Sciencemadness Discussion Board

Separating lead and tin metal?

Draeger - 18-4-2020 at 09:00

So, I found some wire which is tin and lead alloyed together. Is there any way I could separate them, preferably without converting it to anything toxic?

Usually, I'd think of it myself, but for something as potentially toxic as lead without a fume hood, no place outside to do anything and no proper gloves, I'd rather have multiple opinions of more experienced people.

metalresearcher - 18-4-2020 at 09:24

Without fumehood and outdoor experiments it will be difficult. Sn and Pb are very intimately alloyed.

Maybe fractionated crystallization might help to purify one of the metals. But because it is probably an eutectic alloy (60% Sn and 40% Pb) that is also not an option.

I just found another topic on this forum:

http://www.sciencemadness.org/talk/viewthread.php?tid=2277

Texium - 18-4-2020 at 09:57

I think the most effective way, though it doesn't fit your safety bill, would be to heat the wire in a tube while passing chlorine gas over it. This would ultimately produce PbCl2 and SnCl4. The SnCl4 (which has very interesting properties, as a dense, colorless liquid that is truly a covalent compound of tin, and is a useful reagent for synthesizing organotin compounds) can then be distilled off.

This works because while PbCl4 exists, it is not stable at the temperature of the reaction, decomposing into Cl2 and PbCl2.

I haven't tried this with lead/tin solder before, but I have made SnCl4 using pure tin metal and chlorine gas. I didn't have a fumehood at the time, but I was wearing good PPE and had a fan directing all of the fumes out of my garage door.

AgCollector - 18-4-2020 at 11:28

Is your goal the challenge of it? This should be fine. Or is your goal the hope of a cheaper route to tin metal & lead metal versus buying outright? This will be more difficult.

In my opinion it's going to be pretty challenging to do it non-toxicly w/o the need for gloves, etc. But, toxicity is relative, I guess, when you're already starting with lead!

A simple option would be to use as concentrated HCl as you can acquire, cut the wire up into small pieces (or melt & pour into water to make small pieces), then combine and hold it at around 50 or 60 C for several days. You 'll then have PbCl2 as a solid ppt and Sn2+ in solution which you can reduce with zinc metal or iron metal. Based on my own experience this will give you tin that's approximately 98-99% pure with 1-2% lead.

You can also speed up the reaction by adding H2O2 but this will give you Sn4+ instead, which may stay in solution as Sn(Cl)62- if the Cl- concentration is high enough, or it will ppt out as stannic acid which is more problematic.

Lion850 - 18-4-2020 at 13:20

Concentrated nitric acid readily dissolved the common solder I used and I ended up with lead nitrate in solution and tin iv oxide as precipitate. My initial aim was to get to iodides. The lead nitrate was readily converted to lead iodide but the tin dioxide is quite inert and I have yet to find a way to process it further.

S.C. Wack - 18-4-2020 at 14:28

Fusion of Sn with S and carbonate in the absence of air at 450C gives water-soluble sodium tetrathiostannate.
Sn + 3 Na2CO3 + 6 S -> Na4SnS4 + Na2S2O3 + 3 CO2
The lead would be converted to one of its least soluble forms.

PS Same for SnS and SnO2.
SnS + 3 Na2CO3 + 5 S -> Na4SnS4 + Na2S2O3 + 3 CO2
SnO2 + 3 Na2CO3 + 7 S -> Na4SnS4 + Na2S2O3 + 3 CO2 + SO2

[Edited on 19-4-2020 by S.C. Wack]

AgCollector - 19-4-2020 at 08:42

Quote: Originally posted by S.C. Wack  
Fusion of Sn with S and carbonate in the absence of air at 450C gives water-soluble sodium tetrathiostannate.
Sn + 3 Na2CO3 + 6 S -> Na4SnS4 + Na2S2O3 + 3 CO2
The lead would be converted to one of its least soluble forms.

PS Same for SnS and SnO2.
SnS + 3 Na2CO3 + 5 S -> Na4SnS4 + Na2S2O3 + 3 CO2
SnO2 + 3 Na2CO3 + 7 S -> Na4SnS4 + Na2S2O3 + 3 CO2 + SO2

[Edited on 19-4-2020 by S.C. Wack]


Interesting! Have you tried this out?

S.C. Wack - 19-4-2020 at 11:20

Haven't had much reason to. Tin objects have been found on ebay and melted down so I'm good for now. This is from the old analytical lit as an alternative to HNO3 processes and apparently would require some 4 hours. One would want to take some care if sulfur was to sublime and seal a covered crucible, that's probably why PhD candidates use boats.

[Edited on 20-4-2020 by S.C. Wack]

NaK - 19-4-2020 at 17:01

lead is literally one of the cheapest metals you can find as no one really wants it due to its toxicity and therefore replacement. tin not so cheap but there are a lot of old dishes and cups you can easily melt down. Check out ebay and schrottmax

teodor - 20-4-2020 at 07:45

Quote: Originally posted by AgCollector  

....You 'll then have PbCl2 as a solid ppt and Sn2+ in solution which you can reduce with zinc metal or iron metal. Based on my own experience this will give you tin that's approximately 98-99% pure with 1-2% lead.
more problematic.


Yes, I also think HCl is the best way to separate these two. But I wouldn't recommend try to reduce the SnCl2 with Zn because many Sn & Pb alloys contains also Sb and SbCl3 + Zn will give very poisonous SbH3, much more poisonous than any volatile lead compound, so it requires a fume hood as a minimum.

I am not sure about H2O2, because how you can be sure not getting PbCl4? But hot concentrated HCl should work. Well, you need a closed system like distillation setup for this work.

AJKOER - 21-4-2020 at 07:14

An electrochemistry perspective may be valuable here.

My suggestion is to construct a Lead-Acid Battery, which also means in addition to a source of Pb, you will need to prepare some PbO2.

PREPARATION OF PbO2: Dissolve Pb in say a mix of 2 parts 5% vinegar to 1 part 6% NaOCl (chlorine bleach). Add a graphite electrode (from mechanical pencil refill). This is actually a variant of the standard so-called 'Bleach battery' using Pb in place of the Al anode and C in place of Cu, which can be based on either NaOCl or HOCl.

Treat the filtered Pb+ ions with an excess of common chlorine bleach (NaOCl) to await the formation of red PbO2.

For a rendition of my experimenting with Pb leading to PbO2, see this thread http://www.sciencemadness.org/talk/viewthread.php?tid=5490&a... .

So, one can create a Pb | PbO2 cell in dilute H2SO4 (a Lead-Acid battery cell) per the reaction equations:

Pb(s) + HSO−4(aq) → PbSO4(s) + H+(aq) + 2e−

PbO2(s) + HSO−4(aq) + 3H+(aq) + 2e− → PbSO4(s) + 2H2O(l)

Net Cell:
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

And, replacing the source of metal Pb with an impure Pb/Sn in DILUTE H2SO4 also works depositing elemental Sn. Fortunately, per a source:

"With concentrated sulfuric acid, tin reacts very slowly, while it does not enter into a reaction with diluted sulfuric acid at all. ... If tin is heated, it can react with alkalis with the release of hydrogen."

Reference: https://melscience.com/US-en/articles/tin-its-oxidation-stat...

More interesting, use dilute HCl in place of H2SO4 as per a source:

"Tin reacts with dilute acids slower than many other metals and heat is often required to speed up the reaction. This property means tin can be used as a protective coating on other, more reactive, metals."

Now, PbCl2 is not very soluble in dilute solutions, but on warming more soluble. So, use this property to quickly put PbCl2 into solution leaving less reactive Sn.

To isolate Pb, add Al to aqueous PbCl2 and plate out Pb onto part of the Al foil. Add HCl/NaCl to the mix which is now a galvanic cell, and slowly, in time, the anodic Al corrodes forming Al(OH)3 which will dissolve in dilute HCl. Note, here the Pb is the passive cathode.

The above suggested path likely addresses the electrochemical separation of tin and Pb.

Note, a parallel to this process is the anodic corrosion of Al/Si alloy foil in a mix of H2O2/a weak acid with a touch of salt and use a nickel-copper coin for the cathode (microwave to jump salt). Black particles of Si become evident as the aluminum foil dissolves.

[Edited on 21-4-2020 by AJKOER]

[Edited on 21-4-2020 by AJKOER]

AgCollector - 22-4-2020 at 10:11

Quote: Originally posted by teodor  

Yes, I also think HCl is the best way to separate these two. But I wouldn't recommend try to reduce the SnCl2 with Zn because many Sn & Pb alloys contains also Sb and SbCl3 + Zn will give very poisonous SbH3, much more poisonous than any volatile lead compound, so it requires a fume hood as a minimum.


I don't think this would be an issue - while I've read that you can make SbH3 when dissolving an alloy of Sb-Zn into HCl, I have always thought of this as happening because in that alloy, it's almost like Sb is the anion, and Zn is the cation, (that is, existing as Sb3- and Zn2+) so that reaction with HCl via double displacement would give SbH3 and ZnCl2; but, in this case, if it's SbCl3 then the oxidation state of Sb would be 3+ instead of 3- so I wouldn't see how SbH3 would form... even if Zn donated electrons to Sb to produce elemental Sb, I don't believe that reacts with HCl to make stibine.

teodor - 23-4-2020 at 11:27

Quote: Originally posted by AgCollector  
Quote: Originally posted by teodor  

Yes, I also think HCl is the best way to separate these two. But I wouldn't recommend try to reduce the SnCl2 with Zn because many Sn & Pb alloys contains also Sb and SbCl3 + Zn will give very poisonous SbH3, much more poisonous than any volatile lead compound, so it requires a fume hood as a minimum.


I don't think this would be an issue - while I've read that you can make SbH3 when dissolving an alloy of Sb-Zn into HCl, I have always thought of this as happening because in that alloy, it's almost like Sb is the anion, and Zn is the cation, (that is, existing as Sb3- and Zn2+) so that reaction with HCl via double displacement would give SbH3 and ZnCl2; but, in this case, if it's SbCl3 then the oxidation state of Sb would be 3+ instead of 3- so I wouldn't see how SbH3 would form... even if Zn donated electrons to Sb to produce elemental Sb, I don't believe that reacts with HCl to make stibine.


I red this in "Analytical Chemistry" Treadwell, Volume I, 9th English edition, page 163:

... Lead, tin and other methals will also cause the same reaction to take place (deposit of antimony - teodor). Zinc, hovewer, unlike iron, tin and lead, is likely to carry the reduction farther, and stibine gas, SbH3, is formed with a part of the antimony:

2 Sb+++ + 3 Zn -> 3 Zn++ + 2 Sb

Sb+++ + 3 Zn + 3 H+ -> 3 Zn++ + SbH3

I did analysis of several alloys and I would recommend to make at least a primitive fume hood for this purpose - at least a metal box (like for fireplace in a garden), hose and a construction fan. Because very often I felt sick 1 or 2 days after I did experiments with alloys (even dissolving Fe alloy in H2SO4/HCl gives small quantities PH3 and if H2SO4 contains traces of Se you can get traces of H2Se with Zn, and, if I properly remember, my old chemical book (for children) says that H2 from Zn + HCl reaction must be washed from traces of Arsine).
I know only one solvent for alloys which protects your health - HNO3. So, recommendation to add an oxidiser (like H2O2) is wise but a mix of higher valencies can make the separation more complex.
But you should be safe with HCl, just distil it over in a closed system (PbCl2 is soluble in HCl but not very soluble in cold water unlike SnCl2).



[Edited on 23-4-2020 by teodor]