MnO_(4)^(-) ions are areduced in acidic condition to Mn^(2+) ions where they are reduced in neutral conditointo MnO_(2) the oxidation of 25 ml of a solutoin x containing Fe^(2) ions requred in acidic condition 20 ml ofa soluton y containg MnO_(4)^(-) ions what volume of solution Y would be required to oxidise 25 ml of solution x containing Fe^(2+) ions in neutral condition ?

MnO_(4)^(-) ions are areduced in acidic condition to Mn^(2+) ions wher

1.	MnO_(4)^(-) ions are areduced in acidic condition to Mn^(2+) ions where they are reduced in neutral conditointo MnO_(2) the oxidation of 25 ml of a solutoin x containing Fe^(2) ions requred in acidic condition 20 ml ofa soluton y containg MnO_(4)^(-) ions what volume of solution Y would be required to oxidise 25 ml of solution x containing Fe^(2+) ions in neutral condition ?
Answer» 11.4 ml 12.0 ml 33.3 ml 35.0 ml Solution :Redox EQUATION for OXIDATION of `Fe^(2+)` ions in ACIDIC and neutral media respectviely are `MnO_(4)^(-)+8H^(+)+5Fe^(2+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O` `MnO_(4)^(-)+2H_(2)O+3Fe^(2)rarrMnbO_(2)+3Fe^(3+)+4OH` since under acidic condition 1 mole of `MnO_(4)^(-)` OXIDISES 5 moles of `Fe^(2+)` ions but only 3 moles under neutral CONDITIONS therefore volume of `MnO_(4)^(-)` used oxidation of 25 ml of same `Fe^(2+)` solution under neutral condition `(20xx5)/(3)=33.3mL`