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In the electrolysis of $\ce{Na2SO4}$, is it valid to write the anode half-reaction (oxidation) as $$ \ce{OH-} \rightarrow \ce{H+} + \frac{1}{2}\ce{O2} + 2\ce{e-} $$?

My book says it's $$ \ce{H2O} \rightarrow 2\ce{H+} + \frac{1}{2}\ce{O2} + 2\ce{e-}. $$

Other than the additional $\ce{H+}$, there's no difference between the two.

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This depends on the medium in which you do the electrolysis. In the case of a basic medium your equation would be more correct, but in the case of an acidic medium your answer would be wrong, since the concentration of $\ce{OH-}$ ions would be far too small.

Generally: In acidic medium you balance your redox equations using $\ce{H+}$ and $\ce{H2O}$, in basic medium you balance with $\ce{OH-}$ and $\ce{H2O}$.

Note: Since you generate $\ce{H+}$ ions it is possible to argue that starting with a solution at $\text{pH}=7$ gives you an acidic medium, resulting in the equation in the book.

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  • $\begingroup$ The book does not define the medium. It only asks for the half reaction(s). $\endgroup$ – TMOTTM Dec 5 '13 at 13:29
  • $\begingroup$ You generate a base from an acid; while this is not necessarily forbidden, there's a certain ugliness to it in this case, since we're talking about water and proton exchange reactions are very fast. Maybe this lead the authors of the book to that equation. $\endgroup$ – tschoppi Dec 5 '13 at 13:33
  • $\begingroup$ "[...]Base from an acid": Are you referring to OH- -> H+? Would you not mean it the other way around? Ok, so you're saying the ugliness is to write it as if H+ is generated from OH-, right? $\endgroup$ – TMOTTM Dec 5 '13 at 13:48
  • $\begingroup$ Of course, the other way around :) And yes, that's exactly what I meant with that comment. $\endgroup$ – tschoppi Dec 5 '13 at 13:51

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