Given that there is an aqueous solution of sulfuric acid ($\ce{H2SO4}$), $\ce{2H+}$ is reduced in the cathode and $\ce{OH-}$ is reduced in the anode. Why is $\ce{OH-}$ preferentially discharged over $\ce{SO4^2-}$ in the anode?

One explanation I've found is that because $\ce S$ in $\ce{SO4^2-}$ has an oxidation number of $+6$ which is a maximum and thus cannot be oxidized any further. However, can't the $\ce O$ in $\ce{SO4^2-}$ be oxidised instead as it has an oxidation number of $-2$ (just like the case in $\ce{OH-}$ where $\ce{O^2-}$ is oxidised and $\ce{H+}$ is not oxidised according to the half equation $\ce{2OH- <=> 1/2 O2 (g) + H2O (l) + 2e-}$)? Therefore, the argument that $\ce S$ cannot be oxidised because it has an oxidation number of $+6$ doesn't seem to be valid.


1 Answer 1


In aqueous solution the anode will oxidize the water instead, as this reaction can be done at much lower potential than oxidizing the sulfate ion.

Sulfates can be oxidized if you take the water away. Potassium peroxydisulfate can be prepared by electrolyzing a solution of potassium bisulfate ($\ce{KHSO4}$) in sulfuric acid solvent.


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