JEE Chemistry 2025 Question with Solution

Given: The standard electrode potentials are $E_1^\circ = 2.0 \, \text{V}$ for $FeO_4^{2-} \to Fe^{3+}$ and $E_2^\circ = 0.8 \, \text{V}$ for $Fe^{3+} \to Fe^{2+}$.

Find: $E^\circ_{FeO_4^{2-}/Fe^{2+}}$.

For combining electrode potentials, potentials are not added directly unless the electron transfer is the same. Use the relation

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

and add Gibbs free energies.

From the diagram:

• $FeO_4^{2-} \to Fe^{3+}$ involves change of oxidation state from $+6$ to $+3$, so $n_1 = 3$.
• $Fe^{3+} \to Fe^{2+}$ involves $n_2 = 1$.
• Overall, $FeO_4^{2-} \to Fe^{2+}$ involves change from $+6$ to $+2$, so $n = 4$.

Therefore,

$$-nF E^\circ = -n_1 F E_1^\circ - n_2 F E_2^\circ$$

so

$$4E^\circ = 3(2.0) + 1(0.8)$$ $$4E^\circ = 6.8$$ $$E^\circ = \frac{6.8}{4} = 1.7 \, \text{V}$$

Therefore, the value of $E^\circ_{FeO_4^{2-}/Fe^{2+}}$ is $1.7 \, \text{V}$. The correct option is A.

Given: The stepwise potentials are $2.0 \, \text{V}$ and $0.8 \, \text{V}$.

Find: The overall reduction potential from $FeO_4^{2-}$ to $Fe^{2+}$.

A common incorrect approach is to write

$$E^\circ = 2.0 + 0.8 = 2.8 \, \text{V}$$

but this is not valid because standard electrode potentials are intensive quantities. The correct additive quantity is $\Delta G^\circ$.

Using

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

we write

$$\Delta G^\circ_{overall} = \Delta G^\circ_1 + \Delta G^\circ_2$$ $$-4F E^\circ = -3F(2.0) - 1F(0.8)$$ $$4E^\circ = 6.8$$ $$E^\circ = 1.7 \, \text{V}$$

Hence, the overall standard electrode potential is $1.7 \, \text{V}$.