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Reducing agent.

  1. Oct 31, 2008 #1

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    1. The problem statement, all variables and given/known data

    What is the reducing agent in the following spontaneous reaction?

    OCl-(aq) + I2(s) + 2 OH-(aq) --->
    Cl-(aq) + 2 OI-(aq) + H2O(l)



    2. Relevant equations

    The reducing agent is oxidized in the process so it loses electrons.

    3. The attempt at a solution

    Both I2 and OH- lose electrons so im not sure which one it is, any help?
     
  2. jcsd
  3. Oct 31, 2008 #2

    symbolipoint

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    Not as much as that. The hydroxide does not change any of its reduction status.

    The reducing agent reduces some part of another item; the oxidizing agent oxidizes some part of an item. Essentially, the reducing agent becomes oxidized, and the oxidizing agent becomes reduced. Most of the time, you want to check reduction status of some atom other than oxygens.
     
  4. Oct 31, 2008 #3

    a.a

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    so the I2 is the reducing agent?
     
  5. Oct 31, 2008 #4

    symbolipoint

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  6. Oct 31, 2008 #5

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    Thanks..so just so i get why:
    the I2 is the reduceing agent because it reduces another atom and at the same time is oxidized, but the OH- is only oxidized but didnt cause another atom to be reduced?
     
  7. Oct 31, 2008 #6

    symbolipoint

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    You made an interesting interpretation, but the hydroxide is not reduced and is not oxidized in the reaction. The hydroxide is used in the balancing of the equation but does not go through oxidation or reduction. One reason why the hydroxide is needed is that it makes the hypochlorite more stable and available for the redox reaction. The acid form, hypochlorous acid, HOCl, is not very stable and can decompose. The neutralized anion, OCl-, is more stable but needs to be the salt form for stability.

    When you pump chlorine gas into alkaline solution (hydroxide present), you get hypochlorite salt. This is why your question in post #1 starts with hypochlorite in alkaline solution.
     
  8. Oct 31, 2008 #7

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    what would we do in an acidic solution when we need to blance charges?
     
  9. Oct 31, 2008 #8

    symbolipoint

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    My best guess for that reaction in acidic solution is that the hydronium would not be part of any mass or charge balancing; I'm really not so certain in these conditions. Maybe someone else can give better understanding. I could only best imagine showing two hydrogen ions on the left and maybe just two HIO on the right, or just two hydrogen ions (therefore redundant) on the rightside. Like I say, my guess for this condition is not secure and someone else should help respond. It's been a long time.
     
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