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Oxidation Number to balance a redox reaction

  1. May 28, 2012 #1

    agnibho

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    1. The problem statement, all variables and given/known data
    Balance Cl2+NaOH -> NaCl + NaClO3 + H2O
     
    agnibho, May 28, 2012
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  3. May 28, 2012 #2

    chemisttree

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    Start by determining the oxidation numbers of all species.
     
    chemisttree, May 28, 2012
  4. May 31, 2012 #3

    agnibho

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    Cl02 + Na+1O-2H+1 -> Na+1Cl + Na+1Cl-1O3-2 + H2+1O-2

    To equalize the Oxidation Numbers of Cl and Na, 3 atoms of Cl and 6 atoms of Na are required. Thus,
    3Cl2 + 6NaOH -> 5NaCl + NaClO3 + xH2O
    By inspection x = 3
    ∴ 3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O

    I think this is correct, but if I have done something wrong please tell me so, as I'm new to this topic.
    Thanks in advance.
     
    agnibho, May 31, 2012
  5. May 31, 2012 #4

    Borek

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    No idea what you mean by "equalize the ON", specially in the context of sodium - ON for sodium doesn't change, and Na+ is only a spectator.

    You have Cl(0) on the left and Cl(+1) and Cl(-1) on the right. This is disproportionation - teh same element becomes both oxidized and reduced at the same time. Can you tell how many atoms of Cl get oxidized if 1 atom gets reduced?
     
    Borek, May 31, 2012
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