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    Hey, I'm a bit confused on the redox reactions in hydrogen peroxide

    2 H2O2 ----> 2 H2O + O2

    Thats the overall reaction

    Am I correct in saying that

    Oxygen's oxidation numbers
    -1 in hydrogen peroxide
    -2 in water
    0 in oxygen

    and that hydrogen stays the same at +1

    I was wondering how the half equations looked in this reaction?

    would...

    0-1 + e ---> O-2
    0-1 -----> O + e

    be correct? it doesn't seem right to me

    Many thanks
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    Yes, you are right about those oxidation numbers for oxygen.

    The half equations need to be written with the starting species going to the final species with the electrons either being added or produced. Like this

    Fe^{3+} + e \rightarrow Fe^{2+}

    see if you can work out the half reactions now...
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    (Original post by Plato's Trousers)
    Yes, you are right about those oxidation numbers for oxygen.

    The half equations need to be written with the starting species going to the final species with the electrons either being added or produced. Like this

    Fe^{3+} + e \rightarrow Fe^{2+}

    see if you can work out the half reactions now...
    Thanks Is it not just...

    O^{-1} + e \rightarrow O^{-2}
    O^{-1} \rightarrow O + e

    Those half equations don't seem to make complete sense to me in context of the overall decomposition reaction - but they are the only ones that I can make workable.
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    ok, so this is called a disproportionation reaction. The initial species is both oxidised (to form oxygen) and reduced (to form water) at the same time.

    So you have

    H_2O_2 \rightarrow O_2 + 2H^+ + 2e^- oxidation

    and

    2e^- + 2H^+ + H_2O_2 \rightarrow 2H_2O reduction

    in the first, oxygen goes from -1 to 0 and in the second it goes from -1 to -2

    you see how adding both equations together gives you the overall one?
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    (Original post by Plato's Trousers)
    ok, so this is called a disproportionation reaction. The initial species is both oxidised (to form oxygen) and reduced (to form water) at the same time.

    So you have

    H_2O_2 \rightarrow O_2 + 2H^+ + 2e^- oxidation

    and

    2e^- + 2H^+ + H_2O_2 \rightarrow 2H_2O reduction

    in the first, oxygen goes from -1 to 0 and in the second it goes from -1 to -2

    you see how adding both equations together gives you the overall one?
    That's fantastic, thankyou very much for your help. One lat quick question would it be correct to say half the oxygen is being oxidised and half reduced - or is that a simplistic way of wording it?
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    (Original post by Organ)
    That's fantastic, thankyou very much for your help. One lat quick question would it be correct to say half the oxygen is being oxidised and half reduced - or is that a simplistic way of wording it?
    yes. you can imagine that for every two peroxides, one is being oxidised and is supplying the electrons to reduce the other one.
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    I know this questions was asked awhile ago but i was having problems with this myself, I wanted to add that i found it easier to understand if i though about it as 2 oxygens giving up a hydrogen atom and the other two getting an extra hydrogen atom i.e 2H_2O_2 \rightarrow 2(H+O+H)+(O+H-H)_2 , 2H_2O_2 \rightarrow 2H_2O + O_2 Yh so i might make a phew chemists angry doing it this way, but for me chemical equations can be abit difficult to understand without breaking it down into the maths that im use to. Also if you think about each hydrogen atom as carrying an electron with it that will be donated to the oxygen this can apply to question.

    The OP probably doesn't need help with this anymore so i though it could help anyone who is having difficulty understanding the logic beind this equation as i did.
 
 
 
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