Given E_{Fe^{3+}|Fe}^0 = -0.036 V and E_{Fe^ | vedclass

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(C) For $Fe^{3+} + 3e^{-} ightarrow Fe$,$\Delta G_1 = -nFE^0 = -3 	imes F 	imes (-0.036) = 0.108F$.For $Fe^{2+} + 2e^{-} ightarrow Fe$,$\Delta G

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$0.770$

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(C) For $Fe^{3+} + 3e^{-} ightarrow Fe$,$\Delta G_1 = -nFE^0 = -3 	imes F 	imes (-0.036) = 0.108F$.For $Fe^{2+} + 2e^{-} ightarrow Fe$,$\Delta G_2 = -nFE^0 = -2 	imes F 	imes (-0.439) = 0.878F$.We need the reaction $Fe^{3+} + e^{-} ightarrow Fe^{2+}$.This can be obtained by: $(Fe^{3+} + 3e^{-}$ $ightarrow Fe) - (Fe^{2+} + 2e^{-}$ $ightarrow Fe)$.Therefore,$\Delta G_3 = \Delta G_1 - \Delta G_2 = 0.108F - 0.878F = -0.770F$.Since $\Delta G_3 = -nFE^0_{cell}$ and $n=1$,we have $-1 	imes F 	imes E^0_{cell} = -0.770F$.Thus,$E^0_{cell} = 0.770 \ V$.

Given $E_{Fe^{3+}|Fe}^0 = -0.036 \ V$ and $E_{Fe^{2+}|Fe}^0 = -0.439 \ V$,calculate $E^0_{cell}$ for the reaction $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$.

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