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Balancing Oxidation Numbers

  1. May 12, 2004 #1
    I'm not sure how to balance this equation with regards to the half reaction and electrons, as well as the elements:

    Fe+2 + Cr2O-27--->Fe+3 + Cr+3

    Right now, I've gotten this far, though I don't know if it's the right track:

    Fe+2 + 2Cr+72O-27--->Fe+3 + 4Cr+3 + 7O-22

    What should I do next?
  2. jcsd
  3. May 12, 2004 #2


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    My first post was totally wrong so here's a revamp.

    Ok well here's one part..

    [tex]Fe^{2+} \rightarrow Fe^{3+} + e^-[/tex]

    I can't figure out what happens with the Cr2O7. It usually reacts with acid, like this:

    [tex]Cr_2O_7 ^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O[/tex]

    But there's no acid so that ain't it.

    Another thing is that I'm not sure your second equation works out.

    The charges are not the same on both sides.
    Last edited: May 12, 2004
  4. May 13, 2004 #3
    Maybe it should be 2CrO4(2-)? I noticed that on a website...Still can't figure it out though. It's pissing me off. :mad:
  5. May 13, 2004 #4
    ShawnD, that answered my question exactly. I wasn't sure how the notation worked, as far as just throwing the correct number of H+ and so forth.
    Thanks. :wink:
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