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Balancing question

  1. Oct 2, 2006 #1
    Hi, I have been asked to balance this equation:

    [tex]KOH + K_{4}Fe(CN)_{6} + Ce(NO_{3})_{4} --> Fe(OH)_{3} + Ce(OH)_{3} + K_{2}CO_{3} + KNO_{3} + H_{2}O[/tex]

    Can someone please tell me where to start? I have written down the oxidation states of all the elements in the reaction and can see that iron is getting oxidised from +2 to + 3, carbon in cyanide ion is oxidised to + 4 state, Cerium is being reduced from +4 to +3 state. How do I proceed? Please help.
     
  2. jcsd
  3. Oct 2, 2006 #2
    wow this looks confusing.....hmm where's your Cerium after reacting?
     
  4. Oct 2, 2006 #3
    By starting with the oxidation numbers im assuming that you think its a redox reaction. If this is the case you must follow these steps: http://members.aol.com/profchm/redox.html
     
    Last edited: Oct 2, 2006
  5. Oct 2, 2006 #4

    Borek

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    Staff: Mentor

  6. Oct 4, 2006 #5
    I know the rules of balancing a redox reaction. It is just that this one is too tough. There are too many elements undergoing oxidation/reduction. I am not sure whether to employ the oxidation-reduction method or ion-eletron method. What do I do?

    PS. There should be a [tex]Ce(OH)_{3}[/tex] among the products.
     
  7. Oct 4, 2006 #6
    I tried by placing coefficients it just goes back and forth like a never ending combustion reaction
     
    Last edited: Oct 4, 2006
  8. Oct 5, 2006 #7
    I got the half reactions for the equation but its going to take 3+ pages of work to figure out the coefficients. My IB high level chem teacher said she balanced the "mother of all equations" and it took 3 pages, and she thinks this one is more complicated. I will be working with her on it. She also said that coefficients can and quite possibly make it into the 100's. Ill keep you posted. I hope this wasnt due overnight...
     
  9. Oct 6, 2006 #8
    thanks. I have been trying too without any success. Tell me if you do get it.
     
  10. Oct 6, 2006 #9
    have you learned re-dox (by using ionic equations)?
     
  11. Oct 7, 2006 #10
    yes I have.. but It didnt help. I was still not able to do it.
     
  12. Oct 7, 2006 #11

    Borek

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    These things require systematical approach and algebraic method does wonders.

    "Mother of all equations" was most likely so called Stout equation:

    (Cr(N2H4CO)6)4(Cr(CN)6)3 + KMnO4 + H2SO4 -> K2Cr2O7 + MnSO4 + CO2 + KNO3 + K2SO4 + H2O
     
  13. Oct 7, 2006 #12
    :confused:

    This isnt any redox equation...Didnt you even try it. If you did you'd realize how complicated it is. By the way my chemistry teacher got it but was of by 60. Which isnt much conisdering the coefficients, she said its a simple math error and shes going to go over it again i should have an answer for you Tuesday. :biggrin:
     
  14. Oct 7, 2006 #13
    20 extra creidt points in my class if we balance the daughter, son, father, and mother of all equations. These are the names from the book, im not making them up to be funny.
     
  15. Oct 7, 2006 #14
    no, but it's another approach if regular balancing doesnt work...
     
  16. Oct 7, 2006 #15
    Regular balancing isnt going to work, its just plain obvious, you have to break into half reactions.
     
  17. Oct 7, 2006 #16
    yes sir! i need to stop memorizing ways to do probs...hopefully, this forum can shape me up
     
  18. Oct 8, 2006 #17
    thanks all for your help.
    I have balanced it successfully. It was just a matter of dividing it into the half reactions and balancing them individually. I wouldnt advise using the algebraic method though, it turned out to be quite lengthy.
     
  19. Oct 8, 2006 #18
    Good job! you always have to divide into half reacftions. I am not suprised how legnthy it was. It took my teacher a page and a half so im thinking it would take me a few. Again awesome job that problem was intense:smile:
     
  20. Oct 8, 2006 #19

    Borek

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    Algebraic: 15 minutes, half a page :biggrin:
     
  21. Oct 9, 2006 #20
    But it isnt always reliable. When you are pressed for time especially during a competetive exam or so, it is not always handy. Besides, the redox balancing method didnt take too long either. It was just a matter of writing the half reaction correctly.
     
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