# In an Electrotlytic cell using MgSO4 electrolyte, Cu(OH)2 and Mg(OH)2 are visible products, Where Does SO4 go?

I have an electrolytic cell with $$\ce{MgSO4}$$in an aqueous solution at full saturation, running at a potential of $$3.6v$$ with a copper anode because I need some $$\ce{Cu(OH)2}$$.

Production of $$\ce{Cu(OH)2}$$, $$\ce{CuOH}$$ and $$\ce{Mg(OH)2}$$ are all visible.

There's no gas production at the anode but I expect that the oxygen is remaining in solution due to $$\ce{OH}$$ production.

There's no sign of $$\ce{SO2}$$ or $$\ce{SO3}$$, so what Happens to $$\ce{SO4-}$$?

EDIT
$$\ce{SO4-}$$ eventually rejoins $$\ce{Mg}$$ because the more reactive base will always oxidize a less reactive neutral or acidic salt, not to mention that $$\ce{H2SO4}$$ is a species that will form under these conditions as $$\ce{Mg}$$ and $$\ce{Cu}$$ both become insoluble (which should actually yield $$\ce{H2SO5}$$ because $$\ce{Cu + MgSO4 + H2O -> Cu(OH)2 + Mg(OH)2}$$ @cathode, $$\ce{H2O + SO4- -> H2SO5}$$ @anode, which doesn't hang around too long $$\ce{H2SO5 -> 2H2SO4 + O2(g)}$$ and the obvious reformation of $$\ce{MgSO4}$$ via $$\ce{H2SO4}$$ + $$\ce{Mg(OH)2}$$ and $$\ce{CuSO4}$$ + $$\ce{Mg(OH)2}$$ so I was eventually left with $$\ce{Cu(OH)2}$$ as the only insoluble product.

I have since been able to produce $$\ce{Mg(OH)2}$$, omitting $$\ce{Cu}$$ entirely of course, but coming up with a process wasn't easy with my 3 months of experience. I however have been able to completely separate $$\ce{Mg+}$$ from $$\ce{SO4-}$$, yielding $$\ce{H2SO4}$$and $$\ce{Mg(OH)2}$$using perlite.

It forms $$CuSO_4$$, which is soluble in water. $$Cu(OH)_2, CuOH$$ and $$Mg(OH)_2$$ prefer to precipitate, while $$CuSO_4$$ just floating there, completely invisible. Actually, even when ppl try to obtain $$CuSO_4$$ in such electrolytic cells, they simply collect $$Cu(OH)_2$$, add $$H_2SO_4$$ and crystallize sulfate. So it is there, but there is no way you can obtain it directly.