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(C) The ions $Ti^{2+}, V^{2+},$ and $Cr^{2+}$ are strong reducing agents because their standard electrode potentials $(E^{\circ}_{M^{3+}/M^{2+}})$ are

Vedclass · Accepted Answer

$3$

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(C) The ions $Ti^{2+}, V^{2+},$ and $Cr^{2+}$ are strong reducing agents because their standard electrode potentials $(E^{\circ}_{M^{3+}/M^{2+}})$ are negative.These ions can reduce $H^{+}$ ions in a dilute acid to liberate hydrogen gas.The general reaction is: $2 M^{2+}_{(aq)} + 2 H^{+}_{(aq)} \longrightarrow 2 M^{3+}_{(aq)} + H_{2(g)}$.Since all three ions $(Ti^{2+}, V^{2+}, Cr^{2+})$ possess this property,the total number of such ions is $3$.

The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is. . . . . . . .$Ti^{2+}, Cr^{2+}, V^{2+}$

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