Question

How would I balance H2O2 ---> O2 as an oxidation reaction?

The reduction part of the reaction is: 8H^+: + MnO4^- + 6e- ----> Mn^2+ + 4H2O

Answer 1

Well, simply assign oxidation numbers....

Explanation:

We gots…

`stackrel(+I)H-stackrel(-I)O-stackrel(-I)O-stackrel(+I)H`

As always, the oxidation number is the charge left on the atom of interest when the TWO electrons that constitute the atom-atom bond are BROKEN with the charge assigned to the most electronegative atom... And when we do this for `H_2O` we gets...`2xxH(+I) + O(-II)`; when we do this for `F_2O` we gets...`2xxF(-I) + O(+II)`; for `HNO_3` we gets `3xxO(-II) + H(+I)+N(+V)`. Now of course this is a FORMALISM, a useful fiction, but this is a very useful fiction when we use the oxidation states to balance chemical equations...

But when we do this for an element-element bond, the two atoms are of equal electronegativity...and so `HO-OH` gives NEUTRAL `2xxdotOH`...i.e. a FORMAL `O(-I)` oxidation state. Do you follow?

And so reduction half equation...

`underbrace(MnO_4^(-)+8H^+ +5e^(-) rarr Mn^(2+) + 4H_2O(l))_"deep-red to colourless"` `(i)`

Oxidation half equation...

`H_2O_2 rarr O_2 + 2H^+ + 2e^(-)` `(ii)`

And so we add `2xx(i) + 5xx(ii)` to retire the electrons, virtual particles of convenience...

`2MnO_4^(-)+16H^+ +10e^(-) +5H_2O_2rarr 2Mn^(2+) + 8H_2O(l)+5O_2 + 10H^+ + 10e^(-)`

And after cancellation we gets...

`2MnO_4^(-)+6H^+ +5H_2O_2rarr 2Mn^(2+) + 8H_2O(l)+5O_2 uarr`

Is this balanced with respect to mass and charge? Don't trust my 'rithmetic...