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len1
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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]
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woelen
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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.
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PHILOU Zrealone
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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?
PH Z (PHILOU Zrealone)
"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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woelen
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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).
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len1
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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]
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woelen
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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  !
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len1
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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]
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woelen
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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]
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Ephoton
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excellent
I just aquired quite a few of those u tubes never new them to be so usefull.
e3500 console login: root
bash-2.05#
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Ephoton
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Mood: trying to figure out why I need a dark room retreat when I live in a forest of wattle.
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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
e3500 console login: root
bash-2.05#
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