Sciencemadness Discussion Board - Precise electrolysis with a U-tube

Precise electrolysis with a U-tube

woelen - 22-2-2008 at 11:23

Two weeks ago, I have become the lucky owner of a simple, yet very nice piece of glasswork, which allows many new things to be investigated. I first did the funny good old banging with detonating gas:

http://woelen.homescience.net/science/chem/exps/precision_el...

After that, I did some experiments with less-known reactions. The results are surprising (at least to me). I had some clue in the direction of these reactions, but never was able to precisely determine what happens. With this apparatus I could do so experimentally.

http://woelen.homescience.net/science/chem/exps/precision_el...

My next project with this is the electrolysis of monochloroacetic acid/monochloroacetate. If the analogy with acetate holds, then this could be a way to make 1,2 dichloroethane.

EDIT: Changed links, so that they work again

[Edited on 29-12-12 by woelen]

12AX7 - 22-2-2008 at 12:53

Another excellent, detailed experiment!

Tim

chemoleo - 22-2-2008 at 13:34

Very very nice.
I guess you realised, but just in case you didn't, you just rewalked the steps of Adolph Wilhelm Hermann Kolbe, with his famous Kolbe electrolysis!
And yes, the gas is definitely ethane!

I always thought one could make some interesting things with it, imagine i.e. hexachloroethane (insoluble in H2O) from trichloroacetic acid, benzoic acid forming...what exactly (Phe-Phe?) (although unsaturated systems don't work apparently, not sure about phenyl though), citric acid forming ???, succinic acid, etc etc...

Have fun!

monochloroacetate electrolysis

chloric1 - 22-2-2008 at 15:41

With free radicals and halogen compounds foating around, I wonder if you just would get a mix of several dozen materials. Or maybe a chain reaction will just escalate into a polymeric sludge.

Then again the free radicals will just neutralize each other before any halogens have the chance to break free. Free radicals to tend to annihilate themselves, especially in politics

not_important - 22-2-2008 at 18:57

Chain polymerisations occur with unsaturated compounds, where a radical adds to an unsaturated bond, increasing in size and continuing to be a radical.

A concentration of radical does often give dimerisation, as in the Kolbe, but that consumes the radicals.

With free halogens an radicals, you can get chain halogenation, with a radical taking one halogen atom from the dihalogen, leaving the other as a radical that abstracts a hydrogen forming a new radical.

The Kolbe reaction has been used to make halocarbons. Electron withdrawing groups on the alpha carbon of the acid generally enhance the dimer formation, -OH, -NH2, -OR, and to some extent phenyl, tend to decrease yields.

I've not yet found a good general description of Kolbe condition products for benzoic acids. With plain benzoic acid benzene is a major product, biphenyl is formed in small amounts, as is phenol. But the write-up implied others without listing them.

I noticed that the "making beneze" threads look at Kolbe but did not chose to pursue it, I'm not sure why.

woelen - 23-2-2008 at 11:05

@chemoleo: I did not know the Kolbe reaction. Good that you mentioned it and that I now can put my experiment in a broader perspective. I certainly will try different ones, the first one will be monochloroacetate. I unfortunately don't have trichloroacetic acid or trichloroacetates.

@not_important: The production of benzene from benzoic acid is puzzling me. How could that be? This only can happen through a chain of reactions:

C6H5COO(-) - e --> C6H5COO•
C6H5COO• --> C6H5• + CO2

Now I am stuck. C6H5• could react with water:

C6H5• + H2O --> C6H6 + HO•

If this is the mechanism, then I would expect for each molecule of H2 produced at the cathode, two molecules of CO2 at the anode, plus half a molecule of O2 at the anode, and 2 molecules of benzene.

Chloroacetate electrolysis -- unexpected result!

woelen - 23-2-2008 at 16:22

I tried the experiment with chloroacetic acid, CH2ClCOOH. The result is totally unexpected. It is not anything I would expect

.

I dissolved 3 grams or so of solid chloroacetic acid in 10 ml of water, and added NaOH to this and let all of this dissolve. I took so much NaOH, such that the pH ended up between 3 and 4 (according to my pH test strips). I added more water, such that i had 15 ml of solution. This solution is colorless.

I put the solution in the U-tube and allowed the thing to run for 15 minutes before I collected any gases (allowing air to be driven out). After that, I have done 3 experiments. The results of these experiments are quite repeatable. I have the following observations.

1) Volume of collected anode gas is appr. 0.75 times the volume of cathode gas. However, when I count bubbles, then I find a near 1 : 1 volume ratio (120 bubbles at anode for 117 bubbles at cathode). I assume that the difference between collected gas, and bubbled gas is due to dissolving of carbon dioxide in water.
2) When NaOH is brought in contact with the collected anode gas, then appr. 70% of this dissolves. Some gas does remain though, this does not dissolve.
3) The remaining anode gas is flammable. When it is mixed with air (55 parts air with 45 parts gas), then it burns with a peculiar flame (see pics below).
4) The liquid around the anode becomes very light yellow/green. After opening up the apparatus at the end of the experiments, a fairly strong smell of elemental chlorine could be recognized at the anode. However, it is not expected that much of this chlorine made it into the collected gas. Bubbles of the collected gas, going through water, are odorless.

Here follows a sequence of 6 small images, just before igniting the gas mix, until all of it is used up:

The light from the images is rather yellow, this is because I used a setting at the camera, such that it shows real light color. Tungsten light is rather yellow, although we don't perceive it is such. The flames in the pics now have true colors, which they have in reality. With color correction, the color of the flames would become unreliable.
As the pics show, the flame, especially the bottom part, has a very peculiar color. After burning of the gas, no particular smell can be observed, not the sharp odour, which normally remains after burning chlorinated hydrocarbons. The flame also is not sooty, not at all.

From my (rough) measurements and assuming that the difference between bubbled gas and collected gas is dissolved carbon dioxide, I conclude that somewhere between 20% and 25% of the anode gas is the flammable gas, the rest is carbon dioxide.

I do not (yet) have a clue what the precise products of electrolyis could be. One thing, however, I can say is that the analogy with normal acetic acid does not hold. If that analogy existed, then I would expect much more CO2 (two times as much bubbles of anode gas as what I have now) and no flammable gas (1,2 dichloroethane is a liquid, which boils at 83 C).

EDIT: Changed links such that pictures appear in post again.

[Edited on 29-12-12 by woelen]

not_important - 23-2-2008 at 17:20

Quote:

Resource Relation Sov. Electrochem. (Engl. Transl.) ; Vol/Issue: 22:8; Translated from Elektrokhimiya; 22: No. 8, 1029-1033(Aug 1986)

It was shown by direct comparison of kinetic and adsorption data that at platinum in solutions of mono- and trichloroacetic acid, the anodic processes are entirely analogous in character to the processes occurring in acetate solutions.^The lack of Kolbe electrosynthesis products in the electrolysis of chloroacetic acids is to the special reaction features of the secondary chloroalkyl radicals.

http://www.osti.gov/energycitations/product.biblio.jsp?osti_...

And see attached

Attachment: ELECTROLYTIC CONDENSATION OF CARBOXYLIC ACIDS.pdf (868kB)
This file has been downloaded 1185 times

JustMe - 23-2-2008 at 21:04

Interesting stuff. Going back to an old, old, OLD project of mine... I would be really interested in the results of electrolyzing a sodium saccharin solution in a U-tube with a gelatin plug on the bottom separating the cathode and anode.

Look up the saccharin anion...

(Taking off the hydrogen from the nitrogen in this picture: http://en.wikipedia.org/wiki/Sodium_saccharin)

I wonder what would happen?

My old time project (mentioned here before), was the study of transitional metal saccharin compounds with the hypothesis that the saccharin anion would act as a bidentate chelating anion. I made some interesting compounds and grew some interesting crystals. The copper compound(s) (yes, plural) were very interesting. I was able to grown crystals up to 4 mm or so of the hydrate, and it formed a very interesting green crystaline compound when dissolved in pure acetone.

My dream (to this day, even though I don't do chemistry anymore), was to react a solution of sodium saccharin with Chromous Sulfate or Chloride (NOT CHROMIC, the acidity of that solution precipitated free Saccharin) to see if I would get a red compound akin in structure to Chromous Acetate. But alas, I could never get any Chromous compounds, and the synthesis was too complicated. (Obviously it would have to be in boiled, oxygen-free water under an inert gas and done quickly, of course.)

Anyway, sorry to go off on a tangent. Looking at the structure of the Saccharin Anion, what do you think would happen if it were electrolysized? Might be an interesting experiment. Or, to avoid the alkaline of the Sodium, perhaps using a transitional metal compound like Copper Saccharin (precipitates easily on mixing Cupric Sulfate and Sodium Saccharin... super easy to recrystalize). Hint, hint, LOL.

Enjoy your U-tube in many more experiments!!!

len1 - 23-2-2008 at 21:48

Thank you for being one of the few people to continue to post interesting work in this forum! I was quite surprised by Kolbes reaction which you rediscovered I was quite unaware of it myself. Certainly wouldnt have expected decarboxylation, and impressed you deduced the mechanism of something Kolbe didnt himself.

Also surprised by the present results. Apparently chloroacetic acids dechlorinate on electrolysis yielding acetic acid. http://www.freepatentsonline.com/4707226.html. Although many patents are rubbish this apparently is not.

I conclude thereupon the acetic acid must electrolyse as before to give CO2 and C2H6 which is what burns - in the now oxygen-rich gas mixture. But the process competes with the dechlorination, which generates oxygen with the chlorine staying in solution as HCl.

So of the competing processes one generates 1/2 mole of gas at anode per mole at cathode. The other 2 moles per mole at cathode. A ratio of 0.75 lies between these and gives ratio for the two processes. Dissolution of anode is another competing process

[Edited on 24-2-2008 by len1]

Nicodem - 24-2-2008 at 02:48

Very nice experiments, Woelen!
I'm especially fascinated by formate electrolysis since I always wondered what happens to it. Your results indicate that formates/formic acid might be useful as a sacrificial electrolyte in organic electroreductions where a diaphragm must be used. Such reductions could potentially work without a diaphragm if an sacrificial electrolyte which does not yield reactive/oxidative species is used.

You could try a preparative aproach of the Kolbe coupling of benzoate and other such carboxylates that would give you liquid or solid products. For example, you could try getting the paraffin from electrolysis of soap or biphenyl from sodium benzoate…

Quote:

Originally posted by JustMe
Interesting stuff. Going back to an old, old, OLD project of mine... I would be really interested in the results of electrolyzing a sodium saccharin solution in a U-tube with a gelatin plug on the bottom separating the cathode and anode.

This should theoretically yield N-hydroxysaccharin or its nitroxyl, but maybe also something completely different. Seems an interesting experiment. No diaphragm or gelatin plug is needed since sodium saccharin is an anionic species and thus only reacts at the anode.

[Edited on 24/2/2008 by Nicodem]

not_important - 24-2-2008 at 03:46

Quote:

Originally posted by Nicodem
...
I'm especially fascinated by formate electrolysis since I always wondered what happens to it. Your results indicate that formates/formic acid might be useful as a sacrificial electrolyte in organic electroreductions where a diaphragm must be used. Such reductions could potentially work without a diaphragm if an sacrificial electrolyte which does not yield reactive/oxidative species is used.

Related to that, from the document I included above

Quote:

Only hydrogen, oxygen, carbon dioxide , and in some instances glycolic acid could be detected in the
electrolysis products of oxalic acid.

Quote:

You could try a preparative aproach of the Kolbe coupling of benzoate and other such carboxylates that would give you liquid or solid products. For example, you could try getting the paraffin from electrolysis of soap or biphenyl from sodium benzoate…

Again, from that document, it takes special conditions to get Kolbe dimers from benzoic acids

Quote:

For a long time it was assumed that salts of aromatic acids, unlike those of aliphatic acids, did not enter into the Kolbe reaction.
Experiments on the electrolytic condensation of solutions
of salts of benzoic acid in water, sulphuric acid, and absolute methanol, and the electrolysis of fused benzoic acid were indeed unsuccessful. The reaction products were, in general, benzene, free benzoic acid, oxygen, etc. The same was true of the electrolytic condensation of substituted benzoic acids such as ο-nitrobenzoic, p-nitrobenzoic, p-toluic, opianic, salicylic, m-hydroxybenzoic, and p-hydroxybenzoic acids, 3,5-dimethyl-and 2,4-dimethylbenzoic acids, and ethyl phthalate.

In the 1930's Fichter et al. refuted the conclusion that aromatic acids do not undergo the Kolbe reaction...aromatic acids can be electrolysed successfully by using a methanol-pyridine mixture as solvent.
Thus, benzoic, phenylacetic, /3-phenylpropionic, and phenoxyacetic acids yielded the corresponding diphenyl, dibenzyl (50% yield), 1,4-diphenylbutane (37% yield), and
the diphenyl ester of ethylene glycol (45% yield).

woelen - 24-2-2008 at 08:37

@not_important: Thanks for that great PDF. This is an interesting survey of many different reactions. With the Kolbe-synthesis, a complete new world has opened up to me

. The article only briefly covers the electrolysis of chloroacetic acid, it tells that chlorine is split off. I will have to think further about this and try to analyse the gas in more detail. One of the things I intend to do is mixing the gas with dilute bromine vapor. If it is unsaturated, then the color of the bromine vapor will disappear at once.

@len1: The patent you supply is not applicable to my situation. It describes a cathode reaction, mainly being:

CH2ClCOOH + H(+) + 2e --> CH3COOH + Cl(-)

What I am investigating is the anode reaction. I already got flammable gas and CO2 from the start of the experiment, before any acetic acid could have formed at the cathode. I also think that the cathode reaction hardly occured in my situation. I measured a current of 150 mA and with this small current it took almost 25 minutes to fill one test tube of 22 ml with hydrogen gas. This is fairly consistent with formation of almost onloy hydrogen at the cathode.

BromicAcid - 24-2-2008 at 08:59

Here is something interesting of note with relation to Kolbe reactions. The radicals produced can attack the metal electrodes to form organometallic compounds. I found this out when looking through some of the papers on diisopropyl lead. The thread that I made on the subject can be found here:

Easily preparable organometallics and their properties.

This might be interesting to you since the products for lead at least are sometimes brightly colored and separation with something like a U-Tube is essential to prevent their oxidation.

More information can be found in this PDF which I have hosted on my site. It is from the ACS and is a review of methods of synthesis for organometallic compounds and has it's own section devoted to radical reactions.

Anyway, figured this might be an interesting extension to what you have already done.

Great work and good luck!

garage chemist - 24-2-2008 at 10:13

Very nice, woelen! Your extremely good pictures are always something special, as well as your projects.

While it is said that Kolbe electrolysis results in dimerization of the alkyl radicals, this is not always true. The first example is your chloroacetic acid.
Also, organic chemistry books say that Kolbe electrolysis is unsuitable for aromatic carboxylic acids. This can give very different products than what one might expect, and also mixtures of products.

Kolbe electrolysis is used industrially to produce sebacic acid (decanedioic acid, the dicarboxylic acid with 10 carbons) from the monomethyl ester of adipic acid (adipic acid: hexanedioic acid, 6 carbons). The ester group serves as protection, as you can see.

len1 - 24-2-2008 at 18:20

Yes I know the patent I mentioned has the reaction at the cathode, still what you are getting is essentially as I outlined before. H2 at the cathode and mainly O2 at the anode, due to the ratio of cathode to anode gases. You are unlikely to have anything unsaturated in the anode gases because the anode is an oxidative process - no reduction there. The only possibility for a flamable gas is C2H6 plus traces of haloalkanes, made more so because of the excess O2 present. So the major anode reaction

H2O -> 2H+ + 1/2O2 + 2e-

minor

CH2ClCOO- + H2O -> CO2 + 1/2O2 + HCl + 1/2C2H6 + e-

The article not important lists does not provide much illucidation unfortunately on the actual details (common problem of review articles) I shall see if I cna find confirmation elsewhere

[Edited on 25-2-2008 by len1]

PHILOU Zrealone - 25-2-2008 at 08:34

Electrochemistry is very surprising indeed...I would have thought you would end up with CH3-Cl or CH3-CH2Cl

but it is not the case.
Good to know anyway that it is useless to waste good chloracetic salts where cheap acetate provides the same result

woelen - 25-2-2008 at 13:52

Len1, the main reaction I observed at the anode is not production of O2. The gas mix I obtained consists of appr. 80% CO2 (75% to 80% of its volume is absorbed by a solution of NaOH). The remaining part is a flammable gas, as shown above. Of course, it might contain some O2, but if it contains O2, this only is a minor part of the anode gas.

I more and more am thinking it is CH3Cl, but I have to think and read a little bit more on this. Further tests also are needed. I certainly will come back on this, one of these days.

len1 - 25-2-2008 at 16:32

That cant be right, even just stoichiometrically

CH2ClCOO- + H+ -> CH3Cl + CO2

So you cant have 80% CO2 if CH3Cl is the only other major product.

Plus the reviews say you get mainly oxidation products and the Cl- splits off. So if you have a significant quantity of CH3Cl youve got publishable material here!

Also you are not interpreting what is major and minor reactions correctly. The relative contribution of a reaction does not correspond to the volume of gas generated. Note my first reaction gives 1/2 mole of gas per 2 moles of electrons, whereas the second gives 4 moles of gas for the same 2 moles of electrons - > i.e. it generates 8 times more gas.

Because of this ratio of gases the ratio of the reactions is very sensitive and cant be judged from a rough estimate of volumes. So I still stand by my original interpretation.

The only alternative to my reaction 1 would be to consider oxidation of the alkane at the anode by some of the oxygen, instead of it being liberated. That would correspond to what we know from literature - we get oxidation products.

From the abstract we have that the basic reaction is the same as in Kolbe synthesis with acetate

CH2ClCOO- -> CH2Cl. + CO2 + e-

Now we are told we dont get the usual Kolbe reaction produts, which would be CH3Cl and C2H4Cl2 because the Cl- splits off and the radical oxidizes. Choices are

CH2O - oxidation 0
C2H4O2 0
CH2O2 +2
C2H2O4 +3

Formaldehyde is much more soluble than CO2 and gives

CH2ClCOO- + H2O -> CH2O + Cl- + 2H+ + CO2 + 2e-

so it can reduce the ratio of cathode:anode gases to 1:1 but gives no flamable gas - so its a possible alternative to the generation of pure oxygen -> You can test for formaldehyde in the water, but you must still have my reaction 2.

Acetic acid formation would lead to some of it being immediately oxidized further, overall reaction as in my reaction 2.

Further oxidation gives formic acid

CH2ClOO- + 2H2O -> HCOOH + Cl- + CO2 + 4H+ + 4e-

this reduces the ratio of cathode:anode gases to 2:1, same as my reaction 1. Can be tested by presence of formic acid, but you must still have my reaction 2.

Your suggestion

CH2Cl. + H+ + e- -> CH3Cl

is a reduction, which I think is very unlikely at the anode, still you can test for the presence of HCl in the burnt gases with moist indicator paper (stronger effect than H2CO3) or - a spatula with a drop of NH3 solution held in the exhaust gases will fume strongly. Even if by some miracle this reaction were to occur, it can only be a minor reaction as I pointed out above. Add to that that it liberates no net electrons, and we know something must oxidize at the anode, at the rate at which H2 is reduced at the cathode.

I cant unfortunately get to the first article not important posted because my institution doesnt receive it.

[Edited on 26-2-2008 by len1]

woelen - 26-2-2008 at 03:58

I have been thinking about this, and the following could come quite close to my observations:

First reaction: 2CH2ClCOO(-) - 2e --> 2CH2ClCOO•

At the cathode, we then have a single molecule of H2.

Now suppose the radical CH2ClCOO• decomposes: 2CH2ClCOO• --> 2CH2Cl• + 2CO2

Now for 1 molecule of H2 we have two molecules of CO2.

The radicals CH2Cl• could react as follows: 2CH2Cl• + H2O --> CH3Cl + CH2Cl(OH).

The latter compound is unstable and would immediately split off HCl, which goes into solution as hydrated H(+) and Cl(-). The organic compound, which remains behind is CH2O.

CH2O is a strong reductor. At the anode it would be oxidized as follows (net reaction, no mechanism presented here):

CH2O + H2O - 2e --> HCOOH + 2H(+).

So, at the cathode there is no gas for this reaction, at the anode we have another molecule of H2. Now we have 2H2 at the cathode for 2CO2 + CH3Cl at the anode.

Things can go further (see my webpage on oxidation of HCOOH at the start of this thread):

HCOOH - 2e --> CO2 + 2H(+)

No we have another molecule of CO2 at the anode and again a molecule of H2 at the cathode.

The reason why I believe these reactions could occur is that the compounds CH2O and HCOOH are formed directly at the anode. So, they are at the right place for immediate further oxidation. This is not the case for some compounds, formed at the cathode, which first need time to move to the anode.

This makes for a ratio of gases as follows: cathode : anode = 3H2 : (3CO2 + CH3Cl), which makes a volume ratio Vcathode : Vanode = 3 : 4.

I also noticed a strong smell at the anode of chlorine. The liquid become green/yellow, but hardly any chlorine gas made it into the gas phase (it dissolves quite well).

The net reaction for one Cl(-) is:

Cl(-) - e --> 1/2 Cl2 (which is assumed to remain dissolved almost 100%). This makes the volume ratio of gases Vcathode : Vanode = 3.5 : 4

Besides the reaction, mentioned above, there undoubtedly also will be formation of O2 at the anode. That would bring the volume of anode gas, compared to cathode gas even further down. I can come close to the observed approximate 1 : 1 ratio of bubbled gas. Dissolving of CO2 in water can explain the lower amount of collected anode gas.

So, what I have written down is just some theory, and I do not present it as the truth. I'll do the test, suggested by len1. I again make some of the flammable gas and let it burn and see whether it produced smoke with 25% NH3 around, near the rim of the test tube. That indeed is a very sensitive test for HCl, which would be formed by burning CH3Cl. If this test is negative, then my theory does not hold anymore and further searching is required. If the test is positive, then it does not prove my theory, but it again is evidence, supporting it.

YT2095 - 26-2-2008 at 04:12

I`m wondering if perhaps a soln of acidic silver nitrate added to the gas tube (before ignition) and then shaken around a bit may also show the presence of Cl?

len1 - 26-2-2008 at 05:11

Combining what you wrote is completely equivalent to my oxidation to formic acid above

CH2ClOO- + 2H2O -> HCOOH + Cl- + CO2 + 4H+ + 4e-

together with the CH3Cl generation equation

CH2ClCOO- + H+ -> CH3Cl + CO2

Your provide a plausibility argument for linking these two reactions in a 1:1 ratio, the link lies in you combining two alkyl chloride radicals in the one formula as you did

2CH2Cl• + H2O --> CH3Cl + CH2Cl(OH)

so that one molecule goes the first (oxidative way) the other the second (reductive way). This is where the essence of what you are saying lies.

Because the second reaction is local, you can add any amount of it to the real oxidation correponding to the first reaction and still have a valid cell. Experiment has to decide, but I still think that CH3Cl is a reduction which seems unlikely, and it goes against what the literature says. IF your tests show thats what you actually get, youve got publishable material I think.

I dont think shaking a chloroalkane with AgNO3 will work well as the Cl is covalently bonded and so the concetration of free Cl- is likely to be well below that required by the AgCl solubility product.

[Edited on 26-2-2008 by len1]

YT2095 - 26-2-2008 at 08:11

Quote:

Originally posted by len1

I dont think shaking a chloroalkane with AgNO3 will work well as the Cl is covalently bonded and so the concetration of free Cl- is likely to be well below that required by the AgCl solubility product.

[Edited on 26-2-2008 by len1]

it may do Post Ignition though!

woelen - 26-2-2008 at 09:51

I have done more tests now with a new cell and fresh materials.

I have done a test in order to obtain more precise data on the volume ratio in which gases are produced. I have done this, by first letting the electrolysis run for half an hour or so, then blowing out all gas mixes from the tubing, getting only air in the apparatus. The bubbles of gas from the cathode and anode now are air bubbles and there is much less error due to dissolving of the gases. With this test, I found a volume ratio as follows:

Vcathode : Vanode = 90 : 125 (which is very close to 3 : 4)

Another test was done in order to see whether a cathode side reaction occurs, besides the production of hydrogen gas.

I had two experiments:

23.5 ml of gas was produced in 1110 seconds, the current was 0.24 A, the air temperature was 14 C, the test tube with gas was wet and probably was slightly colder due to evaporation of water. Let's say 10 C.

Total amount of gas, produced per second is 0.24/96485/2*(283/300)*22400 ml = 0.0263 ml

Filling a test tube of 23.5 ml with this gas would take 23.5/0.0263 = 893 seconds. Measured time is 1110 seconds. So, appr. 20% of all electrons go into another reaction, than the reaction of making H2.

After running the cell for more than an hour, the liquid becomes warm, and at that time I did another test. Now the current was 0.30 A. With that current one would expect the same test tube to be filled in 715 seconds. Measured time now is 770 seconds. Now, more than 90% of all electrons from the cathode go into production of H2 and the side reaction is much less pronounced.

The ratio of flammable gas to CO2 remains the same for all experiments, it is appr. 1 : 4, so the anode reactions seem to be fairly constant during the experimenting. The only difference is that after some time, a faint green color of Cl2 can be observed.

This is the quantitative part. Now the qualitative part. The flammable gas is not an olefin, in the later stages of the experiment, the bubbles of this gas have a faint smell of elemental chlorine. This would be absolutely impossible if there were unsaturated C=C bonds in the gas mix.

I also did tests about the chlorine contents of the gas mix. First, it is shaken with ammonia, in order to destroy any traces of elemental chlorine from the gas mix and to absorb CO2 from the mix. The ammonia contents in the test tube is appr. 3% and then a gas-air mix (1 : 1) is ignited. No traces of ammonium chloride could be observed, nothing at all.
In another test, the gas mix is cleaned with a solution of NaOH and then it is mixed with air again and the gas mix is ignited, with 15% ammonia just a few cm from the opening of the test tube. Again, no smoke of NH4Cl could be observed. I now am convinced, the gas does not contain chlorine, and hence it is not CH3Cl. Len1, you were right, at least with this aspect of the reaction.

Time to think more about this.....

The only flammable gases I can imagine as plausible alternatives now arre CH4 and CH3CH3.

[Edited on 26-2-08 by woelen]

woelen - 26-2-2008 at 10:25

The appearance of the flame is quite similar to the pics when made in daylight. The preceived color is more blue instead of rose. This probably is due to the white background, instead of the yellow background light.

When the gas was cleaned with ammonia, then the flame appeared different. No orange flame above the blue flame. I doubt that this is due to the ammonia. More likely, this sample has a little bit more air mixed in, but for completeness I'll post this picture as well.

EDIT: The flame color and shape looks a little bit like the flame, produced by CO. I made some CO by mixing 85% formic acid with concentrated sulphuric acid and them mixed this with air and ignited. The result looks somewhat like the flame, shown in the ammonia picture, without the orange tinge.

[CO] could be produced according to the following formal model (only formal, specifying level of oxidation of C):

CH2ClCOO(-) - e --> CH2ClCOO•

Half a molecule of H2 is produced at cathode.

Splitting up of radical, with splitting off of the chlorine as HCl.

CH2ClCOO• --> [CH] + HCl + CO2

Here [CH] stands for a formal species, C has oxidation state -1.

[CH] - e --> [C] + H(+)

Another half molecule of H2 is produced at the cathode.

[C] + H2O - 2e --> CO + 2H(+)

Another molecule of H2 is produced at the cathode.

An alternative reaction is

[C] + 2H2O - 4e --> CO2 + 4H(+)

With this, two molecules of H2 are produced at the cathode.

Using this formal model, I can explain a mix of CO and CO2 as anode gases, and a Vanode : Vcathode ratio between 1 : 1 and 1 : 1.5. Pure CO2 would result in a 1 : 1.5 volume ratio, a mix of equal volumes of CO and CO2 would result in a 1 : 1 volume ratio.

The observed ratio of CO and CO2 is somewhere between 1 : 3 and 1 : 4, this would result in a ratio Vanode : Vcathode ≈ 1 : 1.2.

If one takes into account that the cathode reaction also is not 100%, but between 80 and 90%, then the ratio will be closer to 1 : 1 again.

This could be an explanation for the reaction. More testing now is needed. Is there an easy to apply test for CO?

EDIT: Changed links, so that pictures appear in post again.

[Edited on 24-6-13 by woelen]

len1 - 26-2-2008 at 19:07

OK the results are good in the sense that we have ruled something out, bad in another - with no CH3Cl there goes your publication, unfortunately I thought as much.

What you wrote can be made much simpler. You just have oxidation to formic acid as I wrote before

CH2ClOO- + 2H2O -> HCOOH + Cl- + CO2 + 4H+ + 4e-

with this time formic acid acid decomposing to give you your flamable gas

HCOOH -> CO + H2O

Overall your reaction is

CH2ClCOO- +H2O -> CO + Cl- + CO2 + 4H+ + 4e-

However the docomposition of HCOOH is local - which means you can have any amount of the HCOOH decomposing - from all to none and still have a valid redox cell.

There is a problem with all this . You cant choose all HCOOH decomposing because then you have a 1:1 CO:CO2 ratio which disagrees with your stated 80% CO2. To get the gas to be mostly CO2 you would have to have a lot of the HCOOH not decomposing - which goes against your cathode:anode gas ratio. You can not satisfy what you observe wtih these two formulas.

You still have to use the formula for generation of for ethane together with oxidation to formic acid to get all ratios right. Im sure significant amounts of O2 and Cl2 are also produced. There is one more possibility - methane - but that is reduction even further than C2H6, and with some Cl2 around electrons are clearly at a premium, so I find it hard to believe.

However CO does burn with a blue flame, and in case some CO forms, there is, as it happens a test for CO - based on the same property that makes it so poisnous -it has a lone pair and so forms complexes with transition metals.

Here is a test from a set of notes I have been using:

Prepare a solution of 1.6 M CuCl(aq) in 3 M HCl. This solution will react with CO in a 1:1 volume ratio.

The complex formed makes the CuCl solution change colour. To see this

Fill a syringe with 30-mL CO. Suction in 30-mL 0.4 M CuCl. Shake, the CO will dissolve. The complex has a blue-green colour.

(a) syringe with CO;
(b) CuCl is drawn in;
(c) after shaking
(d) colour of Cu(CO)Cl(H2O)2

[Edited on 27-2-2008 by len1]

CO+CuCl.jpg - 12kB

woelen - 28-2-2008 at 08:04

I did the test with copper(I) chloride, and the test was positive. Approximately 50% of the remaining gas is CO. I made the copper(I) chloride by dissolving some red copper(I) oxide in 25% HCl and sucking this (olive green) solution into the syringe, in which the remaining gas was sucked.

I started with 50 ml of gas. Absorption with NaOH left approximately 10 ml of gas.
This 10 ml of gas was treated with CuCl in HCl. Approximately 5 ml remained.
The remaining 5 ml of gas was treated with catechol in NaOH. Approximately 2 ml of gas remained.
The remaining 2 ml of gas could not be ignited. Probably it is remaining nitrogen from air.

I also did more precise measurements and I noticec that the volume of cathode to anode gas is not constant. At the start of the experiment, Vcathode : Vanode = 1 : 1.5
After two hours of electrolysis it has changed to 1 : 0.7

When the current is measured, and based on this, the expected amount of cathode gas is computed, then it becomes clear that near the start of the experiment, a fairly large part of the cathode electrons is not used for making H2. At the end, almost all electrons at the cathode are used for making H2.

Right now, I am working on a write-up of all the experimental results I have. Len1 has contributed a lot to my understanding with his remarks and ideas. With his remarks, I now can understand the observations I made, and all of this will be made into the webpage on acetate electrolysis. All this work is definitely worth a place on my website.

PHILOU Zrealone - 28-2-2008 at 09:14

What about voltage variation?
Different voltage implies different reactions...with 12V-15V you can separate Fr from F...it is maybe to strong a potential?

woelen - 28-2-2008 at 10:46

I used a series resistor of 4.7 Ohm, in order to limit current and let the cell find its own working voltage. With this resistor, the current was appr. 250 mA, which means that approximately 12 V is accross the cell. With lower voltage, the reactions become VERY slow. Probably, such high voltages are required, because of the very thin anode wire (which is 0.25 mm diameter platinum wire).

len1 - 28-2-2008 at 22:37

Very nice! You were right about your deduction of CO being one of the major products, I didnt think it would be. The O2 reaction which I thought was major, clearly has only 10% of the electrons going to it from your results.

Now we have to put this altogether. I suggest:

Initially cathode current efficiency is not 100% because

2H+ + 2e- -> H2

0) is competing with

CH2ClCOOH + H+ 2e- -> CH3COOH + Cl-

which is the cathode reduction of the chloroacetic acid to acetic acid mentioned in the patent. As the concetration of acetic acid and Cl- released at the anode builds up, by Nernst's equation the potential necessary to drive this reaction rises i.e. the equilibrium is left shifted. When the potenatial is well above that of hydrogen generation mostly H2 is evolved.

At the anode we have a greater variety of reaction. They all have in common the decarboxylative first stage

XCOO- -> CO2 + X. + e-

If X = CH3. (from the acetic acid generated at the cathode) we get CH3CH3. If X contains a halide it oxidizes further.

1) Some of the acetic acid can reach the anode and be electrolysed as before

CH3COO- -> CO2 + 1/2 C2H6 + e-

2) CH2ClCOO- is oxidized to formic acid

CH2ClCOO- +2H2O -> HCOOH + CO2 + Cl- + 4H+ + 4e-

3) Some of the formic acid is decomposed

CH2ClCOO- +2H2O -> CO + H2O + CO2 + Cl- + 4H+ + 4e-

4) We can combine 2) and 3), assuming fraction a of formic acid decomposes, we have

CH2ClCOO- +2H2O -> (1-a)HCOOH + a(CO + H2O) + CO2 + Cl- + 4H+ + 4e-

We can explain what happens at the end. All e- go to producing H2 and the cathode, so the gaseous output of the cell for 4 moles of electrons is

2H2 : (CO2+ a CO)

As you say at the end (1+a)/2 = 0.7, it follows a = 0.4

So at the end CO2 forms about 1/1.4 = 70% of the total gas released, which is about right. We have satisfied both the ratios required with just the assumption of equation 4.

Obviously we have 10% of the elctrons released by

5) H2O -> 1/2O2 + 2H+ + 2e-

but that does not change the numbers for reaction 4) by more than experimental error.

Now we have to explain what happens at the beginning. The cathode anode volume ratio changes from 1.5 to 0.7, i.e. by a factor of 2, but you say the current efficiency in H2 at the cathode changes only from 80% to 90-100% so the change in anode to cathode volumes can not be accounted by the change of hydrogen current efficiency alone , and it follows composition of gases at the anode also changes during this time. That seems to imply a stage of CO2 generation at a rate which can not be accounted for by equation 4), which produces 2 moles of H2 for every 1 of CO2.

I still suggest the generation of ethane by reaction 1) which releases the mole of CO2 without consuming many electrons precisely because X is not oxidized but rather forms ethane. Combining reactions 0) 1) and 5) we get the reaction I posted before

CH2ClCOO- + H2O -> 1/2 C2H6 + CO2 + HCl + 1/2O2 + e-

Note this reaction could occur directly at the anode, or in stages with the acetate migrating from the cathode, and would solve the excess CO2 problem at the beggining, because 4 moles of electrons now generate 4 moles of CO2. But to have any credence we must establish presence of either C2H6 or CH4.

I would suggest trying to burn the remnant gases after absorbing all the CO (i.e. before removing the O2 which will help any C2H6 or CH4 in the mixture burn). If they are not present it is hard to explain the excess CO2. The minor reaction generating Cl2 (oxidation of CH2Cl. on the chlorine) does not solve this problem, since even if we assume some of the NaOH soluble gas is chlorine

CH2ClCOO- +2H2O -> CO2 + 1/2Cl2 +HCOOH + 4H+ + 5e-

we still have a 2.5 : 1.5 cathode:anode gas ratio.

Oxidation of the radical further to CO2 doesnt improve things. The root of the problem is the carbon oxidizing to HCOOH (or CO) is a +4 change of oxidation state. We need an alkane on the right so the mole of CO2 is evolved without eating up too many electrons. That will give us a cathode:anode gas ratio of 1 or less.

[Edited on 29-2-2008 by len1]

woelen - 1-3-2008 at 04:29

I tried burning of the remaining gas after absorbing CO. It did not burn. I tried two times, just to be sure, but I can't get it burning, so I'm quite sure that this gas mix does not contain C2H6 or any other flammable gas.

I also noticed a new special effect. Immediately after switching on the apparatus with a fresh solution, I hardly get any gas at the cathode. I neglected this effect, but it might explain the low cathode gas amount which I obtained in my experiments at the start. But this low production of H2 only is for a few minutes, for the first test tube of gas. After this initial transient, the current efficiency for making H2 goes to 80% or so and this then slowly increases. My experiment with 1.5 times as much anode gas as cathode gas is at the start of the experiment.

Len1, right now, I am working on a write-up of all my findings in a new webpage which hopefully will be available somewhere next week. Any comments on that webpage then are welcomed. I will make a reference to this thread as well from that webpage. Thanks for all your input, ideas and contributions, I appreciate it

len1 - 1-3-2008 at 07:34

Hmm. That sounds pretty definitive. So no flammable gases besides CO.

I have no problem accepting the lack of H2 evolution at the start. I can understand it as nearly all electrons being used to reduce the CH2ClCOOH to acetic acid (which also has the effect of using up H+ and so decreases its local concentration for H2 generation), until a local concentration of Cl- and acetic acid has been built up and equilibrium established (from Nernsts equation the cathode potential goes as E0 + neF log[Cl-]*[CH3COOH]) so its low at the beggining.

The lack of any reduced form of carbon, except the highly oxidixed forms CO and CO2 - I find difficult to explain - unless I have misunderstood something.

If formulas 4 and 5 are the only ones for the anode, CO2 and O2 combined must be generated at the rate of 1 mole per 4 moles of electrons, irrespective of whats going on at the cathode. Neglecting O2, at the end when the cathode efficiency is almost 100%, that would correspond to a 1:2 CO2:H2 ratio (which given 25% of the anode gas is CO agrees with your total ratio of gases 0.7 is indeed approximately 1/.75 : 2)

In the initial period, when the cathode efficiency has risen to 80% the same ratio must be 1/.75 : 2*0.8 ~ 0.83. Could that be the case?

So its still less than 1, but maybe the overall ratio of 1 for the cathode:anode volumes is just due to a very low cathode efficiency in the beginning, much less than 80%.

Thank you for your kind words, I have found it a pleasure communicating with you on this experiment, because it seems we have very similar motivations - to understand things and to report things honestly. I have also enjoyed reading you site to. Made me think about one day if I have the time... You are welcome to use anything I posted if its of use to you - you dont have to reference me if its easier.

[Edited on 1-3-2008 by len1]

woelen - 9-3-2008 at 15:08

As promised, here is a write-up of all results in a webpage. I added this information at the end of the acetate/formiate-electrolysis page:

http://woelen.homescience.net/science/chem/exps/precision_el...

EDIT: Changed link, so that it works again.

[Edited on 29-12-12 by woelen]

excellent

Ephoton - 10-3-2008 at 01:09

I just aquired quite a few of those u tubes never new them to be so usefull.

Ephoton - 11-3-2008 at 03:13

do you think it would be possible to use a u tube that is heated to create calcium metal from
a low melting calcium salt. this would make them extreamly useful for the home chemist