Given the standard electrode potentials,

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.

Solution:

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

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