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Author: Subject: Blue-Orange... Or not :/
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[*] posted on 22-1-2015 at 03:28
Blue-Orange... Or not :/


I tried searching for the recent 'What the *** did I just make-thread but I couldn't find it :s.

I tried a reaction today that should involve some nice colour changes.

- I made 120ml of a 1M solution of Seignette Salt(sodium potassium tartrate) and added to that 80ml of 3% hydrogen peroxide.
- I heated the mixture up to 50 degrees Celsius and added 2ml 1M copper sulphate solution to the mix. The mixture started to fizz, turned blue and heated up from 50 to 60 degrees.
- I heated the mixture further to 70 degrees.
According to my book, the solution should turn orange as copper(I) oxide precipitates, but it didn't, it started out blue and halfway it turned green.
- When the temperature hit 70 degrees, I added 80 ml of 3% hydrogen peroxide.
Instead of the anticipated blue and fizzing, the solution turned a dark, muddy brown, which slowly transitioned into a dark green precipitate.
This was not what I expected to happen. I followed the instructions perfectly and it went completely NOT according to plan.
When the precipitate sank to the bottom, the leftover solution was clear like water.

Someone else's youtube video of the attempted reaction. This is what should have happened..

Can anyone tell me what went wrong?

Source: Roesky & Möckel, Chemische Kabinettstucke, p. 67-68.


Green precipitate after adding H2O2.JPG - 3.8MB

PS My own video of the precipitate when I turned off the stirring and it started to cool down. It looked funny :P




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[*] posted on 22-1-2015 at 03:41


I can't help you with your reaction. It looks like a very interesting one by the way.
The thread you were looking for is here: http://www.sciencemadness.org/talk/viewthread.php?tid=33135#...
You might ask for this thread to be merged.
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[*] posted on 22-1-2015 at 03:57


Threads can be merged if a mod sees this :)



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[*] posted on 22-1-2015 at 04:20


Does the book really call for adding hydrogen peroxide to the mix? The nice orange precipitate you mention is hydrous cuprous oxide, Cu2O.nH2O.

I have done this reaction myself with alkaline tartaric salts, alkaline glycerol or alkaline glucose to which I added a solution of copper sulfate. First the solution turns deep blue, one day later it is colorless and at the bottom you have a nice orange/red precipitate of Cu2O.

More spectacular is the reaction with ascorbic acid (vitamin C). Dissolve some vitamin C in water. Dissolve some NaOH in a separate amount of water. Dissolve some copper sulfate in a third amount of water. Add the solution of NaOH to the solution of ascorbic acid and swirl. Next, add the solution of copper sulfate to the other solution. You get a dark green precipitate, which quickly turns beautifully yellow/orange. This yellow/orange precipitate is hydrated Cu2O.

Cu2O is nice stuff. The more hydrated, the more yellow it is. There is no precise stoichiometry of hydration, best description is Cu2O.nH2O, with n ranging from 0 to quite high values (e.g. 5). For n = 0, the material is brick red. For high values of n, the material is bright yellow. Any color between brick red and yellow (all kinds of shades of red, orange, ochre) are possible.

Have a look at this page in the section about copper in oxidation state +1:

http://woelen.homescience.net/science/chem/solutions/cu.html

[Edited on 22-1-15 by woelen]




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[*] posted on 22-1-2015 at 04:52


Quote: Originally posted by woelen  
Does the book really call for adding hydrogen peroxide to the mix? The nice orange precipitate you mention is hydrous cuprous oxide, Cu2O.nH2O.



It suggests so, it says:
Heat up 120 ml potassium sodium tartrate solution and 80ml hydrogen peroxide solution to 50 degrees.
Add 2ml copper sulphate solution and if necessary, heat further to 70 degrees.
Gas will evolve from the solution, which will make it foam. Almost simultaneously the solution will turn orange as copper(I) oxide forms.
Add 80ml hydrogen peroxide again, the solution will become green/blue, and shortly after that it will turn orange again while foaming.
You can keep adding hydrogen peroxide to the mixture to repeat the reaction. The reaction becomes more 'violent' with each addition.

Quite a literal translation from Dutch. I'll add a photo of the book for the Dutch-speaking people that want to read the exact text themselves.




Boek kleurreactie.jpg - 295kB




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[*] posted on 22-1-2015 at 07:44


Hmm... this does not look really good. The experiment is critical and easily fails.
There also is a big error in the book. At the top of the right page the book says that in the alkaline environment copper(I) oxide is formed according to the following equation:

2Cu(2+) + 2OH(-) --> Cu2O + H2O

This is nonsense. From copper(II) you cannot get copper(I) with only water and hydroxide. The equation is not even balanced correctly (at the left there is net charge +2 and at the right there is charge 0). So, the books shown no real understanding of this experiment.

The first reaction equation also looks very dubious. I do not believe that CO2 and OH(-) ions can appear in a reaction in aqueous solution at the same time and that the OH(-) can exist besides CO2 which escapes as gas. You would get carbonate or bicarbonate. In this particular case I would expect bicarbonate to form, because there is excess CO2. Excess CO2 can escape from the warm liquid as gas. This would hardly lead to rise of pH, because excess CO2 is formed.

I can imagine that this experiment works, but only under fairly well controlled conditions (temperature must be fairly precisely controlled). The real idea behind this experiment must be that copper(II) is reduced by the tartrate at high pH, but this reaction is not immediate. Copper(I) on the other hand is oxidized by peroxide very quickly, while tartrate is oxidized more slowly by peroxide. So, when hydrogen peroxide is added to the tartrate with orange copper(I) oxide, then you get immediate oxidation of copper(I) to copper(II), giving the blue/green color. When copper(I) is used up any excess H2O2 oxidizes tartrate, giving CO2 and bicarbonate. When all H2O2 is used up as well, then the tartrate can reduce copper(II) to copper(I) again, leading to the orange color again. The strong foaming is due to catalytic decomposition of H2O2 and only marginally due to formation of CO2.

The temperature is the critical factor in this experiment. It must be such that reduction of the copper(II) is much slower than the reaction between copper(I) and H2O2, but on the other hand, it must not be too slow, otherwise you do not see the quick transition from blue/green to orange. In practice this can be quite tricky. I personally do not like experiments, which are not robust with respect to changes in one of the parameters, controlling the experiment (e.g. concentration, total amount used, temperature).

Another critical issue is that the pH may not rise sufficiently. I expect that the oxidation of tatrate by H2O2 does not lead to rise of pH. You could try the experiment with some NaOH added to the tartrate as well (e.g. 1 M tartrate and 0.5 M hydroxide) at the start of the experiment. For the rest follow the procedure of the book. Maybe this leads to better results.




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[*] posted on 22-1-2015 at 11:16


I know it's not the same, but there is another rather nice color-changing demonstration that also results in yellow cuprous oxide formation. The method is similar to the one woelen described(and I believe he has also done this one too), only the order in which the reactants are mixed is different. Ascorbic acid is added directly to boiling copper sulfate solution, which precipitates a powder of metallic copper while simultaneously forming a green solution if excess CuSO4 is used and a yellow one if excess ascorbic acid is used(I'm not sure of the stoichiometry involved). Adding dilute sodium hydroxide to this resulting solution causes a bright yellow precipitate of hydrous copper(I) oxide to form. If it is not kept under water or if it is heated too greatly, it turns to a dingier color and finally to reddish-brown, losing its water of hydration.

If you're looking for a colorful demonstration for any of your students, this has the advantage of occurring more quickly and there's more to look at during the process.




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[*] posted on 22-1-2015 at 12:45


Quote: Originally posted by woelen  
Hmm... this does not look really good. The experiment is critical and easily fails.
There also is a big error in the book. At the top of the right page the book says that in the alkaline environment copper(I) oxide is formed according to the following equation:

2Cu(2+) + 2OH(-) --> Cu2O + H2O

This is nonsense. From copper(II) you cannot get copper(I) with only water and hydroxide. The equation is not even balanced correctly (at the left there is net charge +2 and at the right there is charge 0). So, the books shown no real understanding of this experiment.

The first reaction equation also looks very dubious. I do not believe that CO2 and OH(-) ions can appear in a reaction in aqueous solution at the same time and that the OH(-) can exist besides CO2 which escapes as gas. You would get carbonate or bicarbonate. In this particular case I would expect bicarbonate to form, because there is excess CO2. Excess CO2 can escape from the warm liquid as gas. This would hardly lead to rise of pH, because excess CO2 is formed.

I can imagine that this experiment works, but only under fairly well controlled conditions (temperature must be fairly precisely controlled). The real idea behind this experiment must be that copper(II) is reduced by the tartrate at high pH, but this reaction is not immediate. Copper(I) on the other hand is oxidized by peroxide very quickly, while tartrate is oxidized more slowly by peroxide. So, when hydrogen peroxide is added to the tartrate with orange copper(I) oxide, then you get immediate oxidation of copper(I) to copper(II), giving the blue/green color. When copper(I) is used up any excess H2O2 oxidizes tartrate, giving CO2 and bicarbonate. When all H2O2 is used up as well, then the tartrate can reduce copper(II) to copper(I) again, leading to the orange color again. The strong foaming is due to catalytic decomposition of H2O2 and only marginally due to formation of CO2.

The temperature is the critical factor in this experiment. It must be such that reduction of the copper(II) is much slower than the reaction between copper(I) and H2O2, but on the other hand, it must not be too slow, otherwise you do not see the quick transition from blue/green to orange. In practice this can be quite tricky. I personally do not like experiments, which are not robust with respect to changes in one of the parameters, controlling the experiment (e.g. concentration, total amount used, temperature).

Another critical issue is that the pH may not rise sufficiently. I expect that the oxidation of tatrate by H2O2 does not lead to rise of pH. You could try the experiment with some NaOH added to the tartrate as well (e.g. 1 M tartrate and 0.5 M hydroxide) at the start of the experiment. For the rest follow the procedure of the book. Maybe this leads to better results.


What you're saying does make sense. Now that I read it, I can see the nonsense part there.
I thought I was quite precise with my temperature. I may have been max. 1 degree off, but I did not use a digital thermometer, so I can imaging the measurement wasn't that precise.

If even I can't do it, then a 15-year-old with practically no experience can't do it for sure (I tested it for a student, so I could explain it to her).
I'll try the ascorbic acid one tomorrow. Thanks for the explanation and advice :)

[Edited on 22-1-2015 by Jylliana]




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