`MnO_(4)^(-)` ions are reduced in acidic condition 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 `Mn_(4)^(-)` ions . What volume of solution Y would be required to oxidise 25 mL of a solution X containing `Fe^(2+)` ions in neutral condition ?

Options

A. a) 11.4 mL
B. b) 12.0 mL
C. c) 33.3 mL
D. d) 35.0mL

Answer

C

Explanation

In acidic medium , `MnO_(4)^(-) "is reduced to" Mn^(2+)overset(+7)(MnO_(4)^(-))toMn^(2+)`
Change in oxidation number `=7-2=5`
Solution X Solution Y
`underset("For"Fe^(2+))(N_(1)V_(1))=underset("For"MnO_(4)^(-))(N_(2)V_(2))`
`Nxx25=5MxxV[because "For" MnO_(4)^(-),N=5M " in acidic medium "]`
`25N=5Mxx20`
`25N=100M` ...(i)
In neutral medium , `MnO_(4)^(-)` is reduced to `MnO_(2)`
`overset(+7)(MnO_(4)^(-))tooverset(+4)(MnO_(2))`
Change in oxidation number `=7-4=3`
Solution X Solution Y
`underset("For"Fe^(2+))(N_(1)V_(1))=underset("For" MnO_(4)^(-))(N_(2)V_(2))`
`25xxN=3MxxV" " [because"For" MnO_(4)^(-),N=3M" in neutral medium "`
`25N=3MxxV` ... (ii) From Eqs (i) and (ii)
`100M=3MxxV`
`V=(100)/(3)=33.3mL`