Melvin

If you're going to distill the oil-ethanol solution, why not just distill the beer straight away?

It would probably just react to give the salt of sodium and aspirin (sodium acetylsalicylate?) NaOH + CH3COOC6H4COOH --> CH3COOC6H4COONa + H2O I would think the NaOH would break apart the ester though (to sodium acetate and sodium salicylate)

It depends on the concentration of the HCl that you added (its not pure HCl, which is a gas).

It doesn't make sense to me why your Al/Cu cell gives such a low voltage (maybe oxidation of copper producing reverse voltage?). I was saying that seawater was slightly alkaline, which would make it more effective than a plain salt solution. I wasn't aware that you wanted a scale that big . The size of the plates only determines the current (bigger plates = faster reaction = more current). The voltage is determined only by what is reacting; in the case of the "4'x8' Mg and a 4'x8' stainless steel plates in the real sea" part of it would be the reaction of Mg (2.4v) and dissolved oxygen (-.397v). Combined gives you a theoretical maximum of 2.797v (note that since oxygen is reduced you negate the voltage). With the reaction using just the dissolved salts in the seawater, the reaction is between Mg (2.4v) and water (-1.23v), giving 3.63v (although the current is much less because the reaction is pretty slow). All of these values came from the electromotive series in the link I posted earlier. In reality, there will be resistance in the system so your actual voltage and current would be much less. The only real way to get exact values is to test it . Hope this helps a bit

The graphite doesn't actually react; it just serves as a "return point" for the electrons. The Mg loses electrons, which flow through your circuit to the graphite, where the electrons react with the oxygen to reduce it to oxide (and probably the MgO reacts with water to make Mg(OH)2) I think the main problem with the Al/Cu cell is that the amperage is very low (because the reaction occurs slowly). I think if you used a more alkaline solution (which I believe seawater ranges from 7.5 to 8.5 in pH) you would get more current, so this could be feasible. The only real reaction here is with the aluminum, which reacts with the water to produce voltage. The more alkaline the solution, the faster this reaction occurs. The other electrode only serves to return the electrons to the water and produce hydrogen. Aluminum's value on the electromotive series is 1.67v (found at http://www.diracdelta.co.uk/science/source/e/l/electromotive%20series/source.html), so a cell like this could reach up to that voltage (although it probably won't due to some resistance). Magnesium dissolves in acid solutions, so if you used a very dilute acid solution with magnesium and carbon electrodes, you could be able to make a crude cell that could just be replenished with acid (and Mg when necessary). You could always make some kind of pump (non-electrical of course ) to disperse air into the water to replenish the oxygen supply.

Na and Cl2 aren't used because such a cell would require a pure sodium anode, which would instantly react with water anyway and would be consumed too fast. It doesn't work with ions; it only would work with the pure elements. Since sodium has to be produced with electricity anyway, there's no sense in doing this. (To be honest, using the Mg doesn't seem great to me either, because Mg needs to be produced electrolytically as well)

That's what I said. NaCl doesn't work because of the solubility of NaNO3. KCl works for KNO3, though.

Hmm, I would think the barium salt would give a green flame (like barium salts typically do). Weird that yours didn't. KCl should give hints of purple with orange. Now, the way I would do it would be to soak strips of cardstock in solutions of different ions (Na, K, Ca, Ba, etc.) and let them dry out, then burn them to get the flame colors. Copper will also give green; copper with chloride gives blue or blue-green. Not sure about the pink, though EDIT: One thing I forgot is boric acid. Boric acid with methanol gives green flames, and boric acid with ethanol (I think) gives light blue flames.

I thought the "aluminum foil in mouth" was electrochemical and requires that you have fillings. The fillings and the foil are electrodes, and saliva is the electrolyte. At least, that's what I remember...

Just make sure you seal the connection well. The "something" was probably II (as in cobalt(II) chloride). That would give off chlorine at the anode (just like table salt would) instead of oxygen. Not sure where the person was going with that one. As long as you use good inert anodes (which you are), MgSO4 shouldn't give you any problems. The only time you would get different results is if you used a divided cell with MgSO4. In that case, you would get insoluble magnesium hydroxide AND hydrogen at the cathode, and sulfuric acid AND oxygen at the anode.

Your reaction is a little off...it's actually NH4NO3 + NaOH --> NaNO3 + NH3 + H2O Be careful, though. As the video shows, the reaction is very exothermic and you could get sprayed by hot NaOH (which would suck). Ammonia is toxic as well.

Thank you Ok, I will. I was more interested in just making it to see if I could. I might try making esters, but that makes me wonder about any impurities I did use splash-proof goggles and gloves, but yeah, I really need to use more safety equipment.

The issue is where the wire (presumably Cu) contacts the graphite. If any copper is touching the solution, it will dissolve away pretty quickly to give insoluble Cu(OH)2. The pencil graphite is mixed with clay so will usually fall apart, so I use electrodes from D-size zinc-carbon (aka "Heavy Duty") batteries. These are nice, big graphite electrodes great for electrolysis. Just make sure the battery does not have the words "alkaline" or "lithium" on it. There are videos on youtube about it so you can search around. (You should still be careful with it and wear gloves when working with those batteries.)

A saturated solution would give the highest conductivity, but as soon as the amount of water drops (is converted to H2/O2) the MgSO4 will start precipitating. Since the solubility of hydrated MgSO4 is around 71g/100mL, I would say use around 15g/100mL. This should give you plenty of conductivity without using too much.

Should be pure...as long as you use an inert anode (best is Pt, graphite is okay)