# Balance the redox reaction and identify what are the oxidizing and reducing agents H2O2 + MnO4- ---> Mn2+ + O2 (g)   <acidic>

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First we need to know which species got oxidized and reduced.

Reduction:

`MnO_4^(-) -> Mn^(2+)`

`MnO_4^(-) -> Mn^(2+) + 4H_2O`

`8 H^(+) + MnO_4^(-) -> Mn^(2+) + 4H_2O`

`5 e^(-) + 8 H^(+) + MnO_4^(-) -> Mn^(2+) + 4H_2O` -> equation 1

Oxidation:

`H_2O_2 -> O_2`

`H_2O_2 -> O_2 + 2H^+`

`H_2O_2 -> O_2 + 2H^+ + 2 e^(-)`

Balance the electrons by multiplying the equation 2 by 5 and the equation 1 by 2:

`(H_2O_2 -> O_2 + 2H^+ + e^- ) * 5`

`5 H_2O_2 -> 5 O_2 + 5 2H^+ + 10 e^-`

`( 5 e^(-) + 8 H^(+) + MnO_4^(-) -> Mn^(2+) + 4H_2O ) * 2`

`10 e^(-) + 16 H^(+) + 2 MnO_4^(-) -> 2 Mn^(2+) + 8 H_2O`

Combine the two resulting equation:

`5 H_2O_2 -> 5 O_2 + 10 H^+ + 10 e^-`

`10 e^- + 16 H^(+) + 2 MnO_4^(-) -> 2 Mn^(2+) + 8 H_2O`
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`5 H_2O_2 + 6H^(+) + 2MnO_4^(-) ->2 Mn^(2+) +5 O_2 +8 H_2O`

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