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 `N_(1), N_(2) " and" N_(3)` be the normalities of `1M KMnO_(4)` solution in acid, neutral and alkaline mediums, repectively.
`20 mL N_(1)-=33.4 mL N_(2)-=100 mL N_(3)`
In acidic medium, the half reaction is :
`MnO_(4)^(-) +8H^(+)+5e^(-)=Mn^(2+)+4H_(2)O`
`1 M KMnO_(4)=5N KMnO_(4)`
Thus, from above relation,
`N_(2)=(20)/(33.4)xxN_(1)=(20)/(33.4)xx5N =3N`
and `N_(3)=(20)/(100)xx N_(1)=(20)/(100)xx5N =1N`
The equations in the three media are :
`MnO_(4)^(-)+5e^(-) overset("Acid")to Mn^(2+)`
`MnO_(4)^(-)+3e^(-) overset("Neutral")to Mn^(4+)`
`MnO_(4)^(-) +e^(-) overset("Alkaline") to Mn^(6+)`
The balanced equations are :
`MnO_(4)^(-)+8H^(+)+5e^(-) overset("Acid")to Mn^(2+)+4H_(2)O`
`MnO_(4)^(-)+2H_(2)O+3e^(-) overset("Neutral")to MnO_(2)+4OH^(-)`
`MnO_(4)&(-)+e^(-) overset("Alkaline") to MnO_(4)^(2-)`
The balanced equation in the case of acidified `K_(2)Cr_(2)O_(7)` solution can be written as :
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 3Cr^(3+)+7H_(2)O`
`1M K_(2)Cr_(2)O_(7)=6 N K_(2)Cr_(2)O_(7)`
The volume required for the titration of the same volume of reducing agent with acidified `K_(2)Cr_(2)O_(7)` solution as follows :
`20 mL 5 N KMnO_(4)-= V 6 N K_(2)Cr_(2)O_(7)`
`V=(20xx5)/(6)=16.66` mL