Given the standard electrode potentials `K^(+)//K=-2.93V,Ag^(+)//Ag=0.80V`, `Hg_(2)^(2+)//Hg=0.79V,Mg^(2+)//Mg=-2.37V,Cr^(2+)//Cr=-0.74V` Arrange these metals in their increasing order of reducing power.

Given the standard electrode potentials `K^(+)//K=-2.93V,Ag^(+)//Ag=0.

1.	Given the standard electrode potentials `K^(+)//K=-2.93V,Ag^(+)//Ag=0.80V`, `Hg_(2)^(2+)//Hg=0.79V,Mg^(2+)//Mg=-2.37V,Cr^(2+)//Cr=-0.74V` Arrange 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 lt Hg lt Cr lt Mg lt K`.

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Given the standard electrode potentials `K^(+)//K=-2.93V,Ag^(+)//Ag=0.80V`, `Hg_(2)^(2+)//Hg=0.79V,Mg^(2+)//Mg=-2.37V,Cr^(2+)//Cr=-0.74V` Arrange these metals in their increasing order of reducing power.

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