JEE Chemistry 2023 Question with Solution

Given: Standard electrode potentials for $M^{3+}/M^{2+}$ are: $V = -0.26 \, \text{V}$, $Cr = -0.41 \, \text{V}$, $Mn = +1.57 \, \text{V}$, and $Co = +1.97 \, \text{V}$.

Find: Which metal ions can liberate $H_2$ from a dilute acid.

A metal ion can liberate $H_2$ from dilute acid if it can reduce $H^+$ to $H_2$. For this, the relevant reduction potential should be negative relative to the standard hydrogen electrode.

From the given values:

$$V^{3+}/V^{2+} = -0.26 \, \text{V}$$ $$Cr^{3+}/Cr^{2+} = -0.41 \, \text{V}$$

Both are negative.

$$Mn^{3+}/Mn^{2+} = +1.57 \, \text{V}$$ $$Co^{3+}/Co^{2+} = +1.97 \, \text{V}$$

These are positive.

Therefore, only $V^{2+}$ and $Cr^{2+}$ can reduce $H^+$ ions to liberate $H_2$ gas.

The correct option is C. The solution lists option A, but its worked conclusion and final answer clearly identify $(3)\; V^{2+} \text{ and } Cr^{2+}$, which corresponds to C.