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.

Answer

`(16.67 mL)`

Explanation

Let the n-factor of `KMnO_(4)` in acid, neutral and alkaline media are `N_(1), N_(2)` and `N_(3)` respectively. Also, same volumes of reducing agent is used everytime, same number of equivalents of `KMnO_(4)` would be required every time.
`rArr 20 N_(1) = (100)/(3)N_(2) = 100N_(3) rArr N_(1) = (5)/(3)N_(2) = 5N_(3)`
Also, n-factors are all integer and greater than or equal to one but less than six, `N_(3)` must be 1.
`rArr N_(1) = 5, N_(2) = 3`
`:.` In acid medium `MnO_(4)^(-) rarr Mn^(2+)`
In neutral medium `MnO_(4)^(-) rarr Mn^(4+)`
In alkaline medium `MnO_(4)^(-) rarr Mn^(6+)`
`rArr` meq of `K_(2)Cr_(2)O_(7)` required `= 100`
`rArr 100 =1 xx 6 xx V` (n-factor `= 6`)
`rArr V = 100//6 = 16.67 mL`