The standard electrode potential (M3+/M2+) for V, Cr, Mn & Co are -0.26\ V,\ -0.41\ V,\ +1.57\ V and +1.97\ V,…

Strategy: A metal ion can liberate $\ce{H2}$ gas from an acid if it is a stronger reducing agent than hydrogen. Thermodynamically, this means the standard reduction potential for the couple ($E°_{\ce{M^{3+}/M^{2+}}}$) must be negative, as $E°_{\ce{H^+/H2}}$ is defined as 0 V.

Step 1: Analyze the reaction thermodynamics The reaction is: $\ce{M^{2+} + H^+ \rightarrow M^{3+} + \frac{1}{2}H_2}$ For spontaneity, $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} > 0$. $0 - (E°_{\ce{M^{3+}/M^{2+}}}) > 0 \implies E°_{\ce{M^{3+}/M^{2+}}} < 0$

Step 2: Check given potential values

• $\ce{V^{3+}/V^{2+}}$: $-0.26\ \mathrm{V}$ (Negative) $\implies$ Liberates $\ce{H2}$
• $\ce{Cr^{3+}/Cr^{2+}}$: $-0.41\ \mathrm{V}$ (Negative) $\implies$ Liberates $\ce{H2}$
• $\ce{Mn^{3+}/Mn^{2+}}$: $+1.57\ \mathrm{V}$ (Positive) $\implies$ Does not liberate
• $\ce{Co^{3+}/Co^{2+}}$: $+1.97\ \mathrm{V}$ (Positive) $\implies$ Does not liberate

$\boxed{\text{Answer: (c) } \ce{V^{2+}} \text{ and } \ce{Cr^{2+}}}$