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Fernald
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Reaction of Cu(II), NH4OH, and H2O2
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?
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Boffis
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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+.
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AJKOER
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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.
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Eddygp
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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.
there may be bugs in gfind
[ˌɛdidʒiˈpiː] IPA pronunciation for my Username |
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AJKOER
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| 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-
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gatosgr
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| 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. |
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- |
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]
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blogfast25
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There's no such thing as 'ammonium hydroxide'.
Ammonia is a weak base (pK<sub>b</sub> = 4.75) that is highly soluble in water.
In solution it is in equilibrium with water, acc.:
NH<sub>3</sub>(aq) + H<sub>2</sub>O(l) < === > NH<sub>4</sub><sup>+</sup>(aq) +
OH<sup>-</sup>(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
(pK<sub>b</sub> + 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]<sup>-</sup> ≈ [NH<sub>4</sub><sup>+</sup>], so [NH<sub>4</sub><sup>+</sup>] =
10<sup>-2.4</sup> = 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
(NH<sub>3</sub>(aq)).
'NH<sub>4</sub>OH' is a very old and misleading representation of an aqueous solution of ammonia.
Get this wrong and any pathways involving 'NH<sub>4</sub>OH' are merely fanciful baloney.
And ascorbic acid doesn't reduce Cu(+2) to Cu(0), IIRW.
[Edited on 5-5-2015 by blogfast25]
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gatosgr
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thanks for reminding me first year formulas 
ascorbic acid does reduce the metal complex though....
[Edited on 5-5-2015 by gatosgr]
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gdflp
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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.
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Sulaiman
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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)
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DraconicAcid
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| 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.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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gatosgr
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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]
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MrHomeScientist
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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:
CuSO<sub>4</sub> + C<sub>6</sub>H<sub>8</sub>O<sub>6</sub> == Cu +
C<sub>6</sub>H<sub>6</sub>O<sub>6</sub> + H<sub>2</sub>SO<sub>4</sub>
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gatosgr
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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]
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blogfast25
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| 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 ?
(please don't hurt me, I'm a noob) |
DraconicAcid is correct.
For instance, the precipitation of insoluble hydroxides, e.g. M(OH)<sub>z</sub> with ammonia solution is best (and most easily) written
as:
M<sup>z+</sup>(aq) + z NH<sub>3</sub>(aq) + z H<sub>2</sub>O(l) === > M(OH)<sub>z</sub>(s) + z
NH<sub>4</sub><sup>+</sup>(aq)
The equilibrium mentioned in my post above explains why amphoteric aluminium hydroxide with ammonia does not from aluminates
(Al(OH)<sub>4</sub><sup>-</sup> , as do stronger bases like
NaOH, KOH.
[Edited on 5-5-2015 by blogfast25]
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papaya
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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.
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gatosgr
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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...
Anyway if you increase the ph up to 12 you can also reduce iron compounds...
[Edited on 5-5-2015 by gatosgr]
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blogfast25
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| 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. 5<sup>10</sup> for instance).
Edit: that formula doesn't make any sense at all.
[Edited on 5-5-2015 by blogfast25]
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gatosgr
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here it is
http://chem.xmu.edu.cn/teach/fxhx/fxhxwenxian/Biochemists%20...
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blogfast25
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So basically at pH = 7, E<sub>red</sub> = + 0.062 V for ascorbic acid. Acc. that paper.
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gatosgr
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What is acc. that paper?
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blargish
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| Quote: Originally posted by blogfast25 |
DraconicAcid is correct.
For instance, the precipitation of insoluble hydroxides, e.g. M(OH)<sub>z</sub> with ammonia solution is best (and most easily) written
as:
M<sup>z+</sup>(aq) + z NH<sub>3</sub>(aq) + z H<sub>2</sub>O(l) === > M(OH)<sub>z</sub>(s) + z
NH<sub>4</sub><sup>+</sup>(aq)
The equilibrium mentioned in my post above explains why amphoteric aluminium hydroxide with ammonia does not from aluminates
(Al(OH)<sub>4</sub><sup>-</sup>, 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<sup>-</sup>(aq)
As you said, it's a very weak dissociation, but the presence of the metal ion M<sup>z+</sup> 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.
BLaRgISH
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gatosgr
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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]
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DraconicAcid
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| Quote: Originally posted by blargish |
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<sup>-</sup>(aq)
As you said, it's a very weak dissociation, but the presence of the metal ion M<sup>z+</sup> forces the precipitation of the insoluble
hydroxide, and forces the ammonia to dissociate further, pushing the above equilibrium to the right. |
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
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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blogfast25
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According to that paper.
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