What is the standard reduction potential ($$E°$$) for $$Fe^{3+} \to Fe$$?
Given that:
$$Fe^{2+} + 2e^- \to Fe$$; $$E°_{Fe^{2+}/Fe} = -0.47$$ V
$$Fe^{3+} + e^- \to Fe^{2+}$$; $$E°_{Fe^{3+}/Fe^{2+}} = +0.77$$ V

We have to find the standard reduction potential for the overall three-electron process

$$Fe^{3+} + 3e^- \rightarrow Fe$$

Two simpler half-reactions and their standard potentials are given:

$$Fe^{3+} + e^- \rightarrow Fe^{2+}, \qquad E^\circ_{Fe^{3+}/Fe^{2+}} = +0.77\;{\rm V}$$

$$Fe^{2+} + 2e^- \rightarrow Fe, \qquad E^\circ_{Fe^{2+}/Fe} = -0.47\;{\rm V}$$

Direct addition of the two $$E^\circ$$ values is not permissible because potentials are not additive. Instead, we first convert each potential into its standard Gibbs free energy change by using the relation

$$\Delta G^\circ = -\,n\,F\,E^\circ$$

where $$n$$ is the number of electrons involved in the half-reaction and $$F$$ is the Faraday constant.

For the first half-reaction $$Fe^{3+} + e^- \rightarrow Fe^{2+}$$ we have $$n_1 = 1$$, so

$$\Delta G_1^\circ = -\,n_1\,F\,E_1^\circ = -(1)F(+0.77) = -0.77\,F$$

For the second half-reaction $$Fe^{2+} + 2e^- \rightarrow Fe$$ we have $$n_2 = 2$$, so

$$\Delta G_2^\circ = -\,n_2\,F\,E_2^\circ = -(2)F(-0.47) = +0.94\,F$$

Now we add the two Gibbs free energies because Gibbs energies are additive for sequential steps:

$$\Delta G_{\text{total}}^\circ = \Delta G_1^\circ + \Delta G_2^\circ = (-0.77\,F) + (+0.94\,F) = +0.17\,F$$

This total Gibbs energy corresponds to the overall three-electron reduction $$Fe^{3+} + 3e^- \rightarrow Fe$$, for which $$n_{\text{total}} = 3$$. Converting back to a potential, we again use $$\Delta G^\circ = -nF E^\circ$$ and solve for $$E^\circ$$:

$$E^\circ_{\text{overall}} = -\dfrac{\Delta G_{\text{total}}^\circ}{n_{\text{total}}\,F} = -\dfrac{(+0.17\,F)}{3\,F} = -0.056666\ldots\;{\rm V}$$

Rounding to two significant figures,

$$E^\circ_{\text{overall}} \approx -0.057\;{\rm V}$$

Hence, the correct answer is Option A.