Given E_{Fe^{+3}/Fe}^{circ} = -0.036 V and E | vedclass

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(A) We are given the following standard reduction potentials:$(1)$ $Fe^{+3} + 3e^{-} \rightarrow Fe$ ; $E_1^{\circ} = -0.036 \ V$,$\Delta G_1^{\circ}

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$0.770$

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(A) We are given the following standard reduction potentials:$(1)$ $Fe^{+3} + 3e^{-} ightarrow Fe$ ; $E_1^{\circ} = -0.036 \ V$,$\Delta G_1^{\circ} = -n_1 F E_1^{\circ} = -3 	imes F 	imes (-0.036) = 0.108 F$$(2)$ $Fe^{+2} + 2e^{-} ightarrow Fe$ ; $E_2^{\circ} = -0.439 \ V$,$\Delta G_2^{\circ} = -n_2 F E_2^{\circ} = -2 	imes F 	imes (-0.439) = 0.878 F$We need the potential for: $Fe^{+3} + e^{-} ightarrow Fe^{+2}$This reaction is obtained by subtracting equation $(2)$ from equation $(1)$:$(Fe^{+3} + 3e^{-}) - (Fe^{+2} + 2e^{-}) ightarrow Fe - Fe$$Fe^{+3} + e^{-} ightarrow Fe^{+2}$For this reaction,$\Delta G_3^{\circ} = \Delta G_1^{\circ} - \Delta G_2^{\circ} = 0.108 F - 0.878 F = -0.770 F$Since $\Delta G_3^{\circ} = -n_3 F E_3^{\circ}$ and $n_3 = 1$:$-0.770 F = -1 	imes F 	imes E_3^{\circ}$$E_3^{\circ} = 0.770 \ V$

Given $E_{Fe^{+3}/Fe}^{\circ} = -0.036 \ V$ and $E_{Fe^{+2}/Fe}^{\circ} = -0.439 \ V$,calculate the standard electrode potential for the reaction: $Fe^{+3}_{(aq)} + e^{-} \rightarrow Fe^{+2}_{(aq)}$ (in $V$)

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