Making Hydrochloric acid + Halogen Chemistry.

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when you say 2 amperes requires less time, does that mean more amperes = faster

Im reading the website you have provided. i will try it tommorow.

Is the load resistor just so that it doesnt go off for like 5minutes if you accidentally short circuit it but instead act like a fuse and blow out if its overloaded for a long time (like an adaptor)?

EDIT: the power supply i have has a button for turning it on, its in a black thick tube and inside that theres4 wires. brown, blue, black, white. and there a naked wire in it which seems to be the green wire and there connected to the powercord hole at one end and at the other end to the switch

EDIT#2: theres only 1 orenge theres no brown wire other then the one which is 240volts coming in to the switch

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The melting point can be brought down to 600C by adding CaCl2 and the electrolysis will still only effect the NaCl since it is easier to separate. 600C, piece of cake. Just aman

when you say 2 amperes requires less time, does that mean more amperes = faster

Of course! More current --> more electrons per unit of time --> more chlorate per unit of time. I don't want to blame you, but this question gives me the feeling that you do not really understand what you are doing.

Do you have basic knowledge of electronics? Without that, it indeed may be quite hard to understand the second part of the page I wrote.

The load resistors have nothing to do with fuses and protection of the power supply. They are there for controlling the electrolysis process. The resistors make the circuit somewhat controlled and provide negative feedback. If the required redox potential over the electrolyte increases, then the current tends to decrease. But, when that happens, the voltage drop over the resistor also decreases and more voltage remains available for the electrolytic cell. Hence, the current is stabilized.

You just want chlorate production without hassle and difficult math? You do not need precise control over your power supply with selectable currents. Then do the following:

Prepare a power supply, as I describe in the webpage in the text before the "Additional wiring and resistor networks" section.

Next, buy yourself 5 ceramic resistors of 22 Ohm/5 Watt and wire them in parallel (this makes appr. 4.4 Ohms of resistance). Put these parallel resistors in series with the 12 V supply and the electrolytic cell. With this plain simple setup you'll have a current between 1.5 A and 2 A. That allows you to make around 100 grams of NaClO3 in 4 days. You'll need parallel electrodes though with such a current, or you need large electrodes. A simple battery rod will erode way too fast with a current of 2 A.

EDIT: the power supply i have has a button for turning it on, its in a black thick tube and inside that theres4 wires. brown, blue, black, white. and there a naked wire in it which seems to be the green wire and there connected to the powercord hole at one end and at the other end to the switch

EDIT#2: theres only 1 orenge theres no brown wire other then the one which is 240volts coming in to the switch

I'm not talking about wires, which connect to the 240 V input. If your PSU is a normal PC power supply, then you'll see a lot of wires coming out, with plastic connectors, which you normally connect to a mainboard. Those are the wires which you need to modify. PLEASE DO NOT FIDDLE WITH THE 240 V WIRES. Know what you are doing!

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Of course! More current --> more electrons per unit of time --> more chlorate per unit of time. I don't want to blame you' date=' but this question gives me the feeling that you do not really understand what you are doing.

Do you have basic knowledge of electronics? Without that, it indeed may be quite hard to understand the second part of the page I wrote.

[/quote']

Uhh, yer, sorry i miss read something.

I'm not talking about wires, which connect to the 240 V input. If your PSU is a normal PC power supply, then you'll see a lot of wires coming out, with plastic connectors, which you normally connect to a mainboard. Those are the wires which you need to modify. PLEASE DO NOT FIDDLE WITH THE 240 V WIRES. Know what you are doing!

No, it made reference to a brown wire and the only brown wire i got in the psu is the 240volt one.

heres a picture:

http://img165.imageshack.us/img165/7803/dsc005335hu.jpg

im not so good with electronics, ive only done 2 years of it as part of school so i wouldnt know as much as you so dont go mad.

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also, rather than fork out alot of cash on these resistors, if you have one of the old type electric Bar heaters, the sort with the exposed NiChrome wire around the ceramic former, they make Fantastic! ballast resistors

all you need to do is clip one lead of a multimeter to the end, set it too the Resistance 200 OHM mark on the meter and then slide the other lead up the wire until you find the resistance you want, put an alligator clip on it and away you go!

most are good upto a KiloWatt

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@[w00t]: Your PSU probably is quite old. Indeed, no wires, as I mention on my website. But... I see yellow output wires, so, you certainly have 12 V output. If your PSU powers up, just by switching on, then you do not need to do all the difficult things, I write on my site. A modern ATX-PSU, however, requires these things to be done, otherwise it does not power up, even if connected and switched on. So, just use your PSU.

@YT2095: Your method may work, but these resistors are not that expensive. If you buy 5 of them, then you'll probably only need US \$3 (GPB 2). Over here, I pay EUR 0.60 for one of them, but mainland Europe is more expensive with these things.

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yer with this i just press the swtich on the other end of the black thick tube and it goes on, ill go out today and pick up them 5 ceramic resistors of 22 Ohm/5 Watt

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i havent attached the resistors yet but i had the carbon rods attached to 12volts (as shown on site) and basiclly it boiling the water, well not as much as boiling but it went from 200ml to 150ml in 4 hours, i can feel hot gas going past my hand as i wave it over the beaker. also sems like the salt got caught on one of the carbon rods, and basiclly the shit has melted right on it

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What did you do precisely? You say you didn't attach the resistors ? If you apply 12 V directly, then you are ruining all your stuff! That really does not work. You can even kill your PSU without proper fans/cooling.

Please be more precise and follow the direction precisely. The circuit should be as follows:

Take your 5 resistors and solder them in parallel. This makes one big combined resistor.

Next, take a wire and connect this to the + of the PSU. At the other end of the wire, connect one side of the combined resistor. At the other end of the combined resistor, connect another wire. At the other end of that second wire, connect the graphite anode.

At the - of the PSU, connect a wire and at the other end of that wire, connect the cathode (which may be a copper plate, iron plate, but also may be graphite).

With this setup, you dip both electrodes into the liquid and let it run for a few days. That should work and should not melt down anything, nor make a shit out of it. Occasional replenishing of water may be necessary. The liquid will become warm, but not really hot. The resistors can become quite hot, but that is no problem as long as they are at least 5 W. They are designed for that. The graphite anode may erose considerably and you'll get black carbon powder in the liquid, but that should not do any real harm.

Remember, the total graphite surface, in contact with the liquid should be at least 20 cm², preferrably even 40 cm². Optimal current density is 50 mA/cm², but 100 mA/cm² still is acceptable, although that will erode your anode somewhat faster. You know how to compute/estimate the total contact area?

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yup, thats Perfect

although Ide still stay with the 5 volt supply and not the 12 volt rail.

btw, Im not surprised it got hot before and you lost water mass, you do anyway even at the correct power, I ran a Perchlorate cell for just over a week (Making Barium Perchlorate) I had to top it up 3 times! and yes it does get a little Warm (not hot).

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The drawing indeed is correct. YT, using 5 V can be done, but then without the resistors. However, as I stated before, you don't have the benefits of current control in that case. In the beginning the current probably will be too high and at the end it slows down too much. With the suggestion I gave, the current can be controlled much better. It will start of somewhere just below 2A and at the end it will still be above 1.5A.

The most ideal thing would be a real current source instead of a voltage source, but that requires more electronics (we discussed that before, e.g. use of an LM317). The resistor network in series with a higher voltage like 12V provides sufficient current stabilization.

Of course, with this setup, there will be somewhat more power consumption, but for a home/hobby setup that is no concern. In an industrial setup I would advice something totally diffierent, where energy consumption is minimized. For a home setup, I, however, advice the setup, which has the most consistent results over extented periods of time and which has the same effect for almost all kinds of electrolytes used.

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you do realise that youll be pulling a potential nearing 33 watts through resistors rated at only 25W?

there is alot to be said for using the Bar heater element!

Or indeed a more reactive load (with visual indication) of a Car headlamp Bulb

that is IF youre insistant on using the 12v rail.

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The power dissipation in the resistors will be much less. Suppose that your cell takes 4 volts (which is quite standard, it may take a little more). Then 8 volts are across the resistor network.

P = v*v/R = 8*8/4.4 = 14.5 Watts.

Your computation only is valid when the resistors are short circuited and the full 12V is applied over the resistor without electrolytic cell, then the dissipation is 12*12/4.4 = 32.7 W.

I do this kind of electrolysis experiments many times and it works like a charm and no, I've never blown out or overstressed resistors . So, [w00t] can safely use this resistor network without fearing overheating of the resistors.

The nice thing is that the electrolytic cell has an electric characteristic, which closely resembles that of a Zener-diode in series with a small linear resistance. The voltage across the cell almost is independent from the current, it does, however, depend on the chemical composition of the electrolyte. At higher chlorate concentration, the cell takes a higher voltage.

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Im a little confused here, when it says connect 3 yellow wires together.. now i know why that is done but why not more then 3?

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I find this strange indeed.

my psu is a 3 to 5 amp, 3 amps constant, 5 amp surge (should never be maintained at this!).

I run with an Ameter in line and also a Voltage meter across the psu terminals so that I may monitor all occurances.

I note that even at 5 volts a cell will draw alot more than the 3 amps, at near a dead short, but at 3 volts it works quite nicely and the current drops slowly over the day or so.

if I even TRY 12v it gets insanely hot! (as I think Woot told us about also).

no cell should exceed 7 volts really and 200ma cm^3 (and thats for a PERchlorate cell).

basicly the cell at 1st presents near as a dead short, its not until much later that the resistance drops, and so if I can pull a good 5 amps at 5 volts across my cell, then I consider this to very close to a short cct. ergo nearly 33w in your system.

'']Im a little confused here, when it says connect 3 yellow wires together.. now i know why that is done but why not more then 3?

you CAN connect more than 3, it just gets a little harder unless you strip them back a good inch

BTW, please be carefull, as Switch Mode PSUs do Not like being run Open-Circuit, Some form of resistive load should be applied at all time, that includes the Unused outputs (the +5 -5 and -12).

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YT, I still disagree with the 33W across the resistors. I assume you agree your cell works nice around 4 volts but draws a LOT of current at 5 V and hardly draws any current at 3 V. This is a common characteristic of electrolysis cells.

Let's do an experiment of thought .

Now think of a zener diode, rated for 4 V. Connect a series resistor between this and a 12 V power supply. What voltage do you expect to be across the resistor? What is the power dissipated by the resistor and what is the power dissipated by the zener diode?

Now, what happens if you apply 5 V across a zener diode, rated for 4 V? The same for 3 V applied across the zener?

An ideal zener acts as an insulator, when the applied voltage is below its rated voltage, it acts as a short circuit, when the applied voltage is above its rated voltage. Such a component, when used in series with a resistor acts by simply taking away a certain voltage. The voltage, available for the remaining circuitry, is the PSU-voltage minus the zener-voltage.

The electrolytic cell has a characteristic, which is close to the zener characteristic and hence it simply takes somewhere around 4 volts, regardless of the current drawn through it. Only at VERY high currents ut takes 5 volts and at very low currents it takes less. However, if a voltage is applied, then the current will become insanely high, when it is above the "cell zener" voltage and it will be almost 0 when below that voltage.

With this explanation you understand why the resistors of [w00t] only will dissipate around 15W instead of 33 W? Really, just try it yourself with real resistors.

@[w00t]: Why 3 wires and not more? It does not really matter. The more wires you strip , the less resistance you have in all your overall wiring (all wires of the same color are simply in parallel). But 3 wires already have a very low resistance, so there is no real need to strip more of them. But if you want to do that, feel free to do so, without consequences.

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Numbers Mine.

1) Now think of a zener diode' date=' rated for 4 V. Connect a series resistor between this and a 12 V power supply. What voltage do you expect to be across the resistor?

2)What is the power dissipated by the resistor

2.5)and what is the power dissipated by the zener diode?

3)Now, what happens if you apply 5 V across a zener diode, rated for 4 V?

4)The same for 3 V applied across the zener?

[/quote']

firstly it would all depend on the diodes Bias, but in Reverse bias, I shall answer (as thats how these are most commonly employed).

1)12 volts - the 4 from the Zenner breakdown, Total 8 across the resistor.

2)3 times that the Zenner (Power is impossible to determine without the resistance value).

2.5) as above but 1 Third of the resistors.

3)you get 4 across the zenner and 1 across the resistor.

4)3v will not pass.

however I STILL get significant power drain in Current at even 3 Volts across an NaCl (saturated) cell. its practicaly a dead short even at that but not quite as bad.

I think maybe the Best thing to do, is to have the Ameter and Vmeter hooked up, and then plot a graph, Voltage on the Y axis and Amps on the X axis, and maybe come to some truth that way, besides, experimental evidence is worth a TONNE of theory (imo).

cant say When Ill be able to set this up as my cell is in use as a Paint-brush holder at the moment

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YT, that is a nice experiment you suggest. I can perform that experiment. I have an adjustable power supply from 0 to 30 V, with a built in Ampere-meter.

I can prepare a solution of 100 grams salt in 500 ml of water and then measure current as function of voltage. This gives a characteristic I(V).

Using that graph, we can determine the voltage across the resistors for a certain concentration. But right now, already, I am quite confident that this voltage will not be anywhere near 12 V. At most 9 V (and then they are still within tolerance) but probably somewhat lower.

This weekend I'll setup such an experiment. Simple to perform and easy to reproduce by other people. I'll come back on this.

What I'll determine precisely in the experiment are the following:

- Concentration of NaCl in grams per liter.

- Distance between cathode and anode in liquid.

- Surface area of the anode and cathode, which are in contact with the liquid.

- Material used for cathode and anode. I'll use graphite rods in all experiments.

I think these are the main parameters, affecting the outcome. If other parameters come to mind, then could you please let me know?

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sounds Fantastic!

Ill do same here also, although youll more than likely have to help me out with the maths of Surface area, the rest I can do easily (I just hate Maths).

beyond that, I can replicate all you suggest, mines also 0 to 30, and I have the graphite carbon.

for the cell though, youll need to take a reading before AND after the resistors, Ill be running My cell at the 5V level, you may do the 12v level

my PSU would FRY resistorless at that voltage.

can we agree upon NaCl at 400ml MAX, as my cell cell wont hold 500ml volume

as for the rest, this looks like it should be an interesting peice of data to have.

for me itll be resistorless (no ballast), but I wish to see in increments of .5v the difference in current passed through a new unreacted cell upto 5 Volts and take note of the Current consumed during each step.

also, can you think of a Quantitive test with which we may establish the Highest Chlorate result when finished?

we should also fix a time limit too, Ill be ok for a few days running anyway, WHEN I get my Cell back

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2 days is over, for my 30gram batch. it was bubling very low this morning aswell..

anyway i boiled it down to a point where there were a lot of crystals visable, i took them out with a spoon and sun dried them, ok well it melts with a match flame so i know its actually not chloride, but doesnt do anywith with Al power, charcoal or sugar

edit: uhh , the filter paper used to remove the carbon crap, i dried that up aswell, i lit it up and it was very slow, there was a lot of crackling sounds coming from it with salt jumping in the air and etc..

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well the 1st visible crystals that you see will be unreacted Salt(NaCl), the liquid would contain the chlorate (depending how far you boiled down to).

unreacted salt Will "pop" when heated, its water expansion traped in small fissures.

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i boiiled it half way and then the liquid got really thick, a paste crap was forming on top and the crystals looked like snake skin.

also, when i was boiling it, i took it off heat, and i swirled it a few times and it started to bubble, fizz like a coke.. what does this mean?

p.s im redoing the experiment again (30grams) ill post a back in 3 days this time

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@[w00t]: Nice to see that you have progress [w00t]. Try to dissolve your stuff (mix of NaCl and NaClO3) in as little as possible hot water and do the same with some KCl and then mix. Probably you'll get nice crystals of fairly pure KClO3 on cooling down, which should give a reaction with charcoal and so on.

@YT2095: I've done the experiment. I'll make a webpage of it for my website and post a link to it later this day. I did a lot of math as well. I used 33 grams of salt, dissolved in 150 ml of water. These amounts were chosen fairly arbitrarily and were most practical for me. I used a 200 ml beaker.

I'll try to determine the exchange current density for production of hydrogen at a carbon rod and for production of chlorine at a carbon rod. I use Tafel's exponential equation for describing the I/V characteristic of the electrode/electrolyte interface. That is a fairly good approximation at the current densities I use.

What I can say now already, is that [w00t]'s configuration with the 4.4 Ohm resistor in series with the cell is perfectly safe. He'll not obtain more than 15 W of power dissipation in the resistor, probably even less. This can perfectly be shown by my experiments.

More on this follows when the webpage is finished....

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@[w00t]: Nice to see that you have progress [w00t]. Try to dissolve your stuff (mix of NaCl and NaClO3) in as little as possible hot water and do the same with some KCl and then mix. Probably you'll get nice crystals of fairly pure KClO3 on cooling down' date=' which should give a reaction with charcoal and so on.

[/quote']

um wouldnt on cooling NaCl also crystalize with the NaClO3. when NaCl has less solublity in cold water, the chances are that more NaCl will recrystalise since NaClO3 is rather highly soluble?

also since im running out of batterys to open up, could i use the grahite from leadpencils? like would it have a bad effect on the solution since pencil graphite is also binded with some sorta clay

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@[w00t]: You should add KCl to the mix of NaCl and NaClO3. KClO3 is much less soluble than any other compound and that certainly will separate well before anything else separates. Of course, you will not get is 100% pure the first time but with a recrystallize step afterwards, it should be pure enough for your purpose.

Using a pencil stick is not a good idea. These are very thin and also contain a lot of crap. They will have a very low contact area and erode like hell.

===============================================

@YT2095: I did the experiment and I put it on my website. I have derived a lovely model of the electrolysis cell, with a few diodes, zeners and a resistor. With this model at hand, you can very well predict what happens when you connect resistors in series with the cell. You can also predict what current will run at a certain voltage and how sensitive it is to changes in voltage.

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

The last part of this page contains a lot of math (the Tafel approximation and the Butlet-Volmer equation), but if you understand that part, then you understand, why I came up with the model, as presented. There is a beautiful resemblence between the overpotential of an electrolysis cell and an ordinary semiconductor diode. This, however, can only be understood, by working out the Tafel approximation for the electrode/electrolyte interface and the understanding of the exponential relationship between voltage and current for a semiconductor diode.

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