Saved Bookmarks
| 1. |
Given the standard electrode potentials K^(+)//K=-2.93V,Ag^(+)//Ag=0.80V, Hg_(2)^(2+)//Hg=0.79V,Mg^(2+)//Mg=-2.37V,Cr^(2+)//Cr=-0.74V Arrange these metals in their increasing order of reducing power. |
| Answer» Solution :Higher the OXIDATION potential, more EASILY it is oxidized and HENCE greater is the reducing POWER. Thus, increasing order of reducing power will be `Ag LT Hg lt Cr lt Mg lt K`. | |