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Balance the oxidation reduction equation in half reaction

PbO2 + KCl => KClO + KPb(OH)3
thank you a lot

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3 answers
  1. In acid solution.?
    2H^+ + ClO^- + 2e ==> Cl^- + H2O

    In basic solution?
    H2O + ClO^- + 2e --> Cl^- + 2OH^-

    The way to learn this is to do them. Looking at what someone else has done won't get it. If you will share what your problem(s) is/are in doing these perhaps I can help you over your "trouble" spots.

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  2. But ?
    but in the main reaction , the ClO^- is on the product site not reactant , as you put it on the reactant site , I mean Cl- is oxidized to ClO^- !

    On the other hand , I can balance it by myself , but I'm having trouble with KPb(OH)3 , I don't know how to separate it into its cation and anion

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  3. You're right. I turned it around. Sorry about. Just reverse everything and you will have that half reacted.
    For KPb(OH)3 the cation is K^+ and the anion is Pb(OH)3^-
    So I would write
    PbO2 ==> Pb(OH)3^-
    I assume this is a basic solution.
    Pb is +4 on the left and +2 on the right.
    PbO2 + 2e + 2H2O ==> Pb(OH)3^- + OH^-
    You may add K^+ to each side to make it read
    PbO2 + 2e + 2H2O + K^+ ==> KPb(OH)3 + OH^-
    but I wouldn't add K^+ until both half reactions have been balanced, multiplied by the appropriate number and added together. Then add positive or negative ions as needed to make compounds and not ions.
    BTW, I found your follow up question about the 2.48 g salt, endothermic and exothermic and responded to that post. If you have trouble finding it I can find it and post it where you like. And while I'm on the subject, that's another very good reason for using the same screen name. Had I known that student was the same as this one I could have taken care of that earlier.

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