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    I'm stuck on the equation

     H_2O_2 \rightarrow O_2

    As the oxidation state of oxygen in hydrogen peroxide is -2 + -2, and for Oxygen is 0 wouldn't I need 4 e^- on the RHS giving

     H_2O_2 \rightarrow O_2 + 4e^- + 2H^+ however this doesn't give an overall oxidation number of 0 on both sides only if I use 2 electrons rather than 4, it does. Could someone explain this please Thanks!
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    You need to remember that in H2O2, the oxidation number of Oxygen is actually -1, it is an exception
    so itll be H2O2---O2+(2H+)+(2E-)
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    (Original post by noodlemon)
    You need to remember that in H2O2, the oxidation number of Oxygen is actually -1, it is an exception
    so itll be H2O2---O2+(2H+)+(2E-)
    Ohh! I thought the oxygen being 2- each meant the hydrogens would be 2+ each, thanks very much!
 
 
 
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