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Redox Help

  1. May 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Balance the equation (Note: the coefficients provided may be wrong):
    3 Co(s) + 8 H+(aq) + 8 NO3-(aq) → 3 Co2+(aq) + 6 NO3-(aq) + 2 NO(g) + 4 H2O(l)

    3. The attempt at a solution
    I started out by doing a half reaction

    Co → Co 2+ +2e- This one I just added electrons to the appropriate side


    I get stuck here, I tried just balancing the elements...
    4e- +4H+ + 2NO3 - → NO3 - + NO + 2H2O

    No matter how i look at it I don't know why i have this answer incorrect

    the real answer is 3 Co(s) + 8 H+(aq) + 8 NO3-(aq) → 3 Co2+(aq) + 6 NO3-(aq) + 2 NO(g) + 4 H2O(l)

    but i don't know how it came to that
     
  2. jcsd
  3. May 22, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    You have NO3- on both sides. Try to leave it only on the left.
     
  4. May 22, 2009 #3
    yeah the "real answer" you have looks rather strange. It has NO3- on both sides.

    Something looks wrong with this though

    "4e- +4H+ + 2NO3 - → NO3 - + NO + 2H2O "

    How did you arrive at 4 electrons?
     
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