`MnO_(4)^(-)` ions are reduced in acidic conditions to `Mn^(2+)` ions whereas they are reduced in neutral condition to `MnO_(2)`. The oxidation of 25 mL of a solution `x` containing `Fe^(2+)` ions required in acidic condition 20 mL of a solution y containing `MnO_(4)` ions. What value of solution y would be required to oxidize 25 mL of solution x containing `Fe^(2+)` ions in neutral condition ?

Answer

33.3

Explanation

In acidic medium,
`overset(+7)MnO_4^(-) to overset(+2) (Mn^(2+))`
Change in oxidation number =5
`underset(Fe^(2+))ubrace(N_1V_1) = underset(MnO_4^(-))ubrace(N_2V_2)`
`N xx 25 = 5M xx 20`
`25 N = 100 M`… (i)
In neutral medium,
`overset(+7)MnO_4^(-) to overset (+4)MnO_2`
Change in oxidation number =3
`underset(Fe^(2+))ubrace(N_1V_1) = underset(MnO_4^(-))ubrace(N_2V_2)`
`25 xx N = 3M xx V`
`25 N = 3MV`...(ii)
(For `MnO_4^(-), N = 3M` in neutral medim)
Equating (i) and (ii)
`100 M = 3M xx V`
or `V = 100/3 = 33.3 mL`.