# Why can't sulphate ion be discharged in the anode of electrolysis?

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.

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.