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Question

(A) To find the standard electrode potential for the reaction $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$, we use the relationship between Gibbs free energy

Vedclass · Accepted Answer

$0.772$

Vedclass · Answer

(A) To find the standard electrode potential for the reaction $Fe^{3+} + e^{-} ightarrow Fe^{2+}$, we use the relationship between Gibbs free energy change $(\Delta G^{\circ})$ and standard electrode potential $(E^{\circ})$, given by $\Delta G^{\circ} = -nFE^{\circ}$. Let the reactions be: $(1) Fe^{3+} + 3e^{-} ightarrow Fe ; \Delta G^{\circ}_{1} = -3 	imes F 	imes (-0.036) = 0.108F$ $(2) Fe^{2+} + 2e^{-} ightarrow Fe ; \Delta G^{\circ}_{2} = -2 	imes F 	imes (-0.44) = 0.88F$ The target reaction is $(1) - (2)$: $Fe^{3+} + e^{-} ightarrow Fe^{2+} ; \Delta G^{\circ}_{3} = \Delta G^{\circ}_{1} - \Delta G^{\circ}_{2} = 0.108F - 0.88F = -0.772F$ Since $\Delta G^{\circ}_{3} = -nFE^{\circ}_{3}$ and $n=1$: $-0.772F = -1 	imes F 	imes E^{\circ}_{3}$ $E^{\circ}_{3} = 0.772 \ V$.

Standard reduction potentials for the following reactions are :
$Fe^{3+} + 3e^{-} \rightarrow Fe ; E^{\circ}_{1} = -0.036 \ V$
$Fe^{2+} + 2e^{-} \rightarrow Fe ; E^{\circ}_{2} = -0.44 \ V$
What will be the standard electrode potential for
$Fe^{3+} + e^{-} \rightarrow Fe^{2+} (in $V$)?$

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Standard reduction potentials for the following reactions are : $Fe^{3+} + 3e^{-} \rightarrow Fe ; E^{\circ}_{1} = -0.036 \ V$ $Fe^{2+} + 2e^{-} \rightarrow Fe ; E^{\circ}_{2} = -0.44 \ V$ What will be the standard electrode potential for $Fe^{3+} + e^{-} \rightarrow Fe^{2+} (in $V$)?$