woelen Posted November 17, 2005 Share Posted November 17, 2005 I have done quite some electrolysis experiments of halogenide solutions (NaCl, NaBr, NaI) with graphite anodes and i have observed the following things: 1) At low voltage, the halogen is mainly formed at the anode. With bromide and iodide, the electrolysis can be totally free of bubbling oxygen from the anode. 2) At higher voltage, a side reaction occurs, in which oxygen is formed. When this occurs, there also is considerable pulverization of the anode. 3) After a long time of electrolysis and mixing, also more oxygen is formed at the anode, also with bromide and iodide. This formation of oxygen is associated with pulverization of the anode. 4) The current depends on the voltage in a highly non-linear way. Above a certain threshold voltage, the current drawn by the circuit increases sharply with increasing voltage. This threshold voltage is highest for chloride and lowest for iodide. I have an idea to use a current source instead of a voltage source. It is easy to make a current source of 1 A with a simple OPAMP and a moderate power transistor. Using standard graphite rods from batteries with a diameter of 7 or 8 mm and a current of 1 A one should not exceed a current density of 0.1 A/cm2, when the electrode is dipped into the liquid for at least 5 cm. This should prevent strong pulverization, according to Raivo. Using a current source has the advantage, that it automatically adjusts at the halogen to be electrolysed. For an iodide solution a lower voltage is 'chosen' by the current source than for a chloride solution. This eliminates the problem of using too high a voltage, resulting in unwanted formation of oxygen and associated pulverization of the anode. When the electrolysis comes to an end and the amount of halogenide is getting low (i.e. lots of halogenate) in solution, then the current source automatically increases voltage in order to keep the current at 1 A. If there is someone with comments on this, I would appreciate that very much. I'm looking for ways to do electrolysis in a more controlled way for many different electrolytes and I have the idea that a current source is better than a voltage source. Link to comment Share on other sites More sharing options...
raivo Posted November 17, 2005 Share Posted November 17, 2005 I think destruction of electrodes is mostly due of low quality of carbon we use. Graphite from pencils will be pulverized very soon, that from batteryes is better but not much. Destruction of electrodes depends also of current densitiy. Thats amount of electrical current flowing through electrolyser divided by surface area of electrode. Professional electrolysers never use more than 0.5 A/cm2 but more often 0.07 ... 0.1 A/cm2. If we use graphite rod with diameter of 7mm thats soaked 5cm into electrolyte then working area is 0.7 * 3.14 * 5 = 11 cm2, aproximately 10 cm2 so 1A of current will give current density 0.1 A/cm2. If you use graphite from pencils then acceptable current may be 100mA or even less. Current source is a must especially if you want deposit metals from solutions. If you have controll over current you can get more use of Faradays law too. And you will have better repeatable results. I made mine using only one IC: LM317 3-pin adjustable voltage regulator that can also be connected as current source. You have to replace reistor R1 with potentiometer (something like 22 ohms is good). Output current of circuit is 1.2 / R1 . Output voltage may change from fractions of volt to almost input voltage ensuring with this that output current stays at the value set by potentiometer even if resistance of load changes. For foolproofing the device i connected 1 ohm reistor to the output line so output resistance will never be lower than 1 ohm even in if electrodes in electrlyser are short-circuited (this resistor can also be used as current sensor). LM317 must be in heatsink and better yet with small fan from old computer processor. Max output current is 1.5A, max input voltage is something over 30V. 24V adapter is good for input. Not-very-high output current is drawback of this circuit. Something like 10A would be much better but this is not very realistic with reistive regulators even when using transistors to increase output power. Swithmode is needed for this but thats not thing i am currently able to design. LM317 data sheet: http://www.national.com/pf/LM/LM317.html Link to comment Share on other sites More sharing options...
YT2095 Posted November 17, 2005 Share Posted November 17, 2005 it`s the LM 317 I use also, but only as a VRef for a 2n3055 output trans. I get 0 to 35 v continuous at 3 amp constant 5 amp surge. IIRC, I also helped someone else on here with the designs also, it might be worth a search Link to comment Share on other sites More sharing options...
woelen Posted November 17, 2005 Author Share Posted November 17, 2005 Good to hear that a current source is better, so my idea was not that bad . A current source is no problem for me to make. I'll see what scrap components I have around. I can make a current source with any opamp. Btw, you do not have to worry and need to add a 1 Ohm resistor in the output line of a current source. That is the nice thing of current sources. When the output is shortcircuited, it simply gives off the current for which it is designed, so having a shortcircuit does no harm when a current source is used. Having an open circuit can do more harm with a badly designed current source, because that will cause the output voltage to be driven high and the components do not operate in their normal mode of operation. Link to comment Share on other sites More sharing options...
YT2095 Posted November 17, 2005 Share Posted November 17, 2005 another tip that I find usefull, is to get a pair of these cheap Digital multimeters, keep one on the 10 amp setting and always inline with the positive and the other on the voltage setting across the pos and neg. that way you`ll always know the current and the voltage you`re drawing at a glance, and these little meters only cost £2.99 each! I`ve actualy stuck both of mine ontop of the power supply for ease. it`s well worth thinking about! for instance, my fresh batch of Barium Chloride is drawing .33 amps at 3.95 volts at this exact moment, it`s a Chlorate cell. Link to comment Share on other sites More sharing options...
raivo Posted November 17, 2005 Share Posted November 17, 2005 Btw' date=' you do not have to worry and need to add a 1 Ohm resistor in the output line of a current source. That is the nice thing of current sources. When the output is shortcircuited, it simply gives off the current for which it is designed, so having a shortcircuit does no harm when a current source is used.[/quote'] You are right. If i started to use device for chemistry then i did not think of electronics any more, just thought automatically that shortcircuits are bad and put one more resistor there. Its interesting to note that small extra resistance in output of current source does almost no harm to the output parameters of the circuit. Its completely different with voltage sources though. YT, can you send circuit diagram or description of your supply? I have quite a lot of power tranistors somewhere so maybe i can make mine more powerfull. Link to comment Share on other sites More sharing options...
YT2095 Posted November 17, 2005 Share Posted November 17, 2005 basicly put the power rail (raw DC) after smoothing and your reservior cap, drives the LM317 as normal. BUT also the + side goes to the Collector of the 2n3055, and the output of the lm317 goes to the Base of it. all voltage is then taken from the 2n3055s Emiter and obviosly your common -. instead of using the fixed resistor values for the 317 as often drawn, use a variable resistor instead. it really is that simple btw, after the Emiter, do NOT use another capacitor! reservior or other, as it`ll hold charge and you`ll lose your instant voltage variability. imagine you`de been running a car radio off it, then decided to run a digital watch off it, that cap would still hold the 13.8 volts! edit: btw, with regards to the anode decay, in a Chlorate cell driven properly, I get very little to decay at all, it`s only when I crank up the power for the perchlorate phase that I`ll get the corrosion, when the anode`s gone, I know it`s time to stop I`ve been doing this for a long time now, and tend to work it by Eye more than maths etc... I adjust the voltage according to what I see happening and what I smell in the air, the Ideal ballance is little Anode fiz with max chlorine evolution, check the anode in an hour, if there`s corrosion, turn the power down a fraction and so forth... I`ve never made a bad batch yet! Link to comment Share on other sites More sharing options...
jdurg Posted November 17, 2005 Share Posted November 17, 2005 With the high levels of current, that would also heat up the anode a great deal. Perhaps what's happening is that some of the graphite is reacting with the forming oxygen to form little bits of CO2. That is causing the physical structure of the graphite to weaken, thus destroying your anode? Link to comment Share on other sites More sharing options...
raivo Posted November 17, 2005 Share Posted November 17, 2005 Just read from one russian book why graphite electrodes are pulverized! Oxygen indeed is guilty of this. It partly reacts with carbon making some CO2. Electrode surface changes to something porous and small particles start to fall apart. Concentration of NaCl in those pores is lower than in other areas of electrolyte and so there tends to go just electrolysis of water that gives even more oxygen. According to this book even high quality graphite is always rather porous. There is also description of process for treating graphite electrodes in vacuum and soaking into linesed oil to get rid of pores. They also tell that amount of oxigen depends on current density and is lowest somewhere near 0.1 A/cm2. There are two real life industrial electrolysers described here - one with current density 0.07 A/cm2 another 0.02 A/cm2 both use graphite anodes. edit: higher current densities are used when making peroxyacids or peroxides. Those use currents of 0.5 ... 0.7 A/cm2 and as a rule platinum anodes. Still higher current densities make even paltinum electrodes unreliable because electrode destruction. Link to comment Share on other sites More sharing options...
YT2095 Posted November 17, 2005 Share Posted November 17, 2005 that would make perfect sense for sure, in fact the whole soln warms up if done properly, the only problem with that idea is that the cathode gets just as warm so I`de doubt it`s a temperature thing. edit, we must have posted at same time my post was in response to JDurgs idea, Ravio, you said "There is also description of process for treating graphite electrodes in vacuum and soaking into linesed oil to get rid of pores" that Might explain Why Battery carbon electrodes work better then, as they are treated with Wax, heat one up and you`ll even see it come out! ) Now we`re Getting somewhere Link to comment Share on other sites More sharing options...
woelen Posted November 17, 2005 Author Share Posted November 17, 2005 Just read from one russian book why graphite electrodes are pulverized! Oxygen indeed is guilty of this. It partly reacts with carbon making some CO2. Electrode surface changes to something porous and small particles start to fall apart.[...] That is good to read. This perfectly is in line with my observations. As soon as I get oxygen production in my bromate or iodate cell, the anode starts to pulverize. Before that, all is working fine without the pulverization. Link to comment Share on other sites More sharing options...
YT2095 Posted November 17, 2005 Share Posted November 17, 2005 and also backs up mine when my electrodes last as long as they do in a Chlorate process, but not in the PERchlorate one, I get anode buubles in Perchlor And the soln gets warmer (probably enough to melt the wax). Link to comment Share on other sites More sharing options...
woelen Posted November 17, 2005 Author Share Posted November 17, 2005 for instance, my fresh batch of Barium Chloride is drawing .33 amps at 3.95 volts at this exact moment, it`s a Chlorate cell. With such a current it must take a LOT of time before you have an acceptable amount of barium chlorate. At 0.33 amps you have 0.33 Coulomb of charge per second. One mole of electrons is approximately 96500 Coulombs. So, for one mole of electrons, your cell needs to run approximately 3*96500 seconds. For making one mole of Ba(ClO3)2, you need 12 moles of electrons (each chloride converted to chlorate takes 6 electrons), so your total time needed for making one mole of Ba(ClO3)2 is 12*3*96500 seconds. This is approximately 40 days. The molecular weight of Ba(ClO3)2 is 304.23. This means that with your cell you'll get approximately 0.75 grams per day (24 h). Isn't that very slow? Probably we must use a whole array of anodes in parallel, with current evenly divided over each anode. Another possibility is to place cells in series. EDIT: The computation is wrong, 0.75 must be 7.5 With that result the conclusion is not valid anymore, see response further below the thread . Link to comment Share on other sites More sharing options...
YT2095 Posted November 17, 2005 Share Posted November 17, 2005 you`re not TOO far off the mark there with your calc I have 7 grams made in 3 days (it`s only a 300ml Cell). since it was only experimental I saw no reason to commit large quantities to this and risk wasteage on a poor result, it turns out that I had a fantastic result and thus I`m making some more Still, I have no desire to make this stuff on a Factory scale, this soln is considerably more conc than the last batch and I`m a patient man and will be happy with about 25 grams to keep and 10 or so grams to use on my Rocket project, after that I`ll be perfectly happy knowing that my Idea Worked Link to comment Share on other sites More sharing options...
woelen Posted November 17, 2005 Author Share Posted November 17, 2005 you`re not TOO far off the mark there with your calc I have 7 grams made in 3 days (it`s only a 300ml Cell). Ay ay even experts do make blunders. In my previous post I mentioned that in 40 days, 1 mole can be made. One mole is just over 300 grams, so this makes 7.5 grams per day and not 0.75 grams per day . So, your yield is perfectly possible. The efficiency of your cell is not that good, but for home applications it is perfectly suitable and indeed your concept works nicely. The losses most notably are due to 1) Formation of oxygen at the anode 2) Back-reduction of chlorate and hypochlorite at the cathode to chloride. 3) Isolation of the chlorate from the brine with chloride salt Link to comment Share on other sites More sharing options...
YT2095 Posted November 17, 2005 Share Posted November 17, 2005 I was only working with 200ml soln and not saturated when I got my 7 grams though I`m using saturated at the full 300ml now and expect a greater yield ultimately. the electrical dynamics of the cell do change over time also, so the current readings of power used Now, will not be the same as that used in the morning, and so I`ll have to adjust my cell power accordingly, I don`t mind that either, It`s nice to take an active role it should also be noted that the beauty of Barium salts as Chlorates/perchlorates is that their solubility Increases, and so I can now start with a saturated chlorIDE and end with a Not saturated (per)Chlorate Link to comment Share on other sites More sharing options...
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