Question:
Given the standard electrode potentials,
$\mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}, \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}$
$\mathrm{Hg}^{2+} / \mathrm{Hg}=0.79 \mathrm{~V}$
$\mathrm{Mg}^{2+} / \mathrm{Mg}=-2.37 \mathrm{~V} . \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.74 \mathrm{~V}$
Arrange these metals in their increasing order of reducing power.
Solution:
The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is Ag $<$ Hg $<$ $\mathrm{Cr}<\mathrm{Mg}<\mathrm{K}$.
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