Hydrogen peroxide solution `(20 mL)` reacts quantitatively with a solution of `KMnO_(4) (20 mL)` acidified with dilute of `H_(2)SO_(4)`. The same volume of the `KMnO_(4)` solution is just decolourised by `10 mL` of `MnSO_(4)` in neutral medium simultaneously forming a dark brown precipitate of hydrated `MnO_2`. The brown precipitate is dissolved in `10 mL` of `0.2 M` sodium oxalate under boiling condition in the presence of dilute `H_(2)SO_(4)`. Write the balanced equations involved in the reactions and calculate the molarity of `H_(2)O_(2)`.

Explanation

In acidic medium `MnO_(4)^(ɵ)` oxidises `H_(2)O_(2)` to `O_(2)`
The involved reaction are:
`MnO_(4)^(ɵ)+8H^(o+)+5e^(-)toMn^(2+)+4H_(2)O]xx2`
`underline(H_(2)O_(2)toO_(2)+2H^(+)+2e^(-)]xx5)`
`underline(2MnO_(4)^(ɵ)+5H_(2)O_(2)+6H^(o+)to2Mn^(2+)+5O_(2)+8H_(2)O)`
IN neutral medium the reactions of `MnO^(4-)` with `Mn^(2+)` are as follows:
`MnO_(4)^(ɵ)+2H_(2)O+3e^(-)toMnO_(2)+4overset(ɵ)(O)H]xx2`
`underline(Mn^(2+)+4overset(ɵ)(O)HtoMnO_(2)+2H_(2)O+2e^(-)]xx3)`
`underline(2MnO_(4)^(ɵ)+3Mn^(2+)+4overset(ɵ)(O)Hto5MnO_(2)+2H_(2)O)`
The dissolution of `MnO_(2)` is sodium oxalate in acidic medium involes the following reactions:
`MnO_(2)+4H^(o+)+2e^(-)toMn^(2+)+2H_(2)O`
`underline(C_(2)O_(4)^(2-)to2CO_(2)+2e^(-))`
`underline(MnO_(2)+C_(2)O_(4)^(2-)+4H^(o+)toMn^(2+)+2CO_(2)+2H_2O)`
Amount of oxalate used in dissolving `MnO_(2)` is
`n_(1)V_(1)M_(1)=(10mL)(0.2 M)`
`=(10mL)((0.2mol)/(1000mL))=2xx10^(-3)mol`
From the reactions given above, we conclude that
`1 mol C_(2)O_(4)^(2-)-=1 mol MnO_(2)`
`5 mol MnO_(2)-=2 mol MnO_(4)^(ɵ)`
`2 mol MnO_(4)^(ɵ)-=5 mol H_(2)O_(2)`
Hence, `2xx10^(-3)molC_(2)O_(4)^(2-)`
`-=2xx10^(-3) mol MnO_(2)-=(2)/(5)(2xx10^(-3) mol MnO_(4)^(ɵ))`
`-=(5)/(2)[(2)/(5)(2xx10^(-3) mol H_(2)O_(2))]`
That is `2xx10^(-3)` " mol of "`H_(2)O_(2)` was present in 20 " mL of " hydrogen peroxide solution Hence its molarity is
`M=(n)/(V)=(2xx10^(-3))/(20xx10^(-3)L)=0.1molL^(-1)`