Question: why does copper not dissolve in sulfuric cell?

I was wondering if somebody knows the answer to this.

Let me explain what I know about the chemistry of the sulfuric cell.

In the sufuric cell we use two elektrodes, both submerged in concentrated sulfuric, the elektrodes are connected to a power source. The power source deliveres elektrons to the negative elektrode (lead) and pulls elektrons away from the positive elektrodes (copper mesh).
When te current flows we see bubbles appear at both elektrodes but most of them at the negative elektrode.
As far as I know this is due to the elektrolysis of the small amount of water that is present in sulfuric (the remaining 4%? Concetrated sulfuric is mostly 96% pure). So the bubbles are oxygen at the positive and hydrogen at the negative elektrode right?
Also another compound is being formed at the positive elektrode, wich I think is persulfate to dissolve the gold. When the dissolved gold is moving away from the positive eleketrode it wil gain elektrons due to its highly oxidative nature and snow out as small black gold metal particles.
I know this is not the whole story and maybe the wrong or partial wrong story. Who can explain me the whole story, why is only gold attaked and not copper?? Please help..

I don't know but I'm guessing that the base metal is oxidized and reacts with the oxygen into for example copper oxide and that forms a protective shield with oxides that doesn't dissolve into the concentrated sulfuric acid. Even if there is some copper sulfate formed it doesn't dissolve since there isn't any free water, the small amounts of water present is dissociated by the strong sulfuric acid and can probably not dissolve any significant amounts of metal salts.

In other words, the strong sulfuric acid is passivating any base metal while precious metal is not protected by an oxide layer.

That is at least my theory of the process.

Göran

Thanks Göran, that was also what I was guessing but I thought that these oxides would be dissolved by the sulphuric acid so I skipped that theory.

As I am brainstorming I think I understand why copper is left untached and the noble gold is being dissolved. I think both gold and copper are oxidized by the persulfate. Positive gold ions are attrackted to the negative lead elektrode and will, as fast as possible, gain elektrons to becom the very stable gold metal atoms. The copper is reduced to copperoxide wich is, as Göran mentioned, passive so it wil not react.
The fact that it does not react is probably because there are almost no protons (H+ ions) in the liquid. Sulfuric acid is H2SO4(l) in liquid state so there are no ions. But when water is present it becomes H+(aq) + SO42-(aq). You can see it for yourself when you measure the conductivity (or resistance) of concentrated sulfuric. This would be zero. But when water is added the conductivity will go sky high because of the ions that are formed.

Conclusion:

persulfate does oxidize both gold and copper, it creates goldpersulfate and copperoxide
Pure sulfuric does not dissolve oxides as there are hydrogen ions needed, so the oxide passivates the copper.

Questions still remain are:
1 where does the oxide that is attached to the copper come from?
2 what are the chemical equations for the oxidation and for the formation of persulfate?

Well, in answer to your first question, (if this is indeed what is happening)
Sulfuric acid: 2 hydrogen atoms, 1 sulfur and 4 oxygen atoms.

Yes I know, but what about the 4% water that is present?

Ok then yes, you also have the oxygen from the water.
Not sure what you are getting at.

Platdigger said:

Not sure what you are getting at.

Click to expand...

You mean what my point is?

Well the point here is that I want to know what is happening in the sulfuric cell because it interests me, maybe I can learn something from it. I hope that somebody can confirm my idea of what is happening or that somebody can push me in the right direction when I am wrong.

I already found the reaction for the formation of persulfate.

It is formed at the positive elektrode. The nett its reaction is:

2H₂SO₄ --> H₂S₂O₈ + H₂(g)

The bubbles you see at the negative lead elektrode are H₂(g)

The reaction is a net result of two half reactions:

Positive elektrode: 2SO₄^2- <--> S₂O₈^2- + 2e- (oxidation)
Negative elekrode: 2H⁺ + 2e- <--> H₂(g) (reduction)

And out of persulfate should Caro's acid (H₂SO₅) being formed wich can oxidate gold, am I right?

It is unlikely that gold dissolves and then auto-precipitates, due to negative Ec - Ea:
2Au+++ + 6e- = 2Au Ec = 1.5v
6SO4= = 3S2O8= + 6e- Ea = 1.96v . However S2O8= is unprobable at H2SO4 of more than 40%. More probable H2O2 forms through
SO4= + 2H2O = H2O2 + H2SO4 + 2e-
And then, more likely, gold, probably with oxide skin simply falls or reduced at the cathode which has a passive surface to electroplating. H+ to dissolve oxides is in minority because of vanishing as H2 (as archeonist said). Hence copper oxide, which forms through H2O2, dissolution.