The standard electrode potential ( E^ ) for the half-cell reaction...

The given half-cell reactions are:

$\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}$, $E^\circ_1 = -0.04$ V
$\Delta G^\circ_1 = -n_1FE^\circ_1 = -3 \times F \times (-0.04) = 0.12F$

$\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$, $E^\circ_2 = -0.44$ V
$\Delta G^\circ_2 = -n_2FE^\circ_2 = -2 \times F \times (-0.44) = 0.88F$

The required half-cell reaction is:
$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$

This reaction can be obtained by subtracting the second reaction from the first reaction:
$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$

$-1 \times F \times E^\circ_3 = 0.12F - 0.88F$

$-FE^\circ_3 = -0.76F$

$E^\circ_3 = +0.76$ V

Answer: $+0.76$ V