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 is their increasing order of reducing power .

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

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 is their increasing order of reducing power .
Answer» Solution :`Ag^(+)\|,Ag,Hg^(2+)\|HA, CR^(3+)\|Cr, Mg^(2+)\|Mg, K^(+)\|K`