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 V mL of reducing agent be used for `KMnO_(4)` in different medium which act as oxidant
Acid medium, `Mn^(7+) + n_(2)e^(-) rarr Mn^(a+)`
`n_(1) = 7 - a `
Neutral medium, `Mn^(7+) + n_(2)e^(-) rarr Mn^(b+)`
`n_(2) = 7 - b `
Alkaline medium, `Mn^(7+) + n_(3)e^(-) rarr Mn^(c+)`
`n_(3) = 7 - c `
Meq. of reducing agent = Meq. of `KMnO_(4)` in acid medium = Meq. of `KMnO_(4)` in neutral
= Meq. of `KMnO_(4)` in alkaline = `1 xx n_(1) xx 20 = 1 xx n_(2) xx 33.3 = 1 xx n_(3) xx 100`
Since `n_(1), n_(2), n_(3)` are integers and `n_(1) < 7 `
`:. n_(1) = 5 , n_(2) = 3 and n_(3) = 1`
Therefore, different oxidation states of Mn are :
Acid media `Mn^(7+) + 5e^(-) rarr Mn^(a+) :. a = +2 `Neutral media `Mn^(7+) + 3e^(-) rarr Mn^(b+) :. b = +4`Alkaline media `Mn^(7+) + 1 e^(-) rarr Mn^(c+) :. c = +6 `Now same volume of reducing agent is treated with `K_(2)Cr_(2)O_(7)` and therefore,
Meq. of reducing agent = Meq. of `K_(2)Cr_(2)O_(7)`
`20 xx 5 = 1 xx 6 xx V`
`V = 100/6 = 16.66 mL`
`6e^(-) + Cr_(2)^(12+) rarr 2Cr^(3+) :. 1 M = 6 xx 1N`
It is important to note that the conditions are valid only when Mn in each medium exist as monoatomic, i.e., not as `Mn_(2)`