How to balance the following disproportionation reaction $$\ce {XeF2 + H2O → Xe + XeO3 + HF + O2} $$ My attempt: I tried by making total number of reductions equal to total number of oxidation but I couldn't do because Xe is both oxidised and reduced also the solvent is water so what should I do to balance protons and oxygen on both sides?. Please help by solving it on the basis of same procedure that is making oxidation equal to reduction

  • $\begingroup$ since acid is in the product side balancing should be done as in acidic medium $\endgroup$ – JM97 Feb 17 '16 at 6:49

It can't be balanced unambiguously, for this is not a single reaction, but a mechanical sum of two reactions, and these can be formally mixed in arbitrary proportion.

The first reaction is indeed a disproportionation. To balance it, use the common trick: consider these two xenons different elements for a while. $$\ce{\underbrace{XeF2}_{for\;reduction} + \underbrace{XeF2}_{for\;oxidation} +H2O->Xe + XeO3 + HF}$$ The second is a typical redox: $$\ce{XeF2 + H2O -> Xe + O2 + HF}$$

  • $\begingroup$ Nertin how do you know that it is a sum of two reactions? Could you please suggest any good textbook for it $\endgroup$ – JM97 Feb 17 '16 at 8:25
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    $\begingroup$ This is a sum of two reactions because it has two independent oxidized products: $\ce{XeO3}$ and $\ce{O2}$. As for the book, I don't think there is one; the matter is too trivial to deserve a whole book on its own. $\endgroup$ – Ivan Neretin Feb 17 '16 at 8:32
  • $\begingroup$ Nertin how did you figure out what reactants and products would be in each reaction? $\endgroup$ – JM97 Feb 17 '16 at 8:45
  • $\begingroup$ Why, the OP specified them all in the original question. I just figured that the oxidized products are independent, so there would be only one of them in each reaction. $\endgroup$ – Ivan Neretin Feb 17 '16 at 9:15

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