QJEEM23Chd56f: JEE Main 2023 Chemistry - Class 12 JEEM23Chd56f Solution: In an electrochemical reaction of lead, at standard…

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.