Ac electrolysis

[Turtle]

Electrolyte temperature is 76° F.
The gasses have started to collect in the top of the receiving vessels.
In the classic DC current version of this experiment, one vessel fills with Hydrogen while the other fills with Oxygen. Once full, a lighted splint is introduced into each vessel with the Hydrogen vessel expelling a vigorous flash of flame & popping sound while the lighted splint introduced into the Oxgen vessel increase in brightness & flame size.
I plan to continue the AC electrolysis untill I evacuate the vessels entirely & then introduce a lighted splint. What will the reaction be? How will it differ from the DC experiment results?

[Turtle]

It is now 3 hours of electrolyzing & I have collected about 1/2 teaspoon of gases in each vessel.

Pop Quiz! Take out your #2's.
Question #1) At the current rate of displacement in the vessels, how long will it take to completely evacuate them? hint
Question #2) Given the information already provided, what is the current amperage in this experiment? hint

[Jay-qu]

Drinking glasses approx 250mL, 1/2 teaspoon = 2.5 ml so you have to do 100 times the 3 hours you have been running = 300 hours

V = IR
20 = I(2600)
I = 7.69mA

Excellent work turtle! I propose the experiment be repeated with DC for comparison, what I expect is that the reaction taking place is the same BUT for the DC will be a lot faster. Possibly due to AC current not perfectly lining when it changes direction so that all the atoms that just got oxidised wont be reduced. Hence there would be a mixture of gases at the electrodes for the AC.

[Mercedes Benzene]

Turtle? Where DO you find time for such interesting experiments?

[Turtle]

Originally Posted by Jay-qu

Excellent work turtle! I propose the experiment be repeated with DC...

Egracious. The DC experiment I have done before, so it is worth doing. However, this AC experiment I have never done & I find it quite interesting. I won't grade your test until I give more time for other responses. (plenty of time to check your work subtly )

Originally Posted by Mercedes Benzene

Turtle? Where DO you find time for such interesting experiments?

I'm a recluse. Care to submit your quiz answers?

Speaking of pressure, what about the pressure of the gasses in the vessels?

[Vending]

I would like to chime in a bit here to help explain some of the ideas/questions that have been floating around in this thread...

1) Regarding the rate of electron transfer from electrode to chemical. Typically, if one is using a metal electrode (ie. gold or platinum) and small molecules, then the electron transfer is extremely fast. The transfer step itself can be on the order of femptoseconds. This means that the rate of the process is determined by how fast the chemicals can get to the electrode surface -- it is diffusion limited. Diffusion limited processes usually occur on the nanosecond timescale.

Thus, we see that for ordinary (60 Hz) AC, the frequency has very little to do with the rate of the redox process.

2) Concerning the reversibility of the reaction. The idea has been raised that perhaps once the the AC current completes a 180 degree phase shift then the reactions might be reversed (ie. those things that have been reduced will be re-oxidised). This is not the case. Usually anytime that a electrochemical event gives rise to a chemical reaction the electrochemical event is irreverible. This is especially the case for dissociation reactions and reactions in which gases are formed (electrolysis of water is both). For dissociations reactions, the two products must once again find eachother at the surface of the electrode in order for the reaction to be reversed (thus this is a three body problem and it is extremely rare that it would occur). For reactions in which gas is released there are two reasons why this would not occur. First, the gases would have to diffuse back toward the electrode, but they are bubbling away. And second, the gases are in a different phase without supporting electrolyte (air doesn't conduct electricity well).

3) Concerning the real reason why the AC electrolysis proceeds more slowly. It is simply a concequence of the ossilating current. That is to say that on a DC circut the voltage is sufficient to break water 100% of the time. In an AC circut the voltage is only sufficient for some part of the time. Thus, a slower rate of electrolysis results just because a there is less time durring which the AC cell is applying suffiecient voltage to eletrolyse the water.


I hope this helps clear up a few things .

Oh, one more thing i should say. Since you are using AC, you will be collecting both H2 and O2 in the same vial -- a situation a bit more dangerous than in the DC case when you are seperating the two out into seperate vials.

PS. Great job on the experiment. What an excellent way to answer a question!

[Turtle]

Originally Posted by Vending

I would like to chime in a bit here to help explain some of the ideas/questions that have been floating around in this thread...

I hope this helps clear up a few things .

Oh, one more thing i should say. Since you are using AC, you will be collecting both H2 and O2 in the same vial -- a situation a bit more dangerous than in the DC case when you are seperating the two out into seperate vials.

PS. Great job on the experiment. What an excellent way to answer a question!

___Thanks for chiming Vending! You have cleared up things nicely.
___I agree about the danger of igniting the gasses ; the more I have thought about it, the more I keep thinking that the mix is nothing short of rocket fuel. Given all the information so far, is it possible to calculate how much energy will be released from the 9 ounces? What volume do you think is reasonably safe to ignite?
___I just checked the experiment & the electrodes have accumulated a grainy red/brown encrustation. Even though the electrodes cycle between losing metal & depositing metal, I expect them to end up a spongy mass.
___I also wonder how the reaction may change at the moment the electrolyte in a receiving vessel drops below an electrode & exposes it to the gas mixture? Might it cause the gases to burn?

PS Thanks! Ain't da scientific method da bomb!

[Jay-qu]

Originally Posted by Vending

Usually anytime that a electrochemical event gives rise to a chemical reaction the electrochemical event is irreverible. This is especially the case for dissociation reactions and reactions in which gases are formed (electrolysis of water is both).

Ever heard of a re-chargable battery?

So long as the products stay in contact with the electrode they should be able to be reversed. And in the case of gases, as observed the gases where building up as little bubbles on the electrode so they are still in contact with it in this case.

As for the mini-explosion, the reaction is:


that is 572kJ will be released for every 2mol of Hydrogen reacted with 1 mol of Oxygen. You have 9 ounces, so ideally if you had 2 parts H2 and one part O2, that would be about 90mL. At SLC (standard lab conditions - 101kPa, 293K)
PV = nRT
n = (101*.09)/(8.31*293)
n = 0.00368mol

0.00368*572 = 2.10kJ

Thats enough to raise the temp of one litre of water about half a degree

[Vending]

Originally Posted by Jay-qu

Ever heard of a re-chargable battery?

Ever heard of the word "usually"? (hint: look at my post and you will see i used that word when making the statement to which you are replying) (further hint: look up the word usually in a dictionary) -- Sorry, i am done being snide now. Your response just begged for it though.

Yes, rechargeble batteries exist. yes, some (even most) of them reverse chemical reactions associated with redox events. However, let us remember that rechargeble batteries are designed to be this way. There is a lot of thought put into them. The conditions found within a rechargeble battery are not the conditions found around most redox events at an electrode and they certainly do not parelell the environment found in a AC water electrolysis set-up.

Rechargeble batteries are usually designed so that one of the chemical partners (which, is invovled in the chemical reaction that follows the redox event) is present in vast excess. That way when the other of the partners reaches the electrode surface both are garunteed to be present and the reaction will occur. However, I think it is obvious that this situation is not the case in most instances where there is redox events occuring at an electrode. (perhaps it is not so obvious though?)

So long as the products stay in contact with the electrode they should be able to be reversed.

This is only the case when the chemical process following the redox event is a simple one. In the case of complex (multistep and multi-product) reactions this is not usually true. Furthermore, this can only be the case when all the products are in contact with the electrode AND eachother. This is clearly not the case as far as electrolysis of water goes. Here is why; Electrolysis of water produces H2 and O2, but that is not all...

The process at the anode produces Oxygen, but that is not all that it produces. The reaction at the anode is...

H2O = 4 H+ + 4 e + O2

so we see that not only is oxygen produced, but protons (acid) as well. This acid remains in the aqueous phase while the oxygen moves to the gas phase, thus the two products are not in contact anymore (except for at the gas/liquid interface -- and these only conact the electrode surface in 1 dimension -- so they are, from a practical standpoint, not in contact). This is why you do not see the reverse reaction -- the two products are not incontact with eachother and the electrode in any sort of respectable amount.

Likewise for the cathodic process, we have...

4 H2O + 4 e = 2 H2 + 4 OH-

again, we do make H2, but we also make aqueous hydroxide. Once again (for the same reasons given for the anode) we do not have the two products in contact with eachother and the electrode and so we do not expect to observe the reverse reaction at any measureable rate.

Even worse, in the case of electrolysis, the acid and bases created at the anode and cathode neutralize eachother and so they do not exist any longer, so it is very diffucult to find the reaction partners for H2 and O2.

And in the case of gases, as observed the gases where building up as little bubbles on the electrode so they are still in contact with it in this case.

I have already adressed most of the problems with this statement in the above paragraphs, but i thought i would adress one more problem with your thinking. That is this;

Redox processes at electrodes involve the introduction of electrons or their removal. Thus, in order for charge not to build up at the surface of the eletrode your medium must be able to conduct either electricity or charge (in the form of electrolytes). Gases (especially netural ones like H2 and O2) do not conduct electricity nor do they contain electrolytes. Thus, if you were to carry out a redox process in a gas you would quickly build up a significant amount of charge on the electrodes. This then becomes a drving force against more like charged being introduced to the electrode and the redox process quickly shuts itself down as the energy required to introduce more charges near the electrode become prohibitive.

This is why electrolysis and other electrochemistry is mostly done in liquid in the presense of electrolytes and why you would not expect a quick electrode redox process to occur in a pure H2 and O2 environement.

I hope this clears up most of the confusion here. But feel free to ask anyother questions you might have (or raise any concerns you have).

[Jay-qu]

Nice response, I find terms like usually or possibly a bit of a cop out sometimes - sorry if you took offence, but I like to hear from all case senarios(sp?) before I accept something