woelen

I did some experiments, trying to repeat YT's results. I used all reagent-grade chemicals, except the soldering wire of course. Even the sodium chloride was reagent grade in my experiment in order to rule out all kinds of impurities. Well, if I electrolyse with both the anode and the cathode, made of soldering wire, then a lot of hydrogen gas is produced at the cathode and the anode dissolves. You can really nicely see the difference in refractive index of the tin (II) salt and/or lead (II) salt, sinking to the bottom from the anode, and the solution of the sodium chloride. After a while, when hydroxide, formed at the cathode, diffuses through the entiry liquid, a white precipitate is formed. At the anode, now a white precipitate is formed directly as well. I performed this experiment, until the anode broke into little parts. At this stage, I had a white turbid liquid, with a lot of dark grey stuff lying at the bottom. It looks as if only the tin-part of the anode dissolved, the lead remaining as porous metal, which falls off as dark grey particles. So, indeed, you get two kinds of precipitate: white tin (II) hydroxide, finely divided in the liquid, and metallic lead, lying as a heavy and compact precipitate at the bottom. This most likely is YT's conductive precipitate, which at first was not noticed. I swirled the liquid somewhat and waited a few seconds. This causes the darker particles to sink to the bottom very quickly, the white precipitate remains floating around in the liquid. I decanted the white turbid liquid in a separate glass and added a small amount of hydrochloric acid, such that all of it dissolved. Now I did another electrolysis, with both the anode and the cathode made of soldering wire. Now, again a gas is produced at the cathode, but when the voltage is adjusted carefully (to appr. 1.9 volts), then hardly any gas is produced, but a metal is plated. This metal forms a cohesive mass, which sticks to the cathode. The anode slowly dissolves again. I stopped electrolysing at the point, where the metallic stuff has grown towards the anode almost completely. Next, I took the cathode and anode from the liquid. I made pictures of them and attached them to this message. Look at how shiny the piece of metal is, which is attached to the cathode (the thick blob of metal, which almost reached the anode in the second electrolysis experiment). I took the piece of metal and dissolved this in 50% HNO3. The liquid becomes a little turbid (apparently some SnO2 is formed). I diluted the liquid, such that it is not oxidizing anymore, due to high-concentration HNO3. Next, I added a solution of KI. Only a very weak ultrafine crystalline yellow solid was formed. This indicates that only a very small amount of lead was present in the plated metal. I also included a picture of the very fine yellow precipitate of lead iodide. I'm quite sure, that the metal, plated out at the cathode, contains at most of few 0.1%'s of lead, otherwise I would have had a much thicker and more coarse precipitate. Next, I did the same experiment with the clear liquid, in which I did the electrolysis. This did not give a positive reaction on lead at all! After addition of KI, this liquid remained colorless. I did a counter experiment with just a tiny amount of lead acetate dissolved in a lot of water and adding the KI-solution. Now I get a thick bright yellow precipitate of PbI2. Both the electrolyte and the plated metal hardly had any lead. The lead apparently falls to the bottom as the dark conductive stuff, while the white stuff is Sn(OH)2. When lateron some acid is added (not too much, just enough to dissolve the Sn(OH)2), then it can easily be plated onto the cathode.

In order to understand this you need some quantum mechanics. Electrons, when confined to a limited space, can only exist in discrete states. The electrons around atoms are confined in space (by means of a potential, the positively charged nucleus binds the electron to the atom). Because electrons only are allowed to be in discrete states, it is possible to count all states, starting from a lowest-energy state and going upwards. In doing so, one has to introduce the concept of shells, and inside each shell, there is a number of different states. The shell, closest to the nucleus only allows 2 states, the shell, next to it, allows 8 states, the shell above that (nr. 3) allows 18 states. Two states are paired, electrons in the same pair of states are said to be in the same orbital. Shell number 2 for instance has four orbitals, one at a lower level, called the s orbital and three at the same higher level, called p orbitals. Shell three also has d-orbitals (there are 5 of these beasts). Now, if atoms are combined to form molecules, then it is possible that orbitals of different levels inside the same shell combine. E.g. a very common form of combining is that the s-orbital and the three p-orbitals combine into four orbitals of the same energy level and each of them looking the same. So, a carbon atom, which normally has a single s orbital in shell 1 (which is fully occupied with 2 electrons) and a single s orbital in shell 2, and three p orbitals in shell 2 (which are only partially filled, as there are only 6 electrons in carbon), can be transformed, such that at shell 2 it has 4 orbitals, which all are the same. The s-orbital and the three p-orbitals 'merge' and blend into four similar orbitals. These orbitals are called hybridized orbitals and are written as sp3 (one s and 3 p's are merged). In a molecule, like methane, all 4 hydrogen atoms are bound to the C-atom in the same way. The C-atom uses a sp3 orbital and the H-atom uses an s-orbital for the chemical bond. If no hybridization would occur, then the molecule CH4 would have one atom, which would have another type of bonding than the other 3, because it would be bonded through the C-atom's s-orbital and the other three through the p-orbitals. Hybridization also can occur between 2 p-orbitals and one s-orbital. In that case the C-atom has one p-orbital, and three sp2 orbitals. In a molecule like H2C=CH2, the H-atoms are bonded through an sp2-hybridized orbital of the C-atom and an s-orbital from the H-atom. One of the bonds between the two C-atoms also it through sp2 orbitals (from both C-atoms) and the other bond is through the remaining p-orbitals. So, in this molecule, the two bonds between the C-atoms actually are different! In more general terms, hybridization can occur between any orbitals and this occurs, as soon as that is energetically more favorable. Most of the chemical compounds we know are due to some form of hybridization of orbitals. In order to have complete understanding of this, have a look at the concepts of quantum mechanics and molecular orbital theory. These, however, certainly are not the easiest concepts to grasp!

We are not doing your homework! I'll give you some guidelines, but you have to do the computations yourself. How many moles is 15.000 grams of PbO2? You have the equation 2PbO2 --> 2PbO + O2 This means that for each mole of oxygen you get two moles of PbO So, you can compute how many moles of PbO you have after the reaction. The number of moles of PbO formed equals the number of moles of PbO2 which disappeared, so you can compute the number of moles of PbO2 remaining. So, after having done these computations you know how many moles of PbO2 you have left and how many moles of PbO you have. Now you can compute the mass of the formed PbO and you can compute the total mass easily This gives you the percentage. Success!

If you use that, then you'll get oxygen and acid at the anode, also some peroxodisulfate, S2O8(2-). At the cathode, you still get hydrogen and the white precipitate of Mg(OH)2. With vigorous stirring, that precipitate may redissolve again, because of the acid, produced at the anode. If no peroxodisulfate is formed, then the acid at the anode precisely balances the hydroxide, formed at the cathode. You have no need to replace your aluminium cathode, only the anode must be a carbon rod.

YT, The next weekend I have some time to experiment with the tin-stuff. I'll try to perform an experiment in a very controlled environment at a micro-scale and see what happens. Especially the conductive layer under the white stuff is really intriguing. I'll see if I can manage to create a conductive bridge between the two electrodes. I've made such a thing once before from a solution of lead acetate. Really spectacular to see it grow. The black stuff at the anode also intrigues me. I almost have the idea that the polarity of the electrodes was reversed for a while, so I'll certainly check out that. The black stuff certainly can be metal. Think of a precipitate of silver metal. It also is deep black. Look at B/W pictures, the black image is due to small silver particles. Many metals, especially if spongy, have a black appearance (I have tried and confirmed this with Pd, Ag, Ru, Co and Ni). As soon as I have some interesting results I'll let you know.

At the negative electrode you get the following primary reaction: 2H2O + 2e ---> H2 + 2OH(-) This in turn causes the following secundary reaction: 2OH(-) + Mg(2+) ---> Mg(OH)2 The bubbles are H2 gas. The white stuff is the Mg(OH)2, which precipitates with the hydroxide ions, formed at the negative electrode. The negatve electrode itself is not corroded. As you see, this is not the way to get magnesium metal! At the anode, the following reaction occurs: Al --> Al(3+) + 3e The electrode slowly corrodes away. This would give rise to colorless aluminium ions and not to a precipitate. Most likely there is some impurity in the aluminium, which causes another metal ion to go in solution. You also get a side reaction: 2SO4(2-) ---> S2O8(2-) + 2e The persulfate ion is formed. It might be that this forms a yellowish precipitate with aluminium ions. It can be yellowish under certain conditions, but I do not know the precise nature of this yellow compound. You definitely do not get sulphur at the anode. Remember, the anode is strongly oxidizing and in the sulpate ion the sulphur is oxidized already to its max. oxidation state. If you electrolyse a solution of a sulfide instead of a sulfate (e.g. Na2S), then you probably will get a yellow precipitate of sulphur at the anode, but in your case this most likely is an impure aluminium-compound.

For people, who are fascinated by strange and obscure compounds, have a look at this reaction with very common chemicals, but with a really strange result. http://woelen.scheikunde.net/science/chem/riddles/thiocyanate+nitrite/index.html I have had some discussion about this reaction recently and it appears that a compound, called nitrosyl isothiocyanate (ON-NCS) is formed in this reaction. This stuff is supposed to form a polymeric species, giving rise to the dark brown color in the liquid. This stuff also is supposed to escape from the liquid as a gas, where it decomposes as follows: ON-NCS ---> ONN + CS The compound with structure ONN is the well-known mild anaesthetic dinitrogen monoxide (laughing gas), the compound CS is a very unstable radical, which reacts with other molecules in the reaction mix, giving rise to the white smoke. All of this is just theory and is not (yet) verified. If some of you have ideas about the actual reaction products and mechanisms or if some of you have ideas on how to analyse the outcome of this reaction further, then I would be very pleased. The really nice thing about this is that there is a very uncommon reaction with very common reagents.

Most metals are shiny, because of the fact that there are free electrons in a conducting band. Metals have a special form of bonding, with atoms in a lattice and electrons freely flowing through the lattice. These electrons, which are in a conduction band, cannot be said to belong to a particular atom or molecule. This 'free cloud' of electrons is very reflective for light of most wave lengths. Why this is reflective can only be well understood by means of quantum mechanics. This same 'free cloud' of electrons also explains why metals are conductive. If a potential is applied over a piece of metal, then the electrons are attracted to the anode and they can move freely to that and at the same time, electrons are sourced by the cathode.

I don't make my HCl you can buy it almost everywhere in any hardware store and it is dirt cheap.

Probably molten lead nitrate is not possible. It will decompose before it melts. But the answer can be given for molten NaNO3. At the cathode you get the metal. At the anode you'll get oxygen and nitrogen oxides (most likely NO2, but NO also may be a possibility): 2NO3(-) --> 2NO2 + O2 + 2e (electrons absorbed by anode) When molten PbSO4 or Na2SO4 is electrolysed, then at the cathode again you get the metal. At the anode, oxygen is produced, together with SO3. The latter will react with the sulfate, forming pyrosulfate, but eventually it will also escape from the liquid. 2SO4(2-) ---> 2SO3 + O2 + 4e (electrons absorbed by anode) In general, when oxoanions are electrolysed in melts, then you may expect that oxygen is produced and the corresponding anhydride, which may or may not escape from the melt, depending on the actual anhydride.

Now some counterinfo: How do you explain that the parcel I received is from the UK and not from the USA? On the scimad forum, someone else (from Australia) also mentioned that he ordered some chems from kno3.com. This person also wrote that the parcel is from the UK, and NOT from the USA. How do you explain this site ? Same name as the owner of kno3.com, same style of writing, this is a UK-based site. Looks somewhat sloppy, but definitely does not look like a police-run site from the USA. I agree with you that it is suspicious that the site of kno3.com is Texas-based, but this does not necessarily mean that the company and the person also are in Texas. For example, many dutch sites (especially the somewhat more obscure ones) are not in the Netherlands, but in the USA for instance. They, however, have a domainname, ending in .nl. Maybe kno3.com is too obscure for UK-laws and they feel better if their site is not hosted in the UK? Yet another thing. How do you explain that this company ships worldwide? This is including the RP, as I also have some now for my element collection. If this were a police-run site in the USA, then why would they ship such an "evil chemical" like RP woldwide? If I were the police and wanted to use this as a trap, I would certainly limit the shipping to USA residents only. If the US-police allowed shipping of potential dangerous chems like RP, but also KClO3 etc. worldwide, then I'm quite sure that there are countries in the world who would not be amused with that. Could you explain this to me? What do you want to say with this? I think this is fairly expensive, but the RP even is more expensive at GBP 19.50 per 100 gram. You want to say that the other chems on the site besides the RP only are there to make the site look less like a police-trap? Indeed the RP is by far the most interesting chem from this site, but the other chems are not fake. They really can be ordered (in fact I ordered and received two rolls of magnesium ribbon at a discount price of just GBP 2.50 per roll). Kerry-An Shanks also was an eBay seller till the end of last year, but he is not an active member anymore. Most of the chems from their site also were sold through eBay, but eBay is quite strict with regards to selling chems, so that may be the reason that he stopped selling chems on eBay. If this really were a police-run company, then why not have a deal with eBay and sell some RP over there? I'm still not convinced and I think it is not OK to introduce fear and anger among fellow hobby-chemists. Of course it is good to warn people, but you bring this as a FACT and doing that simply is not fair. Not towards other people out there, but also not towards kno3.com. What if this really is a company, trying to make some money with selling chems? It is an extra source and if we as hobby chemists are acting paranoia on new sources, then soon no company will exist which sells to the general public. I write this out of some frustration, because we now have a similar fear and paranoia discussion about a small Dutch company, which just started a few weeks ago, selling chems online to the general public (on the Dutch market only), including nice stuff as KClO3, KNO3, CS2, CH3COCH3, conc. HNO3, conc. H2SO4. Many hobby-chemists think this is a police-trap and so, they only get few orders. If things are going on like that, then the company will not survive, due to lack of orders, and a new potentially interesting source of chems is killed already, before it even can grow significantly.

Yet some other nice experiments with copper sulfate: Only Al-foil, dilute HCl and table salt needed besides copper sulfate: http://woelen.scheikunde.net/science/chem/exps/cu+al/index.html Another experiment, which requires more chems, but for the somewhat better equipped home lab, it also is feasible: http://woelen.scheikunde.net/science/chem/exps/cu-redox/index.html If you have sodium thiosulfate (hypo, photo fixer) at hand, then the following may be very interesting. If someone can explain the observations, please let me know : http://woelen.scheikunde.net/science/chem/riddles/copper+thiosulfate/index.html Finally, if you have conc. HCl and plain copper wire (electricity wire) besides copper sulfate, then you can do the following (the CuCl2 can be replaced by CuSO4 in most cases without any problem): http://woelen.scheikunde.net/science/chem/riddles/copperI+copperII/index.html OK, for now there are enough interesting copper sulfate experiments. If you want more, I have tons of experiments with copper compounds. This is one of my favorite elements, because it has a really remarkable and very rich chemistry, which can be explored by people with even very moderate resources.

Even better, synthesise yourself TACS (tetra ammine copper sulfate). I have done this synthesis. Look at the beautiful crystals of the blue complex mentioned by YT2095. http://woelen.scheikunde.net/science/chem/compounds/copper_tetrammine_sulfate.html This can be done by adding concentrated NH3 (25%) to a concentrated solution of copper sulfate, until all precipitate, which is formed initially has redissolved again. You get a really beautiful dark royal blue liquid. To the clear liquid, add half the volume of ethanol (denatured is OK) and then let stand for a night. Beautiful blue crystals are formed. Rinse with a cold mix of ethanol and water, 1 : 1. Then rinse with cold ethanol (96%). Finally rinse with diethyl ether and let dry. If you don't have the latter, then just let it dry after the final alcohol rinse, but in that case, you may loose some ammonia during the slower drying process and the crystals become covered by some light blue/cyan powdery stuff. As soon as the crystals are made, store them in an air-tight container in order to prevent loss of ammonia. DON'T do this with copper nitrate, or you'll have a serious risk to blow off your hands! The compound formed is TACN and that is REALLY bad stuff, which you don't want to have in your house. The sulfate, TACS, however, is safe to make and safe to keep. Of course, you have to be careful with the NH3 fumes and the blue compound itself, as it is quite corrosive.

Could you summarize in short, step by step, what you did? Start with the NaCl solution and the molten electrodes of tin/lead solder. From that point I would like a step by step description. I have quite some idea what you did, but in order to repeat what you've observed, I think we must be very precise in describing what happened (e.g. voltages, concentration of solutions, precise order of actions, precise description of precipitates and also at which steps you obtained them, etc. etc.). I'm in for trying this also. So, if you make a step by step description, then I'll try to repeat it next weekend, then I have some hours for such a lengthy experiment. My experience is that even with simple "garage-chemistry" people can discover VERY interesting phenomena. I already have puzzled professional chemists a few times with simple plain reactions, just have a look at the riddles section of my website: http://woelen.scheikunde.net/science/chem/riddles I asked these questions at Usenet/sci.chem, at a Dutch forum of acedemic chemists and in personal communications with two university employees. Only the sulfite/iodide riddle is resolved so far, the others still are open more or less. So, with your tin experiments, there may also be something interesting, which cannot be explained with basic highschool chemistry or even beyond highschool chemistry. The appealing thing of these riddles is that they can be repeated with such simple equipment and simple reagents. Maybe your experiment can be added to this? Even better would be an explanation. I'm looking forward for your as precise as possible reproduction of all steps taken by you.

The original method I gave was with NaCl-solution. With the NaOH I indeed can imagine that you get a lot of white stuff (being insoluble Pb(OH)2 and SnO2) from the anode. I however, cannot explain how you ever can end up with grey metallic particles plated at the cathode if you use NaOH. With NaCl of course, this can be explained quite well, because Pb goes in solution as Pb(2+) ions, but in NaOH solution things are totally different. Can you give a little more details about your setup with the NaOH-solution? Maybe it forms plumbate ions? But these are anionic, so how would these be reduced at the cathode?

Is this also the case with RuO4? I made some of this stuff in my home lab. It can be made remarkably easily, simply by adding a solution of a ruthenate (VI) salt to dilute hydrochloric acid, adding some persulfate and heating. A mix of Cl2-gas and yellow RuO4-vapor leave the liquid and the RuO4 can be collected higher up on the glass.

Yes, I think that this will be mainly tin. As I understand, you added dilute H2SO4 to the liquid with HCl and H2O2. This means that you precipitated almost all lead with the sulfate ions from the H2SO4. However, before adding the H2SO4, you almost precipitated most of the tin also with the H2O2, by oxidizing it to SnO2. So, I can explain quite well why you hardly get any metal at the cathode and why you get gas at both electrodes. This is, because there hardly is any metal left in the clear liquid. If you have added sufficient H2SO4, then you removed virtually all lead (PbSO4 is highly insoluble) and apparently a small amount of tin remained in the liquid, so you get a very thin layer of tin, but mainly hydrogen gas at the cathode. The yield of this method, of course is terrible.

Indeed, the first person, who actually made elemental bromine, thought that what he had made was iodine (I) chloride, which was known at that time already. So, the person, who officially is called the discoverer of bromine (Antoine Je´rome Balard) was not the first one to prepare it (in 1826). Years before, Justus von Liebig, a German chemist, had prepared elemental bromine. Jdurg's argument that the interhalogens form compounds, with properties intermediate between the properties of the parent halogens, only holds for Cl, Br, I. Interhalogen compounds, containing fluorine (e.g. ClF3, IF7) are colorless gases and do not resemble the parent halogens. The element fluorine is quite an anomaly in the set of halogens anyway. BTW, I made a little more of the ICl (appr. 1 ml) and kept it in a small vial, but this stuff is really corrosive. It has eaten a hard plastic cap, just within a few days. I "disposed of" the stuff now, by reacting the whole lot with fine aluminium powder, which gave a beautiful red plume of gas, almost like my avatar (just slightly more purple), and a little white smoke, again only after some water was added. The red gas probably is a mix of gaseous I2 and ICl. Quite spectacular....

I may have found a solution for separating tin and lead. I did not try it personally, but it looks reasonably. I take a quote from another forum: What this person achieved was making tin (II) chloride. I did not try this personally, I just wanted to mention it at this place. No special chems required and no special equipment required. You might give it a try. BTW: the source of this information is science madness, search for 'tin chloride' in the forum's search page and the link will pop up.

At the cathode you get hydrogen and hydroxide. 2H2O + 2e ---> H2 + 2OH(-) The 2e are supplied by the voltage source and the hydroxide ions OH(-) remain dissolved in the liquid. With the Fe(2+) ions, formed at the anode, they react as follows: Fe(2+) + 2OH(-) ----> Fe(OH)2 Fe(OH)2 is an off-white solid, which becomes green when even the slightest trace of Fe(OH)3 is contained in it. Lateron it becomes brown, due to aerial oxidation: 4Fe(OH)2 + O2 + 2H2O --> 4Fe(OH)3 The latter is brown/red. The stuff you see at the surface most likely is due to very finely suspended iron hydroxide or other small particles, which remain floating in the water. Due to these particles, bubbles do not break apart at once when they reach the surface, but remain floating on the surface for a while.

It is tin oxide, but with tin in the +4 oxidation state. Remember, oxides at high oxidation states usually are not basic and need not react with acids. Tin (IV) oxide is a notorious example of a rather inert oxide. It can be dissolved in strong bases, however, where it acts as an acid and in solution it forms stannate (IV) anions. Other examples of metal oxides which are acidic are chromium (VI) oxide, CrO3, vanadium pentoxide, V2O5 and molybdenum oxide, MoO3. If tin (IV) oxide dissolves in HCl, then that is not due to an acid/base reaction, but due to a coordination reaction, where chloride ligands replace the oxo-ligands. A nice example, where this reaction is strong is with mercury (II) oxide. This dissolves in an excess solution of KI, forming HgI4(2-) ions and the solution becomes alkaline. Unfortunately this effect is not strong with tin (IV) oxide.

Are you sure that tin reacts with chlorine forming tin (II) chloride? I'm afraid you'll get a tin (IV) species (probably some basic oxochloride stuff) or relatively pure SnCl4, which with the slightest contact with water is changed to SnO2 and HCl. The problem with tin is that the +2 oxidation state is quite unstable and very easily goes to the +4 oxidation state. In fact, I have some SnCl2, but over the months it already is severely oxidized to some tin (IV) compound , containing chloride and oxide (or hydroxide).

Almost... At the anode you do not even get any O2 at first. The iron is oxidized immediately, apparently to Fe(2+) ions, which forms the green precipitate with hydroxide formed at the cathode. Indeed, after some time, the iron (II) species is oxidized to an iron (III) species by oxygen from the air. So, the green color goes to a brown color.