JEE Main 2025 — 24 January, Evening Shift — Question 24

A strong reducing agent has a low reduction potential (or high oxidation potential). Given: $E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = +1.33 \text{ V}$, $E^\circ_{\text{Cl}_2/\text{Cl}^-} = +1.36 \text{ V}$. For Cr metal: Oxidation half-reaction is $\text{Cr} \to \text{Cr}^{3+} + 3e^-$. The reduction potential for $\text{Cr}^{3+}/\text{Cr}$ is $-0.74 \text{ V}$ (standard value). For Mn²⁺: $E^\circ_{\text{MnO}_4^-/\text{Mn}^{2+}} = +1.51 \text{ V}$, so oxidation potential of Mn²⁺ is $-1.51 \text{ V}$. For MnO₄⁻: It is a strong oxidizing agent with high reduction potential (+1.51 V). For Cl⁻: Oxidation potential is $-1.36 \text{ V}$ (since $E^\circ_{\text{Cl}_2/\text{Cl}^-} = +1.36 \text{ V}$). Comparing oxidation potentials: Cr ($+0.74 \text{ V}$ for oxidation) > Cl⁻ ($-1.36 \text{ V}$) > Mn²⁺ ($-1.51 \text{ V}$). Higher oxidation potential means easier oxidation, so Cr is the strongest reducing agent.