The reduction potential ($$E^0$$, in V) of $$MnO_4^-$$(aq)/Mn(s) is ______.

[Given : $$E^0_{MnO_4^-(aq)/MnO_2(s)} = 1.68$$ V ; $$E^0_{MnO_2(s)/Mn^{2+}(aq)} = 1.21$$ V ; $$E^0_{Mn^{2+}(aq)/Mn(s)} = -1.03$$ V]

For any redox couple, the relation between the standard Gibbs free-energy change and the standard reduction potential is
$$\Delta G^\circ = -nF E^\circ$$  $$-(1)$$
where $$n$$ is the number of electrons and $$F$$ is the Faraday constant.

To find $$E^\circ_{MnO_4^-(aq)/Mn(s)}$$ we must combine the three given half-reactions so that their electron counts add up.

Step 1: $$MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$$  $$n_1 = 3$$, $$E_1^\circ = 1.68\ \text{V}$$

Step 2: $$MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O$$  $$n_2 = 2$$, $$E_2^\circ = 1.21\ \text{V}$$

Step 3: $$Mn^{2+} + 2e^- \rightarrow Mn(s)$$  $$n_3 = 2$$, $$E_3^\circ = -1.03\ \text{V}$$

The overall balanced reduction is obtained by adding the three steps:
$$MnO_4^- + 8H^+ + 7e^- \rightarrow Mn(s) + 4H_2O$$
Hence $$n_{\text{total}} = 7$$ electrons.

Using $$-(1)$$, the overall Gibbs free-energy change is the sum of the three individual changes:

$$\Delta G^\circ_{\text{overall}} = -F\,(n_1E_1^\circ + n_2E_2^\circ + n_3E_3^\circ)$$

Therefore the required potential is
$$E^\circ_{\text{overall}} = -\dfrac{\Delta G^\circ_{\text{overall}}}{n_{\text{total}}F} = \dfrac{n_1E_1^\circ + n_2E_2^\circ + n_3E_3^\circ}{n_{\text{total}}}$$

Substituting the numbers:

$$\begin{aligned} E^\circ_{\text{overall}} &= \dfrac{3(1.68) + 2(1.21) + 2(-1.03)}{7} \\ &= \dfrac{5.04 + 2.42 - 2.06}{7} \\ &= \dfrac{5.40}{7} \\ &\approx 0.77\ \text{V} \end{aligned}$$

Thus, the standard reduction potential of the couple $$MnO_4^-(aq)/Mn(s)$$ is $$0.77\ \text{V}$$.