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Hydrogen Peroxide oxidation state

  1. Feb 14, 2005 #1
    In a redox reaction I found, Hydrogen Peroxide (H202) was taken as having an oxidation state of zero

    However my chemistry teacher keeps telling me that oxygen ALWAYS has an oxidation state of -2 so that would mean the hydrogen in H202 must have an oxidation state of +2 to keep the molecule neutral. Since this +2 can only come from the loss of 2 electrons how is it possible for hydrogen, which when neutral has 1 electron, to lose 2 electrons and become +2?

    Please forgive my ignorance on all chemistry matters :confused:
  2. jcsd
  3. Feb 14, 2005 #2
    Oxidation states are used to name compounds, balance equations and for ionic equations. There are rules that say you must go by certain elements first. One is that oxygen is on of the ones that almost always takes the charge of -2. So in your example the overall charge of oxygen is -4 (-2 x 2). This means, as you said, that hydrogen must have an overall charge of +4 (+2 for each atom). I don't think that it has anything to do with the electrons in this case.

    The Bob (2004 ©)
  4. Feb 14, 2005 #3
    i still don't understand how something with only one proton (providing +1) can have a charge of +2, surely its just not possible
  5. Feb 14, 2005 #4


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    1. Oxidation states are a tool for solving problems.

    2. What problem are you trying to solve here ?

    3. Treat oxygen as -1 in this case. (exception to the rule, if you may)
  6. Feb 14, 2005 #5


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    As Gokul indicated, treat the oxygen as -1 state, and hydrogen as +1.

    The oxygen atoms share a bond between each other.

    In general H2O2 --> H2O and O. Hydrogen peroxide is a powerful oxidizing agent. The O in H2O has a -2 valence state, and the free oxygen (O) is neutral until it reacts with another element (atom), and it will most likely take a -2 state.
  7. Feb 14, 2005 #6
    Remember too that the bond structure for H202 is H-O-O-H, so essentally though the O's are -2 each (each have 6 electrons), they share a valence electron this making the sub-molecule O-O have a total -2 valence state(in the bohr view). Then the 2 H's (each with only one electron) fill those slots.
  8. Feb 14, 2005 #7


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    The self bonding (encountered especially for C in organic compounds) always alters the ON.For oxygen,it is -1.I advise you to compute these numberds for di,tri and tetrationats of Natrium,for example...Or peroxodisulphuric acid [itex] H_{2}S_{2}O_{8} [/itex] You'll be surprised...

    These numbers are essential when discussing redox reactions...

  9. Feb 14, 2005 #8
    ok so assuming Oxygen in H2O2 is -1 and the hydrogen +1:
    H2O2 --> H2O + O
    The oxidation state of oxygen in the H20 is now -2, so would you say that one oxygen has been reduced and the other oxidised?
  10. Feb 14, 2005 #9


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  11. Feb 14, 2005 #10


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    Hint: Hydrogen peroxide. Try reading up on the section of polyatomic ions in your text, it should describe to you the oxidation state of oxygen in this compound.
  12. Feb 19, 2005 #11
    I learned the reason as being peroxide is a polyatomic, [itex]O_2^{-2}[/itex], so the bond stil works out. It's sort of like mercury(I) being [itex]Hg_2^{+2}[/itex] and what not.
  13. Feb 19, 2005 #12


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    What do you mean " the bond still works out"...?

  14. Feb 20, 2005 #13
    That's correct. Hydrogen is almost always +1. The only exception is when it's -1, as in a hydride (eg, LiH). Oxygen is usually -2. In peroxides, it's taken as -1. There are also superoxides (like NaO2), where it's -0.5. These oxidation numbers don't necessarily refer to the electrons in any individual atom, but are used for calculating overall electron transfers in the half-reactions.
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