Titration of KMnO4 and H2O2 under Acidic Conditions

[watermarked]

Homework Statement

A redox titration is carried out by adding Purple KMnO₄ solution from a burette to a solution of H₂O₂ in a flask. Which of the following would occur in the flask before the equivalence point is reached?

The answer is: The solution is colorless and O₂ is formed.
The other options were that the solution would remain purple and O₂ gas would form, that the solution would remain purple and H₂ gas would form, and the solution would be colourless and O₂ gas would form.

Homework Equations

So my table says:
H₂O₂ + 2H⁺ + 2e^- <=> 2 H₂0 has a reduction potential of +1.78
And Mn04 + 8H= + 5e- <=> Mn2⁺ + 4H20 has a reduction potential of +1.51

The Attempt at a Solution

Shouldn't H₂O₂ be the oxidizing agent, forming H₂O?

 

[Borek]

First of all, potentials you listed are for standard conditions and they are strongly pH dependent. This titration takes place in a very low pH, have you tried to estimate formal potentials in the solution?

Shouldn't H2O2 be the oxidizing agent, forming H2O?

And what would get oxidized in the process?

But the most important part is, there is another reaction taking place at 0.70V.

 

[watermarked]

No -- since the question only mentions that it takes place under "acidic conditions", I assumed it was under standard acidic concentrations. I guess that assumption was wrong, since it isn't one of the options.

Thanks for your help.