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" n_(1)e+Mn^(7+)rarrMn^(alpha+)`
` :. n_(1)=7-a`
`"Neutral Medium"n_(2)e+Mn^(7+)rarrMn^(b+)`
`:. n_(2)=7-b`
`"Alkaline Medium" n_(3)e+Mn^(7+)rarrMn^(c+)`
`:. n_(2)=7-c`
`:.` Meq.of reducing agent`=` Meq.of `KMnO_(4)` in acid `=` Meq.of `KMnO_(4)` in neutral `=` Meq.of `KMnO_(4)` in alkali
`=1xxn_(1)xx20xx1xxn_(2)xx33.3=1xxn_(3)xx100`
`n_(1)=1.665n_(2)=5n_(3)`
`because n_(1),n_(2),n_(3)` are integers and `n_(1)gt7`.
`:. n_(3)=1`
`:. n_(1)=5,n_(2)=3` and `n_(3)=1`
Therefore, different oxidation states of `Mn` are `:`
`{:("Acid Media",5e+,Mn^(7+)rarr,Mn^(a+),:.a=+2),("Neutral Media",3e+,Mn^(7+)rarr,Mn^(b+),:.b=+4),("Alkaline Media",1e+,Mn^(c+)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)`
`20xx5=1xx6xxV`
`therefore V=(100)/(6)=16.67 mL`
`( therefore 6e+Cr_(2)^(6+)rarr2Cr^(3+) therefore 1M=6xx1N)`
The conditions are valid only when `Mn` is each medium exist as atom, i.e., not as `Mn_(z)`.