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Finding the oxidized and reduced element

  1. Sep 24, 2013 #1
    1. I need to find the element being oxidized and reduced for:

    CO2 + C = 2CO

    I don't know how to do it for this one.





    3. I know how to do it for ones like this:

    2Al + 3Cl2 = 2Al(3+) 6Cl(-)

    it would be

    zero for elemental Al and Cl

    and the oxidation number for the product would be its charge.

    but how do I do it for ones like this.

    CO2 + C = 2CO
     
  2. jcsd
  3. Sep 24, 2013 #2

    Borek

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    Staff: Mentor

    Have you tried to calculate oxidation numbers for all atoms present in the equation?
     
  4. Sep 24, 2013 #3
    Would it be

    for the left side:

    C=+4, O =+2 for CO2

    and elemental C = 0

    and for the right side:

    C=+2, O =+2 for 2CO
     
  5. Sep 24, 2013 #4

    Borek

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    Staff: Mentor

    So far, so good. Which atoms oxidation numbers changed?
     
  6. Sep 24, 2013 #5
    If I add the C atom and subtract from the other side would this be correct:

    (4+0) - 2 = 2

    So, C would be reduced.
     
  7. Sep 24, 2013 #6

    Borek

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    Staff: Mentor

    I am not sure what you are doing, but your conclusion is partially right. Carbon is being reduced, but you can't have reduction without oxidation.

    What is being oxidized?

    Hint: I was referring to atoms (not elements) for a purpose.
     
  8. Sep 24, 2013 #7
    Oxygen is being oxidized, but how much is the oxidation number increased by?
     
  9. Sep 24, 2013 #8

    Borek

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    Staff: Mentor

    No, oxygen oxidation number doesn't change - it was -2 before and it is -2 after.

    Sigh, I just realized you calculated oxidation numbers for oxygen wrong - you wrote them as +2. No, it was -2 in all compounds.
     
  10. Sep 24, 2013 #9
    yes I noticed that as well
     
  11. Sep 24, 2013 #10

    Borek

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    Staff: Mentor

    So, what is being oxidized?
     
  12. Sep 24, 2013 #11
  13. Sep 24, 2013 #12

    Borek

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    Staff: Mentor

    Yep. You have two different carbon atoms on the left - one is getting oxidized, the other is getting reduced.

    Sometimes it can get even more surprising, when you have two identical atoms on the left and they get reduced and oxidized at the same time - it is called disproportionation. The simplest example is probably

    Cl2 + 2OH- → Cl- + OCl- + H2O

    (try to assign oxidation numbers and see what is happening here).
     
  14. Sep 24, 2013 #13
    that is surprising. I did not know about that.

    So, is the other C going from 0 to 2?
     
  15. Sep 25, 2013 #14

    Borek

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    Staff: Mentor

    Yes, one C is going from +4 to +2, the other one from 0 to +2. First is reduced, the other is oxidized.
     
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