For the decolorization of `1` mol of `KMnO_4`, the moles of `H_2O_2` requiered are .

Options

A. `1//2`
B. `3//2`
C. `5//2`
D. `7//2`

Answer

C

Explanation

`Mn^(2+)` ions are colourless, so ` MnO_4^(-)` will reduce to `Mn^(2+)` ions in acidic solution
`overset(+7)(MnO_(4)^(-))rarrMn^(2+)`
`MnO_(4)^(-)+8H^(+)+5e^(-) rarrMn^(2+) +4H_2O " "...(i)`
`H_2O_2` will oxidize to `O_2`
`overset(-1)(H_2O_(2))rarroverset(0)(O_2)`
`H_2O_2 rarrO_2 +2H^(+) +2e^(-) " "...(ii)`
Applying: eq. (i) `xx2 ` + Eq. (ii) `xx 5`
`2MnO_4^(-) +5H_2O_2 + 6H^(+) rarr2Mn^(2+) +8H_2O +5O_2`
So 1 mole of `KMnO_4` requires 5 / 2 moles of `H_2O_2` .