Question

Given the standard electrode potentials,

K⁺/K = −2.93V, Ag⁺/Ag = 0.80V,

Hg²⁺/Hg = 0.79V

Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = − 0.74V

Arrange these metals in their increasing order of reducing power.

Answer 1

The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K⁺/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag.

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K