Arrange these metals in their increasing order of reducing power

Given the standard electrode potentials,
${{K}^{+}}$/K =- 2.93 V, ${{Ag}^{+}}$/Ag= 0.80 V ${{Hg}^{2+}}$/ Hg= 0.79 V, ${{Mg}^{2+}}$/Mg=- 2.37 V,
${{Cr}^{3+}}$/Cr=-0.74 V
Arrange these metals in their increasing order of reducing power.

The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials are in the order
${{K}^{+}}$/K < ${{Mg}^{2+}}$/Mg < ${{Cr}^{3+}}$/Cr < ${{Hg}^{2+}}$ /Hg < ${{Ag}^{+}}$/ Ag Therefore, these metals arranged in the increasing reducing power will be
Ag < Hg < Cr < Mg < K