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(D) We are given:$(1) \ Fe^{3+} + 3e^{-} \rightarrow Fe \ ; \ E^o_1 = -0.036 \ V \ ; \ n_1 = 3$$(2) \ Fe^{2+} + 2e^{-} \rightarrow Fe \ ; \ E^o_2 = -0

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$+0.772$

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(D) We are given:$(1) \ Fe^{3+} + 3e^{-} ightarrow Fe \ ; \ E^o_1 = -0.036 \ V \ ; \ n_1 = 3$$(2) \ Fe^{2+} + 2e^{-} ightarrow Fe \ ; \ E^o_2 = -0.440 \ V \ ; \ n_2 = 2$We want to find $E^o_3$ for $Fe^{3+} + e^{-} ightarrow Fe^{2+}$.This can be obtained by $(1) - (2)$:$Fe^{3+} + 3e^{-} - (Fe^{2+} + 2e^{-}) ightarrow Fe - Fe$$Fe^{3+} + e^{-} ightarrow Fe^{2+}$The Gibbs free energy change is $\Delta G^o = -nFE^o$.$\Delta G^o_3 = \Delta G^o_1 - \Delta G^o_2$$-n_3 F E^o_3 = -n_1 F E^o_1 - (-n_2 F E^o_2)$$-1 	imes E^o_3 = -(3 	imes -0.036) - (2 	imes -0.440)$$-E^o_3 = 0.108 + 0.880$$E^o_3 = 0.108 - 0.880 = -0.772 \ V$Wait,checking the calculation: $\Delta G^o_3 = \Delta G^o_1 - \Delta G^o_2 = (-3 	imes F 	imes -0.036) - (-2 	imes F 	imes -0.440) = 0.108F - 0.880F = -0.772F$.Since $\Delta G^o_3 = -n_3 F E^o_3$ and $n_3 = 1$,we have $-1 	imes F 	imes E^o_3 = -0.772F$,so $E^o_3 = +0.772 \ V$.

Given the standard electrode potentials,what is the standard electrode potential for the reaction $Fe^{3+}_{(aq)} + e^{-} \rightarrow Fe^{2+}_{(aq)}$ in $V$?
$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.036 \ V$
$Fe^{2+}_{(aq)} + 2e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.440 \ V$

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Given the standard electrode potentials,what is the standard electrode potential for the reaction $Fe^{3+}_{(aq)} + e^{-} \rightarrow Fe^{2+}_{(aq)}$ in $V$?$Fe^{3+}_{(aq)} + 3e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.036 \ V$$Fe^{2+}_{(aq)} + 2e^{-} \rightarrow Fe_{(s)}$ ; $E^o = -0.440 \ V$