Hydrogen Peroxide and Ozone

Three days ago, while I was reading an article about Oxygenated Water (H₂O₂) and Ozone (O₃) chemistry, I saw 3 different reactions between the same substances mentioned above, all three possible. The reactions were:

1. 2H₂O₂ + 4O₃ → 2H₂O + 7O₂
2. H₂O₂ + O₃ → H₂O + 2O₂
3. H₂O₂ + 3O₃ → H₂O + 5O₂

I let the book down for a minute and asked myself why: "Why all three reactions are possible?". Immediately, I realised that the first one is the sum of the last two; so, all I had to do is to prove the last 2 reactions.

1. H₂O₂ + O₃ → H₂O + 2O₂
• RedOx Method

Firstly everything went wrong. I thought that the oxygen atoms in the H₂O₂ molecule have the oxidation state number -2, but after a little bit of documentation I had found that, due to it's allotropic state, oxygen in H₂O₂ has an N.O. of -1 .

OK. This wasn't a problem any more. Then, i noticed that O atoms are both the oxidizers and the reducers in the reaction (Nothing new, of course. There are such reactions, in nature, where the same element is both reducer and oxidizer; for example H₂SO₄ + S → SO₂ + H₂). In conclusion the partial ionization reactions are:

O^-1_(H2O2) —( -1e^- )→ O⁰_(O2) O⁰_(O3) —( +2e^- )→ O^-2_{(H2^δ+O^δ-)}

Reaction Possible.
• Separate Decomposition Equations Method
In a more simple way, I could solve this by taking the reactions separately, making an equation system, multiplying each reaction by the coefficients resulted and finally, adding everything up.

So we have:

H₂O₂ → H₂O + 1/2 O₂

O₃ ↔ 3/2O₂

————————————(+)

H₂O₂ + O₃ → H₂O + 2O₂ (→TRUE)

2. H₂O₂ + 3O₃ → H₂O + 5O₂
We observe that this reaction is just the anterior one with one more O₃ ↔ 3/2O₂ added. That means that in an athmosphere of O3 in exces this is perfectly possible. (→TRUE)

From these two we deduce that the first one is also plausible. And that's it: All three are true!