In this problem, we're considering three different half-cell reactions involving lead ions.

1) The reduction of $$\mathrm{Pb}^{2+}$$ ions to lead metal :

$$\mathrm{Pb}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$$

The Gibbs free energy change for this process can be written as $$\Delta \mathrm{G}_1^0=-2 \mathrm{FE}_1^0$$, where $$\mathrm{E}_1^0$$ is the standard cell potential, F is Faraday's constant, and the factor of 2 is because two electrons are involved in the reaction.

2) The reduction of $$\mathrm{Pb}^{4+}$$ ions to lead metal :

$$\mathrm{Pb}^{4+} + 4 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$$

This reaction has a Gibbs free energy change of $$\Delta \mathrm{G}_2^0=-4 \mathrm{FE}_2^0$$.

3) The oxidation of $$\mathrm{Pb}^{2+}$$ ions to $$\mathrm{Pb}^{4+}$$ ions :

$$\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+} + 2 \mathrm{e}^{-}$$

The Gibbs free energy change for this process is $$\Delta \mathrm{G}_3^0=-2 \mathrm{FE}_3^0$$.

We can write the third reaction as the difference between the first two reactions (i.e., reaction 2 - reaction 1). This implies that the Gibbs free energy changes for these reactions should add up accordingly :

$$\Delta \mathrm{G}_3^0=\Delta \mathrm{G}_2^0 - \Delta \mathrm{G}_1^0$$

Substituting the expressions for the Gibbs free energy changes from the half-cell reactions into this equation, we get :

$$-2 \mathrm{FE}_3^0 = -4 \mathrm{FE}_2^0 - (-2 \mathrm{FE}_1^0) = 2F (2n - m)$$

Solving this equation for $$E_3^0$$ gives :

$$E_3^0 = m - 2n$$

However, the problem statement tells us that $$E_3^0$$ can also be written as $$m - xn$$. Comparing these two expressions for $$E_3^0$$, we see that $$x$$ must be equal to 2.

So, the value of $$x$$ is 2.