Given the standard electrode potentialsK+/K = – 2.93 V, Ag+/Ag = 0.80 VHg22+/Hg = 0.79V, Mg2+/Mg = – 237V, Cr2+/Cr = – 0.74 VArrange these metals in their increasing order of reducing power.

Given the standard electrode potentialsK+/K = – 2.93 V, Ag+/Ag = 0.8

1.	Given the standard electrode potentialsK+/K = – 2.93 V, Ag+/Ag = 0.80 VHg22+/Hg = 0.79V, Mg2+/Mg = – 237V, Cr2+/Cr = – 0.74 VArrange these metals in their increasing order of reducing power.
Answer» Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus increasing order of reducing power will be Ag < Hg < Cr < Mg

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Given the standard electrode potentialsK+/K = – 2.93 V, Ag+/Ag = 0.80 VHg22+/Hg = 0.79V, Mg2+/Mg = – 237V, Cr2+/Cr = – 0.74 VArrange these metals in their increasing order of reducing power.

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