# Electrolysis Anode equation

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.

## 1 Answer

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.

• The book does not define the medium. It only asks for the half reaction(s). Dec 5 '13 at 13:29
• 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. Dec 5 '13 at 13:33
• "[...]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? Dec 5 '13 at 13:48
• Of course, the other way around :) And yes, that's exactly what I meant with that comment. Dec 5 '13 at 13:51