An equal volume of reducing agent is titrated separately with `1 M KMnO_(4)` in acid, neutral and alkaline medium. The volumes of `KMnO_(4)` required are `20 mL, 33.3 mL` and `100 mL` in acid, neutral and alkaline medium respectively. Find out oxidation state of `Mn` in each reaction product. Give balance equation. Find the volume of `1 M K_(2)Cr_(2)O_(7)` consumed if same volume of reductant is titrated in acid medium.

Explanation

Let the normality of `KMnO_4` in acidic, neutral, and basic mediua be `N_1,N_2`, and `N_3` respectively.
m" Eq of "different solution for the same quantity of reducing agent.
`20N_1=33.4,N_2=100N_3`
`MnO_4^(ɵ)+8H^(o+)+5e^(-)toMn^(2+)+4H_2O`
`MnO_4^(ɵ)` is reduced to `+2` oxidation state, so
`1M=5N_1`
`N_2=(20xx5N)/(33.4)=3N`
The fall in oxidation number is 3. Hence oxidation state in the reduced state is `+4`, oxidation i.e., it should be `MnO_2`.
`MnO_4^(ɵ)+2H_2O+3e^(-)toMnO_2+4overset(ɵ)(O)H`
`N_3=(20xx5N)/(100)=1N`
Fall in oxidation sate in the reduced product is `+6`, i.e., it should be `MnO_4^(2-)`
`MnO_4^(ɵ)+e^(-)toMnO_4^(2-)`
`K_2Cr_2O_7` reacts in acidic medium as
`K_2Cr_2O_7+14H^(o+)+6e^(-)to2Cr^(3+)+7H_2O`
`1 M K_2Cr_2O_7` as reducing agent `=20mL` of `5 N KMnO_4`.
`V=(20xx5N)/(6N)=16.7mL`