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 $$\ce{Cu + MgSO4 + H2O -&gt; Cu(OH)2 + Mg(OH)2}$$ @cathode, $$\ce{H2O + SO4- -&gt; H2SO5}$$ @anode, which doesn't hang around too long $$\ce{H2SO5 -&gt; 2H2SO4 + O2(g)}$$ 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. 

[remark] You can also try to crystallize it from solution, since it's there. But maybe you won't succeed.