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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)
Detailed Solution
(1) $Fe^{3+}+3e^-\rightarrow Fe$; (2) $Fe^{2+}+2e^-\rightarrow Fe$; (3) $Fe^{3+}+e^-\rightarrow Fe^{2+}$. Eq. (3) = Eq. (1) - Eq. (2), so $\Delta G_3^\circ=\Delta G_1^\circ-\Delta G_2^\circ \Rightarrow -1\cdot FE_3^\circ=-3FE_1^\circ-(-2FE_2^\circ) \Rightarrow E_3^\circ=3E_1^\circ-2E_2^\circ=3(-0.04)-2(-0.44)=-0.12+0.88=+0.76$ V.