`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 ?

Options

A. 11.4 mL
B. 12.0 mL
C. 33.3 mL
D. 35.0 mL

Answer

C

Explanation

`MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2+)+4H_(2)O`
`([Fe^(2+)rarrFe^(3+)+e^(-)]xx5)/(MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarrMn^(2+)underset(("in acidic medium"))(+5Fe^(3+)+4H_(2)O))`
`MnO_(4)^(-)+4H^(+)+3e^(-)rarrMnO_(2)+2H_(2)O`
`([Fe^(2+)rarr Fe^(3+)+e^(-)]xx3)/(MnO_(4)^(-)+4H^(+)+3Fe^(2+)rarr MnO_(2)underset(("in neutral medium"))(+3Fe^(3+)+2H_(2)O))`
In acidic medium
5 vol. of `Fe^(2+)` requires 1 vol. of `MnO_(4)^(-)` in acidic medium.
`:.` 25 vol. of `Fe^(2+)` requires `1/5 xx 25` vol of `MnO(4)^(-)` in acidic medium.
`1/5xx15` vol. of `MnO_(4)^(-)~=20` vol. or 20 mL
In neutral medium.
3 vol. of `Fe^(2+)` requires 1 vol. of `MnO_(4)^(-)` in neutral medium
Then 25 vol. of `Fe^(2+)` requires vol. of `1/3 xx25 MnO(4)^(-)` in neutral medium
`1/2xx25 ` vol of `MnO_(4)^(-)~=20` vol.
`:.1.3xx15` vol. of `MnO_(4)^(-)~~ (20)/(5)xx(25)/(3)` vol. of `MnO_(4)^(-)`
`=33.3` vol. of mL