MnO_(4)^(-) ions are reduced in acidic condition to Mn^(2+) ions whereas they are reduced in neutral condition to MnO_(2). The oxidataion of 25 mL of a solution X containing Fe^(2+) ions required in acidic medium 20 mL of a solutionY containing MnO_(4)^(-) ions. What volume of solution Y would be required to oxidise 25 mL of solution Y containing Fe^(2+) ions in neutral condition ?

MnO_(4)^(-) ions are reduced in acidic condition to Mn^(2+) ions where

1.	MnO_(4)^(-) ions are reduced in acidic condition to Mn^(2+) ions whereas they are reduced in neutral condition to MnO_(2). The oxidataion of 25 mL of a solution X containing Fe^(2+) ions required in acidic medium 20 mL of a solutionY containing MnO_(4)^(-) ions. What volume of solution Y would be required to oxidise 25 mL of solution Y containing Fe^(2+) ions in neutral condition ?
Answer» 11.4 mL 12.0 mL 33.3 mL 25.0 mL Solution :In acidic medium, `overset(+7)MnO_(4)^(-) to overset(+2)Mn^(2+)` Change in oxidation number = 5 `underset(Fe^(2+))underbrace(N_(1)V_(1))= underset(MnO_(4)^(-))underbrace(N_(2)V_(2))` `N xx 25 = 5 M xx 20 "" …(i)` 25 N = 100 M IN neutral medium, `overset(+7)MnO_(4)^(-) to overset(+4)MnO_(2)` Change in oxidation number = 3 `underset(Fe^(2+))underbrace(N_(1)V_(1))= underset(MnO_(4)^(-))underbrace(N_(2)V_(2))` `25 xxN = 3 M xx V` 25N = 3MV Equating (i) and (ii) 100 M = 3 MV or `V = (100)/(3) = 33.3` mL