Jump to content

Water electrolysis


speakerguy

Recommended Posts

Hi,

 

I would like to produce a small flame, perhaps about the size of a golf ball or a little smaller, by burning hydrogen and oxygen. I understand that electrolysis of water can be done more efficiently by adding something like sodium sulfate to increase the conductivity. I need to know if it is feasible to electrolyze enough water into H2 and O2 to produce the size of flame I need (well, actually two flames this size) given the power I can get out of a wall outlet. Please let me know the required voltage and amperage and I can figure out what kind of transformer/rectification is necessary (is DC really needed if I am just going to be mixing the two gases, or can I hook up AC?). Thanks in advance,

 

-speakerguy

Link to comment
Share on other sites

If I understood you properly, what you are trying to do is not feasable, the oxygen and hydrogen combust at a much higher rate than the electrolysis that produces the gases. Now if you want to store them and burn them later thats another story.

 

"given the power I can get out of a wall outlet."

 

Electrolyzing using a wall outlet is not a good idea since its very dangerous and you run a risk of getting electricuted. There is no specific requirement in terms of amps or volage, but the time it takes to electrolyze a given amount is directly related to the amps supplied. and using the faraday constant you can calculate how much time it will take to generate n moles of gas.

Link to comment
Share on other sites

I am an EE so no worry about electricty, I am more scared by the kW of lights over my reef tank :) now that is dangerious. also electrolysis is easier since my only other choice is to ionize air to get the plasma i need == ozone, or have bulky helium tanks laying around so i can ionize it and not get o3.

 

I can use a 2kVA step down trafo to get the voltage down to the absolute min required for electrolysis and get as much amperage out of it as possible. I can also calculate moles of gas produced but I have no idea of how many moles of gas are consumed by a burning flame of the required size.

Link to comment
Share on other sites

You ever watch a video of old hydrogen filled blimps go 'boom' ? Pretty damn fast, I don't know if such values exist (combustion rate of hydrogen & oxygen) but i can safetly say that it will burn faster than you can generate the two gases at any rate. I tried something similar with Acetylene buy 'dunking' chunks of calcium carbide in water and lighting the end from which the gas escapes. It burned out quickly.. stayed lit for a few seconds only. As for your comparison between the size of the flame to a golf-ball. If you are trying to produce a circular shaped flame, that is only possible in microgravity. Unless you can recreate such conditions, you are stuck with the standard conic type flame you usually see.

Link to comment
Share on other sites

Ha, I see that you have quite a lot of knowledge of electronics. Then I would say, build yourself a current source of 1 A (this can be done with an LM317) and use that for the electrolysis. You definitely do not want to use AC, as that is extremely dangerous and using wall outlet voltages (115/230) is straightout idiot.

 

If you want things really simple, then use a 5 V power supply and a fairly concentrated solution of Na2SO4 or dilute H2SO4.

 

Another cheap but much better way of doing electrolysis is using an approximate current source. I have made this from a 12V power supply (old PC PSU) and a series resistance. See

 

http://woelen.scheikunde.net/science/chem/misc/psu.html

 

This page you should definitely understand. I've written this with people in mind, who absolutely have no knowledge of electronics, and only have knowledge of chemistry.

 

You need to store the H2 and O2. Even with 2 A you'll net get sufficient H2 / O2 at once. Maybe you could use a higher voltage source and put cells in a series circuit and adjust the current to 1 ... 2 A through all different cells. The cell voltage is somewhere from 4 ... 6 volts at normal operation. I've investigated that as well. Maybe this characteristic can help you determine the type of power supply you need.

 

http://woelen.scheikunde.net/science/chem/exps/electrolysis/index.html

Link to comment
Share on other sites

OK so it looks like I am going to be wanting a DC current source based on your links. My education was mainly in computer engineering and DSP so remembering some of this stuff is from a ways back. Could I get an isolation transformer (or hell even hook a step down trafo in reverse so long as I watch my VA ratings to get mega voltage), full wave rectifier for the isolated 120VAC, craploads of filter capacitance (like 50kuF or more at 200VDC), appropriate fuses, and power resistors to set the current drive? It wouldn't be regulated but it would be decent enough I think. I have access to 50-100-200w alu heatsinked power resistors for audio amplifer bench testing, 4 and 8 ohms, with quad resistors I could have anywhere from 1-32 ohms source resistance. Is there a formula for impedance of an electrolyte solution based on ion molarities (isn't there another term for that? genchem is way back too) or should i just treat it as a short. I could check just how 'regulated' this current source would be based on load impedance fluctuation and voltage drop across the load to get currents at best and worst cases. I know this sounds dangerous but I have a healthy respect for electricity, heard too many bad stories about 600V-1kV B+ supplies getting someone when they got a little lax with their tube amps open.

 

Re: acid, I think I would prefer a neutral salt like Na2SO4 if it would be equivalently good. This is already getting a little scary, mega amperage through aqueous sol's, h2 and o2, flame, high freq/voltage RF. So strong diprotic acid = not really my thing. plus my dog might decide to go drink it. but man would this be an awesome technology showcase.

 

The second link is quite interesting. I never would have though the IV curve would be nonlinear. Would it also be nonlinear for H2SO4 or Na2SO4? I assume they are both strong electrolytes. Can i find those IV curves for different solutes or predict them with an eq or must it be experimentally determined?

 

Thank you for your post, you have been quite helpful!

Link to comment
Share on other sites

have you considered using a rod welding transformer for isolation?

Plenty of power from those puppies :)

and almost harmless voltages at very high current.

If you go the direct mains route, you`ll need significant ballast in line, a bar heater or maybe even a popup toaster should do, although I`me quite against using mains directly, even with ballast or variac.

 

the key here will be to keep the voltage safely low, the current High and have lots of electrode surface area rockwool sheeting inbetween the plates is a good idea too, prevents potentialy lethal mishaps for accidental plate shorting.

 

just a few extra ideas :)

Link to comment
Share on other sites

The second link is quite interesting. I never would have though the IV curve would be nonlinear. Would it also be nonlinear for H2SO4 or Na2SO4? I assume they are both strong electrolytes. Can i find those IV curves for different solutes or predict them with an eq or must it be experimentally determined?

The I/V curve actually is very similar for these solutes. The model, which I derived on that page is valid for any ionic solute, the only difference being the parameters. The forward voltage drop and the redox potential can be somewhat different. For the solutes H2SO4 and Na2SO4 you can take a redox potential of appr. 2 V and a forward voltage drop of 500 mV to 1 V.

 

the key here will be to keep the voltage safely low

As YT pointed out, you need only volts for electrolysis, not hundreds of volts or kilovolts. Use high current and low voltage, but keep in mind that the current density at your electrodes does not exceed 100 mA/cm². With an oxygen producing cell, it would even be better to keep the current density as low as 50 mA/cm². So, you need large electrodes.

 

Of course, if you have a high current high voltage supply, then you can put cells in series, but collecting the gas from all different cells separately will be quite difficult and require a lot of apparatus. Never mix O2 and H2 in more than ml quantities. That definitely will lead to heavy explosions and accidents.

Link to comment
Share on other sites

I want to add that one of most serious problems you will encounter when using electrolysis is destruction of electrodes, especially anode. So you indeed need to use low current densities and anode must be made from graphite and better yet from platinum.

 

In my opinion NaOH is better to use than acid or salt electrolytes as this gives you another choice of electrodes - you can use iron or stainless steel and this wil not corrode much if electrolyte is NaOH solution (aprox. 10%).

 

Indeed you need to use low voltage power supply. This is a must because even if you get cell resistance high enough to use, say 220V then excess voltage (everything that is over some volts) just gets converted to heat. So if you put 10A @ 220V through your cell something like 3V x 10A = 30W does your electroolysis and 217V x 10A = 2.17 KW starts to heat your electrolyte. So you get boiler instead of electrolyser.

Link to comment
Share on other sites

You really shouldn't EVER use platinum when you are working with hydrogen and oxygen gas. Platinum is an incredible catalyst and if for some reason you get some oxygen and hydrogen mixed together, the platinum will instantly ignite it and you could have some trouble. At room temperature, platinum metal will catalyze the combustion of hydrogen and oxygen. I've done this myself by taking a balloon that I filled with 2 parts hydrogen and one part oxygen and popped it with a tiny piece of platinum wire attached to the end of a meter stick. As soon as the balloon popped, the H2/O2 mixture ignited with a VERY loud report.

Link to comment
Share on other sites

First of all I should mention cost is no restriction in this system, my speaker systems routinely cost $1000+ for the high end designs that I make. This being an engineering exercise which I hope to publish in an attempt to land myself a job at a loudspeaker complany, it has no cost limits since it could gain me future employment.

 

Second, about the high voltage, I am not talking about applying the high voltage to the electrolysis setup itself - i'm not putting 100+ volts into the cell. I am thinking about supplying it with a current source. An ideal current source has infinitely high output impedance - this means that current delivered to the load will be entirely independent of load impedance. Since infinite impedance is impossibly, the best way to get an approximation is with very very high source voltage, and very high source resistance. Say I had a 10V source and a 10ohm resistor in series to give a 1amp current delivered to the cell. Let's said the cell resistance varied from 2 ohm to .5 ohm depending on concentration of H2SO4 as the H2 and O2 evolved from the electrolysis setup - I am just pulling these numbers out of thin air, not sure if they are accurate. Anyway, system current would vary from 10/(10+2) = .83 amp to 10/(10.5) = 0.95 amp. If I instead use a 168V source (rectified isolated 120V) with a 168 ohm resistor, and the same 2 ohm to .5 ohm resistance of solution based on varying concentration of H2SO4, current variance is only .988 amps to .997 amps - giving a more stable evolution of H2 and O2 as the solution is electrolysed. I think the actual voltage across the electrolysis setup would be only that required to electrolyse the solution based on it's I/V curve like what was discussed above. Is my thinking on that wrong?

 

Also regarding the electrodes (anode in particular) - is there any solute that would make it possible to use anodized aluminum? I have access to very large, very high surface area aluminum plates in the form of old power amplifier heatsinks. I could have custom iron or stainless steel electrodes machined but that would be quite a bit more expensive (I use emachineshop.com for that kind of stuff, minimum setup fee looks to be about $100 for any type of milled part in single unit quantities).

 

Again I thank you guys for the very helpful information, I am primarily an EE guy but I find this kind of stuff very interesting. I think it is sad that in my 4 years as an electrical engineer we never even learned how batteries worked (2nd sem genchem wasn't required :( )

 

Thank you very much again!

Link to comment
Share on other sites

to speakerguy: Anodized aluminium is not suitable with any kind of electrolyte as far as i know. It gets oxidised by acid or salt electrolytes and imepedance of cell rises to kiloohms or even more. This is in fact how common electrolytic capacitors are made. In alcaline electrolytes Al just dissolves with evolution of extra hydrogen.

 

Stainless steel may be best you can get. In this case NaOH must be used as electrlyte. Another way is to find some large graphite plates. These are used in industrial electrolysers and might be available. In case of graphite you can use acid or salt electrlytes as well but current density must be lower. Otherwise graphite my pulverise with fast rate.

 

Your computations on resistor based current controll are ok but more practical way is just to get some power supply that has low output voltage and high current rating. Even 3V should be enough and amperage limit should be as high as possible. In industries they use kiloampers for single cell. That of course means that electrode surface must be 1m2 or more. Power supply does not need to be precisely current controlled (although this is good) but it should surely be current limited so that you do not need to worry when cell resistance drops to unexpectedly low level. In industries they use switchmode PSU-s specially designed for this purpose. Powerfull transformes (similar to those that are used for electrical welding) can also be used if suitable recitifier (4 power diodes) is aded. Hobby chemists usually use computer PSU-s or car battery chargers. But as i said there must be series resitors or some other tricks to protect PSU if cell wants to draw dangerously high current.

 

to jdurg: good that you pointed this out. I have not played with platinum myself and did not think of its catalytic properties. It still can be used in hydrogen-oxygen producing cells (and is sometimes used) but it seems that cell must be carefully designed to prevent any mixing of oxygen and hydrogen.

Link to comment
Share on other sites

Using a very high voltage source and a high resistor gives indeed a good current source approximation, but that definitely is not necessary. I would not use a plain voltage source, but the setup with the PSU, which I put on my website, gives good enough current control. Electrolysis is not very sensitive.

 

Just some calculations. Assume the electrolysis cell operates at 4V and 1.5A. With my setup, there is 8V across the series resistor. Now, suppose that due to temperature changes, changes of chemical composition or whatever cause, the voltage across the cell changes to 5 V (which is a LOT), then still, with the same resistor, you only have a 12.5% change of current. That is not a problem at all. Even a variation of 20% is no problem at all. You still get H2 and O2, the amount per unit of time being proportional to the current. You get appr. 5*10^(-6) mol of H2 per second per ampere of current. This is appr. 0.11 ml H2 gas per second per ampere and 0.055 ml O2 gas per second per ampere at room temperature.

 

So, using coarse approximate current control is more than sufficient. So, I would suggest you to use a 10 .. 15V power supply and use a series resistor. In that way you have sufficient current control, while keeping dissipation at an acceptable level. When you use a 100+ V source, then you'll dissipate enormous amounts of energy in the resistor and that it totally useless, unless you want a room heater at the same time.

 

As Raivo stated, NaOH indeed is a nice choice of electrolyte. But, you have to be very careful with this. It is more corrosive than dilute sulphuric acid. It slowly dissolves your skin and almost instantly blinds you when it comes into the eye. So, be very careful with this, especially if you use larger quantities. Never dunk your hands in a solution of this, your skin will become very slippery, due to the dissolving powers of this solute. This stuff, however, is very nice on electrodes. Even plain copper wire, as used in electronics, is very suitable. It does not dissolve, nor erode. Graphite rods erode (pulverize), when used as anode.

Using copper or iron anodes with acid or salts solutions does not work. Under those conditions, the anode dissolves and no oxygen is formed. The cathode is less critical. You can use a copper cathode without problem in all situations and an iron cathode can be used as long as the liquid is not acidic.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.