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Given the standard electrode potentials,...

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

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More negative the reduction potential, more easily it is oxidised 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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