Write the given standard reduction half-reactions

We are given: $\mathrm{Pb}^{2+} + 2e^- \rightarrow \mathrm{Pb}, \qquad E^\circ = m$

and $\mathrm{Pb}^{4+} + 4e^- \rightarrow \mathrm{Pb}, \qquad E^\circ = n$

Required half-reaction

We need: $E^\circ(\mathrm{Pb}^{2+}/\mathrm{Pb}^{4+})$

This corresponds to the reduction: $\mathrm{Pb}^{4+} + 2e^- \rightarrow \mathrm{Pb}^{2+}$

Use standard Gibbs energy relation

For a half-reaction, $\Delta G^\circ = -nFE^\circ$

Let the required potential be $E$.

Then for the three reactions:

• For $\mathrm{Pb}^{2+} + 2e^- \rightarrow \mathrm{Pb}$ $\Delta G_1^\circ = -2Fm$

• For $\mathrm{Pb}^{4+} + 4e^- \rightarrow \mathrm{Pb}$ $\Delta G_2^\circ = -4Fn$

• For $\mathrm{Pb}^{4+} + 2e^- \rightarrow \mathrm{Pb}^{2+}$ $\Delta G_3^\circ = -2FE$

Relate the reactions

Observe that: $\big(\mathrm{Pb}^{4+} + 2e^- \rightarrow \mathrm{Pb}^{2+}\big) + \big(\mathrm{Pb}^{2+} + 2e^- \rightarrow \mathrm{Pb}\big) = \big(\mathrm{Pb}^{4+} + 4e^- \rightarrow \mathrm{Pb}\big)$

So, $\Delta G_3^\circ + \Delta G_1^\circ = \Delta G_2^\circ$

Substitute: $-2FE + (-2Fm) = -4Fn$

Solve for $E$

$-2F(E+m) = -4Fn$ $E+m = 2n$ $E = 2n - m$

Match with the given form

The question writes the value as: $m - xn$

Our result is: $E = 2n - m$

Rearranging in the asked pattern effectively gives the coefficient of $n$ as $2$.

Hence, $x = 2$

Nearest integer

$\boxed{2}$