Electrolysis Problem

[Theophrastus]

Greetings. Not long ago, I was conducting a simple electrolysis experiment, performing electrolysis of water, using a voltage of approximately 13- 15 volts, in a saturated aqueous salt solution. I chose to use a magnesium anode, and a copper cathode, to my surprise, a substance, faintly green in colouration, began to rise from one of the electrodes.

Fearing this was chlorine, I quickly disbanded this setup, building a different one, in which the reaction would take place in an erlenmeyer flask, with a holed stopper. Through the hole in the stopper, I placed a curved glass tube, which would release any gases made in the reaction, into a separate vessel. The wires connecting to the cathode and anode, were also inserted into the hole.

Soon, the reaction ensued, as a stream of bubbles vigorously rose from the anode. As time passed by, I quickly watched the solution change colour from a pale green- yellow, to a soft golden yellow.

However, upon my return to the room, the next time, the solution had become discolored and opaque, due to the presence of a dark brown precipitate. I waited a while for the solution to settle, to find two precipitates of varying density, layered upon each other, at the bottom of the flask. The lower one was a dark brown, while the upper one a dull orange. What could have gone wrong? What are these precipitates? Can they be of any use? Help with these questions would be most appreciated!

 

,Theophrastus

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In retrospect, I believe that the black substance, may be copper (II) oxide, from the anode, however I am unsure, and uncertain. Any ideas??

[cplusplus]

Hello,

 

I have had that happen to me also. I think that it is the decomposition of the copper as the electricity flows through it along with the water creating copper oxide.

[salter]

the yellow is CuCl4 2- obviously the 4 is supposed to be a subscript and the 2- a superscript but i can't figure out how to do that in this... NaCl dissasociates into Na+ Cl-. So at your + end you get OH- and Cl-. the OH- - 4e = 2H20 and 2 O2. Theoretically you would get CuCl2 (or Cl2) from the copper from the probe and Cl-; however, the abundance of electrons and additional Cl-'s can create CuCl3 - and CuCl4 2- as well. CuCl4 2- has a yellow color. CuCl3 - has a redish color, and CuCl2 has a bluish color. The ratio for yellow:red:blue should be dependant upon current and voltage

(number of electrons in the area... or perhaps surface area and concentration of salt water). when i did this with a car battery, i got a deep orange that looked like mango juice. here is a link to a wiki about copper chlorides

http://en.wikipedia.org/wiki/Copper(II)_chloride

 

also, The black is not copper oxide since, if you do the experement with both the anode and the diode as copper, all you get is the yellow/orange. i have done this myself and if you allow the experiment to run longer using solid copper rods, you will see that the copper erodes; therefore the orange/yellow must be a copper chloride. i would guess magnisium oxide for the black.

[UC]

salter, use the chem notations: [ce] CuCl4^2^- [/ce] for example, delivers [ce] CuCl4^2^- [/ce]

 

Magnesium hydroxide would be formed in aqueous solution, not the oxide, and it is white colored. The black may be nonreactive contaminants in the Mg though.

 

I also think you mean cathode instead of diode Very different things.

[Theophrastus]

To Salter: I doubt that the black chemical is magnesium oxide, due to the fact that magnesium oxide is white, however, the idea you gave me about the copper chloride, was certainly interesting, as I think that you were right in saying that the yellow colour of the solution was due to CuCl2, and I think the green colour, I noticed, earlier on, was CuCl. However, I doubt that my precipitate was CuCl or CuCl2, as most chlorides, (with the rare exceptions of Pb 2+, Ag+, Hg2 2+, etc.) are in fact soluble. Thanks for the earlier not though; copper(II) chloride has its uses, and I suppose I could extract it, were I to perform a similar experiment again, however, it seems I've sadly returned to the start. (Insert frowning face)

I still believe that the black is copper(II) oxide. (correct me if I'm wrong.)

(Sudden thought!) Hold on... couldn't one of the precipitates be a chlorine- oxygen compound, like a chlorate, hypochlorite, etc. No, but since I was using copper in its usual valency (II), wouldn't it be dichlorate, dichlorite. Personally, I know little about these compounds. Also, none of the precipitates can be magnesium, as aren't most magnesium compounds white? I also examined the magnesium anode, more carefully, and decided to finally note a thin layer of crusty, white powder upon it. Examining it thoroughly with a magnifying glass, I also found miniscule traces of a brownish gold residue. Could this perhaps be traces of my upper precipitate, that was left inscribed upon the anode? Any thoughts?!?

Edited March 20, 2009 by Theophrastus
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[salter]

to UC: did i seriously say diode? wow. I meant anode and cathode.

 

to Theo: you should tell us which one was positive and which was negative and what ends gave what.

 

If the copper was positive then Cl- and OH- gather around it, and that with the copper limits your possibilities. Assuming that the yellow is [ce]CuCl4^2^-[/ce] and the black comes from the magnisium end (since when i used both copper probes i only got the yellow/orange), the only possibilities of what the black one could be are combinations of H+, Na+, and whatever is in the magnesium. if the copper was negative and the magnesium was positive there is a whole range of other possibilities... you should try reversing the charges and tell us what you get then.

 

Since I did the experement with both copper probes and got no black... i would assume that somehow the black has to be related to the magnesium or addatives in salt or impurities in your magnesium.

but I could be wrong so lets go with it...

if the black was copper (II) oxide (again assuming that the copper was + since thats the only way you might get copper (II) oxide), then the Cl- goes unused and would make [ce]Cl2[/ce], correct? and you would have 2OH- + 2Cu - 2e = 2CuO + [ce]H2[/ce] in addition to the chlorine gas. So if you feel sick when you do this experiment... lol. in theory you would then get chlorine gas, hydrogen gas, and copper (II) oxide off of the +... leaving the yellow to the other side. you have Na+, H+ and Mg (assuming the magnesium is pure) on the - side and I don't see any possibilites... correct me if i'm wrong, but this rules out the copper (II) oxide idea... unless both the black and the yellow come off of the copper and I missed that somehow when I did it... you should perform the experiment again and take note of where they come off of. If you can't within the next three days or so i'll experiment a little to try to figure this out since i'll be home from vacation.

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also the white powder... I suppose it's possible that it was created before you applied the charges and is magnesium hydroxide. Once you apply a negative charge to the Mg you repel the OH-.

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Hmm... looking at the possibilities if you reversed the charges... hmm you could get chlorine gas and MgOH if the Mg was +, or you could get [ce]MgCl2(H2O)x[/ce] http://en.wikipedia.org/wiki/Magnesium_chloride and oxygen gas (4 OH- - 4e = 2[ce]H2O[/ce] + [ce]O2[/ce]). On the - (copper) you have Na+, H+ and Cu... hydrogen gas. doesn't seem to fit.

Edited March 20, 2009 by salter
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[Theophrastus]

My apologies, for ommitting the fact, but I am confident that I used a magnesium anode, and a copper cathode for the experiment. Any thoughts? Thanks, for the input, though! Quite interesting!

Edited March 20, 2009 by Theophrastus
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[salter]

All things considered, i think the most likely answer is that both copper (II) oxide and copper chlorides are coming off of the cathode, accounting for both colors. I would assume that the Mg doesn't actually react with anything so you just get [ce]H2[/ce] from the anode and the Na+ makes NaOH with an OH-. hmm... I would still say that the white powder is magnesium hydroxide... perhaps you could do a Ph test or something.

[Theophrastus]

Excellent idea, and you're right that it does seem strange that a white powder is laden upon the "anode." My only explanation is that either it is salt from the solution, or else magnesium hydroxide, as the excited magnesium might react with the water: Mg* + H2O > MgOH + O. I'll do a PH test tonight, if I can, then share my results tomorrow.

[salter]

If the salt water solution was concentrated it is possible that the white is salt (but that should be easy to identify) or what would make more sense to me would be just NaOH on it. Or a combination... the more I think about the MgOH idea the less likely i think it is. If you do find that it is basic, it just means it isn't salt not that it's MgOH since it could be NaOH as well. thinking about it a little more... i think the OH-'s would be more likely to just create [ce]H2O[/ce] and [ce]O2[/ce] than bond with the Mg; however, Na+ ions should be left over from the Cl-'s usage and want to bond. Then an H+ would be floating around and create additional hydrogen gas with little additional use of just 2e... The simplest way to solve this would be just see if the Mg has eroded at all. Maybe run the experiment longer to see. That coupled with the Ph would be pretty strong evidence one way or another.

 

If i was you I would go crazy with this and build an apparatus that gathered gasses seperately (this is what I did when I used both copper probes) using clear tubing and hot glue, then take the amounts of gasses produced at each end and figure out exactly how much of each thing is being produced with respect to each other and time. Could be a pain but... fun to attempt . and after that you could even figure out how battery power affects the rates and amounts of different compounds produced relative to each other and amounts of different copper (II) chlorides etc.

 

Also... there is a guy i know who goes by the name of Theo and speaks in the same style that you do. Do you live in Flagstaff, AZ by chance?

[Theophrastus]

I've conducted my PH test, this evening, reaching the conclusion, that it was in fact magnesium hydroxide, layered upon the magnesium electrode. Further proof of a reaction, was after the completion of the titration, I poured out my indicator solution, getting rid of any remaining salt, upon my metal, with acetic acid, I found that all across the magnesium anode, there were tiny pockmarks, where the ionized magnesium, reacted with the water, forming magnesium hydroxide. Once the salt was neutralized, the magnesium, lost in the reaction, was evident.

Good idea, with the gases! (the collection of them, I mean) I could give it a go this weekend, if I have enough free time on my hands. Thanks for the corrspondence to my question. I think, that if I'm unable to figure out how to separate the two precipitates, by the week's end, I'll simply filter both out, and start some testing. Flame tests, seeing if they react with particular compounds, if I get any clues towards what they may be. If I find out, I'll post my resolve, on this forum.

 

Ps: Sadly, I do not live in arizona, or even USA. Think further north, up in Canada. Its really quite a shame that there are so many limitations, in obtaining chemicals up here. Cheers!

Edited March 20, 2009 by Theophrastus
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[salter]

problem: Mg and Acetic acid react.

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Hmm... I suppose that what you found must be MgOH... considering that NaOH and Mg react . That rules out the NaOH idea... but the MgOH is likely from a reaction with the NaOH rather than the electrolysis itself.

[Theophrastus]

Of the first, yes, it is true that vinegar and acetic acid react, to produce hydrogen, and magnesium acetate, however, the time I placed it into the acid, was relatively short, and as such, the contamination would only be minor. And secondly, if what you say is true, there is a serious flaw to my theory, as magnesium, being less reactive, will not displace NaOH. Even though, this reaction, theoretically occurs, due to the magnesium ion's "excited" state, Na, being far more reactive, will following the creation of Magnesium Hydroxide, displace it. Also, I doubt a large amount of NaOH formed, as, I checked the PH of the solution itself, following my experiment, to find its PH to be between 7, and 8. Just in case, following the filtration process, I'll boil down the solution, and see, whether a change occurs.

[salter]

I looked it up and Mg and NaOH do react. wiki.answers.com/Q/What_happens_when_magnesium_reacts_with_sodium_hydroxide and http://www.jtbaker.com/msds/englishhtml/s4034.htm

 

This would create Magnesium Oxide... the idea that it comes from a reaction between the NaOH and the Mg makes more sense to me because then there isn't any problem because of the - charge repelling the OH-.

And the Ph would be much lower because much of the NaOH would have reacted with the Mg.

 

If i recall correctly... I think I did do a Ph test on my version with only copper probes and it came out to a 9 or 10.

 

You could test this because it releases additional [ce]H2[/ce] gas.

Edited March 21, 2009 by salter
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[Theophrastus]

Ah, perfect! Thanks for the input. Thus, it can be safe to conclude, that it was mostly magnesium oxide, caused by the reaction between Mg and NaOH, as so, I believe:

 

2Mg + 2NaOH > 2Na + 2MgO + H2

 

That explains the violent bubbling!

as such, the vinegar would then react with it, to form magnesium acetate, releasing water, in the process. That, said, this implies that both precipitates would have to come off of the cathode, not the anode. The hydroxide, used up, there is only Copper, Oxygen and Chlorine, to speak of. As such I believe the lower precipitate is copper oxide, while the upper one... well that I'm unsure of. The only way it could be copper chloride, is if the solution is already saturated with CuCl2, however, as the solution remains clear, this is not the case. On an alternative note, can you think of any way to separate the two precipitates, from one another; I'm clueless. If I were to boil it off, would they not crystalize upon the beaker, and then, with like, clinging to like, could I not extract the crystals separately. Cleaning off any excess, stuck to the walls of the beaker, with acetic acid solution, or perhaps, a heavy- duty drain cleaner.