + 2e^- 	o Fe$, the standard Gibbs free energy change is $\Delta G^\circ_1 = -n_1 F E^\circ_{Fe^{2+}/Fe} = -2F X$.

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$3Y - 2X$

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(B) $1$. For the reaction $Fe^{2+} + 2e^- 	o Fe$, the standard Gibbs free energy change is $\Delta G^\circ_1 = -n_1 F E^\circ_{Fe^{2+}/Fe} = -2F X$. $2$. For the reaction $Fe^{3+} + 3e^- 	o Fe$, the standard Gibbs free energy change is $\Delta G^\circ_2 = -n_2 F E^\circ_{Fe^{3+}/Fe} = -3F Y$. $3$. To find the potential for $Fe^{3+} + e^- 	o Fe^{2+}$, we subtract the first reaction from the second: $(Fe^{3+} + 3e^- 	o Fe) - (Fe^{2+} + 2e^- 	o Fe) \implies Fe^{3+} + e^- 	o Fe^{2+}$. $4$. The change in Gibbs free energy for this reaction is $\Delta G^\circ_3 = \Delta G^\circ_2 - \Delta G^\circ_1 = -3FY - (-2FX) = 2FX - 3FY$. $5$. Since $\Delta G^\circ_3 = -n_3 F E^\circ_{Fe^{3+}/Fe^{2+}}$ where $n_3 = 1$, we have $-1 \cdot F \cdot E^\circ_{Fe^{3+}/Fe^{2+}} = 2FX - 3FY$. $6$. Therefore, $E^\circ_{Fe^{3+}/Fe^{2+}} = 3Y - 2X$.

Given at $298 \ K$: $E^\ominus_{Fe^{2+}/Fe} = X \ V$; $E^\ominus_{Fe^{3+}/Fe} = Y \ V$. The $E^\ominus_{Fe^{3+}/Fe^{2+}}$ in Volt at $298 \ K$ is given by:

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