Question:
Given the standard electrode potentials,

K⁺/K = -2.93V, Ag⁺/Ag = 0.80V,

Hg²⁺/ Hg = 0.79V

Mg²⁺/ Mg = -2.37V. Cr³⁺/ Cr = -0.74V

Arrange these metals in their increasing order of reducing power.

Answer:

The lesser the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is 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 = -2.37V. 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 = -2.37V. 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 = -2.37V. Cr³⁺/ Cr = -0.74V

Arrange these metals in their increasing order of reducing power.