JEE Main 2023 — Electrochemistry Question with Solution

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 :

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

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 :

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

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 :

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

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.