woelen

Indeed, [w00t], good work . One word of caution though. Did you use the batch with the copper contamination? You cannot see that contamination, it is very pale blue, in low concentrations just white, but it is lurking there, ready to set things off when you don't expect it. The copper is not there as solid brown/red particles, but as ionic impurity, incorporated in the crystal lattice of the NaClO3. So, be careful when you experiment with this. You have to be careful anyway, chlorates are sensitive, too sensitive to my opinion. If you want a really good product, you can dissolve KCl in water (make a concentrated warm solution) and you dissolve your impure NaClO3 in an as little as possible volume of warm water. Next, you mix the two solutions and you let things cool down slowly. Take approximately equal volumes of the two concentrated hot solutions. The KClO3 will settle at the bottom, the NaCl and possibly excess KCl will remain in solution. The crystalline stuff has to be separated from the liquid and needs to be rinsed briefly with some ice cold water (not too much and not too long, otherwise you loose KClO3 again). That gives you a nice product, which also remains nice on storage (KClO3 is not hygroscopic and remains good, also over extended periods of time, while NaClO3 is prone to deterioration). Now, your product is like KNO3, but if it is purified, it is much more energetic. A glowing match, put in a heap of pure KClO3 will cause a very violent reaction with flames and smoke, and not just eating away of the wood. KClO3 is very energetic, much more so than KNO3. Try the following thing to assess the purity of your NaClO3. If you have access to concentrated hydrochloric acid, then do the following experiment: http://woelen.scheikunde.net/science/chem/exps/clo2/index.html If you also get the bright and intensely colored yellow gas, then you have NaClO3 at quite high purity. Beware, do this experiment outside, but not in bright sunlight. ClO2 is a very sensitive gas and may explode, when hit by sunlight, so be careful and use small quantities. If you do not have conc. HCl, but you have H2SO4 of 30% or better, then use that (dilute to 30% if more concentrated), combined with a lot of table salt to make in-situ HCl and then add your NaClO3 to that. That gives the same results if your NaClO3 is sufficiently pure.

The lamps use sodium in the gas phase. There is some sodium in these lamps, but it is in microscopic quantities. Not useful for experiments, it would be an insanely expensive source of sodium and you would have to put quite some effort in it to isolate it from the bulb. Before you have done that, it already reacted with humidity and oxygen from the air.

Nope, filtering is not an option, the copper is inside the crystals. Recrystallizing might be a better option, but that requires quite some skills and a lot of patience and it introduces quite some losses. Throw away the batch and make a new one. That is my advice. I hope you think your fingers are worth more than 3 days of lost time . Yes, [w00t], real science is not easy and sometimes it is frustrating. I've been working as a researcher in university for almost 5 years and I once had to redo work, on which I spent three months instead of 3 days. Whoooaaa..., but it's all part of the game.

Don't use a copper electrode when electrolysing for chlorate production. Even if it is a cathode, there will always be traces of copper ions in solution. Copper ions, however, form a VERY dangerous mix with chlorate ions. Copper ion is a good catalyst for decomposition of chlorate. Any composition made of the chlorate becomes a suicide bomb, when the chlorate is contaminated with copper. If you value your fingers (or even more), then consider using another electrode for that copper place. A stainless steel cathode is much safer.

You'd better say, there are very few common polyatomic cationic species, which are used on high schools. In reality there are many more, even quite common ones: vanadyl: VO(2+) hydrazinium: NH2NH3(+) and (+)NH3NH3(+) all kinds of substituted ammonium ions with 1,2,3 or even 4 substituted H's. nitrosyl: NO(+) nitryl: NO2(+) ...

I'm not saying that you are a k3wl, because you want KClO3. I also want/have that stuff. My problem is the 'borrowing' (stealing) of materials not the material itself. A real scientist does not steal, that person is honest. K3wls do steal if that makes things easier for them. I personally like your attempts to make KClO3/NaClO3 and I really hope you can make some of it succesfully. I am also willing to help you with that (as you may have noticed already), but if you start stealing things, then my part is over. Over here in the Netherlands, chemicals also are hard to obtain, but in due time, you will learn to know your sources.

The latter problem is not an issue at 1000 V. I understand that it is at 6.3 kV as in your apparatus, but with 1000 V I would not worry too much about the static fields. As long as wires are kept well-separated (at least 1 cm) there is no risk of unintended breakdown. However, the other warnings remain valid for 100%. High-voltage cascade circuits are no childrens toys!

Why all these dangerous and difficult things? Making a spark periodically is very easily done, especially if sparking already starts at 700 V. Have a look at the following page: http://www.oberlin.edu/physics/catalog/demonstrations/em/neonosc.html The idea is that a capacitor is charged to a certain voltage, until a breakdown of the gas/air occurs. Then a flash is produced (discharge of capacitor) and next the capacitor is recharged again. The principle behind this is that a certain breakdown voltage Vb is needed to start the sparking, but once the spark exists, a conductive (ionized) channel exists, which can sustain a current while the voltage across the channel becomes much lower than Vb. This example with the neon bulb can easily be made with a rectifying circuit, a small capacitor and a resistor from the 220 mains line. Actually, I have one of these circuits lying around here, just for fun. When you don't use a neon bulb, you need to build a cascade circuit, which multiplies the rectified voltage several times. With three cascade steps you can make 1000 V DC from 220 V AC. http://www.kronjaeger.com/hv/hv/src/mul/ A word of warning here: When you build a rectifying circuit or a cascade circuit, which connects to the wall outlet directly, ALWAYS ALWAYS ALWAYS connect resistors in series with the capacitors. Use 100 nF capacitors in series with 100 K resistors, which can withstand the high voltages. Failing to use the resistors will result in exceedingly dangerous situations! Another danger is that the cascade circuit is galvanically connected to the wall outlet. That may give an electrical shock on touching. That problem can easily be overcome by using a separation 1 : 1 transformer (these are used in many hotels etc. for making a shaver's power outlet). They are easy to obtain and it does not need to be expensive, because the transformer does not need to provide more than a few hundreds of mW of power. At the n-th stage of the cascade circuit, you need to connect an additional resistor (e.g. 1 MOhm, and a capacitor of 100 nF / 1500 V). Using that, you will have a spark every second or so (probably somewhat more frequently). Using a larger capacity gives a more powerful spark, but less frequently. Again, be very careful that you do not touch both ends of the capacitor, charged to 1000 V. Although the capacitance is small, I assure you, 100 nF charged to 1000 V can give a REALLY nasty shock.

Really ???? I cannot stop you, but don't expect any input from me anymore on this subject. Really, honestly obtained materials provide more fun! If you really are interested in science and not a k3wl, then please also take some effort and/or money in obtaining your chemicals and materials. It will make you much more respected and noone has to blame you for anything.

'Borrow' ??? What do you mean with this ? Ask them kindly and if they don't want to give you a few of those rods, then accept that and try to buy some! Really, honestly obtained materials provide much more fun . They can be found in hardware stores (welding rods), and you could also ask people around whether they have old 1.5 V batteries around (you need the zinc-ones).

If it is an ionic compound, then in front of the name of the anion, the name of the cation is mentioned. SO4(2-) : Sulfate Na2SO4 : Sodium sulfate (here the cation is Na(+)). You need to memorize the names of anions (sulfate, nitrate, sulfile, chlorate, ....). You also need to memorize the names of cations (sodium, potassium, ammonium, ....). When you see a combination of such names, then it is an ionic compound. When you only see the name of the anion, then it is an ion. There also are compounds like SO3, ClO2, NO2. These of course are not called sulfite, chlorite and nitrite, these are neutral oxides.

Jdurg, yes it does, provided the power source is capable of handling this. My experiments have shown that. The overpotential needed for obtaining a certain current is reduced by a small amount (appr. 17 mV) when the surface area is doubled, but when the same overpotential is applied at double the surface area, then twice as much current is flowing. Only the resistive effect in the solution suppresses this somewhat, but still, the effect will be quite noticeable. Look at the effect, described in my webpage. In one of the experiments, the surface area was halved, leaving all other settings the same. This resulted in a drop of current from 1.2A to 0.65A. So, [w00t], try adding additional rods and you'll see that you get more current. You of course also have to double the resistors (you need to connect more in parallel), otherwise the voltage drop over these resistors increases and the effect also is gone. Another approach, which works equally well, is to buy another set of 5 resistors and hook up another totally independent cell. They only share the PSU, but a PC-PSU easily can handle two such cells in parallel.

@[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.

Depends on you compound. Please give more information.

@[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....

I'm not going to give you all answers, but with my hints you should be able to answer your questions. 1) In both cases, hydrogen peroxide acts as a reductor. Oxygen goes from oxidation state -1 to 0. Try to determine the half-reaction for hydrogen peroxide, acting as reductor. One of the products is oxygen. 2) In acidic environments, the permanganate is converted to colorless Mn(2+) ions. What is the half-reaction for permanganate to Mn(2+) in acidic environments? 3) In neutral or weakly alkaline environments, the permanganate is converted to brown MnO2. What is the half-reaction for permanganate to MnO2 in neutral/alkaline environments?

This question has been posed many times before. Making sodium at home in a safe and responsible way is almost impossible at home with the limited equipment you have available. You need to work in an inert gas atmosphere (argon, maybe nitrogen also works) and you'll have large difficulties isolating the sodium. As soon as it comes in contact with air, it ignites at a temperature of 300 C. So, I would say, forget about sodium making at home, unless you have really good equipment and can cope with the severe safety issues very well. The average home chemist simply cannot do this.

I've been thinking about the formula at the second link. This formula for pressure drop seems to be for a stationary situation, where there is a constant gas flow. I'm not an expert on this type of systems, but I have the idea that you cannot easily derive an equation from this for the type of problem you want to solve, with a step-shaped pressure at the input.

Ah, I see the problem. How long is your pipe? If the pipe is very long, then I'm afraid that you cannot do anymore with an ordinary differential equation and that you really have to resort to partial differential equations. Then you need to determine P(l, t), where l is the length along the pipe and t is the time. This differential equation then will contain operators ∂/∂l and d/dt. Have a look at this link. It may help you. http://www.le.ac.uk/eg/ar45/eg7029/eg7029w/node10.html However, be prepared that analytical solutions may become very hard to obtain and you may have to resort to numerical processing. If the pipe is thin, compared to its length, then you can simplify the equations by only using 1 dimension instead of 3 dimensions. Your current model, however, is 0-dimensional, it regards the pipe as a single lumped object without any physical dimensions. That kind of models is very nice for electronics (almost all electronic devices can very well be described by means of lumped physical models) but for mechanical and fluid/gas computations that approach may be too limited. Another approach may be to model the system as a lumped system, together with a time delay. If the lumped system can be described as a linear system, then you can determine a polynomial Laplace transform. The time delay (time needed to flow from one end to the other end) can also be Laplace transformed. It is an exponential of s in the Laplace-domain (i.e. transcendental).

As stated in the other thread, are you sure that this equation describes what you want? Besides that, it is not a linear equation, so you cannot derive a Laplace transform. With linear equations you can do so. For finite-order systems, simply replace d/dt by a factor s and you can derive Laplace transforms. When distribution in space needs to be taken into account as well, then the only proper description is a partial differential equation. If this is linear, then you still can find Laplace transforms, but these are not polynomial in s anymore. They contain transcendental functions of s and can be regarded as infinite order lumped systems. A nice example of such a thing is a long coaxial cable with a certain resistance per meter and a certain capacity per meter. One can see this as an infinite chain of RC-combo's switched after each other, with infinitesimal R and C values. This leads to an infinite dimensional linear system, with a trancendental Laplace transfer function. Before proceeding with your problem, first try to find a better model for your tank. If the model is non-linear, but you only use that model around a certain pressure and do not deviate too much from it, then you can derive an approximating linear model, which is valid for a small area around that certain pressure. From that linear model you can derive a Laplace transform.

If this differential equation is your model, then I can say immediately that it is not a good model. Indeed, the pressure will go to its final value asymptotically. This differential equation does not, so it cannot be a good description.

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?

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.

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.