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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 .

A

`Ag lt Hg lt Cr lt Mg lt K`

B

`Ag lt Hg lt Mg lt Cr lt K`

C

`Mg lt K lt Ag lt Hg lt Cr`

D

`Ag lt Mg lt Hg lt Cr lt K`

Text Solution

Verified by Experts

The correct Answer is:
1
Higher the oxidation potential, more easily it is oxidized and hence greater is the reducting power. Thus, increasing order of reducing power will be `Ag lt Hg lt Cr lt Mg lt K`.
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