The standard electrode potential $\mathrm{(M^{3+}/M^{2+})}$ for V, Cr, Mn & Co are $-$0.26 V, $-$0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate $\mathrm{H_2}$ from a dilute acid are :
Solution
Metal cation with $(-)$ value of reduction potential $\left(\mathrm{M}^{+3} / \mathrm{M}^{+2}\right)$ or with $(+)$ value of oxidation potential $\left(\mathrm{M}^{+2} / \mathrm{M}^{+3}\right)$ will liberate $\mathrm{H}_{2}$<br/><br/> Therefore they will reduce $\mathrm{H}^{+}$ i. $\mathrm{e~V}^{+2}$ and $\mathrm{Cr}^{+2}$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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