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Balancing (in acidic medium)

  1. Sep 1, 2009 #1
    Cr2O7(-2) + HNO2 + H(+) --> Cr(+3) + NO3(-) + H2O

    The question asks me to balance the above rxn, and then say what the coefficient in front of H2O is.
    The answer key says it's 4, but I've tried three times and I keep getting 6.
    I've always had trouble with these, so chances are I'm doing it wrong. Here was my attempt, using the half-reaction method.

    Cr2O7(-2) --> Cr(+3)

    HNO2 --> NO3(-)

    1. I balanced Cr and N
    2. Added H2O to balance O
    3. Added H(+) to balance H
    4. Added e- to balance charge
    5. Combined both half-rxns into one and cancelled terms that were on both sides.

    So I got this result:

    4e- + 11H(+) + Cr2O7(-2) + HNO2--> 2Cr(+3) + NO3(-) + 6H2O

    If anyone could help me on this I'd really appreciate it! Thanks ahead of time for any responses.
     
  2. jcsd
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