The standard electrode potential (E^circ) for | vedclass

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(B) To find the standard electrode potential for the reaction $Fe^{3+} + e^- \rightarrow Fe^{2+}$, we use the Gibbs free energy change $(\Delta G^\cir

Vedclass · Accepted Answer

$+0.76 V$

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(B) To find the standard electrode potential for the reaction $Fe^{3+} + e^- ightarrow Fe^{2+}$, we use the Gibbs free energy change $(\Delta G^\circ)$.For reaction $(1): Fe^{3+} + 3e^- ightarrow Fe$, $\Delta G_1^\circ = -n_1 F E_1^\circ = -3 	imes F 	imes (-0.04) = 0.12 F$.For reaction $(2): Fe^{2+} + 2e^- ightarrow Fe$, $\Delta G_2^\circ = -n_2 F E_2^\circ = -2 	imes F 	imes (-0.44) = 0.88 F$.We want the reaction: $Fe^{3+} + e^- ightarrow Fe^{2+}$, which is reaction $(1) - (2)$.Therefore, $\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ = 0.12 F - 0.88 F = -0.76 F$.Since $\Delta G_3^\circ = -n_3 F E_3^\circ$ and $n_3 = 1$, we have $-0.76 F = -1 	imes F 	imes E_3^\circ$.Thus, $E_3^\circ = +0.77 V$ (approximately $+0.76 V$ based on given values).

The standard electrode potential $(E^\circ)$ for the half-cell reaction $Fe^{3+} + e^- \rightarrow Fe^{2+}$ at $298 K$ is (Given: $E^\circ(Fe^{3+}/Fe) = -0.04 V$ and $E^\circ(Fe^{2+}/Fe) = -0.44 V$ at $298 K$)

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