How many ions liberate H_2 gas with dilute ac | vedclass

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(A) For an ion to liberate $H_2$ gas from dilute acids,it must act as a strong reducing agent such that its standard reduction potential is more negat

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$1$

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(A) For an ion to liberate $H_2$ gas from dilute acids,it must act as a strong reducing agent such that its standard reduction potential is more negative than that of the hydrogen electrode $(E^{\circ}_{H^+/H_2} = 0.00 \ V)$.Among the given species,only $Cr^{2+}$ has a standard reduction potential $(E^{\circ}_{Cr^{3+}/Cr^{2+}} = -0.41 \ V)$ which is lower than $0.00 \ V$,allowing it to reduce $H^+$ to $H_2$.The reaction is: $Cr^{2+} + H^+ 	o Cr^{3+} + \frac{1}{2}H_2 \uparrow$.Therefore,only $1$ ion $(Cr^{2+})$ liberates $H_2$ gas.

How many ions liberate $H_2$ gas with dilute acids among the given species:
$Mn^{2+}, Cr^{3+}, Cr^{2+}, Zn^{2+}, Cu^{2+}, Sc^{3+}, Hg^{2+}$

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How many ions liberate $H_2$ gas with dilute acids among the given species:$Mn^{2+}, Cr^{3+}, Cr^{2+}, Zn^{2+}, Cu^{2+}, Sc^{3+}, Hg^{2+}$