In an electrochemical reaction of lead, at standard temperature, if $\mathrm{E}^{0}\left(\mathrm{~Pb}^{2+} / \mathrm{Pb}\right)=\mathrm{m}$ Volt and $\mathrm{E}^{0}\left(\mathrm{~Pb}^{4+} / \mathrm{Pb}\right)=\mathrm{n}$ Volt, then the value of $\mathrm{E}^{0}\left(\mathrm{~Pb}^{2+} / \mathrm{Pb}^{4+}\right)$ is given by $\mathrm{m-x n}$. The value of $\mathrm{x}$ is ___________. (Nearest integer)

In this problem, we're considering three different half-cell reactions involving lead ions. <br/><br/>1) The reduction of $\mathrm{Pb}^{2+}$ ions to lead metal : <br/><br/>$\mathrm{Pb}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$ <br/><br/>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. <br/><br/>2) The reduction of $\mathrm{Pb}^{4+}$ ions to lead metal : <br/><br/>$\mathrm{Pb}^{4+} + 4 \mathrm{e}^{-} \rightarrow \mathrm{Pb}$ <br/><br/>This reaction has a Gibbs free energy change of $\Delta \mathrm{G}_2^0=-4 \mathrm{FE}_2^0$. <br/><br/>3) The oxidation of $\mathrm{Pb}^{2+}$ ions to $\mathrm{Pb}^{4+}$ ions : <br/><br/>$\mathrm{Pb}^{2+} \rightarrow \mathrm{Pb}^{4+} + 2 \mathrm{e}^{-}$ <br/><br/>The Gibbs free energy change for this process is $\Delta \mathrm{G}_3^0=-2 \mathrm{FE}_3^0$. <br/><br/>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 : <br/><br/>$\Delta \mathrm{G}_3^0=\Delta \mathrm{G}_2^0 - \Delta \mathrm{G}_1^0$ <br/><br/>Substituting the expressions for the Gibbs free energy changes from the half-cell reactions into this equation, we get : <br/><br/>$-2 \mathrm{FE}_3^0 = -4 \mathrm{FE}_2^0 - (-2 \mathrm{FE}_1^0) = 2F (2n - m)$ <br/><br/>Solving this equation for $E_3^0$ gives : <br/><br/>$E_3^0 = m - 2n$ <br/><br/>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. <br/><br/>So, the value of $x$ is 2.