JEE MAIN - Chemistry (2023 - 11th April Morning Shift - No. 24)
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)
Answer
2
Explanation
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.
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.
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