#### Question

Given the standard electrode potentials,

K^{+}/K = −2.93V, Ag^{+}/Ag = 0.80V,

Hg^{2+}/Hg = 0.79V

Mg^{2+}/Mg = −2.37 V, Cr^{3+}/Cr = − 0.74V

Arrange these metals in their increasing order of reducing power.

#### Solution 1

The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K^{+}/K < Mg^{2+}/Mg < Cr^{3+}/Cr < Hg^{2+}/Hg < Ag^{+}/Ag.

Hence, the reducing power of the given metals increases in the following order: Ag < Hg < Cr < Mg < K

#### Solution 2

Higher the oxidation potential more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be Ag<Hg<Cr<Mg<K.

Is there an error in this question or solution?

Solution Given the Standard Electrode Potentials Concept: Galvanic Cells - Introduction.