Sciencemadness Discussion Board

Electrolysis with glassy carbon

smeesh - 2-6-2012 at 13:50

hello there sirs.
im working on a setup that is as follows:
12 volt DC current
Glassy carbon electrodes
sodium bicarbonate or sodium chloride electrolyte

the objective is to produce hydrogen and oxygen with limited unwanted bi-products.

is this feasible?
depending on the electrolyte, what other gases might be produced?
is there any way to reduce the likelyhood of producing unwanted gases?
is pure water electrolysis possible using such a low voltage?

any info would be greatly appreciated.
thanks guys

Hexavalent - 2-6-2012 at 14:16

What do you intend to do with the gases produced? How much of them do you want to produce? What kind of power supply are you using (what current can it give)?

Usually, 12V is sufficient to make small amounts of hydrogen and oxygen from water.

Sodium chloride will produce chlorine at the anode when electrolysed, so if you want the pure hydrogen and oxygen then look for another electrolyte.






[Edited on 2-6-2012 by Hexavalent]

smeesh - 2-6-2012 at 14:34

thank you hex.
one of the things i was quite interested in was where the sodium went when salt is the electrolyte.
baking soda would produce CO2 instead of oxygen, would it not?

Hexavalent - 2-6-2012 at 14:51

As you can see in the diagram, the sodium would go on to form sodium hydroxide.

Back to your original question, providing that the voltage is above approximately 1.5VDC current will flow in tap water (needed for the ion content) and electrolysis will begin. The electron flow begins to split the water molecules into positive Hydrogen ions (H) and negative Hydroxide ions (OH).

The positively charged Hydrogen ion is attracted to the negative cathode where it regains an electron to become neutral again. This Hydrogen atom joins up with another Hydrogen atom and becomes a Hydrogen molecule (H2), and this molecule bubbles to the surface.

H + H ---> H2

The negatively charged Hydroxide ion is attracted to the anode where it gives up its extra electron and becomes neutral again. The Hydroxide molecule combines with three others to form one Oxygen molecule (O2)and two water molecules(H2O). The Oxygen molecule bubbles to the surface.

4OH ---> O2 + 2H2O

So we now know that Hydrogen gas forms at the cathode and Oxygen gas at the anode. The amount of gas produced is proportional to the amount of current.

As for your second question in that last post, very simply, no - not quite. Assuming you mean aqueous baking soda (sodium bicarbonate) hydrogen and oxygen production would continue - the Na+ and HCO3- ions would just help the current flow through the cell.

smeesh - 2-6-2012 at 15:49

that is excellent to know. so, sodium bicarbonate would indeed be a good electrolyte for production of oxygen in addition to the hydrogen?

Endimion17 - 2-6-2012 at 16:55

No, it's not a good electrolyte because it has low solubility. Sodium sulphate, potassium nitrate, sulphuric acid, they make good electrolytes.

elementcollector1 - 2-6-2012 at 16:55

Keep in mind that for the salt solution, any NaOH formed would react with dissolved chlorine unless a two-cell setup for electrolysis was used. It makes bleach! :D
If you want just the gases hydrogen and oxygen, then I would recommend using a dilute solution of salt / baking soda, as the concentrated solution produces chlorine (as seen above) or carbon dioxide. (I know it's carbon dioxide just because of the vigorous bubbling and the gas will *not* relight a splint.) If you want to trap the two gases, inverted test tubes full of the dilute solution should be placed over the two electrodes.
I hope this helps!

[Edited on 3-6-2012 by elementcollector1]

woelen - 3-6-2012 at 00:16

In reality, no sodium is formed at all in aqueous solution. What really happens is that there are competing possible reactions at the cathode and competing possible reactions at the anode. I'll try to explain in detail and let's take NaCl as example with carbon electrodes.

In solution we have the following species:
- Na(+) ions from the sodium chloride
- Cl(-) ions from the sodium chloride
- water

At the cathode, electrons are 'pushed' into the solution, at the anode, electrons are drawn from the solution. The electrons which are 'pushed' into the solution must be absorbed by one of the species present near/on the cathode. We have the following candidates for absorbing electrons:
- sodium ions
- chloride ions
- water molecules
- carbon atoms from the cathode
If we look at the ease at which these absorb electrons, then we see that water molecules do so easiest, sodium ions are much more reluctant to accept electrons and pushing another electron on the chloride ion is nearly impossible. The same is true for the carbon. So, easiest is pushing the electrons on the water molecules. The following reaction occurs. Water molecules are charged, but these ions are unstable and immediately break down to hydroxide ions and hydrogen atoms.

H2O + e --> "H2O(-)" --> OH(-) + H.

Two hydrogen atoms in turn combine to hydrogen molecules, which escapes as gas:

2H. --> H2

When the electrolysis has been carried out for some time, hydroxide ions are present as well near the cathode, but these do not accept electrons, hence the the above reaction remains the main reaction.




At the anode, electrons are drawn away. The following are candidates for providing electrons:
- sodium ions
- chloride ions
- water molecules
- carbon atoms

Here, the electrons can easiest be drawn from the chloride ions, but water molecules also can provide electrons. Sodium ions and carbon atoms do not provide any electrons. So, at the anode we have the following reactions:

Main reaction: Cl(-) - e --> Cl.
Two Cl atoms combine to a molecule: 2Cl. --> Cl2

Another reaction which also occurs somewhat: H2O - e --> "H2O(+)" --> H(+) + OH. The H2O(+) ion is very unstable and as soon as the water molecule gives up an electron, it at once breaks down to H(+) and hydroxyl, that's why I write the " around the formula of the ion. Hydroxyl in turn is extremely reactive and usually reacts as follows:

2OH. --> H2O + O.
The O. atoms in turn react to form O2.
Some of the hydroxyl may react as follows: 2OH. --> H2O2
The latter reaction only occurs at very high current density and low temperature and it certainly is not favorable under ordinary conditions.

So, at the anode you get a mix of Cl2 and O2, mainly Cl2, but the amount of O2 cannot be neglected.

The ratios at which hydrogen and chlorine plus oxygen are formed is fixed. For each electron, which is pushed in solution at the cathode, there is one electron which is taken from the anode.


--------------------------------------------------------------------------------------------------------

Now suppose you use copper electrodes instead.

At the cathode instead of carbon atoms now there are copper atoms, which are candidates for acceptance of an electron. Copper atoms, however, do not want electrons and hence, at the cathode things do not change. It is still the water molecules which accept electrons.

At the anode things become really different. Copper atoms more easily provide electrons than either chloride ions or water molecules. So, with copper anode you get the following primary reaction at the anode:

Cu - e --> Cu(+)

The Cu(+) in turn reacts with chloride to form CuCl. Two Cu(+) ions also can combine to form Cu + Cu(2+). When a solution of NaCl is electrolysed, then the main product formed at the anode is a copper(I) species, most likely impure CuCl, but also hydrous Cu2O. This copper(I) containing material is contaminated with copper(II) species. You'll see this as a brown/mustard slurry, falling from the anode to the bottom. When a solution of Na2SO4 is electrolysed, then also Cu-metal provides electrons easiest and in that case, the Cu(+) ions combine to Cu and Cu(2+). The Cu(2+) ions give a blue color to the solution.

smeesh - 3-6-2012 at 08:39

this is really excellent stuff guys. i greatly appreciate all the info. most of the information i received prior to this was from hho websites...often times, they dont know a great deal about chemistry.

vmelkon - 4-6-2012 at 05:29

Quote: Originally posted by woelen  
If we look at the ease at which these absorb electrons, then we see that water molecules do so easiest, sodium ions are much more reluctant to accept electrons and pushing another electron on the chloride ion is nearly impossible. The same is true for the carbon. So, easiest is pushing the electrons on the water molecules. The following reaction occurs. Water molecules are charged, but these ions are unstable and immediately break down to hydroxide ions and hydrogen atoms.

H2O + e --> "H2O(-)" --> OH(-) + H.


Is there some scientific paper that discusses this?
Because I don't understand how the chloralkali process works then. Sodium forms and it dissolves in the mercury and the mercury is just dumped into distilled water to make some NaOH.
Therefore, I think that no matter what type of electrode you use for the cathode (graphite, copper, iron), sodium still forms and reacts with water and produces hydrogen.


[Edited on 4-6-2012 by vmelkon]

woelen - 4-6-2012 at 06:22

I understand your point. I also read about the formation of sodium amalgam.

But on the other hand, when you look at the ease at which compounds are reduced, then you see that water is reduced much more easily than sodium, so why would first sodium be formed and then water react with sodium. If this were true for all kinds of cathodes, then you would have tremendous formation of heat during electrolysis processes and a lot of loss of energy, because the energy, required to make sodium is much higher than the energy required to make hydrogen from water.

I think (but I must admit this is my personal guess and it is not backed up by any reference) that the mercury surface changes the redox properties quite a lot. It is known that different electrode materials can have a strong influence on the ease at which reactions occur. E.g. with MMO anodes one can make chlorate, but no perchlorate. The latter requires platinum anodes. With MMO, electrolysis of a chlorate solution simply gives oxygen at the anode (oxidation of water, as described in my previous post), with platinum, perchlorate ion is formed. Most likely a similar change occurs at the cathode when other materials are used.

All together, you brought up a good point and it may be worthwile to investigate this further. It also demonstrates that true understanding of electrolysis is more demanding than many people think. Actually, really understanding electrolysis at full detail is hard, because the electrons are pushed (or drawn) from ions and atoms one by one, but we observe net reactions in which multiple electrons are involved. Finding the precise mechanistic pathway for the reactions is a real challenge, especially for such complicated things like chlorate cells and perchlorate cells in which a multitude of chlorine-species exists near the anode.


[Edited on 4-6-12 by woelen]

vmelkon - 4-6-2012 at 09:08

What do you mean by water is reduced much more easily? What numbers are you looking at?

Look at this example, in the case of galvanic cells, if you have Mg/Mg++ in one cell and Zn/Zn++ in the other cell,

Mg -> Mg2+ + 2e (2.37 V)
Zn2+ + 2e -> Zn (-0.76)

we would expect a cell potential of 1.61 V and that zinc would be plated out.
Plating out is something that we know happens : chrome plating is a well known example in the automotive industry.

Yes, I agree that it would be a very inefficient way to make hydrogen if first it is making sodium and then it reacts with the water.
I guess the only way to test that scientifically is to do it in a calorimeter and try NaOH, compare that with KOH, compare that with LiOH and also CsOH and RbOH and finally H2SO4 (or some other acid).

I would expect for electrolysis to be more efficient with acids than alkali metals.

[Edited on 4-6-2012 by vmelkon]

barley81 - 4-6-2012 at 10:53

2H2O + 2e- -> 2OH- + H2 E0 = -0.83V

Na+ + e- -> Na E0 = -2.71V

To reduce water in a cell requires less voltage than to reduce sodium ions if the oxidation half-reaction is the same.

[Edited on 4-6-2012 by barley81]

woelen - 4-6-2012 at 12:32

At 3 volts there already is a noticeable current in a cell with NaCl.

Reduction of water requires 0.83 volts.
Oxidation of chloride to chlorine requires 1.36 volts
So, in theory one needs 2.19 volts. In practice, appr. 3 volts is needed (I tried this myself with a real cell, using carbon electrodes), and at 3.5 volts there already is quite some current.

If the reduction at the cathode would go through sodium, then one would at least need 4 volts according to theory and well over 4 volts in a practical cell. So, I indeed am inclined to conclude that no sodium is formed as intermediate, but that water is reduced instead.

[Edited on 4-6-12 by woelen]

vmelkon - 6-6-2012 at 05:02

Thanks for the replies. Yes, electrochemistry isn't straightforward.

Doesn't all that mean that in order to stay efficient, you should keep the voltage at 3 V? I would imagine if you raise it enough (1.36 V (chlorine) + 2.71 V (sodium)) 4.07 V, it might produce sodium.