Sciencemadness Discussion Board - Reaction of Cu(II), NH4OH, and H2O2

Reaction of Cu(II), NH4OH, and H2O2

Fernald - 19-7-2013 at 15:16

Hello everyone. First post to these forums.

A solution of 3% H2O2 and household NH4OH is prepared in a 10:1 ratio. Then a few drops of copper acetate are added. Immediately the blue copper solution turns dark reddish brown. After about five minutes the solution has bubbled up about 3cm above the liquid surface. The solution is now full of a cloudy brown precipitate.

What has happened here? I'm thinking that the H2O2 is releasing O2 gas, the bubbles. But what has happened to the copper and the ammonia? Is it becoming a Cu(I) oxide?

Boffis - 19-7-2013 at 16:21

Yes, the H2O2 is acting as a reducing agent to reduce the copper ammonia complex to cuprous oxide and the ammonia is liberated again. H2O2 also acts as a reducing agent with substances like potassium permanganate too. It is interesting to contrast the different reactions between copper and cobalt under the same condition, the later is oxidized to Co3+.

AJKOER - 30-7-2013 at 17:52

Add more NH3 and you should form a royal blue copper ammine complex: [Cu(NH4)4(H2O)2]++

What is also interesting is that by adding Ascorbic acid (Vitamin C) to a solution at a pH 6~7 between 60 to 70 C, the solution will form a very fine Cu/Cu2O suspension.

Eddygp - 3-8-2013 at 07:15

Quote: Originally posted by AJKOER

Add more NH3 and you should form a royal blue copper ammine complex: [Cu(NH4)4(H2O)2]++

What is also interesting is that by adding Ascorbic acid (Vitamin C) to a solution at a pH 6~7 between 60 to 70 C, the solution will form a very fine Cu/Cu2O suspension.

Yes, when I read the title I thought that this was what happened. Actually, you need no H2O2 for the tetraammine complex. The H2O2 anionthe (if the complex had formed) to form NH3 and water.

AJKOER - 3-8-2013 at 13:52

Quote: Originally posted by Eddygp

....
Yes, when I read the title I thought that this was what happened. Actually, you need no H2O2 for the tetraammine complex. The H2O2 anion the (if the complex had formed) to form NH3 and water.

I know of one pathway that definitely requires the presence of free oxygen (or H2O2). Now, O2 is not only available from exposure to air, it is also dissolved in water. Here is reference "Kinetics and Mechanism of Copper Dissolution In Aqueous Ammonia", available at http://academia.edu/292096/Kinetics_and_Mechanism_of_Copper_... the dissolution mechanism is, in my opinion, best described as electrochemical in nature. To quote from the paper:

"The generally accepted theory on the corrosion of a metal (Evans[18]), is that when a metal comes into contact with an aqueous salt solution to which oxygen is accessible, oxygen takes up electrons at one part of the surface (the cathodic zone) while the metal gives it up at another (the anodic zone). In this way the attack of the metal proceeds at an appreciable rate at room temperature. These principles are well established and they were successfully demonstrated in many cases, e.g. the dissolution of zinc in sodium chloride solution in contact with air, or gold in a cyanide solution saturated with air, Thompson [19]"

Here are some of the actual equations cited by the author occurring in the overall electrochemical reaction:

1. 1/2 O2 + H2O + 2 e- ---> 2 OH- (Cathodic reduction of O2 at surface of the Copper)

2. Cu + 4 NH3 ---> [Cu(NH3)4]2+ + 2 e- (Anodic dissolution of Cu by a complexing agent)

Overall:

Cu + 4 NH3 + 1/2 O2 + H2O ---> [Cu(NH3)4]2+ + 2 OH-

gatosgr - 5-5-2015 at 05:44

Quote: Originally posted by AJKOER

Quote: Originally posted by Eddygp

....
Yes, when I read the title I thought that this was what happened. Actually, you need no H2O2 for the tetraammine complex. The H2O2 anion the (if the complex had formed) to form NH3 and water.

Here is schweizer reagent

Cu(CH3COO)2+2OH-=Cu(OH)2+2CH3COO

Cu(OH)2+4NH4OH-> [Cu(NH3)4(H2O)2](OH)2+2H2O

if that's the case what is the preferred pathway for this reaction? amonnium hydroxide has all the oxygens needed for the complexing reaction to happen.

When you add ascorbic acid the copper percipitates BUT are you sure it forms CuO too? excess H2O2 could oxidize copper...
The reaction is like this I think:

[Cu(NH3)4(H2O)2](OH)2 -> [Cu(NH3)4(H2O)2](2+) + 2OH-

H2A+[Cu(NH3)4(H2O)2](2+) + 2OH- -> Cu + 4NH3 + A + 6H2O

is this inner sphere or outer sphere?? if it was outer sphere then it should work with sugars as well which leads me to think it's inner sphere

[Edited on 5-5-2015 by gatosgr]

[Edited on 5-5-2015 by gatosgr]

blogfast25 - 5-5-2015 at 08:55

Quote: Originally posted by gatosgr

amonnium hydroxide has all the oxygens needed for the complexing reaction to happen.

There's no such thing as 'ammonium hydroxide'.

Ammonia is a weak base (pKb = 4.75) that is highly soluble in water.

In solution it is in equilibrium with water, acc.:

NH3(aq) + H2O(l) < === > NH4+(aq) + OH-(aq)

Kb = [NH4+][OH-]/[NH3]. pKb = - log Kb

Use the simplified Hasselbalch formula to get an idea of the degree of protonation of the ammonia. For a weak and fairly dilute base: pOH ≈ 1/2 (pKb + pC) (C = molar concentration of base).

For e.g. for C = 1 mol/L (1 M), pC = 0, so pOH = 0.5 x 4.75 = 2.4

With pH + pOH = 14, pH = 14 - 2.4 = 11.6.

Also, [OH]- ≈ [NH4+], so [NH4+] = 10-2.4 = 0.004 M.

So only about 0.004/1 x 100 % = 0.4 % off the dissolved ammonia has been protonated, the rest is present as 'free' ammonia (NH3(aq)).

'NH4OH' is a very old and misleading representation of an aqueous solution of ammonia.

Get this wrong and any pathways involving 'NH4OH' are merely fanciful baloney.

And ascorbic acid doesn't reduce Cu(+2) to Cu(0), IIRW.

[Edited on 5-5-2015 by blogfast25]

gatosgr - 5-5-2015 at 09:12

thanks for reminding me first year formulas

ascorbic acid does reduce the metal complex though....

[Edited on 5-5-2015 by gatosgr]

gdflp - 5-5-2015 at 09:16

Quote: Originally posted by blogfast25

And ascorbic acid doesn't reduce Cu(+2) to Cu(0), IIRW.

Actually, it does http://www.sciencemadness.org/talk/viewthread.php?tid=2654 I've found personally that it needs some heat, but it certainly does work.

Sulaiman - 5-5-2015 at 09:53

blogfast25,
from your post above, you clearly know much more about this than me ... but ...
is it likely that reactions occur mainly with the 0.4% 'ammonium hydroxide'
which by Le Chatalier's principle is rapidly replaced with the less reactive NH3(aq) ?
so 'ammonium hydroxide' should be in the chemical equation rather than the less reactive ammonia ?

(please don't hurt me, I'm a noob)

DraconicAcid - 5-5-2015 at 10:03

Quote: Originally posted by Sulaiman

blogfast25,
from your post above, you clearly know much more about this than me ... but ...
is it likely that reactions occur mainly with the 0.4% 'ammonium hydroxide'
which by Le Chatalier's principle is rapidly replaced with the less reactive NH3(aq) ?
so 'ammonium hydroxide' should be in the chemical equation rather than the less reactive ammonia ?

No. Ammonia is perfectly capable of acting as a base, and doesn't need to react with water first to form hydroxide. That kind of thinking is a remnant of Arrhenius' theory of acid-base chemistry, where he said that only hydroxide could act as a base. This is why Arrhenius' theory is only used in beginning chemistry classes, and never by real chemists.

gatosgr - 5-5-2015 at 10:04

Increasing the ionic strength of the solution and the available hydrogen ions will increase the rate of the of ascorbic acid oxidation reaction.

Stop spamming with NH4OH.

[Edited on 5-5-2015 by gatosgr]

MrHomeScientist - 5-5-2015 at 10:33

Yes I've done the ascorbic acid reduction myself a number of times, it does certainly produce a copper-looking powder. I'll admit to not having tested this to definitively prove it is copper, though. Simply add a solution of Vitamin C to a solution of copper sulfate, and the mixture instantly turns emerald green. Heating to near boiling causes a fine coppery-colored precipitate to quickly fall out. The green color in the solution still remains, interestingly. The equation I've been going by is:
CuSO4 + C6H8O6 == Cu + C6H6O6 + H2SO4

gatosgr - 5-5-2015 at 10:45

Well bring out a litmus paper and measure the ph, you can work out if H2OS4 is formed like this. The redox potential of AA is 0.062V at ph 7 which is very little compared to 0.39V so I though about make a buffer solution to keep the ph constant at 7.

[Edited on 5-5-2015 by gatosgr]

blogfast25 - 5-5-2015 at 12:39

Quote: Originally posted by Sulaiman

DraconicAcid is correct.

For instance, the precipitation of insoluble hydroxides, e.g. M(OH)z with ammonia solution is best (and most easily) written as:

Mz+(aq) + z NH3(aq) + z H2O(l) === > M(OH)z(s) + z NH4+(aq)

The equilibrium mentioned in my post above explains why amphoteric aluminium hydroxide with ammonia does not from aluminates (Al(OH)4-

, as do stronger bases like NaOH, KOH.

[Edited on 5-5-2015 by blogfast25]

papaya - 5-5-2015 at 12:39

Here's an article about catalytic H2O2 decomposition with Cu(II)/NH3, where formation of brown precipitate is also discussed.
https://www.jstage.jst.go.jp/article/bcsj1926/47/5/47_5_1162...
However I understood not much from there as what is the composition of that precipitate, anyone can explain? (seem not an oxide).
In my own experiments (if I remember correctly, too long ago) brown precipitate forms when you add too much peroxide at once, and doesn't form (correct if wrong) if you add peroxide slowly.

gatosgr - 5-5-2015 at 14:00

H2O2 should oxidize copper.

Well if you want to reduce a metal with lower redox potential than copper you can use this equation to find the ph dependent standard redox potential of ascorbic acid which is a diprotic acid

0.39+0.05916/2*log((10^(-ph)^2+10^(-4.1)*10^(-ph)+10^(-4.1)*10^(-11.79))

I still think about using a buffer solution has anyone used a buffer for electrochemical reactions before? I dont have a digital ph meter so I'll have to try a buffer tablet or smthing... :mad:

Anyway if you increase the ph up to 12 you can also reduce iron compounds...

[Edited on 5-5-2015 by gatosgr]

blogfast25 - 5-5-2015 at 14:08

Quote: Originally posted by gatosgr

I still think about using a buffer solution has anyone used a buffer for electrochemical reactions before?

Yes. What pH are you looking for?

Where did that equation come from? (Try and use the superscript tags, it makes it more readable. 510 for instance).

Edit: that formula doesn't make any sense at all.

[Edited on 5-5-2015 by blogfast25]

gatosgr - 5-5-2015 at 14:10

here it is

http://chem.xmu.edu.cn/teach/fxhx/fxhxwenxian/Biochemists%20...

blogfast25 - 5-5-2015 at 14:22

So basically at pH = 7, Ered = + 0.062 V for ascorbic acid. Acc. that paper.

gatosgr - 5-5-2015 at 14:25

What is acc. that paper?

blargish - 5-5-2015 at 14:55

Quote: Originally posted by blogfast25

, as do stronger bases like NaOH, KOH.

[Edited on 5-5-2015 by blogfast25]

Wait, correct me if I'm wrong, but in the case you mention, isn't the dissociation of the ammonia integral to the proceeding of the reaction? So the ammonia dissociates in water via:

NH₃ + H₂O <=> NH₄⁺(aq) + OH-(aq)

As you said, it's a very weak dissociation, but the presence of the metal ion Mz+ forces the precipitation of the insoluble hydroxide, and forces the ammonia to dissociate further, pushing the above equilibrium to the right.

Technically, would it not be the so-called "ammonium hydroxide" doing the work? I understand that ammonia by itself acts as a base (ie the formation of ammonium chloride smoke when exposed to HCl), but I would guess for such reactions with ammonia in aqueous solution the formation of "ammonium hydroxide", if you want to put it that way, is part of the mechanism.

As for the case with aluminates, I would assume that no precipitation is driving the ammonia dissociation forward as in the case of the insoluble hydroxide, thus the ammonia remains undissociated and no reaction occurs.

gatosgr - 5-5-2015 at 15:12

the complexing ion also has an effect on standard reduction potential for the oxidizing metal ion with it's formation constant similar to AA's

[Edited on 5-5-2015 by gatosgr]

DraconicAcid - 5-5-2015 at 15:20

Quote: Originally posted by blargish

No, because metal ions in solution are hydrated. A generic metal(II) ion in water is better represented as [M(H2O)6]2+.

[M(H2O)6]2+ + 2 NH3 -> M(OH)2 + 2 NH4(+) + 4 H2O

blogfast25 - 5-5-2015 at 16:49

Quote: Originally posted by gatosgr

What is acc. that paper?

According to that paper.

blogfast25 - 5-5-2015 at 16:57

Quote: Originally posted by DraconicAcid

No, because metal ions in solution are hydrated. A generic metal(II) ion in water is better represented as [M(H2O)6]2+.

[M(H2O)6]2+ + 2 NH3 -> M(OH)2 + 2 NH4(+) + 4 H2O

The solvated metal cation as a weak acid reacting with the weak base NH₃? Interesting. I've always seen it more as a case of ligand exchange (H2O < == > OH-

. Evidence for your pathway?

[Edited on 6-5-2015 by blogfast25]

gatosgr - 6-5-2015 at 01:54

So if that's the case the reduction potential for Cu(NH3)4 for 1 molar solution and 0.1 M for ammonia

Cu2+ +4NH3-> Cu(NH3)4(2+)

stability constant of formation for metal complex Cu(NH3)4(2+)
β=[Cu(NH3)4]/[Cu2+][NH3]^⁴

http://www.vaxasoftware.com/doc_eduen/qui/kfcomplejos.pdf

β=1.1x10^13

Reduction of Cu

Cu2+ +2e -> Cu E0=0.34V

Nernst:

E=E0-RT/2Flog(αCu/(αCu2+)(ae2-)^2)<=>E=E0-RT/2Flog(1/(αCu2+))

[Cu2+]=β[NH3]^4/[Cu(NH3)4]

E=0.39+0.0592/2log([Cu(NH3)4]/(β[NH3]^4))=0.11V

Is that right?

[Edited on 6-5-2015 by gatosgr]

DraconicAcid - 6-5-2015 at 02:09

Quote: Originally posted by blogfast25

Quote: Originally posted by DraconicAcid

No, because metal ions in solution are hydrated. A generic metal(II) ion in water is better represented as [M(H2O)6]2+.

[M(H2O)6]2+ + 2 NH3 -> M(OH)2 + 2 NH4(+) + 4 H2O

The solvated metal cation as a weak acid reacting with the weak base NH₃? Interesting. I've always seen it more as a case of ligand exchange (H2O < == > OH-

. Evidence for your pathway?

[Edited on 6-5-2015 by blogfast25]

Not off of the top of my head, but it is well-known that hydrated metal cations are acidic because the metal cation withdraws electron density from the water ligands.

gatosgr - 6-5-2015 at 02:25

Some metal salts are acidic some are basic according to lewis theory... ammonia doesnt need water according to bronsted theory.. I was taught these first year along with arhenius theory.. they're ok for beginners I guess.. according to lewis theory the amount of protonation depends on electronegativity of the metal.

That chinese paper is weird it doesnt cancel the ratios... http://chem.xmu.edu.cn/teach/fxhx/fxhxwenxian/Biochemists%20...

https://ch301.cm.utexas.edu/help/ch302/ab/fracspecies.pdf

The fraction of ascorbic acid and dehydroascorbic isnt that the same as α=γ(c/c0)??

-0.05916/2log(fH2A/D) why is this here it's still dependent on H+ ... how is this solved ?

nobody knows?

I found the graph for ascorbic acid:

if you read in the literature ascorbic acid is used for iron and cobalt compounds as well..

So this is why you needed to boil the water for the AA to reduce Cu2+ ΔG=ΔH-ΤΔS.. now you can try with adjusting the ph, ALTHOUGH some metal complexes have ph dependent E0 as well.

[Edited on 6-5-2015 by gatosgr]

blogfast25 - 6-5-2015 at 06:51

Quote: Originally posted by DraconicAcid

Not off of the top of my head, but it is well-known that hydrated metal cations are acidic because the metal cation withdraws electron density from the water ligands.

I was also taught that some solvated cations behave like weak acids because of the repulsion exerted by the central electrical force on the protons:

[M(H2O)6]z+ + H2O < === > [M(H2O)5(OH)](z-1)+ + H3O+

Especially for z = 2, 3, ... and small ionic radius.

Quote: Originally posted by gatosgr

Some metal salts are acidic some are basic according to lewis theory... ammonia doesnt need water according to bronsted theory.. I was taught these first year along with arhenius theory.. they're ok for beginners I guess.. according to lewis theory the amount of protonation depends on electronegativity of the metal.

That chinese paper is weird it doesnt cancel the ratios... http://chem.xmu.edu.cn/teach/fxhx/fxhxwenxian/Biochemists%20...

https://ch301.cm.utexas.edu/help/ch302/ab/fracspecies.pdf

The fraction of ascorbic acid and dehydroascorbic isnt that the same as α=γ(c/c0)??

-0.05916/2log(fH2A/D) why is this here it's still dependent on H+ ... how is this solved ?

Does it have to be solved? You have an expression for Ered = f(pH) anyway.

Can it be expressed only in concentrations and equilibrium constants? I think so but it’s not worth the effort, I think. It's likely to lead to a polynomial [H+]3 + etc. Not neat. That's why they keep it implicit, IMO.

[Edited on 6-5-2015 by blogfast25]