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:

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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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.

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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.

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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.