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