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Balancing Redox Equation

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data
    For those who have mastered the balancing of any type o redox equation, an extra challenge is provided below. Balance each of the following equations. Note: all of the reactions take place in aqueous solution.

    2. Relevant equations

    HIO3 + FeI2 + HCl ====> FeCl3 + ICl + HOH

    3. The attempt at a solution

    The oxidation numbers that is got in order of equation

    reactants
    H +1
    I +5
    O3 -6 (-2each)
    Fe + 2
    I -2 (-1each)
    H +1
    Cl -1

    Products
    Fe +3
    Cl3 -3 (-1each)
    I +1
    Cl -1
    H +1
    O -2
    H +1
    The two half reactions (I'm not sure about these, though this is only thing that comes to my head)

    1. FeI2 ===> FeCl3 (I dont get how i would balance these) do i add HCl to the right side and ICl and HOH to the left?
    2. HIO3 ===> ICl Do I add HCl to the right and HOH to the left?


    i asked all my friends but none seem to know how to do these. i have more like this but if i know how to do this i would get the rest.

    Do i balance it first? 5HIO3 + 4FeI2 + 25HCl ==> 4FeCl + 13 ICl + 15HOH
    All help Appreciated Thanks!
     
    Last edited: Oct 16, 2007
  2. jcsd
  3. Oct 16, 2007 #2
    Well first you want to balance the equation. Then you carrry out the two half reactions. Which is seperated into the substance being reduced and the substance thats being oxidized. Think you can do it now? Also the only way you balance in half reactions is by adding electrons or H+ or H20
     
  4. Oct 16, 2007 #3
    i dont get what the two half reactions would be like the fe loses an electron and the iodine gains an electron.
    so i got the two half reactions a
    FeI2 ==> FeCl3
    HIO3 ===> ICl
    i dont get how to balance these because in the first one where does the iodine go.
    so far in class all we did was when there are two reactants and two products so im stumped
     
    Last edited: Oct 16, 2007
  5. Oct 16, 2007 #4
    What is the question specifically asking?
     
  6. Oct 16, 2007 #5
    its asking to balance the redox equation
    HIO3 + FeI2 + HCl ====> FeCl3 + ICl + HOH
     
  7. Oct 16, 2007 #6
    Well the only way i can see you doing it is this way FeI2 +HCl====> FeCl3 + H2O as one reaction and FeI2 ==> FeCl3. Balance using acid, bases, and electrons. If you get stuck just ask. You don't need to balance the equation into this 5HIO3 + 4FeI2 + 25HCl ==> 4FeCl + 13 ICl + 15HOH. The whole point is that redox will balance it.

    Edit: I misread part of you problem. Guy below me did it right
     
    Last edited: Oct 16, 2007
  8. Oct 16, 2007 #7
    KEY STEP: To determine the half reactions, look at the oxidation states and what is losing and what is gaining electrons

    HIO3 (I = +5) --> ICl (I = +1) reduction
    FeI2 (Fe=+2) (I = -1) ---> ICl (I = +1) + FeCl3 (Fe = +3) oxidation;

    balance Cl by adding HCl, H+ for H and H2O for O

    you will need to multiply both reactions to get the electrons to cancel out

    but when they do, and you add the two half reactions and cancelling H+ where needed, you do get back the balanced equation

    25HCl + 4FeI2 + 5HIO3 ---> 13 ICl + 15H2O + 4 FeCl3

    that IS a good problem
     
    Last edited: Oct 16, 2007
  9. Oct 16, 2007 #8
    Your not supposed to do it for him just help him get to it himself. And it's not a good problem because your not really allowed to add acids like hcl. You can acids as H+ and bases as OH-
     
    Last edited: Oct 16, 2007
  10. Oct 16, 2007 #9
    what do you think H+ is in the form of? HCl can be put in the half reactions as well.

    with the other half reactions suggested, not sure how the I would have been balanced

    the key step is to look at the oxidation states, well, yes maybe I did give a little too much info..
     
    Last edited: Oct 16, 2007
  11. Oct 16, 2007 #10
    nvm, I misread part of his problem. I didn't see the charge of I also changed
     
  12. Oct 16, 2007 #11
    :smile:come on, you have to admit, it was GOOD problem
     
  13. Oct 16, 2007 #12
    It wasn't as good as I thought at first. Now it's just a regular redox equation. Before I thought it was something I've never seen.

    Edit: I wonder if he understood
     
  14. Oct 17, 2007 #13
    i do understand it now. the problem was i didnt know you could add hcl. that helps alot. thank you both for your help and effort.
     
  15. Oct 17, 2007 #14
    Adding HCl to balance redox titration is not normal procedure but it works in this case because Cl- charge does not change, essentially a spectator ion, so its going to be there anyway in that form. Just some different thinking - I haven't done many of these. You did assign the oxidation states well, keep them in mind before and after.
     
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