We need to find the standard electrode potential for the reaction: $$Fe^{3+} + e^{-} \rightarrow Fe^{2+}$$ Given: (1) $Fe^{3+} + 3e^{-} \rightarrow Fe$ ($E^{\circ}_1 = -0.041$ V) (2) $Fe^{2+} + 2e^{-} \rightarrow Fe$ ($E^{\circ}_2 = -0.44$ V)
Since electrode potential is not additive, we must use the Gibbs free energy change ($\Delta G^{\circ} = -nFE^{\circ}$): $$\Delta G^{\circ}_3 = \Delta G^{\circ}_1 - \Delta G^{\circ}_2$$ $$-n_3FE^{\circ}_3 = (-n_1FE^{\circ}_1) - (-n_2FE^{\circ}_2)$$
For reaction (3): $n_3 = 1$ For reaction (1): $n_1 = 3$ For reaction (2): $n_2 = 2$ $$-1 \cdot F \cdot E^{\circ}_3 = -(3 \cdot F \cdot -0.041) - (-2 \cdot F \cdot -0.44)$$ $$-E^{\circ}_3 = 3(0.041) - 2(0.44)$$ $$-E^{\circ}_3 = 0.123 - 0.88$$ $$-E^{\circ}_3 = -0.757 \text{ V}$$ $$E^{\circ}_3 = +0.77 \text{ V (approx +0.76 V)}$$
Final Answer: +0.76 V
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