JEE Chemistry 2023 Question with Solution

Given:

• $\mathrm{FeO}_4^{2-} + 3e^- \rightarrow \mathrm{Fe}^{3+}$ with $E^\circ = 2.2 \, \text{V}$
• $\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}$ with $E^\circ = 0.7 \, \text{V}$

Find: $E^\circ_{\mathrm{FeO}_4^{2-}/\mathrm{Fe}^{2+}} = x \times 10^{-3} \, \text{V}$ and hence the value of $x$.

Use the Gibbs free energy relation:

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

For the overall reaction,

$$\mathrm{FeO}_4^{2-} + 4e^- \rightarrow \mathrm{Fe}^{2+}$$

we have

$$\Delta G_3 = \Delta G_1 + \Delta G_2$$

Substituting $\Delta G = -nFE^\circ$,

$$-4FE^\circ_3 = -3F(2.2) + (-1F)(0.7)$$

Therefore,

$$4E^\circ_3 = 6.6 + 0.7 = 7.3$$

So,

$$E^\circ_3 = \frac{7.3}{4} = 1.825 \, \text{V}$$

Expressing it as $x \times 10^{-3} \, \text{V}$,

$$1.825 \, \text{V} = 1825 \times 10^{-3} \, \text{V}$$

Therefore, the value of $x$ is $1825$.

Given: The reduction from $\mathrm{FeO}_4^{2-}$ to $\mathrm{Fe}^{2+}$ occurs through two steps shown in the solution.

Find: The numerical value of $x$.

Standard reduction potentials cannot be added directly when electron counts differ. Instead, convert each half-reaction into Gibbs free energy terms and then add them.

Step 1:

$$\mathrm{FeO}_4^{2-} + 3e^- \rightarrow \mathrm{Fe}^{3+}, \qquad \Delta G_1 = -3F(2.2)$$

Step 2:

$$\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}, \qquad \Delta G_2 = -1F(0.7)$$

Overall:

$$\mathrm{FeO}_4^{2-} + 4e^- \rightarrow \mathrm{Fe}^{2+}$$

so

$$\Delta G_3 = -4FE^\circ_3$$

Now add the free energies:

$$\Delta G_3 = \Delta G_1 + \Delta G_2$$ $$-4FE^\circ_3 = -3F(2.2) - F(0.7)$$ $$-4FE^\circ_3 = -6.6F - 0.7F = -7.3F$$

Cancelling $-F$ from both sides,

$$4E^\circ_3 = 7.3$$ $$E^\circ_3 = 1.825 \, \text{V}$$

Now compare with the required form:

$$E^\circ_3 = x \times 10^{-3} \, \text{V}$$

Hence,

$$x = 1825$$

Therefore, the required answer is $1825$.