Given the standard electrode potentials L^(+)//K =-2.93V,Ag^(+)//Ag=0.80 V ,Hg^(2+)//Hg=0.79V,Mg^(2+)//Mg=-2.37V Cr^(3+)//Cr=-0.74 V Arrange these metals in increasing order of their reducing power

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

1.	Given the standard electrode potentials L^(+)//K =-2.93V,Ag^(+)//Ag=0.80 V ,Hg^(2+)//Hg=0.79V,Mg^(2+)//Mg=-2.37V Cr^(3+)//Cr=-0.74 V Arrange these metals in increasing order of their reducing power
Answer» Solution :Lower the electrode potentialbetter is thereducing agent since the electrode POTENTIAL increase in the order `K^(+)//K(-2.93 V)MG^(2+)//Mg(-2.37 V),Cr^(3+)//Cr(-074 V),F^(2+)//Hg(0.79V ),Ag^(+)//Ag(0.80 V)`therefore oreducing power of METALS decrases in the same order i.e k, Mg , Cr Hg ,Ag