COMEDK 2025 Afternoon Solved Paper

Concept:The ease of oxidation from $+2$ to $+3$ state is determined by the oxidation potential $E^\circ_{\text{ox}}$.Explanation:The given $E^\circ_{M^{3+}/M^{2+}}$ values are reduction potentials for $M^{3+} + e^- \to M^{2+}$.To find the ease of oxidation $M^{2+} \to M^{3+} + e^-$, we reverse the reaction and take $E^\circ_{\text{ox}} = -E^\circ_{\text{red}}$.The more positive $E^\circ_{\text{ox}}$, the easier the oxidation.For Cr: $E^\circ_{\text{red}} = -0.41\ \mathrm{V}$, so $E^\circ_{\text{ox}} = +0.41\ \mathrm{V}$.For Mn: $E^\circ_{\text{red}} = +1.57\ \mathrm{V}$, so $E^\circ_{\text{ox}} = -1.57\ \mathrm{V}$.For Fe: $E^\circ_{\text{red}} = +0.77\ \mathrm{V}$, so $E^\circ_{\text{ox}} = -0.77\ \mathrm{V}$.For Co: $E^\circ_{\text{red}} = +1.97\ \mathrm{V}$, so $E^\circ_{\text{ox}} = -1.97\ \mathrm{V}$.Cr has the highest (most positive) $E^\circ_{\text{ox}}$ value of $+0.41\ \mathrm{V}$, indicating the greatest tendency to lose an electron.Thus, oxidation from $+2$ to $+3$ is easiest for Cr.Answer:Option C, Cr.