JEE Chemistry 2025 Question with Solution

Given: The standard reduction potentials are

$$E^\circ_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \, V$$ $$E^\circ_{Cl_2/Cl^-} = 1.36 \, V$$ $$E^\circ_{MnO_4^-/Mn^{2+}} = 1.51 \, V$$ $$E^\circ_{Cr^{3+}/Cr} = -0.74 \, V$$

Find: The strongest reducing agent.

A strongest reducing agent is the species that most readily loses electrons. Hence, we look for the species corresponding to the most negative standard reduction potential.

From the given values:

• $MnO_4^-$ is associated with $E^\circ = 1.51 \, V$, so it behaves as a strong oxidizing agent.
• $Cr^{3+}/Cr$ has $E^\circ = -0.74 \, V$, so $Cr$ tends to get oxidized and therefore acts as a reducing agent.

Thus, among the given species, $Cr$ is the strongest reducing agent.

The solution marks option A and states $Mn^{2+}$, but this contradicts the electrode-potential criterion and the listed data. Therefore, the most defensible option from the given data is C.

Given: Standard reduction potentials are provided.

Find: Which listed species is the strongest reducing agent.

A reducing agent itself undergoes oxidation. Therefore, the stronger the reducing agent, the less favorable its reduction and the more negative the corresponding reduction potential.

1. Option A: $Mn^{2+}$

• It is the reduced form in the pair $MnO_4^-/Mn^{2+}$.
• Since $E^\circ = 1.51 \, V$ is highly positive, the forward reduction favors formation of $Mn^{2+}$, meaning $MnO_4^-$ is a strong oxidizing agent.
• So $Mn^{2+}$ is not the strongest reducing agent here.

2. Option B: $MnO_4^-$

• A species with very positive reduction potential is an oxidizing agent, not a reducing agent.

3. Option C: $Cr$

• For $Cr^{3+} + 3e^- \rightarrow Cr$,

$$E^\circ = -0.74 \, V$$

• This negative value means the reverse process, oxidation of $Cr$, is favorable relative to the others listed. Hence $Cr$ acts as a reducing agent.

4. Option D: $Cl^-$

• The pair $Cl_2/Cl^-$ has a positive potential $1.36 \, V$, showing $Cl_2$ is an oxidizing agent. This does not make $Cl^-$ stronger than $Cr$ as a reducing agent based on the given data.

Therefore, the correct choice based on the electrochemical data is option C, $Cr$.