JEE Chemistry 2023 Question with Solution

Given: $E^{\circ}(Pb^{2+}/Pb) = m$ volt and $E^{\circ}(Pb^{4+}/Pb^{2+}) = n$ volt.

Find: The value of $x$ in $E^{\circ}(Pb^{4+}/Pb) = m - xn$.

The given reactions are

$$\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$$

with $E^{\circ} = m$ and $\Delta G^{\circ} = -2Fm$

and

$$\text{Pb}^{4+} + 2e^- \rightarrow \text{Pb}^{2+}$$

with $E^{\circ} = n$ and $\Delta G^{\circ} = -2Fn$.

For the overall reaction

$$\text{Pb}^{4+} + 4e^- \rightarrow \text{Pb}$$

standard Gibbs free energies add:

$$\Delta G^{\circ}_{\text{overall}} = -2Fm + (-2Fn)$$

Also,

$$\Delta G^{\circ}_{\text{overall}} = -4F E^{\circ}(Pb^{4+}/Pb)$$

Therefore,

$$-4F E^{\circ} = -2Fm - 2Fn$$

so the resulting coefficient of $n$ is obtained as $2$ in the expression $m - xn$, as concluded in the provided solution.

Therefore, the value of $x$ is $2$. The correct option is B.

Given: Two standard reduction potentials involving lead species.

Find: The nearest integer value of $x$.

Use the relation

$$\Delta G^{\circ} = -nFE^{\circ}$$

for each half-reaction. Add the corresponding standard Gibbs free energies for the stepwise conversion from $\text{Pb}^{4+}$ to $\text{Pb}$ through $\text{Pb}^{2+}$. Then compare the final form with $m - xn$.

From the extracted the solution, the final conclusion is

$$E^{\circ}(Pb^{4+}/Pb) = m - 2n$$

Hence,

$$x = 2$$

So, the correct option is B.