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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 the reducing power .

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Verified by Experts

The correct Answer is:
`Ag lt Hg lt Cr lt Mg lt K`
Reducing power increases with decrease in reduction potential.
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Given that the standard electrode (E^(@)) of metals are : K^(+)//K=- 2.93 V, Ag^(+)//Ag = 0.80 V, Mg^(2+)//Mg =- 2.37 V, Cr^(3+)// Cr =- 0.74 V, Hg^(2+)//Hg = 0.79 V . Arrange these metals in an increasing order of their reducing power.

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