Given the standard electrode potentials,...

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

Text Solution

Verified by Experts

As reduction potential is more positive, then it is strong reducing agent. Increasing reduction potential series is : `K^(+)//K(-293V),Mg^(+2)//Mg(-2.37V),Cr^(3+)//Cr(-0.74V),Hg^(+2)//Hg(0.79V),Ag^(+)//Ag(0.80V)`
Increasing order of reducing power is as under :
Ag, Hg, Cr, Mg, k

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

Text Solution

Topper's Solved these Questions

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Given the standard electrode potentials, `K^(+)//K=-2.93V,Ag^(+)//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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KUMAR PRAKASHAN-REDOX REACTIONS-SECTION-D (NCERT EXEMPLAR SOLUTION LONG ANSWER TYPE)

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