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

Using Gibbs energy additivity: $\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ \implies -n_3 F E_3^\circ = -n_1 F E_1^\circ - (-n_2 F E_2^\circ)$. Here $n_3 = 1, n_1 = 3, n_2 = 2$. Thus, $\Delta G_1^\circ = -3F(-0.04) = 0.12F$ and $\Delta G_2^\circ = -2F(-0.44) = 0.88F$. Therefore $\Delta G_3^\circ = 0.12F - 0.88F = -0.76F$. Since $\Delta G_3^\circ = -n_3 F E_3^\circ$, we have $-1 \times F \times E_3^\circ = -0.76F \implies E_3^\circ = +0.76\text{ V}$. Hence, Option 2 is correct.