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.

Given the standard electrode potentials, K^(+)//K=-2.93V,Ag^(+)//Ag=0

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 :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