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Given standard electrode potentials K^...

Given standard electrode potentials
`K^(o+)|K=-2.93V, Ag^(o+)|Ag=0.80V`,
`Hg^(2+)|Hg=0.79V`
`Mg^(2+)|Mg=-2.37V,Cr^(3)|Cr=-0.74V`
Arrange these metals in their increasing order of reducing power.

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Higher the oxidation potential more easily it is oxidised and hence stronger is the reducing power. Therefore order is `Ag lt Cu lt Fe lt Mg lt K`.
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Given the standard electrode potentials, K^(+) | K = -2.93 V , Ag^(+) | Ag = 0.80 V , Hg_(2)^(2+) | Hg = 0.79 V , Mg_(2)^(2+) | Mg = -2.73 V , Cr_(2)^(3+) | Cr = -0.74 V . Arrange these metals in increasing order of their reducing power.

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