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

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

1.	Given the standard electrode potentials , K^(+)//K = - 2.93 V , Ag^(+) // Ag = 0.80 V , Hg^(2+) // Hg = 0.79 V Mg^(2+)//Mg = -2.37 V. Cr^(3+)//Cr = - 0.74 V arrange these metals in their increasing order of the reducing power .
Answer» SOLUTION :`AgltHgltCrltMgltK`