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Redox equations help

I'm stuck on the equation

H2O2O2 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 ee^- on the RHS giving

H2O2O2+4e+2H+ 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 :biggrin: Thanks!
Reply 1
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-)
Reply 2
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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