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MrHomeScientist
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Renewed Interest
After a very long hiatus, I uncovered some information that is extremely interesting and makes me want to have a go at this project once more. I'll
get to that in a later post. In the meantime, I tried out the neodymium double-sulfate separation method and it looks like it works great. I read
through the whole thread again and I guess I never tried this route myself before.
I started with 15mL of "magnet soup" leftover from the last time I pursued this experiment. It's been sitting on the shelf for a year or so (!), in a
flask covered with parafilm. As you recall from earlier posts, this soup consists of iron(II), iron(III), and neodymium sulfate from dissolving
magnets in sulfuric acid. Not having any potassium sulfate handy, I tried out sodium sulfate instead.
I made up a near-saturated solution of Na2SO4 by dissolving 7.4g of it into 25mL distilled water. Some very fine powder was left
over that refused to dissolve, so I let that settle and pipetted the liquid above it for the following procedure.
The sodium sulfate solution is on the left, and "magnet soup" on the right.
I added about half of the Na2SO4 solution to the magnet liquor. Nothing happened immediately, but after a few minutes a light
pink sandy precipitate slowly appeared and settled fairly quickly. I left this to settle/react overnight.
The next day, a layer of brown gel had formed on top of the Nd precipitate!
The liquid tested as a pH of 5 to pH paper, as I suspected it might. I added a few drops of ~50% H2SO4 to bring pH back down to
1, stirred vigorously for a few minutes, and allowed to settle. This effectively dissolved the brown goo (iron hydrolysis, most likely).
Here it is after adding the acid.
This was filtered off and rinsed with several small portions of acidified (with sulfuric acid) Na2SO4 solution, then a final
rinse with cold distilled water. I collected 0.6g of the neodymium double salt, Nd2(SO4)3 *
Na2SO4 * 3H2O . Unfortunately, I don't have any thiocyanate to test for iron in the precipitate, but it looks quite
nice.
Next, I treated this powder with sodium hydroxide to convert the neodymium to the hydroxide. I believe this goes via:
Nd2(SO4)3*Na2SO4*3H2O + 6NaOH --> 2Nd(OH)3 +
4Na2SO4 + 3H2O
Accordingly, 0.6g of the double salt should require 0.2g of NaOH, and produce 0.32g Nd(OH)3 and 0.47g Na2SO4. Based
on the solubility of the sodium sulfate and required amount of sodium hydroxide, I made up a solution of 0.8g NaOH in 12mL of distilled water. This is
a large excess of hydroxide, to ensure full conversion of the neodymium.
I poured this solution directly onto the dry double-sulfate powder, and stirred vigorously for ~1 hour. I then allowed this to settle for 2 hours. The
precipitate had lightened into a white color with a hint of pink. I filtered this off and rinsed with a small portion of hot water, followed by a few
rinses with cold water. After drying, I recovered 0.3g of Nd(OH)3 - a near-stoichiometric yield! (My scale is only accurate to 0.1g,
though)
Here's a photo of the hydroxide:
So in conclusion, I concur with blogfast: this appears to be an excellent way to separate the iron from the neodymium! I should be getting some
thiocyanate in the near future, and when I do I'll make a more quantitative determination.
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Brain&Force
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I'm going to try making some neodymium salts soon from some neodymium magnets I was careless with and shattered. I also have access to calcium,
magnesium, and sodium fluoride so I may be able to do a Nd thermite (which someone has already done) - that is, if I can find a place to do it.
Why not make some unusual Nd salt? I haven't seen much on the acetate, citrate, benzoate, perchlorate, bromide, iodide, or carbonate.
Other than surface area reasons, is removing the nickel coating really necessary? NiSO4 is soluble in water, so it shouldn't interfere with
Nd precipitation.
At the end of the day, simulating atoms doesn't beat working with the real things...
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MrHomeScientist
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Wow that's a godawful video. No description, terrible focus, no idea what's going on. He could have filmed just about anything and claimed it was an
Nd thermite...
The problem with the fluorides you listed is they are basically insoluble in water, so you can't really use them to arrive at NdF3. You
need a soluble fluoride salt; in my case, ammonium bifluoride. (EXTREMELY DANGEROUS!)
You need to remove at least some of the nickel coating because nickel doesn't react with sulfuric acid in these conditions (maybe in hot concentrated
acid). So you need to expose the magnet material itself to allow the reaction to proceed. If some did dissolve, then you're correct in that it
shouldn't interfere.
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Brain&Force
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By calcium and magnesium I meant the metals, not the fluorides. And I thought sodium fluoride was soluble in water.
All my magnets are chipped and corroding at the minimum, so I should be OK with just leaving them in acid.
At the end of the day, simulating atoms doesn't beat working with the real things...
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blogfast25
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That won't work, trust me.
Reduction of the fluoride NdF<sub>3</sub> with lithium might just about work at elevated temperatures, going by the Heats of
Formation.
MrHS already tried Nd trifluoride + Mg w/o much success.
The RE oxides are almost impossible to reduce chemically. Aluminum certainly does NOT do it. The guy in the video is a plain old LIAR: the heats of
formation of Nd2O3 and Al2O3 simply do NOT allow such a reduction to take place.
NaF: a fairly useless chemical, max. solubility at RT about 1 M. You really need ammonium bifluoride. Quite a dangerous chemical, read up and suit up
before use.
@Mr HS: that Nd hydroxide look very much the real deal! Lovely pink. Never got mine quite that good...
[Edited on 20-1-2014 by blogfast25]
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Brain&Force
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Quote: Originally posted by MrHomeScientist  | | I added about half of the Na2SO4 solution to the magnet liquor. Nothing happened immediately, but after a few minutes a light
pink sandy precipitate slowly appeared and settled fairly quickly. I left this to settle/react overnight. |
This same exact thing happened to me with the terbium extraction (but with potassium sulfate). It seems the double sulfates have a tendency to
supersaturate. Conveniently, most of the K-Tb sulfate came in a single flat bed of crystals. Did this happen with the Na-Nd sulfate?
This thread seriously needs to be stickied.
blogfast25 - then I guess thermite is out of the question. I'll just try to make some of the salts mentioned previously.
At the end of the day, simulating atoms doesn't beat working with the real things...
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blogfast25
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A few words on the metallothermic reduction of NdF3 with Li.
The Standard Heat of Formation of NdF3 is – 1657 kJ/mol (Wolfram alpha), a real whopper!
For LiF NIST gives – 616.93 kJ/mol.
So for NdF3 + 3 Li === > Nd + 3 LiF3 the Standard Heat of Reaction would be – 194 kJ/mol. Ignoring any entropic effects, it’s safe to say this
should proceed, if you can get it to light.
But – 194 kJ/mol is almost certainly not enough to heat the reaction products to above the MP of Nd (1024 C, the MP of LiF is 845 C), needed to
allow metal and slag to separate out, thermite style. So external heating would be needed to ensure the end temperature is, say, 1100 C or higher.
NdF3 is easy to prepare, both Mr HS and me have done so: it’s water insoluble.
Lithium is easy to obtain, but expensive and hard to handle.
A graphite crucible loaded with Li metal on the bottom and finely powdered, dry NdF3 on top, perhaps rammed a bit to reduce air, then heated quickly
to about 600 C, may well give the desired result. Heavy, lump Nd metal covered with LiF slag is what you’d want here, post cooling.
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MrHomeScientist
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I think it's still worthwhile to pursue the magnesium thermite route, just with a larger charge than I did previously. I now have a charcoal furnace
capable of melting aluminum, so I could also get to a higher temperature than before. Metallothermic reaction with lithium metal is a little scary to
me - molten lithium is just about one of the most aggressive things there is, so your choice of crucible would need to be pretty careful.
Another really interesting method I came across uses a setup that refluxes magnesium metal onto magnet chips directly - the molten metal being able to
dissolve Nd but not Fe or B! It collects in a crucible at the bottom of the apparatus, where Mg boils off and Nd remains. Really cool, but quite
likely out of our league.
Attachment: Direct Extraction and Recovery of Neodymium Metal from Magnet Scrap.pdf (257kB)
This file has been downloaded 806 times
=====================================
=====================================
Now, for the idea that renewed my interest in this project. Completely unrelated to neodymium, I was browsing around for new things to try. I found a
thread here on deep eutectic solvents, and that sent me on a quest of discovery that I'll sum up for you.
Deep Eutectic Solvent (DES) thread here on SM:
http://www.sciencemadness.org/talk/viewthread.php?tid=10529
I thought it'd be neat to make one of these room temperature "ionic liquids," specifically choline chloride & urea. In the reference link in the
first post of that thread, it states that many metal oxides are soluble in these ionic liquids. Scanning through other links in that thread, I found
the attached PDF that shows the electrodeposition of aluminum from AlCl3 dissolved in such a liquid.
Attachment: Conductivities of AlCl3 in Ionic Liquid Systems and Their Application in Electrodeposition of Aluminum.pdf (191kB)
This file has been downloaded 1620 times
Further, ShadowWarrior4444 (the author of the DES thread) claimed in this thread that "as for Ionic Liquids, one of them should be able to dissolve" NdF3.
All of these bits of information got me very excited. I thought it just might be possible to electrodeposit neodymium metal from a solution of
NdF3 in a choline chloride/urea DES.
Now, I don't have much experience with electrolysis. Looking at the standard electrode potentials, here's what I think would happen using Al
electrodes. (starting with Al because that's what they used in the PDF above)
Nd3+ + 3e- <---> Nd ........... -2.323V
Al <---> Al3+ + 3e- ............... 1.676V
Overall Cell Potential: Ecell = Ecathode - Eanode = -2.323 - 1.676 = -3.999V
So the cell would need to be run at 4V, using Al electrodes. This metal is usually unsuitable for electrolysis, though, so I'd prefer to use graphite
rods as I've done in the past. I'm not sure what the anode reaction would be in that case, though.
My concern is that I don't really know the reactivity of this solvent system, so I'm not sure if Nd will react with it in some way. I already received
my choline chloride (smells awful, reminds me of triethylamine), and the urea is on the way. I can't wait to start experimenting with this.
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DraconicAcid
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| Quote: Originally posted by MrHomeScientist |
Now, I don't have much experience with electrolysis. Looking at the standard electrode potentials, here's what I think would happen using Al
electrodes. (starting with Al because that's what they used in the PDF above)
Nd3+ + 3e- <---> Nd ........... -2.323V
Al <---> Al3+ + 3e- ............... 1.676V
Overall Cell Potential: Ecell = Ecathode - Eanode = -2.323 - 1.676 = -3.999V
So the cell would need to be run at 4V, using Al electrodes. This metal is usually unsuitable for electrolysis, though, so I'd prefer to use graphite
rods as I've done in the past. I'm not sure what the anode reaction would be in that case, though.
My concern is that I don't really know the reactivity of this solvent system, so I'm not sure if Nd will react with it in some way. I already received
my choline chloride (smells awful, reminds me of triethylamine), and the urea is on the way. I can't wait to start experimenting with this.
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You're mixing conventions here. If you're going to subtract one half-reaction potential from the other, then they either both have to be reductions,
or both have to be oxidations. If you've written one as an oxidation and the other as a reduction, then their signs are different, and you have to
add them, giving something like -0.6 V.
As you've written the half-reaction, the Nd one does not want to go, but the Al one does. The overall reaction will be somewhere in the middle.
Those standard electrode potentials- are those the ones from aqueous solution? They won't apply at all in a different solvent.
[Edited on 21-1-2014 by DraconicAcid]
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MrHomeScientist
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Bah you're right - I got them from a table of standard potentials here, which is probably aqueous. I guess that means I'll have to slowly raise the voltage until I see something start to happen?
As for the equations, they were both listed in the table as reductions. I flipped the aluminum equation and changed the sign on the potential before
inserting them into the equation. I was under the impression you needed to do this to form a complete chemical equation for the cell, before you could
plug numbers into the potential equation. At the cathode Nd is plated out, while the anode dissolves into solution.
I appreciate the help. Like I said, I'm a real beginner at electrochemistry. I haven't been able to find a decent tutorial on the subject online
either, so if anyone knows of a good one I'd appreciate it.
[Edited on 1-21-2014 by MrHomeScientist]
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DraconicAcid
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| Quote: | | Bah you're right - I got them from a table of standard potentials here, which is probably aqueous. I guess that means I'll have to slowly raise the voltage until I see something start to happen?
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Yes.
| Quote: | | As for the equations, they were both listed in the table as reductions. I flipped the aluminum equation and changed the sign on the potential before
inserting them into the equation. I was under the impression you needed to do this to form a complete chemical equation for the cell, before you could
plug numbers into the potential equation. At the cathode Nd is plated out, while the anode dissolves into solution. |
That's the trouble- there's two ways of looking at it. The electrochemist's view is that reduction potentials are the only thing anyone would ever
look at, so you subtract one from the other. The thermodynamicist's viewpoint is that you flip one half-reaction, add the equations, and *add* the
half-potentials (just like you would add enthalpy changes, Gibb's Free Energy changes, etc.). Both will give you exactly the same answer (and exactly
the same "you're doing it wrong" sneer from the other group), but if you change the sign *and* subtract, you'll get the wrong answer.
Consider this:
A -> A+ + e E = +1.0 V
A+ + e --> A E = -1.0 V
If you add these up, you get a null reaction (A -> A, nothing happens), which must have E = 0 V, right? You can get this by a) adding the
reduction potential to the oxidation potential (+1.0 V + -1.0 V), or b) subtracting one reduction potential from the other (+1.0 V - +1.0 V). Either
way will give you the correct result. BUT changing the sign of one reduction potential (to make it negative) and then subtracting will give you 2.0
V, which is wrong.
[Edited on 21-1-2014 by DraconicAcid]
[Edited on 21-1-2014 by DraconicAcid]
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blogfast25
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Mr HS:
I don't think lithiothermic reactions are any scarier than magnesiothermic ones.
Remember though that I made a mistake calculating the thermochemistry for NdF3 + Mg. On correction we found that basically ΔH = 0 for that
reduction. There's no getting around that.
A graphite crucible (see gold smelting) should resist molten Li well, bearing in mind also that there won't be any Li for very long. It's quite a
'common' reduction agent for higher fluorides (tri or tetra).
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blogfast25
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Well, well, well. I ran the numbers for the adiabatic end temperature of NdF3 + 3 Li === > Nd + 3 LiF and got a fairly crude estimate of 1100 to
1200 C. For a thermite that would be abysmal but here it's fairly close to target...
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MrHomeScientist
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Hmmm...
I do have some small graphite crucibles, and a strip of Li foil from a lithium battery. Do you think it would be sufficient to cram the foil into the
bottom of the crucible, pour powdered (dry) NdF3 on top, and heat with a propane torch until some kind of ignition? My Li is stored under
oil, though, and surely I'd want to get rid of that somehow.
The crucible should be totally dry and free of any contaminants for safety. I'd read elsewhere that molten lithium can rip the oxygen right out of
glass!
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DraconicAcid
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I want to hear about what happens in the choline-urea melt....
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MrHomeScientist
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Me too! 
I received my urea in the mail yesterday, so I'm set to start down that road as well. First I'd like to dry my choline chloride, though - it came as
pretty damp white crystals. I'll put some in a dessicator bag with calcium chloride (Damp Rid) tonight.
One thing that bothers me, though: numerous sources list choline chloride as completely nontoxic (indeed, it's an additive in animal feed), yet the
MSDS from Sigma gives it a 2 health hazard. My sample has a pretty 'scary' smell, which reminds me of tiethylamine. Might there be some contamination
in my sample? I hadn't seen anything referencing it having a smell.
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blogfast25
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| Quote: Originally posted by MrHomeScientist | Hmmm...
I do have some small graphite crucibles, and a strip of Li foil from a lithium battery. Do you think it would be sufficient to cram the foil into the
bottom of the crucible, pour powdered (dry) NdF3 on top, and heat with a propane torch until some kind of ignition? My Li is stored under
oil, though, and surely I'd want to get rid of that somehow.
The crucible should be totally dry and free of any contaminants for safety. I'd read elsewhere that molten lithium can rip the oxygen right out of
glass! |
My guess is that that should suffice. Ram the NdF3 powder down onto the Li to reduce oxygen availability to the Li. And propane should be enough: I
doubt if ignition doesn't start from about 500 C or so.
Re. liquid lithium and glass, yeah, ok. A bit extreme though, isn't it? And I'd still like to see proof of it: lithium isn't used often in oxidic
reductions...
The deep-eutectic idea is of course wonderful but unchartered territory (as far as I know). Be a trail blazer!  By electrolysis, which anion will you expect to be decharged?
[Edited on 22-1-2014 by blogfast25]
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DraconicAcid
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My sample also has a bit of a smell. I didn't try drying it.
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blogfast25
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Have you prepared a deep eutectic with it?
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DraconicAcid
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I tried a choline chloride-copper(II) chloride eutectic, which didn't work properly.
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Brain&Force
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Why not try electrolysis of a soluble neodymium salt in pyridine? Apparently lithium can be made with that method, so Nd should be possible in theory.
However, the salt would have to be perfectly dry for it to work properly, and lanthanide salts are pretty much impossible to dry without thionyl
chloride or heating in a HCl atmosphere.
Interesting 1919 reference that states NdCl3 is soluble in pyridine: http://books.google.com/books?id=W2zxN_FLQm8C&pg=PA116&a... There's a lot of useful info in these older references.
At the end of the day, simulating atoms doesn't beat working with the real things...
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blogfast25
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@B&F:
I think you need anhydrous NdCl<sub>3</sub> for that. The anhydrous form is tricky to prepare from the hexahydrate. Once I tried
with NH<sub>4</sub>Cl and only got a mess.
Edit: oops, you said so.
[Edited on 23-1-2014 by blogfast25]
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MrHomeScientist
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I started my experiments with the deep eutectic choline chloride / urea mixture last night. I managed to make the liquid, but haven't done anything
else so far. Tough to do stuff during the work week, especially when its been so cold outside lately!
Until I have something more relevant to neodymium come out of that route, I'll be posting my eutectic experiments in that thread instead of this one.
Here's the link: http://www.sciencemadness.org/talk/viewthread.php?tid=10529&...
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Brain&Force
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If I dissolve hydrous NdCl3 in pyridine, will I be able to dry it with calcium oxide? I can't find any info on calcium hydroxide's
solubility in pyridine though.
Unfortunately I cannot acquire pyridine; even if I am able to, I can't use it in my school - the substance is banned. Very odd, because I've heard
that there was this one teacher somewhere who would denature alcohol with a drop of pyridine.
NdCl3 is soluble in alcohol (44.5g/dL), but I don't see how one could produce Nd metal in such solution.
At the end of the day, simulating atoms doesn't beat working with the real things...
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MrHomeScientist
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Finally got my potassium sulfate in the mail, but it's highly impure. Off white color with black specks throughout; turns out it's fertilizer grade. I
guess that's what I get for ordering "organic sulfate of potash." At least it was cheap - I'm recrystallizing some now and then I'll be able to
proceed.
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