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Titration of KMnO4 and H2O2 under Acidic Conditions

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  1. Jan 24, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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

    3. The attempt at a solution
    Shouldn't H2O2 be the oxidizing agent, forming H2O?
     
  2. jcsd
  3. Jan 25, 2016 #2

    Borek

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    Staff: Mentor

    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?

    And what would get oxidized in the process?

    But the most important part is, there is another reaction taking place at 0.70V.
     
  4. Jan 25, 2016 #3
    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.
     
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