When we use Copperas How is wait for gold
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if using ferrous sulfate ( copperous) to precipitate gold should see it reacting in your solution fairly soon after adding it,and smell the sulfur gas bubbling , if fairly pure gold solution (Yellow) it will start to turn brown after several minutes, stir well.
I let my solution settle for 24 hours, then test solution with stannous chloride to see if any gold still left disolved in the acid solution.
I let my solution settle for 24 hours, then test solution with stannous chloride to see if any gold still left disolved in the acid solution.
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I'd like to add that if you have not eliminated free nitric acid, the reaction may be delayed, or not occur at all. When you pour your ferrous sulfate solution in the gold chloride solution and you have evaporated properly, the reaction is instantaneous. If your solution is fairly concentrated, you'll see a thick cloud of gold form instantly.
The advice to allow the material to settle well is very good. It isn't uncommon for a fine layer to take as much as a day to settle.
Be certain you test the remaining solution with stannous chloride, and to stir well so you don't have any gold chloride solution trapped under the layer of gold. It's not common, but it does happen.
Harold
The advice to allow the material to settle well is very good. It isn't uncommon for a fine layer to take as much as a day to settle.
Be certain you test the remaining solution with stannous chloride, and to stir well so you don't have any gold chloride solution trapped under the layer of gold. It's not common, but it does happen.
Harold
Harold said the Oxidizer you used must be eliminated (nitric acid, peroxide, or bleach) they are what helped to make the gold to dissolve to form an ion solution of chlorides, and if they are not eliminated the gold will just keep redissolving in your solution and will not precipitate.
I use by weight ferrous sulfate if I think I have an ounce of gold will use ounce of ferrous sulfate. a small excess if not too much is ok.
Stannous chloride test is very important, it lets you know if have gold left in your solution.
I use by weight ferrous sulfate if I think I have an ounce of gold will use ounce of ferrous sulfate. a small excess if not too much is ok.
Stannous chloride test is very important, it lets you know if have gold left in your solution.
Here's the copperas Reduction formula:
3FeSO4*7H2O + AuCl3 --> Fe2(SO4)3 + FeCl3 + Au + 7H2O
So yes, the iron is oxidized and replaces the gold, the gold is therefore reduced.
Steve
3FeSO4*7H2O + AuCl3 --> Fe2(SO4)3 + FeCl3 + Au + 7H2O
So yes, the iron is oxidized and replaces the gold, the gold is therefore reduced.
Steve
thanks steve and harold, but maybe I am missing something here, if it was just Iron replacing the gold then why couldnt we just use Iron? it will precipitate the metals above it on the activity series of metals, of coarse that is not many, but still I would not expect the purity of gold if it was only the Iron doing the precipitation reaction.
I understand to precipitate the gold must get back its missing electron. but I thought that chemically the sulfate helped to keep the other metals lower than Iron in solution.
if not then why wouldn't Iron chloride work as well? I know enough of it would precipitate gold, but it is not used like FeSO4
can someone do more explanation on this.
I understand to precipitate the gold must get back its missing electron. but I thought that chemically the sulfate helped to keep the other metals lower than Iron in solution.
if not then why wouldn't Iron chloride work as well? I know enough of it would precipitate gold, but it is not used like FeSO4
can someone do more explanation on this.
Metallic iron would obviously precipitate several other metals with your gold (cobalt, nickel, tin, lead, copper) which would defeat the purpose of refining the gold.
Iron chloride would already have three chlorines attached to it (FeCl3) and therefore it would not displace the chlorines from the AuCl3.
It is the sequential exchange of electrons between the three ferric sulfate molecules and the auric chloride that drives the reaction. Notice how the sulfate anions and iron cations rearrange in the three FeSO4 molecules to Fe2(SO4)3.
Steve
Iron chloride would already have three chlorines attached to it (FeCl3) and therefore it would not displace the chlorines from the AuCl3.
It is the sequential exchange of electrons between the three ferric sulfate molecules and the auric chloride that drives the reaction. Notice how the sulfate anions and iron cations rearrange in the three FeSO4 molecules to Fe2(SO4)3.
Steve
yes I seen how they rearanged in the formula earlier , and see Iron would precipitate the other metals, which ferrous sulfate will not. so your saying we are displacing the chlorine. and that besides the Iron giving away its electron it has no other purpose? and really except for this solution the sulfate has no purpose except to transporter of iron, and also I assume changes other metals to sulphides or sulphates Cu,Sn etc. soluble.
this still semms a bit strange. not to have you give me chemistry lessons here, but what would be the difference in using copper sulfate
it could displace clorine ,exchange electrons, saturate solution and change it to a sulfide .
but it does the make a selective precipitant that ferrous sulphate does.
this still semms a bit strange. not to have you give me chemistry lessons here, but what would be the difference in using copper sulfate
it could displace clorine ,exchange electrons, saturate solution and change it to a sulfide .
but it does the make a selective precipitant that ferrous sulphate does.
Butcher,
There are a few reasons why copper sulfate won't work.
It has to do mainly with the oxidation states that copper normally exists in. There are two normal forms of iron sulfate, FeSO4 and Fe2(SO4)3, from Fe II and Fe III respectively. Copper only forms CuSO4 from Cu II, and the higher oxidation states of copper (3 and 4) only occur in very rare circumstances with copper compounds, so only Cu I and II normally are seen.
With the above in mind, recall that three chlorines must be removed or displaced from the gold to fully reduce it to elemental metal. Less reactive copper (as compared to iron) cannot achieve this, because it doesn't oxidize to Cu III under normal conditions. Secondly, recall that CuCl (copper I chloride) is insoluble in aqueous solutions. This means that CuCl would precipitate once the first chlorine was displaced. If the acid levels were high enough the copper could reach the second level of oxidation, CuCl2, which is soluble, but that would still leave a remaining chlorine attached to the gold.
Think about the way AP behaves when saturated and this will be more clear. It's very common to find solid CuCl in your gold foils after filtering the AP. This is also a common salt in Poormans AR reactions. Luckily CuCl dissolves in more acid.
Steve
There are a few reasons why copper sulfate won't work.
It has to do mainly with the oxidation states that copper normally exists in. There are two normal forms of iron sulfate, FeSO4 and Fe2(SO4)3, from Fe II and Fe III respectively. Copper only forms CuSO4 from Cu II, and the higher oxidation states of copper (3 and 4) only occur in very rare circumstances with copper compounds, so only Cu I and II normally are seen.
With the above in mind, recall that three chlorines must be removed or displaced from the gold to fully reduce it to elemental metal. Less reactive copper (as compared to iron) cannot achieve this, because it doesn't oxidize to Cu III under normal conditions. Secondly, recall that CuCl (copper I chloride) is insoluble in aqueous solutions. This means that CuCl would precipitate once the first chlorine was displaced. If the acid levels were high enough the copper could reach the second level of oxidation, CuCl2, which is soluble, but that would still leave a remaining chlorine attached to the gold.
Think about the way AP behaves when saturated and this will be more clear. It's very common to find solid CuCl in your gold foils after filtering the AP. This is also a common salt in Poormans AR reactions. Luckily CuCl dissolves in more acid.
Steve
Thank you Steve.
That gives me a little more homwork to do on studying the oxidation states.
You explained that so well, how do you do that?
I can understand or see how things work but have a hard time explaining them.
THANKS :wink:
That gives me a little more homwork to do on studying the oxidation states.
You explained that so well, how do you do that?
I can understand or see how things work but have a hard time explaining them.
THANKS :wink:
