JEE Chemistry 2025 Question with Solution

Given: The standard reduction potential values of the given species are provided.

Find: The species with the strongest oxidising capacity.

A stronger oxidising agent has a greater tendency to gain electrons. Therefore, the species with the highest standard reduction potential is the strongest oxidising agent.

The given values are:

• $E^\circ_{\text{I}^- / \text{I}_2} = +1.26 \, \text{V}$
• $E^\circ_{\text{Al}^{3+} / \text{Al}} = -1.66 \, \text{V}$
• $E^\circ_{\text{Pb}^{4+} / \text{Pb}^{2+}} = +1.67 \, \text{V}$
• $E^\circ_{\text{Sn}^{4+} / \text{Sn}^{2+}} = +1.15 \, \text{V}$

Comparing these values, $E^\circ_{\text{Pb}^{4+} / \text{Pb}^{2+}} = +1.67 \, \text{V}$ is the highest.

Therefore, $\text{Pb}^{4+}$ has the strongest oxidising capacity because it most readily accepts electrons to get reduced to $\text{Pb}^{2+}$.

The correct option is C.

Given: Standard reduction potentials of four p-block ion systems are listed.

Find: Which species has the strongest oxidising power.

Step 1: Understand oxidising capacity. An oxidising agent causes oxidation of another species by itself getting reduced. Hence, stronger oxidising capacity means stronger tendency to gain electrons.

Step 2: Use the relation between oxidising strength and reduction potential. A species with a more positive standard reduction potential has a greater tendency to undergo reduction and therefore acts as a stronger oxidising agent.

Step 3: Analyse the values. Among the given data,

which is higher than

• $E^\circ_{\text{I}^- / \text{I}_2} = +1.26 \, \text{V}$
• $E^\circ_{\text{Sn}^{4+} / \text{Sn}^{2+}} = +1.15 \, \text{V}$
• $E^\circ_{\text{Al}^{3+} / \text{Al}} = -1.66 \, \text{V}$

Step 4: Conclude. Since $+1.67 \, \text{V}$ is the highest value, $\text{Pb}^{4+}$ is the strongest oxidising agent among the given species.

Therefore, the correct option is C, corresponding to $E^\circ \, \text{Pb}^{4+} / \text{Pb}^{2+} = +1.67 \, \text{V}$.