SpudFiles

I've been trying to produce hydrogen as a propellant gas, and I think I'm finally making progress, but I'm a little lost as to the physics and chemistry combination I ended up with.

I've prepared a saturated solution of phosphoric acid, along with two graphite electrodes with leads. I connected a 1.5V D-cell battery, and no bubbles. So I just grabbed the 9V battery, and plenty of hydrogen poured off the cathode, but to my dismay, some oxygen was produced as well. With two 1.5V batteries in series, there is a tiny bit of hydrogen produced, and no apparent production of oxygen.

Why does the reaction require so much voltage to proceed?

This is based on how a fuel cell works. Hydrogen and Oxygen have an electrical bond. They need a higher voltage than the bond voltage to separate them. This bond voltage in reverse is used in a fuel cell to get the voltage and current back when they combine to produce water.

Try 2 or 3 D batteries. It is lower than 9 volts. 1.2 volts should start the process, but at a very slow rate. 1.5 volts is barely above the minimum. A low battery will stop it.
http://www.crscientific.com/electrolysis.html

Electrolysis of water will begin around a minimum of 1.2 volts and will increase in rate as the voltage is increased. Typically, the electrolysis is carried out around 6 volts.

A higher potential difference (voltage) applied to the cell means the cathode will have more energy to bring about reduction, and the anode will have more energy to bring about oxidation. Higher potential difference enables the electrolytic cell to oxidize or reduce energetically more "difficult" compounds.

Higher voltage typically decreases the efficiency by causing the water to heat up and the electrodes to corrode. Large flat plates, closely spaced, are probably your best bet since they offer a better connection than, say grapite rods.

Don't mean to be rude Tech, but your post is to me mostly patronizing. The breakdown voltage of water is 1.34V, based on the SHE. And in any case, I'm not trying to break down water (I think), I'm trying to reduce the H⁺ that's dissociated from the H₃PO₄. That's where I'm confused. That should occur at lower voltages... right?

Well in any case, it seems I've proved my own theory wrong, and it's impossible to produce hydrogen by electrolysis without evolving a gas on the other electrode Back to the "drawing board"

Might as well close down discussion on this thread, awaiting a final product thread.

If you're reducing the dissociated hydrogen ions, doesn't that mean that you're going to have to have an oxidation reaction to maintain the net neutral charge of the solution? My chemistry knowledge isn't brilliant, but what's happening to your phosphate ion if that is the case? Something has to happen to maintain the ionic constant of the water!

I'm pretty sure what you're seeing is simple breakdown of water as in standard electrolysis, with the acid simply acting as an electrolyte.

saefroch wrote:Don't mean to be rude Tech, but your post is to me mostly patronizing. The breakdown voltage of water is 1.34V, based on the SHE. And in any case, I'm not trying to break down water (I think), I'm trying to reduce the H⁺ that's dissociated from the H₃PO₄. That's where I'm confused. That should occur at lower voltages... right?

Well in any case, it seems I've proved my own theory wrong, and it's impossible to produce hydrogen by electrolysis without evolving a gas on the other electrode Back to the "drawing board"

Might as well close down discussion on this thread, awaiting a final product thread.

Breaking down H+ from the acid is identical to breaking down H2O. Indeed, there is no such things as "H+" in a water solution, it is present as H3O+.

Like tech said, you need a bit of over voltage since the SHE or SEP are the minimum voltages, not the practical voltages. You get higher production rates at higher voltages at the cost of efficiency and selectivity. (Selectivity being that other ions in solution may start to participate in REDOX reactions.)

Was Tech patronizing? Perhaps, but then again you are sufficiently uninformed to not know that oxidation and reduction reactions always happen together. If you generate H2 gas then you also generate O2 at the other electrode. If you could actually just convert 2H+ to H2 using your setup the reaction would explode since there would be a buildup of negative ions. Positive and negatively charged ions are alway balanced in chemistry, if you remove H+ then you must also remove some negatively charged species.

To produce a pure gas you need to separate the two electrodes. Fnord's suggestion is how you get maximum production of gas with minimal losses but you get a stoichiometric mixture of H2 and O2 since with closely spaced electrodes you can't keep the gases from mixing.'

Perhaps the most practical, and efficient, setup is to use 5 or 6 cells in series powered by say a 12V car battery or car battery charger. The multiple series electrolysis cells mean you only drop 2 or so volts across each cell and the huge power content of the battery (or big ass charger) can provide the needed current. A fresh 9V battery will source perhaps 1 or 2 amps for a short period of time. 6 good AA batteries in series will source perhaps 10A for a short time. A car battery will source 10A for hours, and the charger will probably do more like 15V at 30A or so indefinitely.

To be honest, I was kind of hoping to initiate some sort of other oxidation reaction other than that of oxygen to compensate, but I just discovered that that's rather impossible.

Since the discussion has turned so far away from the original question, I'll just continue with what I'm planning, in a somewhat vain attempt to show that not all of my chemistry knowledge has leaked out of my brain so close to AP exams.

jimmy101 wrote: Indeed, there is no such things as "H+" in a water solution, it is present as H3O+

Yeah, I might as well have mentioned it as the hydronium ion, but when writing net ionic equations it's just easier to write it has H⁺. I'm lazy.

jimmy101 wrote:To produce a pure gas you need to separate the two electrodes. Fnord's suggestion is how you get maximum production of gas with minimal losses but you get a stoichiometric mixture of H2 and O2 since with closely spaced electrodes you can't keep the gases from mixing.

Wrong, actually. After I concluded that it was impossible to oxidize another species in the solution, I did a few experiments and found a barrier that I can make an airtight seal to and will allow charge to flow, but not any appreciable amount of oxygen or hydrogen.

As to power sources: I was hoping to get the resistance down far enough to use a whole lot of 1.5V batteries in parallel, since I'm too cheap to design something with multiple cells.

I'm planning to use nickel electrodes, since those are often touted as better than stainless steel, but I'd feel more comfortable about purchasing a few if I was sure they wouldn't corrode quickly. Does anyone have evidence?

Don't mean to be rude Tech, but your post is to me mostly patronizing.

I did not intend to be that way, but I'll be first to admit sometimes my posts in trying to give the info tend to be that way. If you think I'm being patronizing simply PM me. its is less embarrassing when others see it and comment on it.

My intent on the post was simply from my limited knowledge of of the subject. Your intended reaction was beyond my knowlege so I simply followed the thread for a while trying to learn what I as missing. I try to be helpful, but I am the first to admit that I don't know it all. I'm in my 50's and still learning.

Ah. I understand the lack of knowledge business. I'm not even out of high school yet. I'll be more courteous in the future, but to be honest I think I demonstrated a greater lack of knowledge than you did.

In any case, I think it's just difficult to establish on a web forum like this what you know and don't know about a subject. It just sort of bothers me (and I have NO DOUBT I've done this to other members) that sometimes I'm looking for... higher-level? information about an object, and instead sometimes have to go through a page of posts of people explaining things to me that I already know before my question gets answered.

*End rant*

No hard feelings. We all make mistakes. Keep going in your research and good luck.

Rather than use (and pay for) a bunch of 1.2V batteries, could you get a computer PSU and use the +3.3 rail with two cells in series? In my area, 500W supplies can be had for under $20.00. They should be able to provide a great deal of current.

you can use H2O and NaCL

The bad stuff is the CL.. People have used NaCL, but Chlorine is still nasty stuff. HCL or salt, both are not recommended for the same reason. NaOH is a good substitute. It is caustic.

I was planning on NaOH or KOH for an electrolyte, since I've actually tried NaCl, and I can tell you the results are not good.

ramses wrote:Rather than use (and pay for) a bunch of 1.2V batteries, could you get a computer PSU and use the +3.3 rail with two cells in series?

Well I was actually planning on using rechargeable batteries, and just recharging them (since I already have a bunch). So far as I know, I can't acquire a computer PSU, but I'll look around.

I have a battery charger that can charge 4 AA or AAA batteries pretty quickly... could I hook up the connections like they're in parallel to use it as a power source and not fry it?

Should work fine as long as you avoid direct contact between the electrodes. I've fried a couple small transformers when my ghetto electrolysis rigs got moved and the electrodes shorted.
You can put a resistor in series to avoid this.