Non-Chemical cellphone scraps

[ericrm]

makes sense to me (at least with Cl₂), but in excess oxidizer you can evt. still not test them, driving off the oxidizer will again let them react with reducers, I would think.....

no your missing the point the oxidiser react with the precipitant to make another salt ,it wont be be precipitant anymore... stannous chloride to stannic chloride, ferrous chloride to ferric chloride....

What good is it to put the colloidal gold back into solution if you cannot separate the problem that causes the colloid to form, (where it will form again)?

i have added stannous chloride in enuf quantity in gold bearing AR to make sure it wont dissolve back, than i added nitric acid until everything was back to normal... even at room temperature after theyr was no sign of coloid , it is not because the oxidiser keep the gold dissolved it is because the oxidiser react with the precipitant....

 

[solar_plasma]

Sn + 2HCl --> SnCl2 + H₂O

BUT we use HCl/chlorox or AR, so:

SnCl₂+Cl₂ --> SnCl4
Sn+2Cl₂ --> SnCl₄
SnCl₂+HCl+HNO₃ --heating--> SnCl4 (equation not complete)

after dillution:
SnCl₄+2H₂O -exotherm--> SnO₂ + 4HCl
at the end:
SnO₂+HCl --> H₂[SnCl₆] + 2H₂O

H₂[SnCl₆] crystalizes out of the solution or forms metal compounds like M⁺₂[SnCl₆] or M⁺⁺[SnCl₆]

(sources: wiki eng/ger, Römpp Chemielexikon, Inorganic Chemistry - Holleman/Wiberg)

This I would say is the reason why we never get purple gold colloid from a dirty solution, unless we take a drop of the solution and test it with a pervert excess of SnCl₂ from our stannous test solution. Nevertheless the gold will be bound to complexes.

 

[butcher]

Sn + 2HCl --> SnCl2 + H2O
Stannous chloride:
Sn(s) + 2HCl(aq) --> SnCl2(aq) + H2
(g)

BUT we use HCl/chlorox or AR, so:

SnCl2+Cl2 --> SnCl4
Sn+2Cl2 --> SnCl4
SnCl2+HCl+HNO3 --heating--> SnCl4 (equation not complete)
In aqua regia tin forms stannic chloride:
3Sn + 4HNO3 + 12HCl --> NO(g) + 8H3O + SnCl4
(when excess HCl present it can also form chlorostannic acid H2SnCl6 in solution)


after dillution:
SnCl4+2H2O -exotherm--> SnO2 + 4HCl
at the end:
SnO2+HCl --> H2[SnCl6] + 2H2O

Here I have to disagree, stannic oxide (SnO2)
would not form in the acid environment, after dilution of the aqua regia solution. I believe first after denoxing the solution the reaction above moves back towards the left:
SnCl4 --> SnCl2 + Cl2
then on dilution the stannous chloride in solution can form an hydroxide of tin with water (along with the stannous chloride, as we are still in a highly acidic environment), but as acidic as our solution is I believe this reaction would stay pretty much to the left as stannous chloride in solution:
4SnCl2 +2H2O --> Sn(OH)4 + 4HCl
and
SnCl2 + H2O --> Sn(OH)Cl +HCl
for this to occur I think is basically impossible in this highly acidic environment of even dilute aqua regia, so I believe the reaction would stay to the left leaving us with stannous chloride in our diluted aqua regia ( if gold was involved in the solution, the stannous would reduce the gold to colloidal gold, re-oxidizing could put the gold back into solution, but would not get rid of the tin or its problems, later we would be right back here; trying to get out gold back out of this colloidal solution).

(with air or oxygen upon dilution and standing over time the stannous chloride can also form this hydroxide along with the stannic chloride):
6SnCl2 + O2 + 2H2O --> 2SnCl4 + 4Sn(OH)Cl

For us to form the insoluble stannic oxide SnO2, we would have to evaporate the solution to powders, and oxidize the tin powders.

 

[solar_plasma]

Thank you Butcher!

The first one has been a print error - ofcourse I meant H₂

the HCl+HNO3 line I could only find incomplete on wiki - thanks again

Here I have to disagree, stannic oxide (SnO2)
would not form in the acid environment, after dilution of the aqua regia solution. I believe first after denoxing the solution the reaction above moves back towards the left:
SnCl4 --> SnCl2 + Cl2
then on dilution the stannous chloride in solution can form an hydroxide of tin with water (along with the stannous chloride, as we are still in a highly acidic environment), but as acidic as our solution is I believe this reaction would stay pretty much to the left as stannous chloride in solution:
4SnCl2 +2H2O --> Sn(OH)4 + 4HCl
and
SnCl2 + H2O --> Sn(OH)Cl +HCl
for this to occur I think is basically impossible in this highly acidic environment of even dilute aqua regia, so I believe the reaction would stay to the left leaving us with stannous chloride in our diluted aqua regia ( if gold was involved in the solution, the stannous would reduce the gold to colloidal gold, re-oxidizing could put the gold back into solution, but would not get rid of the tin or its problems, later we would be right back here; trying to get out gold back out of this colloidal solution).

(with air or oxygen upon dilution and standing over time the stannous chloride can also form this hydroxide along with the stannic chloride):
6SnCl2 + O2 + 2H2O --> 2SnCl4 + 4Sn(OH)Cl

For us to form the insoluble stannic oxide SnO2, we would have to evaporate the solution to powders, and oxidize the tin powders.

So, I guessed right with what I wrote before: a lot of equilibria

 

[arthur kierski]

JM-Johnson Matthey----have a material(kind of fibre),is called smopex----this product recuperates precious metals from solutions---probably even colloidal gold------------

if someone is interested,please enter JM site-----very interesting
regards
Arthur

 

[niteliteone]

JM-Johnson Matthey----have a material(kind of fibre),is called smopex----this product recuperates precious metals from solutions---probably even colloidal gold------------

if someone is interested,please enter JM site-----very interesting
regards
Arthur

Here is a write up on the "Sompex" material. It sounds interesting.
http://www.scavengingtechnologies.com/userfiles/files/SMOPEX-301%28TM%29%20an%20effective%20replacement%20for%20Triphenylphosphine.pdf

It has also been mentioned here on the forum by the manufacturer.
http://goldrefiningforum.com/~goldrefi/phpBB3/viewtopic.php?f=84&t=16948&start=0