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Half equations for the decomposition of hydrogen peroxide Watch

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    So for some reason I am confusing myself with this reaction

    2H2O2(aq) -> 2H2O(l) + O2

    Oxygen in this reaction goes from 2O2 ----> 2O + O2.
    We know that the element O2 has no charge, but 2O has a a charge of -4. This means that it has gained 4 electrons from somewhere. Should we include those electrons in the half equation?
    2O2 + 4e- ----> 2O + O2

    Hydrogen in the reaction is 2H2 ----> 2H2 so I don't see where the oxygen has gained it's 4 electrons from...

    ....... i'm being really stupid here
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    (Original post by Cinamon)
    So for some reason I am confusing myself with this reaction

    2H2O2(aq) -> 2H2O(l) + O2

    Oxygen in this reaction goes from 2O2 ----> 2O + O2.
    We know that the element O2 has no charge, but 2O has a a charge of -4. This means that it has gained 4 electrons from somewhere. Should we include those electrons in the half equation?
    2O2 + 4e- ----> 2O + O2

    Hydrogen in the reaction is 2H2 ----> 2H2 so I don't see where the oxygen has gained it's 4 electrons from...

    ....... i'm being really stupid here
    No, you're not being stupid...

    ... the reaction is not as straighforward as it seems. Although it is described as simply decomposition, it is effectively a disproportionation reaction.

    half equation 1 (reduction): H2O2 + 2e --> 2OH-

    here the oxidation state of the oxygen atom changes from -1 to -2

    half equation 2 (oxidation): H2O2 --> 2H+ + O2 + 2e

    here the oxidation state of the oxygen atom changes from -1 to 0

    Add the two equations together:

    H2O2 + 2e --> 2OH-
    H2O2 --> 2H+ + O2 + 2e
    -----------------------------------------------------------------------------------------
    2H2O2 --> 2H+ + 2OH- + O2


    ...which is the same as writing:

    2H2O2 --> 2H2O + O2
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    (Original post by charco)
    No, you're not being stupid...

    ... the reaction is not as straighforward as it seems. Although it is described as simply decomposition, it is effectively a disproportionation reaction.

    half equation 1 (reduction): H2O2 + 2e --> 2OH-

    here the oxidation state of the oxygen atom changes from -1 to -2

    half equation 2 (oxidation): H2O2 --> 2H+ + O2 + 2e

    here the oxidation state of the oxygen atom changes from -1 to 0

    Add the two equations together:

    H2O2 + 2e --> 2OH-
    H2O2 --> 2H+ + O2 + 2e
    -----------------------------------------------------------------------------------------
    2H2O2 --> 2H+ + 2OH- + O2


    ...which is the same as writing:

    2H2O2 --> 2H2O + O2
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