# 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.

## 1 Answer

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.

• That's what I thought, if not elemental sulfur, which I thought I saw in the top layer. If I centrifuge (I made one) the solution to separate the insoluble Copper and Magnesium would it be possible to plate the copper out of solution to obtain H2SO4? – user14828 Dec 1 '18 at 13:25
• Check this website. quora.com/… It shows various ideas about this process. – Kelly Shepphard Dec 1 '18 at 13:27
• I originally started this process because I thought it would be a cheap source of H2SO4 while being able to produce Cu(OH)2 for a reaction with ammonia to produce Copper Nitrite [Cu(NO2)2] Is that where all of the sulfate anions go?(I've only been playing with chemistry for a month or so.) – user14828 Dec 1 '18 at 13:44
• It's not really a good idea. I've found youtube video for you, it's like instruction for doing this, but I don't recommend trying anything dangerous at home, without teacher or supervisor who can help you. But at least now you have CuSO4 solution, which is needed (still it's better to use not home-made solution, but as you wish). Promise that you will be careful! Don't work without proper personal protective equipment. youtube.com/watch?v=t5nHi5_Obd8 – Kelly Shepphard Dec 1 '18 at 14:02
• Lol, I'm a quick learner. I've already HCl by two methods, H2SO4 + NaCl and via Cl oxidation from NaCl. I'm currently designing the circuit for a high power O3 generator to make pure NaClO for a more efficient source for chloroform synthesis. I'm guessing I can synthesize H2O2 from water if I bubble the ozone through it. I'm trying to synthesize H2O2 electrochemically using carbon fiber electrodes (far superior to regular carbon or lead dioxide, no exfoliation), but I'm having purification issues. How do you take the H2SO4 out and how can I avoid making peroxymonosulfuric acid as a byproduct? – user14828 Dec 1 '18 at 14:39