1.

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.

Answer»

Solution :* The LOWER the reduction potential, the higher is the reducing power.
The GIVEN standard electrode potentials increase in the order of
`(E_(K^(+)|K)^(Theta)=-2.93V) lt (E_(Mg^(2+)|Mg)^(Theta)=-2.37V)lt(E_(Cr^(3+)|Cr)^(Theta)=-0.74V) lt (E_(Hg^(2+)|Hg)^(Theta)=0.79V) lt (E_(AG^(+)|Ag)^(Theta)=0.80V)`
The metal which has low standard reduction potential will work as strong reducing agent.
* HENCE, the reducing power of the given metals increases in the following order:
`Ag lt Hg lt Cr lt Mg lt K`.


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