The standard electrode potential (E^{circ}_{M | vedclass

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(C) metal ion $M^{2+}$ can liberate $H_2$ from a dilute acid if it can reduce $H^{+}$ to $H_2$ while itself getting oxidized to $M^{3+}$.This requires

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$V^{2+}$ and $Cr^{2+}$

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(C) metal ion $M^{2+}$ can liberate $H_2$ from a dilute acid if it can reduce $H^{+}$ to $H_2$ while itself getting oxidized to $M^{3+}$.This requires the reduction potential of the $M^{3+}/M^{2+}$ couple to be lower than the reduction potential of $H^{+}/H_2$ $(0.00 \ V)$.Alternatively,the oxidation potential of $M^{2+}/M^{3+}$ must be positive.Given $E^{\circ}(V^{3+}/V^{2+}) = -0.26 \ V$ and $E^{\circ}(Cr^{3+}/Cr^{2+}) = -0.41 \ V$,both are negative,meaning $V^{2+}$ and $Cr^{2+}$ are strong reducing agents that can reduce $H^{+}$ to $H_2$.$Mn^{2+}$ and $Co^{2+}$ have positive reduction potentials,meaning they are stable and cannot reduce $H^{+}$ to $H_2$.

The standard electrode potential $(E^{\circ}_{M^{3+}/M^{2+}})$ for $V$,$Cr$,$Mn$,and $Co$ are $-0.26 \ V$,$-0.41 \ V$,$+1.57 \ V$,and $+1.97 \ V$,respectively. The metal ions which can liberate $H_2$ from a dilute acid are

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Reduction of species is dependent on its reduction potential value.

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