JEE MAIN - Chemistry (2023 - 29th January Morning Shift - No. 11)
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 :
$$\mathrm{Mn^{2+}}$$ and $$\mathrm{Co^{2+}}$$
$$\mathrm{V^{2+}}$$ and $$\mathrm{Mn^{2+}}$$
$$\mathrm{V^{2+}}$$ and $$\mathrm{Cr^{2+}}$$
$$\mathrm{Cr^{2+}}$$ and $$\mathrm{Co^{2+}}$$
Explanation
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 $\mathrm{H}_{2}$
Therefore they will reduce $\mathrm{H}^{+}$ i. $\mathrm{e~V}^{+2}$ and $\mathrm{Cr}^{+2}$
Therefore they will reduce $\mathrm{H}^{+}$ i. $\mathrm{e~V}^{+2}$ and $\mathrm{Cr}^{+2}$
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