Question

Given the standard electrode potentials,

 K +/K = −2.93V, Ag+/Ag = 0.80V,

 Hg2+/Hg = 0.79V

 Mg²⁺/Mg = −2.37 V, Cr 3+/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