Rhodium from the beginning.

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HAuCl4

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I got this "sketchy" procedure from wikipedia:

Rhodium (Greek rhodon (ῥόδον) meaning "rose") was discovered in 1803 by William Hyde Wollaston,[3][4] soon after his discovery of palladium.[5][6][7] He used crude platinum ore presumably obtained from South America.[8] His procedure involved dissolving the ore in aqua regia and neutralizing the acid with sodium hydroxide (NaOH). He then precipitated the platinum as ammonium chloroplatinate by adding ammonium chloride, NH4Cl. Most other metals like copper, lead, palladium and rhodium were precipitated with zinc. Diluted nitric acid dissolved all but palladium and rhodium, which were dissolved in aqua regia, and the rhodium was precipitated by the addition of sodium chloride as Na3[RhCl6]·nH2O. After being washed with ethanol, the rose-red precipitate was reacted with zinc, which displaced the rhodium in the ionic compound and thereby released the rhodium as free metal.[9]

Please feel free to add "simplifications, and/or variations and/or improvements" over this basic and 200+ year old process, with the aim of obtaining extremely pure Rhodium. There are a few obvious ones, and I'm sure several "secret" ones. :p
 
Rhodium does not dissolve well in aqua regia, if boiling some of the very finely dissolved powders can dissolve, but the metal will not dissolve easily into solution, it will normally be left as powders after dissolving other metals into the aqua regia solution, these powder can contain other metal powders depending on what was processed and what insoluble powder could possibly remain, like silver chloride, and iridium.
Treatment of these insoluble powders I would have to consider starting materials and proportions of volume of the metals involved to decide how to proceed.
Silver chloride could be dissolved from the powders using ammonium hydroxide, and then silver precipitated again as silver chloride using HCl (leaving a solution of ammonium chloride);

Or the insoluble powders from the aqua regia solution could be washed in dilute NaOH, rinsed in boiling water to remove the salt (silver chloride to silver oxide), wrap powders in fine lead and cupel and sell to someone who can collect enough of the material to refine.

Also can take the powders remaining from the aqua regia solution and wash well,
This residue can be fused with sodium bisulfate and then powders boiled in water making the rhodium soluble as Rh2(SO4)3, at this point we can separate it from other metals, or insoluble materials(that may have been in these powders that will not dissolve in aqua regia like iridium)

Once we have a soluble solution of Rhodium we can treat it several ways depending on if we wanted it in another salt of Rh or we could precipitate or cement it out of solution.
 
It never ceases to amaze me the great results that these old researchers obtained with limited tools, chemicals, and knowledge.

And also how good one can duplicate or even better those results by changing a few little things, like scavenging the traces of palladium from the nearly pure rhodium (obtained after reducing the pink rose rhodium salt) with DMG, OR elliminating all base metals first, etc. IMO much better than the endless gymnastics that one can observe in some books and papers, some written centuries later.
 
HAuCl4 said:
It never ceases to amaze me the great results that these old researchers obtained with limited tools, chemicals, and knowledge.

And also how good one can duplicate or even better those results by changing a few little things, like scavenging the traces of palladium from the nearly pure rhodium (obtained after reducing the pink rose rhodium salt) with DMG, OR elliminating all base metals first, etc. IMO much better than the endless gymnastics that one can observe in some books and papers, some written centuries later.

It's amazing how much time we spend re-inventing the Wheel.

Dissolving Platinum ore in AR is a bit of a simplification. I think it was a bit more complicated than that.
 
Please feel free to add "simplifications, and/or variations and/or improvements" over this basic and 200+ year old process, with the aim of obtaining extremely pure Rhodium. There are a few obvious ones, and I'm sure several "secret" ones.
An elegantly formulated experiment in phishing for a secret piece of the rhodium-refining-cake?
 
freechemist said:
Please feel free to add "simplifications, and/or variations and/or improvements" over this basic and 200+ year old process, with the aim of obtaining extremely pure Rhodium. There are a few obvious ones, and I'm sure several "secret" ones.
An elegantly formulated experiment in phishing for a secret piece of the rhodium-refining-cake?
HeHe. It could be... save from the fact that THERE IS NO RHODIUM!. Except for small experiments... :lol:
 
Please note that the method is only to remove platinum. What do you do when iridium (III) is there?


Lou
 
IrCl6 (-2) can be easily converted to IrCl6 (-3) and viceversa.

Procedure supposedly in: http://onlinelibrary.wiley.com/doi/10.1002/14356007.a21_075/abstract, which I've never attempted.

So... oxidize to IrCl6 (-2), and then TBP+PE extract along with the PtCl6 (-2), leaving the Rh behind. ?. Piece of cake. :lol:

IMHO all (but traces?) of Ir and/or Ru should stay in crystalline form after doing lead fusion as per Gilchrist modified procedure. (Paper posted by jimdoc in another thread). Pt and Rh and any added gold should dissolve, Pd too if any. (hint-hint :!: )
 
@Lou: "What do you do when iridium (III) is there?"

Oxidize it to IrCl6(2-) followed by extraction, together with PtCl6(2-), and eventually PdCl6(2-), if also present. Even simpler, and equally effective, is absorption on a strong-base (quaternary ammonium-function) anion-exchange resin.

"Ir will do everything rhodium does when it comes to aqua regia."

Not in every case. An alloy, e.g. made up by melting Pt/Rh20% together with ca. 10-15% of its weight Cu will dissolve completely, quite easily in aqua regia, whereas a similar alloy, prepared from Pt/Ir20% will dissolve only partially, resulting in a Pt(Cu)-solution, containing also significant amounts of dissolved Ir (mainly Ir(IV)), and a solid residue consisting mainly of Ir and significant amounts of Pt.


@HAuCl4: "IrCl6 (-2) can be easily converted to IrCl6 (-3) and viceversa. So... oxidize to IrCl6 (-2), and then TBP+PE extract along with the PtCl6 (-2), leaving the Rh behind. ?"

It's possible. A different procedure would be, to precipitate yellow, hydrated Rh(OH)3 and redissolve it in aqueous H2SO4. Chances are, that Ir will stay undissolved, as dark, hydrated IrO2 (really a bitch to be dissolved completely with acid).


freechemist
 
Yes, it is well known that iridium must be oxidized to (IV) state or it will not co extract into the TBP. To clarify, Pt in its highest common oxidation state, Ir should be trivalent if one is separating the platinum out with a ternary or quaternary ammine.

Thanks for pointing out my switch freechemist, you know I know better.
 
It is abso-freakin'-lutely AMAZING The level of, as well as the practical application of knowledge found on this forum every single day.
some say "Those who can't do, teach." A bit of oversimplification, perhaps, but a man who fishes to eat isn't concerned with showing off his knowledge of aquatic sea life. Thus those whose livelihood is not dependent on the efficacy of their processes, such as the authors of the complicated texts mentioned above, simply can not hold a candle to the wisdom evident here daily.
thank you all for sharing.
'Nuff said
 
I certainly need to fish to eat.


The basic premise of rhodium refining is to remove as much crap as possible from it then put it into solution and treat it. There is a selective organic agent that can be used to bind just rhodium (well, iridium will go to, but that hopefully has been S/X'd out). Iridium must tetravalent to leave Rh and trivalent to be ignored when separating from platinum (or mostly ignored, at least).

Mostly, we plate it out of the sulfate after dissolving the yellow, fluffy oxide produced by treatment with caustic soda in 35% room temperature sulfuric.

We modulate the Ir(IV)--> Ir(III) with SO2 before going after Pt and back with Cl2. The Pt is still treated hydrolytically to remove traces of Rh and and Ir as their hydrous oxides.


Also,

I have some different experience with 50% Rh 50 % Pt added to 2X it weight in copper--the rhodium stays behind as a black cake and is only extracted into the lixiviant minimally.

On the flip side, I run 90% Pt, 10% Ir jewelry alloy that is diluted to 70% PGM basis in copper and invariably, at least 15-20% of the iridium will co-dissolve.

I think it has much to do with how much of the Ir or Rh in solution. Strange phenomena in any case.
 
Hi Lou,

IMHO Your statement "Yes, it is well known that iridium must be oxidized to (IV) state or it will coextract into the TBP" must have been infected by a printing mistake, and should read as ".....or it will not coextract into the TBP".

"Strange phenomena in any case": Thank you for sharing. Your experiences with Pt/Ir are similar to mine. Your 50% Rh-case is new to me. I never worked with Pt/Rh-alloys with such a high Rh-content.


freechemist
 
I've recently provided some 999 Rh to a customer who was casting various items of it--they alloyed it as far down as 50% Pt 50% Rh to improve ductility. It was sent back as credit for more Rh.
 
Thanks for the discussion gentlemen. 8)

freechemist: My TBP discussion was in the context of "cleaning up" or scavenging for different contaminants a mostly 98%+ pure Rhodium solution, but I appreciate your alternate hydroxide/sulphuric acid procedure and thank you for it.

"The basic premise of rhodium refining is to remove as much crap as possible from it then put it into solution and treat it."

I think Lou's observation can be applied to all metals, and in doing so the steps needed can be broken down to sequential, sometimes iterative, techniques that can be implemented by operators with limited knowledge, with the seldom need for the expert, supervising chemist. This may not be the optimal way to save on chemicals, but the extra expense on chemicals is compensated by the reduced cost of labor, and sometimes time savings as many processes can run in parallel, whilst one expert chemist only has 2 hands to work with. :idea:
 
The hydroxide treatment is really only applicable after most of the base metals have been removed. Iron, for instance, will contaminate it if one is not careful. Much of the garbage can be eliminated by proper treatment with sodium sulfide (removes other PGMs as well)...
 
HAuCl4 said:
IrCl6 (-2) can be easily converted to IrCl6 (-3) and viceversa.

Procedure supposedly in: http://onlinelibrary.wiley.com/doi/10.1002/14356007.a21_075/abstract, which I've never attempted.

So... oxidize to IrCl6 (-2), and then TBP+PE extract along with the PtCl6 (-2), leaving the Rh behind. ?. Piece of cake. :lol:

IMHO all (but traces?) of Ir and/or Ru should stay in crystalline form after doing lead fusion as per Gilchrist modified procedure. (Paper posted by jimdoc in another thread). Pt and Rh and any added gold should dissolve, Pd too if any. (hint-hint :!: )
Do you have any other resources of mentioned documents. Its no free and i cannot pay outside country especially in $. Thanks
 
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