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    yo

    please see attatched

    will give rep if you help me ..that is if I'm not all out for today ..if I am I can always give it tomorrow though
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    (Original post by CasualSoul)
    yo

    please see attatched

    will give rep if you help me ..that is if I'm not all out for today ..if I am I can always give it tomorrow though
    Not the best source material to work from!

    Sodium is not included as it is a spectator ion, i.e. nothing happens to it in the course of the reaction.

    You are dealing with a reaction in which the chlorine is disproportionating.
    You have not identified the first half equation correctly. It should be:

    Cl2 + 2e --> 2Cl-

    The other is:

    Cl2 + 6OH- --> 2ClO3- + 6H+ + 10e

    To balance the two half equations you must multiply the first by 5 and then add them together...
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    (Original post by charco)
    Not the best source material to work from!

    Sodium is not included as it is a spectator ion, i.e. nothing happens to it in the course of the reaction.

    You are dealing with a reaction in which the chlorine is disproportionating.
    You have not identified the first half equation correctly. It should be:

    Cl2 + 2e --> 2Cl-


    The other is:

    Cl2 + 6OH- --> 2ClO3- + 6H+ + 10e

    To balance the two half equations you must multiply the first by 5 and then add them together...

    Hey that';s what I thought and what I wrote originally but then in the book they gave the answer as how I've written it in the attatchment..is it because they have done it for the Cl2 going to the Cl in the NaClO- when you are meant to do it for the Cl2 going to the Cl in the Nacl-?


    anyway thanks this has helped lot ..yeah I may refrain from doing these summary questions and just do the exam style instead as they just end up confusing me even more :cool:
 
 
 
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