# Balance the oxidation reduction equation in half reaction

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

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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^-