Sciencemadness Discussion Board - Copper sulphate preparation question

Copper sulphate preparation question

reactofurnace - 28-8-2020 at 10:34

I'm working on a collaboration with another channel and he asked if I could explain the reason of adding H2O2 for the reaction of Cu and H2SO4 to progress. I thought of it and researched it but couldn't find any convincing answers. I know that H2O2 acts as an oxidizer, presumably oxidizing the copper to Cu2+.
I have a slight hunch that H2O2 decreases the Gibbs free energy of the system making the reaction more thermodynamically spontaneous. However, until I've calculated it I'm not sure if that would be correct. Any ideas on how this reaction works:
Cu + H2SO4 + H2O2 ---> CuSO4 + 2 H2O
and why this reaction is slow:
Cu + 2 H2SO4 ---> CuSO4 + 2 H2O + SO2

[Edited on 28-8-2020 by reactofurnace]

unionised - 28-8-2020 at 10:53

An important part of the reason is
SO2 + 2 H2O2 --> H2SO4 + H2O
Which means you use less acid (because some of it is regenerated from the SO2
And you don't have to take care of lots of SO2.

Bedlasky - 28-8-2020 at 11:57

Cu + H2O2 --> CuO + H2O

CuO + H2SO4 --> CuSO4 + H2O

This is what actually happen. Copper can't be dissolve in nonoxidizing acids because nascent hydrogen reduce Cu(II) back to its metalic state. However you can dissolve it in conc. H2SO4 (>90%), because this acid have oxidizing properties.

Cu + 2 H2SO4 ---> CuSO4 + 2 H2O + SO2

But this reaction take place only with concentrated sulfuric acid, not with dilute. Btw. this reaction is simplified, in reality is much more complex, look at this:

http://pubs.sciepub.com/wjce/2/2/1/

Fulmen - 28-8-2020 at 12:02

Acids attack less noble metals by a replacement reaction where it switches place with hydrogen. For iron the reaction will be something like:
Fe + H2SO4 => FeSO4 + H2
But this only happens if the metal is more reactive than hydrogen (https://en.wikipedia.org/wiki/Reactivity_series). Copper is more noble, so acids do not attack it directly.
It is however readily oxidized into CuO by both H2O2 (and air for that matter). Once oxidized the acid have no problem attacking it through a different reaction:
CuO + H2SO4 => CuSO4 + H2O

Unionseed: What are you talking about???

unionised - 28-8-2020 at 12:28

Quote: Originally posted by Fulmen

Unionseed: What are you talking about???

Which bit don't you understand.

Fulmen - 28-8-2020 at 12:42

@Unionseed: Everything. As Bedlasky points out H2SO4 isn't oxidizing except for in extreme cases. Also, you only presented part of the reaction. Where does the SO2 come from? The starting ingredients are H2SO4, Cu and H2O2, any intermediate steps should be included in a well presented argument.

macckone - 28-8-2020 at 13:35

Fulmen's analysis is correct.
Copper is sufficiently inert toward sulfuric acid it won't displace hydrogen.
H2O2 will react with sulfuric acid forming caro's acid aka Piranha solution which is a stronger oxidizer.
That reacts with copper forming copper(II) oxide which then reacts forming copper(II) sulfate.
H2O2 and Sulfuric acid are very very slow to attack copper individually.
Peracetic acid does a similar reaction.

[Edited on 28-8-2020 by macckone]

DraconicAcid - 28-8-2020 at 13:47

I don't think it's correct to say that the hydrogen peroxide reacts with copper to form copper(II) oxide, which then reacts with the acid to form copper(II) sulphate. Acidic hydrogen peroxide will react with copper metal to give water and copper(II) ions.

2 H+ + H2O2 + Cu ==> 2 H2O + Cu(2+)

The sulphate is a spectator ion.

macckone - 28-8-2020 at 15:24

DraconicAcid,
You are correct technically once the copper (ii) oxide is attacked by the hydronium ions to yield copper ions and water.
In this reaction the sulfate ions are first converted to monopersulfate, so they aren't exactly spectators.

DraconicAcid - 28-8-2020 at 16:10

Quote: Originally posted by macckone

DraconicAcid,
You are correct technically once the copper (ii) oxide is attacked by the hydronium ions to yield copper ions and water.
In this reaction the sulfate ions are first converted to monopersulfate, so they aren't exactly spectators.

You'd have to have some hard evidence to convince me that copper(II) oxide is formed as an intermediate at all in this reaction.

The conversion of sulphate to a persulphate is believable, but I'm not convinced that it's a required step. Has it been shown to be involved in this reaction?

draculic acid69 - 28-8-2020 at 18:50

Quote: Originally posted by Bedlasky

Cu + H2O2 --> CuO + H2O

CuO + H2SO4 --> CuSO4 + H2O

Cu + 2 H2SO4 ---> CuSO4 + 2 H2O + SO2

http://pubs.sciepub.com/wjce/2/2/1/

I
Bedlasky is correct.the copper has to react to CuO first. the peroxide will help speed this up and if it stops the rxn from giving off so2 that's a good thing.also don't start with copper pipe.use headphone cords or some other thin electric wire as boiling sulfuric acid with copper pipe would take a week to dissolve.thin wire dissolves in a day or less.use a wash bottle with caustic soda to dissolve any fumes

macckone - 28-8-2020 at 21:15

Here is the link to a 1962 paper on it:
https://www.nrcresearchpress.com/doi/pdf/10.1139/v62-227

It has to form a copper oxide before it will react with the sulfuric acid.
But neither sulfuric acid nor hydrogen peroxide by itself will oxidize copper in a timely manner.
Therefore the intermediate monopersulfuric acid has to be present.
Bubbling oxygen into sulfuric acid is known to form monopersulfuric acid which is detected as peroxide in solution.
Also in the provided paper.

As I said previously peracetic acid does the same type of oxidation.
This is true for copper and other metals below hydrogen on the reactivity scale.
In the case of lead, the sulfate is insoluble so you have to go the peracetic route.

macckone - 28-8-2020 at 21:34

Here is one that goes more into persulfates:
Simpkins, D. C. (1979). Hydrogen Peroxide/Sulphuric Acid Etching Systems. Circuit World, 6(1), 54–57. doi:10.1108/eb043607

It is available from you favorite hub of scientific papers.

DraconicAcid - 28-8-2020 at 21:38

If you read the end of the paper where they talk about the proposed mechanism, they don't suggest that copper oxide gets formed- just copper(I) and copper(II) ions, the former of which gets oxidized by hydrogen peroxide.

I can accept that neither hydrogen peroxide nor sulphuric acid alone will oxidize copper at a reasonable rate. The fact that they do so together can be adequately explained by the fact that acidic hydrogen peroxide is a stronger and more rapid oxidizing agent than neutral hydrogen peroxide, without postulating the involvement of peracids or the formation of copper(II) oxide.

As far as I know, hydrogen peroxide will corrode copper quite well in the presence of *any* acid. Hydrochloric acid certainly won't form an oxyacid with hydrogen peroxide, and will react with hydrogen peroxide and copper. Anyone have any fluoboric acid to test the reaction of that with copper and hydrogen peroxide? It won't form a peroxyacid, neither will it coordinate the copper ions.

Bedlasky - 28-8-2020 at 22:17

Quote: Originally posted by draculic acid69

Quote: Originally posted by Bedlasky

Cu + H2O2 --> CuO + H2O

CuO + H2SO4 --> CuSO4 + H2O

Cu + 2 H2SO4 ---> CuSO4 + 2 H2O + SO2

http://pubs.sciepub.com/wjce/2/2/1/

I try to say, that if you want to dissolve copper in dilute sulfuric acid you need some oxidizer, because dilute sulfuric acid haven't oxidizing properties. So there isn't SO2 formation at all. Dilute sulfuric acid alone just don't react with copper. This can do only concentrated sulfuric acid which have oxidizing properties (so SO2 is produced as a product of reduction of sulfuric acid). This needs heating for reasonable speed of the reaction. Of course that you can add some peroxide for speeding things up, but it isn't necessary.

Quote: Originally posted by DraconicAcid

Anyone have any fluoboric acid to test the reaction of that with copper and hydrogen peroxide? It won't form a peroxyacid, neither will it coordinate the copper ions.

I think that perchloric acid is more common nonoxidizing and noncoordinating acid, which doesn't form peroxyacid. I have some NaClO4 and I plan make some HClO4 from it so I can try it.

[Edited on 29-8-2020 by Bedlasky]

wg48temp9 - 29-8-2020 at 00:22

Quote: Originally posted by unionised

An important part of the reason is
SO2 + 2 H2O2 --> H2SO4 + H2O
Which means you use less acid (because some of it is regenerated from the SO2
And you don't have to take care of lots of SO2.

If you add sufficient even 50% H2O2 to concentrated sulphuric acid for it to be stoichiometric the 50% H2O of the 50% H2O2 will dilute the acid to about 75% and therefor little or no SO2 formed to be oxidised, assuming that about 75% acid does not oxidise copper to produce SO2.

If 30% H2O2 is used the acid will be diluted to only about 50%.

unionised - 29-8-2020 at 03:22

OK, so I actually did the experiment.
I got some copper , cleaned it up and put it in aqueous hydrogen peroxide overnight.
It is definitely blackened.

So CuO is a possible intermediate.
I can't rule out the idea that Cu2O is the actual intermediate.
The reaction H2SO4 + H2O2 <--> H2SO5 + H2O lies fairly well to the left in aqueous conditions.
I doubt any peroxysulfate is an intermediate.
Not least because it's not obvious why persulphate would do a better job of oxidising copper than peroxide.

In response to Fulmen's question "Where does the SO2 come from?" I suggest that he re-reads the OP to which my reply was a direct response; and which gives the equation for the production of SO2.
To be fair you do need a relatively high concentration of sulphuric acid to produce much SO2 (If I get round to it, I will find out how high), but since the OP was talking about it, I assumed that he had observed it.

S.C. Wack - 29-8-2020 at 07:24

Rather high and quite hot, to get it started. Not for the fearful or careless.

[Edited on 29-8-2020 by S.C. Wack]

reactofurnace - 29-8-2020 at 17:20

Quote: Originally posted by unionised

OK, so I actually did the experiment.
I got some copper , cleaned it up and put it in aqueous hydrogen peroxide overnight.
It is definitely blackened.

So CuO is a possible intermediate.
I can't rule out the idea that Cu2O is the actual intermediate.
The reaction H2SO4 + H2O2 <--> H2SO5 + H2O lies fairly well to the left in aqueous conditions.
I doubt any peroxysulfate is an intermediate.
Not least because it's not obvious why persulphate would do a better job of oxidising copper than peroxide.

In response to Fulmen's question "Where does the SO2 come from?" I suggest that he re-reads the OP to which my reply was a direct response; and which gives the equation for the production of SO2.
To be fair you do need a relatively high concentration of sulphuric acid to produce much SO2 (If I get round to it, I will find out how high), but since the OP was talking about it, I assumed that he had observed it.

Interesting result. Copper (II) oxide could have formed but unfortunately doesn't rule out formation of persulphuric acid.

(1) H2SO4 + H2O2 <--> H2SO5 + H2O
(2) H2SO5 + Cu ---> CuSO4 + H2O

Fulmen - 30-8-2020 at 00:04

Quote: Originally posted by unionised

I doubt any peroxysulfate is an intermediate.
Not least because it's not obvious why persulphate would do a better job of oxidising copper than peroxide.

That's my conclusion as well. We know that copper is oxidized by H2O2 (and air for that matter), and that acids readily attacks copper oxide.

Quote:

In response to Fulmen's question "Where does the SO2 come from?" I suggest that he re-reads the OP

You're right, my bad. I completely missed that part of the post.

Bedlasky - 30-8-2020 at 02:38

Quote: Originally posted by reactofurnace

Interesting result. Copper (II) oxide could have formed but unfortunately doesn't rule out formation of persulphuric acid.

(1) H2SO4 + H2O2 <--> H2SO5 + H2O
(2) H2SO5 + Cu ---> CuSO4 + H2O

Yes, some very little amount of H2SO5 is formed. So there will be two reactions with copper metal - major with H2O2 and minor with H2SO5.

Arthur Dent - 9-11-2020 at 11:58

Preparing a batch of CuSO4, gently boiling and releasing some ominous NO2 fumes.

Nice blue color that will produce lovely crystals. Started with fine telephone wire with the insulation stripped off, poured 100 ml of sulfuric acid, added very gradually about 20ml of nitric acid, then put the erlenmeyer on a quick and dirty sand bath, heat at medium.

Also added about 15 ml of hydrogen peroxide 10% to give a bit of a boost to the reaction. When I cut the heat and removed the flask, the reaction continued for a while. The resulting solution is a beautiful sky blue and will be filtered, then boiled again to increase concentration.

Such a nice, warm day in November, it was the ideal time for a bit of outdoors chem!