In a titration, `H_(2)O_(2)` is oxidised to `O_(2)` by `MnO_(4)^(-)`. 24 mL of 0.1M `H_(2)O_(2)` requires 16 mL of 0.1M `MnO_(4)^(-)` solution. Hence `MnO_(4)^(-)` changes to :

Explanation

(i) `2MnO_(4)^(-)+16H^(+) + 10 CI^(-) rarr5CI_(2)+2 Mn^(2+) to +2 in Mn^(2+)`
oxidation number of Mn changes form +7 in `MnO_(4)^(-) to + 2 Mn^(2+)`
(ii) `2MnO_(4)^(-) + 6H ^(+) + 5NO_(2)^(-) rarr 5NO_(3)^(-) +3H_(2)O + 2 Nn^(+)`
Oxidatoin number of N changes from +3 in `NO_(3)^(-) "ion to" +5 in NO_(3)^(-)` ion
(iii) `2I^(-) + 4H + 2 NO_(2)^(-) rarr I_(2) + 2NO+2H_(2)O`
oxidation number of N changes form +3 is `NO_(2)^(-)` to +2 in NO
(iv) `3Mn^(2+) +CIO_(3)^(-) +6H^(+)rarr3Mn^(4+)+CI^(-)+3H_(2)O`
oxidation numbr of CI changes from + 5 in `CIO_(3)^(-) "to -1 in" CI^(-)`
(v) `2CrO_(3)^(-)+ h_(2)O_(2)+2OH^(-)rarr2CrO_(4)^(-)+2H_(2)O`
oxidation number of Cr changes from +5 in `CrO_(3)^(-)` to +6 in `CrO_(4)^(2-)]`