Sciencemadness Discussion Board - H2O2 + Cu(OH)2

H2O2 + Cu(OH)2

guy - 26-11-2005 at 13:34

What is the reaction when copper(II) hydroxide is mixed with hydrogen peroxide? The product seems to be CuO (black) and some gas (O2 or H2). It only works when H2O2 is added to a basic precipitate of copper(example CuCO3). I can't figure it out because it doesn't do this to other hydroxides that I have tried, Fe(OH)3.

The_Davster - 26-11-2005 at 13:42

It does that to other hydroxides as well, I have made Fe2O3 by reacting Fe(OH)2 with peroxide, creating a suspension of Fe2O3 which can be precipitated by adding NaOH.

woelen - 26-11-2005 at 14:59

This certainly is not a beginners question. H2O2 forms very peculiar compounds with a whole bunch of metal hydroxides, carbonates and many other precipitates.

I personally think that the peculiar reactions are a combination of redox reactions and complex formation (coordination reactions).

Similar effects occurs with titanium, vanadium, iron, cobalt, cerium (IV) and possibly others. The problem is that most of these are not simple redox reactions where the metal is oxidized to higher oxidation state.

E.g. ferrous ion under neutral conditions seems to form very special iron (IV) species (google "Fenton's reagent" for more info). Copper seems to form peroxo-complexes of varying color. The reaction of cerium (IV) with hydroxide and hydrogen peroxide is a total mystery to me.

BromicAcid - 26-11-2005 at 15:08

Thanks Woelen for bringing up Fenton's Reagent, there is a variation on it that actually uses copper salts in place of iron salts (Searching 'Fenton's Reagent Copper' gives some relevent results), although I thought most of the oxidizing ability of Fenton's Reagent took place under slightly acidic conditions, could it just be that the intermediate oxidizing agents are just more stable under basic conditions so less oxidizing?

woelen - 27-11-2005 at 12:33

Yes, you are right. The conditions must be slightly acidic. I have tested it by dissolving Mohr's salt (ferrous ammonium sulfate) in water and adding H2O2. A solution of Mohr's salt is slightly acidic on its own.

The copper variation of Fenton's reagent indeed is very special. It is a combination of copper (II) ions, chloride ions and ascorbate, which allows hydrogen peroxide or even oxygen from the air to form hydroxyl radicals in solution. I know of its special oxidizing properties, but I do not understand precisely how it works (why does it form hydroxyl radicals, what is the role of the chloride ions?).

I am a fish - 27-11-2005 at 12:55

How hot did the reaction mixture get? Copper(II) hydroxide readily decomposes to the oxide in warm water. It is possible that the Cu(OH)2 merely catalysed the decomposition of the hydrogen peroxide into water and oxygen, and then the resulting rise in temperature (H2O2 decomposition is very exothermic) caused the Cu(OH)2 to decompose.

guy - 27-11-2005 at 20:31

I only used 3% H2O2. The reaction was exothermic and was warm. Not like super warm, but it got warmer. I don't think its warm enough to decompose because I have heated Cu(OH)2 to higher temps without decomposition.

[Edited on 11/28/2005 by guy]

woelen - 28-11-2005 at 12:47

Quote:

Originally posted by I am a fish
How hot did the reaction mixture get? Copper(II) hydroxide readily decomposes to the oxide in warm water. It is possible that the Cu(OH)2 merely catalysed the decomposition of the hydrogen peroxide into water and oxygen, and then the resulting rise in temperature (H2O2 decomposition is very exothermic) caused the Cu(OH)2 to decompose.

I don't think it is heating. I have done this similar experiment with cold 1% H2O2 and then still the precipitate turns dark.

A similar effect is observed, when a solution of Na2S2O8 is added to a precipitate of Cu(OH)2. Apparently the oxidizing capabilities have something to do with this.

guy - 28-11-2005 at 17:38

Maybe this reaction is similar to the reaction of calcium hydroxide added to H2O2. That reaction forms calcium peroxide. So maybe this reaction forms copper (ii) peroxide which ultimatly decomposes to copper (ii) oxide. But somehow the reaction only takes place in a basic solution. Perhaps it is because H2O2 is a very weak acid, which causes it to react and form a peroxide (due to insolubility).

Quote:

A similar effect is observed, when a solution of Na2S2O8

Peroxysulfate hydrolyses to hydrogen peroxide, so that explains the similar effect.

Cu(OH)2(s) + H2O2(aq) ---> CuO2(s) + 2H2O(s)
then
2CuO2(s) ---> 2CuO(s) + O2(g)

Anyways, that's my guess. It seems to make sense.

[Edited on 11/29/2005 by guy]

[Edited on 11/29/2005 by guy]

[Edited on 11/29/2005 by guy]

[Edited on 11/29/2005 by guy]

12AX7 - 28-11-2005 at 19:03

CuO2 as a transient state, or do you mean the (unbalanced equation using) CuO2(2-) cuprate ion? This does exist, in strongly basic conditions.

Tim

guy - 28-11-2005 at 19:15

I meant copper(ii) peroxide. I'm guessing that the formula is CuO2.

woelen - 29-11-2005 at 13:58

Quote:

Originally posted by guy

Quote:

A similar effect is observed, when a solution of Na2S2O8

Peroxysulfate hydrolyses to hydrogen peroxide, so that explains the similar effect.

Are you sure? How can you explain the following observations, assuming your theory is right?

1) Persulfate added to a light green precipitate of Ni(OH)2 forms a black solid, at once.
2) Hydrogen peroxide added to Ni(OH)2 does not give any (visible) reaction.
3) Hydrogen peroxide, added to the black result of Na2S2O8/Ni(OH)2 causes the black stuff to change to light green Ni(OH)2 again.

If persulfate would form hydrogen peroxide, then observations (2) and (3) could not be explained at all.

stygian - 29-11-2005 at 14:26

'persulfate' is a broad term. I don't think theyre like percarbonates, which have H2O2 of hydration, these actually have peroxide bonds on the sulfur.

guy - 29-11-2005 at 19:36

Well, I thought you meant peroxydisulfate ((SO4)2)2- or S2O8 2-. And the peroydisulfate hydrolysizes to H2O2 (it is how H2O2 is produced in the electrolysis of ammonium bisulfate).

S2O82- + 2H2O -----> H2O2 + 2HSO4-

H2O2 only shows its strongest oxidizing abilities in an acidic sloution. Alkaline solutions are needed to form NiOOH(Ni(III)). So I think thats why persulfate works, since it is an oxidizer in alkaline slns.

[Edited on 11/30/2005 by guy]

[Edited on 11/30/2005 by guy]

[Edited on 11/30/2005 by guy]

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[Edited on 12/2/2005 by guy]

woelen - 30-11-2005 at 01:27

Yes I am talking about S2O8(2-), the peroxidisulfate ion. But still, this does not answer my question.

Why is the black nickel (III) (or nickel (IV)) oxide/hydroxide precipitate turned light green by hydrogen peroxide in alkaline solution. In order to be clear, I'll repeat the entire experiment:

1) Add a solution of NiSO4 to a solution of NaOH, such that there is quite some excess NaOH. This gives a light green precipitate of Ni(OH)2.
2) Add a solution of Na2S2O8. This immediately causes the light green precipitate to turn black. I think this is due to formation of either nickel (III) or nickel (IV) oxide/hydroxide. I've seen both explanations.
3) Add some H2O2 to the still very alkaline solution with the black precipitate. Quickly the precipitate becomes light green again. The liquid still is very alkaline.

A second experiment: Add some H2O2 to a green suspension of Ni(OH)2 in an alkaline solution of excess NaOH. No visible change at all can be observed. When more H2O2 is added, it remains the same.

I have the feeling there is something contradictory in stating that S2O8(2-) gives H2O2 in solution and at the same time, adding H2O2 makes the precipitate turn green again.

guy - 30-11-2005 at 18:32

It is because H2O2 is not a strong oxidizing agent in a basic solution, therefore it will not oxidize Ni(II) to Ni(III). H2O2 will exihibit its strong oxidizing abilities in an acid solution.

2H+ + H2O2 --> 2H2O E = 1.77V

And as for turning green, it is because H2O2 is a weak acid and shifts the equilibrium left again.

Ni(OH)2 + 2OH– + S2O8(2-) → NiO2 + 2SO42- + 2H2O

Adding acid, reacts with the Ni(OH)2 and shifts it back. And furthermore the hydrolysis of S2O8(2-) is significantly slower than the reduction of S2O8(2-) in this reaction.

Also, the changing of S+7 to S+6 is stronger than the oxidizing power of hydrogen peroxide. It is +2.42V.
So in this case, the peroxydisulfate is the main oxidizer. In the case otf Cu(OH)2, the H2O2 acid-base reaction is the main component.

[Edited on 12/1/2005 by guy]

[Edited on 12/1/2005 by guy]

woelen - 1-12-2005 at 01:40

I'm very persistant on this, because I'm still not convinced

. This is not to annoy you

, but I simply want to have this issue settled, until I totally understand it.

If I do the experiment with very concentrated NaOH with a little Ni(OH)2 in it, then adding Na2S2O8 makes the precipitate black. Adding a few drops of H2O2 again makes the precipitate green. But after this, the liquid still is very alkaline. I used a LOT of NaOH, so the concentration hardly changed with the small amounts of other reagents mixed in.

So, the argumentation of equilibrium shifting towards green Ni(OH)2 due to shift of pH does not hold.

I also did a counter experiment without adding H2O2, but dripping in HCl. When this is done, then the precipitate remains black, until a point is reached, where the liquid really becomes acidic. At that point the precipitate very slowly dissolves and a smell of chlorine can be observed.

So, with H2O2 a few drops of 10% H2O2 are sufficient to make the black precipitate green again. With 10% HCl a lot of acid was needed before the black precipitate disappears.

I also did another experiment by using a large excess amount of Na2S2O8, relative to the Ni(OH)2. The rationale behind that is that the Ni(OH)2 is quickly converted to the black NiO2/NiO(OH). After this, a lot of Na2S2O8 remains. If this slowly forms H2O2, then the black color should fade again and it should become light green again. Well, I've done this and left the test tube overnight, but it remained black.

Tonight I will do another test experiment. If Na2S2O8 indeed gives H2O2, due to hydrolysis, then an (acidic) solution of Na2S2O8 should turn an acidified solution of K2Cr2O7 deep blue. This is a well-known qualitative test for dichromate or peroxide. I'll keep you updated on this.

Another remark: S in S2O8(2-) does not have oxidation state +7. It still is +6. Only the two oxygens between the two sulphurs in this ion have oxidation state -1 instead of -2.

Are you sure that the redox potential is +2.42V under the conditions I have? I thought it is just over 2 V. But this is not essential for the question about the Ni(OH)2 --> NiO2 conversion

guy - 1-12-2005 at 20:09

Okay
I have another (I hope better) explanation.

To make peroxydisulfuric acid:
H2O2 + 2H+ + 2SO42- <------> S2O82- + 2H2O

If you add H2O2, the equilibrium is shifted back to S2O82-. Adding and acid (HCl for example) will also shift the equilibrium right, but more is needed because it takes 4x as much. I hope that this is a better explation. Also, try adding excess sulfate ions and see if this causes a similar effect.

AND ALSO
going bak to the main topic

, does anyone agree with my acid-base theory for the reaction of Cu(OH)2 with H2O2?
[Edited on 12/2/2005 by guy]

[Edited on 12/2/2005 by guy]

[Edited on 12/2/2005 by guy]

[Edited on 12/2/2005 by guy]

neutrino - 2-12-2005 at 03:28

That should make peroximonosulfufic (Caro's) acid:

H2O2 + H2SO4 --> H2SO5 + H2O

woelen - 2-12-2005 at 14:53

Quote:

Originally posted by guy
Okay
I have another (I hope better) explanation.

To make peroxydisulfuric acid:
H2O2 + 2H+ + 2SO42- <------> S2O82- + 2H2O

If you add H2O2, the equilibrium is shifted back to S2O82-. Adding and acid (HCl for example) will also shift the equilibrium right, but more is needed because it takes 4x as much. I hope that this is a better explation. Also, try adding excess sulfate ions and see if this causes a similar effect.

I have done some more experimenting with persulfate and H2O2.

First the test with acidified dichromate. Addition of persulfate does not result in formation of the deep blue peroxo compound of chromium. Not even a trace of this can be observed. A counter experiment with adding a single drop of 1% H2O2 immediately results in formation of deep blue CrO5. So, we can safely conclude that persulfate does not hydrolyse to hydrogen peroxide in acidic media.

I did the same test in alkaline environment. Adding H2O2 to a yellow solution of K2CrO4 results in formation of a brown complex (CrO8(3-)). This reaction also is quite sensitive. When persulfate is added to an alkaline solution of K2CrO4, then no change of color can be observed, nothing at all, even when a lot of persulfate is added. So, we can also safely conclude that persulfate does not form hydrogen peroxide in alkaline environments.

Besides that, I also read about Caro's acid, as neutrino mentions. This indeed is the equilibrium reaction, not the persulfate. Persulfate is formed electrolytically from sulfate:

SO4(2-) --> SO4(-) + e (at anode)
2SO4(-) --> (-)O3SOOSO3(-), which is persulfate

Finally, a remark on the 4 times strength of H2O2 relative to HCl. Even if this were the case, it cannot explain my observations. With H2O2 (10%) a single drop suffices to remove the black compound, with HCl, several ml are needed. I think a factor of 50 is more close to what I observe than a factor of 4.

Quote:

AND ALSO
going bak to the main topic

, does anyone agree with my acid-base theory for the reaction of Cu(OH)2 with H2O2?

No, I think that the mechanism is different. I agree with that a peroxide may be formed, such as CuO2. I, however, think it is a coordination reaction and not an acid-base reaction. The Cu(OH)2 is not real Cu(OH)2, it is a very complicated macro-structure with Cu-ions, OH(-) ligands and H2O ligands forming a complex structure, with empirical formula Cu(OH)2.nH2O. As in Fenton's reagent, I believe that H2O2 ligands replace H2O or OH(-) ligands, followed by subsequent loss of hydroxyl radicals. This can lead to formation of CuO2.nH2O.

guy - 2-12-2005 at 16:43

From
http://astronautix.com/props/h2o2.htm

Quote:

Hydrogen peroxide is manufactured commercially by several processes. Inorganic processes employ the electrolysis of an aqueous solution of sulphuric acid or acidic ammonium bisulphate, followed by hydrolysis of the peroxydisulfate which is formed.

There are also more sites which support this.

Quote:

That should make peroximonosulfufic (Caro's) acid:

I said

H2O2 + 2H+ + 2SO42- <------> S2O82- + 2H2O

This is reaction is definately valid for the formation of a peroxydisulfate.

And, for the test with persulfate in acidic solution will not form H2O2, because that will the reaction form more S2O8(2-). Alkalne enviroment will cause H2O2 to decompose to O2 and H2O. Try in a neutral enviorment, and see if this causes any change.

Also, try adding excess sulfate ions to see if the NiO2 will redissolve.
*****************************
And on the Cu(OH)2 + H2O2 subject.

You believe the reaction is
Cu(H2O)3(OH)2 + H2O2 ---> Cu(H2O)3(H2O2)(OH)]- ---> ?CuO2 + .OH?

Can you explain more on the mechanism of this reaction?

And can you explain how this reaction is different from Ca(OH)2 + 2H2O2 --> CaO2 + 2H2O?

[Edited on 12/3/2005 by guy]

[Edited on 12/3/2005 by guy]

[Edited on 12/3/2005 by guy]