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

Given that :

where `x_(1),x_(2) and x_(3)` are the oxidation states of Mn in the product in acidic , neutral and alkaline media respectively .Since equal volumes of the reducing agent is used in each titration ,
` :. ` ,.e of reducing agent = m.e of `KMnO_(4)` in acidic medium
= m.e of `KMnO_(4)` in neutral medium
= m.e of `KMnO_(4)` in alkaline medium
or `1xx (7-x_(1)) xx 20 = 1 xx ( 7 - x_(2)) xx 33.4 `
` = 1xx ( 7 - x_(3)) x 100`
[ m.e = `N xx V` (mL) , N = M `xx` change in On]
or ` (7-x_(1))/5 = (7-x_(2))/3 = (7-x_(3))/1 `
On inspection , we see that the equality exists for
`x_(1)= +2 , x_(2) = +4 and x_(3) = + 6 " as " x_(1),x_(2) and x_(3)` can never be greater than 7 .
The balanced chemical equations of all the three half reactions are
`MnO_(4)^(-) +8H^(+) +5e to Mn^(2+) +4H_(2)O` (acidic medium )
`MnO_(4)^(-) 2H_(2)O+3e to MnO_(2)+4OH^(-)` (neutral medium )
`{:(MnO_(4)^(-)+e, to,MnO_(4)^(2-)),(,,+ 6):}`(alkaline medium)
Further , in acidic medium ,
` {:(Cr_(2)O_(7)^(2-) to ,2Cr^(3+),,"Change in ON =6"),(+12,+6,):}`
` :. ` normality of `K_(2)Cr_(2)O_(7)` solution ` = 1 xx 6 = N `
Let the volume of `K_(2)Cr_(2)O_(7)` solution be v mL .
` :. ` m.e of `K_(2)Cr_(2)O_(7)` = m.e of `KMnO_(4)` in acidic medium
` 6 xx v = 5 xx 20 ` ( normality of `KMnO_(4) ` = 5 N)
` v = 16.67 mL `