Sciencemadness Discussion Board - Does more current increase the rate of electrolysis?

Does more current increase the rate of electrolysis?

Jango - 23-11-2010 at 13:17

Hi,

I'm interested in doing various experiments, including electrolysis of various solutions. I'm searching for DC power supplies, and have found some good ones on eBay. I want the electrolysis to be done as fast as possible (i.e: larger amounts of chemical to be formed at each electrode per second).

Should I look for a DC power supply that has a higher voltage and a lower current (e.g: 20V 2A)? Or should I look for one with a lower voltage and a higher current (e.g: 13.8V 15A)?

I've been looking for something that has a constant, reliable supply so that I can leave it running so that the electrolysis finishes relatively quickly (also why I am only electrolysing small quantities).

I can't afford the ones that have a massive voltage and current, and I only want to electrolyse relatively small amounts at once (e.g: 100-200ml of a 1M solution).

Thanks.

[Edited on 23-11-2010 by Jango]

woelen - 23-11-2010 at 13:41

The only thing which counts is current. By means of electrolysis you perform redox reactions at the electrodes. At the cathode you 'push' electrons on the reactants and at the anode you take electrons from the reactants.

Current is a measure of charge per second, which is a measure of number of electrons per second. At a current of 1 A, appr. 1/96500 mole of electrons are pushed on the reactants at the cathode and the same number of electrons is taken at the anode.

As you can see, at a current of 1 A you need to 96500 seconds to have one mole of electrons transferred to/from reactants. If you want to make e.g. chlorate from chloride, then you need 6 electrons for each transformation of chloride to chlorate, so making 1 mol of chlorate requires 6*96500 seconds of electrolysis at 1 A. This is almost one week!

So, you want high current.

The voltage determines how much current is going through your cell. Part of the voltage is required for doing the redox reactions, the remaining part is just consumed as heat. Typical redox potentials are in the order of magnitude of 2 volts for both reactions at anode and cathode summed together. You also need to take into a few tenths of volts for overpotential at the electrodes. Altogether this makes up a minimum of around 3 volts before any reaction occurs.

If you want a decent current, then you also need some voltage for ohmic resistance. If this resistance is e.g. 2 ohm, your overpotential is 1 volt and the total redox potential is 2.5 volts, then you need a voltage of 5.5 volts if you want 1 A of current:

5.5 volts - 2.5 volts redox potential - 1 volt overpotential --> 2 volts remaining
these 2 volts remaining give a current of 1 A if the ohmic resistance is 2 ohm

Summarizing: Take a voltage of 5 volts for many electrolysis experiments (a simple PC powersupply does the job). If you want more versatility, take a voltage of 12 V, combined with a resistor network.

I have written several pages on the subject of electrolysis and making a decent very cheap power supply. The first link tells how to make a good and versatile power supply, suitable for a lot of electrolysis experiments and even suitable for small to medium scale production work if you also keep the 5V. The second link is an example of the use of this power supply and the making of a chemical by means of electrolysis. The third is a more theoretic page, which gives insight in the electrical characteristics of an electrolysis cell.

http://woelen.homescience.net/science/chem/misc/psu.html
http://woelen.homescience.net/science/chem/exps/miniature_ch...
http://woelen.homescience.net/science/chem/exps/electrolysis...

[Edited on 23-11-10 by woelen]

Jango - 23-11-2010 at 14:01

Thank you for your good reply.

I've tried to convert an ATX PSU into a DC power supply before, but it broke. I haven't really got the equipment to build one properly, and I would expect that buying a 15A bench supply would be better, as it would be more reliable.

So, if more current is important, then I will go with the 15A one instead of the 2A.

Is 15A enough to, for example, fully electrolyse 200ml of a 1M solution of NaCl in 12-24 hours? Or would I need more than that for it to be more successful?

watson.fawkes - 23-11-2010 at 16:01

Quote: Originally posted by Jango

Is 15A enough to, for example, fully electrolyse 200ml of a 1M solution of NaCl in 12-24 hours? Or would I need more than that for it to be more successful?

At this point, you might consider asking for help with how to do the computation rather than asking for someone to do it for you. Here's my contribution: look up the Faraday constant.

Jango - 24-11-2010 at 00:08

So, according to Faraday, if I was to do the electrolysis in one second, I would need 96,500A.

But I want to do it in 12 hours (43,200 seconds).

So, the minimum current I would need = 96,500/43200 = 2.234A am I right?

Or is it more complicated than that?

watson.fawkes - 24-11-2010 at 03:44

Quote: Originally posted by Jango

Or is it more complicated than that?

Not much more complicated than that, but your units are off. Yes, I know you put down amperes and seconds as units, but there are more units lurking (and it's not hours). Read up on dimensional analysis and you'll figure out what you're missing.

Jango - 24-11-2010 at 12:19

Sorry, I can't work out what I'm missing.

I thought:

*96,500 coulombs of charge needed per mole of electrons. So, using the I = Q/t equation, if I wanted 1 mole of electrons to be electrolysed in 43,200 seconds (12 hours), I would need a current of: I = 96,500/43,200 = 2.234A.*

I also thought (but I think I'm wrong):

*1 mole of electrons = 6.02 X 10^23 electrons. In 1 molecule of NaCl, there are 28 electrons (10 in Na+, 18 in Cl-). Therefore, I would need 28 times the amount of charge (and thus, 28 times the current) to electrolyse 1 mole of NaCl in a certain period of time. So 2.234 X 28 = 62.552A. I hope I'm wrong. A power supply with that much current would be above my budget!

But looking at Dimensional analysis (on Wikipedia), I didn't see anything about electrolysis on there. Please could you possibly give me a bit more guidance? I don't mean to be lazy, but I'm only a 15-year-old British GCSE student interested in a bit of electroysis, and all of this is above my level (apart from the I = Q/t equation, I learnt that at school).

Thank you for all of your help so far.

[Edited on 24-11-2010 by Jango]

watson.fawkes - 24-11-2010 at 13:15

Quote: Originally posted by Jango

*96,500 coulombs of charge needed per mole of electrons.

"mole" is a unit you didn't use in your previous posting. Dimensional analysis is not electrolysis-specific, it's about making sure all your units are accounted for. The Faraday constant has units, your current has units, etc., and you'll need to always ensure that you don't have any left over.

Quote:

In 1 molecule of NaCl, there are 28 electrons (10 in Na+, 18 in Cl-).

In this context, 28 is wrong, as are both 10 and 18. It's time to go read up on "half reactions".

not_important - 24-11-2010 at 13:28

One ampere-hour equals 3,600 coulombs, thus 26,8 amp-hours equals one mole of electrons; or your 1st answer of 2.34 amps for 12 hours.

The number of electrons involved is just the ionisation ones, Na(1+) Cl(1-), not the total electrons in each atom - it takes a lot more energy to pull of more than one electron from a sodium atom.

Jango - 25-11-2010 at 08:30

Quote: Originally posted by not_important

One ampere-hour equals 3,600 coulombs, thus 26,8 amp-hours equals one mole of electrons; or your 1st answer of 2.34 amps for 12 hours.

So, was I right originally?

Quote: Originally posted by not_important

The number of electrons involved is just the ionisation ones, Na(1+) Cl(1-), not the total electrons in each atom - it takes a lot more energy to pull of more than one electron from a sodium atom.

Oh yes, you're right. I forgot that! As if sodium would lose ALL of its electrons! I obviously wasn't thinking.

[Edited on 25-11-2010 by Jango]

White Yeti - 13-9-2011 at 15:28

If you want to force a lot of current through your cell, you will have to decrease its electrical resistance. One thing you could do is set up multiple cells in parallel. What this will do is decrease the combined resistance.
The general idea.

The more cells you add, the lower the total resistance, the more power you will draw, and the more amps will go through your cells.

bdbstone - 14-9-2011 at 07:51

One way to have high current power supply is to use welding machine.. Max.current goes up to 150A.

White Yeti - 14-9-2011 at 11:28

Quote: Originally posted by bdbstone

One way to have high current power supply is to use welding machine.. Max.current goes up to 150A.

It might be capable to run through a current of dozens of amps, but remember:
V=IR
As resistance goes up, current goes down, no matter how powerful your welding machine might be.

phlogiston - 14-9-2011 at 14:55

More current = faster reaction, but there is are limits. Pushing 95000 amps through a 100 ml cell for one second probably won't work, depending on the reaction you want to try. Why not?

Well (if we ignore the magnetic field that will make your cell explode) some of the intermediate reactions may proceed at a comparatively slow rate. For instance, for making chlorate from chloride, it is necessary for chlorine evolving at the anode to dissolve for it to participate in a subsequent reaction taking place in the bulk of the solution (not at the electrode). The solubility of chlorine is limited and increasing the chlorine evolution (by increasing the current) beyond the rate at which it is consumed by the reaction will only result in losing chlorine gas to the atmosphere.

White Yeti - 15-9-2011 at 12:32

Obviously there are limits, I'm just saying that putting several small cells in parallel will allow you to draw more power and more amps than if you set up a giant tub in line with the power source. One more question remains, why does the OP want to speed up an electrochemical reaction? I would rather use less power and leave the reaction overnight rather than pump hundreds of amps and crossing my fingers that nothing blows up.
Also, one more thing that the OP should consider is that passing more amps will generate tremendous amounts of heat. This can cause thermal runaway, cracking glass etc. etc...

phlogiston - 16-9-2011 at 11:59

White yeti, my previous post was not in response to you but I disagree. There is no difference between using two cells in parallel or using a single big cell with twice the volume and electrode surface area. Not in terms of resistance, yield, etc. anyway, provided the distance between the electrodes remains the same when scaling up the cell. This is easy to see if you imagine a cell and then imagine extending the length of the electrodes by a factor 2. The result is: half the resistance, as it would if you had two cells in parallel of half the size.

The only real difference I see is that a big cell will have a smaller surface to volume ratio, making passive cooling less efficient. Big cells will run in to cooling problems sooner.
There is a clear advantage to multiple cells if you have a fixed voltage power supply that is high enough to feed several cells in -series-. Or if there is a long distnce between the power supply and the celll, so you need to use a higher voltage to minimise losses in the wires runnging to the cell. Then it helps to run cells in -series- from the higher voltage.

Otherwise,you can in principle simply increase the voltage to increase the rate of the reaction in a single big cell. The rate is proportional to the current. The price you pay is reduced power efficiency (more heat).

White Yeti - 16-9-2011 at 12:37

Phlogiston, you were probably asleep in physics, but take a look at this formula:

If you plug in multiple resistances, it turns out that the overall resistance is smaller than your lowest resistance (if arranged in parallel). Going back to good old V=IR, if voltage remains constant, current increases when resistance decreases.

phlogiston - 16-9-2011 at 14:23

I did meet my wife during physics classes so I may not have been paying attention as much as I should

However, electronics being another hobby of mine, I am very familiar with that formula.
Let me introduce another formula, the resistance of a wire:

R=rho*L/A
rho = specific resistance of whatever the wire is made of
L=length
A=cross sectional area

Think of a cell as a wire. For simplicity, imagine a cell consisting of a 1x1x1 cube, two opposing sides of which are the electrodes. Then A=1, L=1, and (let's assume rho=1) therefore R=1. Now take a cell of 2x1x1 instead of 1x1x1, with correspondingly larger electrode. Then A=2, so R=1*L/2 = 0.5. As I said: increase electrodes surface area by 2, and maintain the same distance over the surface area and the resistance will be half.

Putting two cubes in parallel will have exactly the same effect, with the same total electrolyte volume and electrode surface area so you gain nothing by constructing two cells:
Req = 1/(1/1 + 1/1) = 0.5

Same result.

For more complex cell configurations, consider that current can flow from every point on one electrode to every point on the other electrode, via every possible path through the cell. Integrate over all these paths to find the resistance. I assure you the result will be the same, however.

[Edited on 16-9-2011 by phlogiston]

watson.fawkes - 17-9-2011 at 06:23

Quote: Originally posted by phlogiston

Think of a cell as a wire.

This is valid for the bulk resistance of the electrolyte, but is not a full electrical model of a cell, which has to include electrode polarization, amongst other things. For the point you were making, though, it's OK. The right target is specific electrode current, which you can't improve by parallel cells.

White Yeti - 17-9-2011 at 06:47

What you are saying is correct and I like your analogy.
However, if you want to get very technical, you are missing some key factors. The fact that the electrodes are larger does not necessarily mean that current is evenly distributed throughout the electrode. Going to your example of a rectangular prism, 2x1x1, it depends where you place the wires if you want to use all the electrode surface area available to you. You could find wires that have a cross-sectional area as large as your electrode (good luck).

Practically,when your electrodes get big enough, another factor comes into play. When you connect an electrode to your power supply with a wire, (the wire has negligible resistance, unless it's very thin), the cross section is no longer the entire surface area of the electrode. If you think of your electrode as a playing card, your cross sectional area is about the area of the shortest edge of the card, pretty darn small, (if not smaller due to uneven current distribution). Considering the fact that the cross sectional area of your electrode does not increase much when you increase the total electrode surface area, resistance stays about the same, whether your cell is big or small. The only way to increase the cross sectional area is to make your electrode thicker, (a big waste of electrode material). In a practical situation, I still think connecting up several small cells is both more economical, safer and capable of handling higher currents, than a giant cell.

[Edited on 9-17-2011 by White Yeti]

quicksilver - 20-9-2011 at 14:48

Many excellent points have previously been pointed out, yet I do want to underline how many variables complicate this topic.

I have followed this thread with a degree of interest as I have enjoyed per / chlorate electrolysis cell. experimentation for a few years now. From what I (and a friend) have determined is that the amount of variables that INTERACT is somewhat profound. So a simple query such as amount of current in relationship to yield is complicated by solution density, pH (& pH management) spacing of the electrodes, temperature of solution, type of anode/cathode construction, & depth / size of electrode contact area within the electrolytic solution. - That's a great many variables!

I was lucky enough to have teamed up with an experienced friend and we were able to cut some good deals on both MMO/Ti and graphite electrodes. I had been using computer power supplies with success yet was able to find some professional (fan cooled) switching power supplies, one of which is 150amps (@5vdc). There has been some material written which demands construction further from the simplistic designs. Pumping cells, current pulsing, auto pH handlers and even more professional designs that take considerable work to start up. Yields and crystal formation have been fantastic. yet to categorize this higher yield(s) from one element alone is very difficult. pH level itself can alter the efficiency of current usage toward final output yield.

Thus far if one were to have a switching PS such as a computer PSU and to monitor pH carefully I believe those to be the most consistent contributing factors - yet such a thing as solution temp and timing of pH adjustment can easily allow a 30a unit to preform as well as a non-pH adjusted higher current unit. I have had difficulty with containerization beyond 5 gallons due to cost and design issues. However I have found methods of reduction of graphite material and degradation through pH. I encourage people to "branch out" toward varying techniques - but records and notes become extremely important as the variable compound upon one another.

I encourage the use of a digital pH meter. Even an inexpensive one is perhaps one of the more valuable tools I have used. I also found that array and spacing require a great deal of trial and error and solution volume and density should always be recorded in conjunction with this type of experiment. Slight alterations can (at minimum) cause extra work and possibly confuse weeks of well written notes.

Personally I found that larger cells have an advantage in that they can produce visual yields quicker and make the endeavor rewarding in that source current may be utilized with a wider spread.

Chromates for stabilization have also become a worthwhile addition. I also believe that it may be a useful shortcut to purification processes.

[Edited on 20-9-2011 by quicksilver]