1
$\begingroup$

I know Aqua regia is used for purifying gold by dissolving gold in it, but how that dissolved gold is obtained back in the pure state?

I searched everywhere, but I didn't get the answer. So, I am asking here.

| improve this question | |
$\endgroup$
  • 2
    $\begingroup$ Pretty much anything can reduce gold. $\endgroup$ – Ivan Neretin Jul 1 '19 at 4:49
  • 1
    $\begingroup$ The Niels Bohr's story has already been mentioned several times in related questions(1, 2); and there is nothing special about reducing gold, there are numerous agents, both organic and inorganic (e.g. citric acid or tin(II) chloride) that can be utilized. The choice is also dictated by the post-refining process ("Affinage"), if any. $\endgroup$ – andselisk♦ Jul 1 '19 at 5:45
  • $\begingroup$ Gold is not really purified by dissolving it into aqua regia. Most base metals dissolve in aqua regia, too. The actual purification comes from choosing a reducing agent that acts on the gold but leaves the less reducible base metal ions in solution. $\endgroup$ – Oscar Lanzi Jul 1 '19 at 12:25
  • $\begingroup$ There is a historic anecdote about this question. De Hevesy dissolved his gold Noble prize medals in aqua regia during World war II. After the war, he found the solution undisturbed and precipitate d back his gold medals from the solution without any loss in weight. See: en.m.wikipedia.org/wiki/George_de_Hevesy $\endgroup$ – Nilay Ghosh Oct 10 '19 at 7:38
7
$\begingroup$

By electrolysis. In general electrolysis is a very common way of purifying metals if you have them dissolved, since you can selectively control the reaction by adjusting the potential current. Al and Cu are also obtained through electrolysis of their salt solutions.

Electrolysis is a redox reaction that occurs in the non spontaneous direction driven by a external potential. In electrolysis we have two electrodes (a cathode and an anode) submerged in a solution with the salt of the metal we want to purify. As it always happens in electrochemistry, the reduction reaction will occur in the cathode, while the oxidation in the anode.

Since in a gold solution obtained through Aqua Regia we have the gold forming a tetrachloride complex (AuCl4), the oxidation reaction we will have will be the production of chlorine gas from the chloride. So the global redox reaction that will be occurring will be: $$\ce{2[AuCl4]- + 6e- -> 2Au + 8Cl-}$$ $$\ce{6Cl- -> 3Cl2 + 6e-}$$

And if you look at the standard potentials of the two reactions in this Wikipedia webpage you can see that you just need a 0.43 V external potential for the reaction to occur. The negative electrode should be made out of gold and it will grow during the process by the deposit.

Industrially this process is done by having hundreds of electrodes connected in series and submerged into the gold salt solution. In this image you can see a factory for Cupper purification. enter image description here

You can see from the electrodes lifted by the crate that they have copper lamina that grow during the process. Electrolysis is normally used when you want very high purity in your product, like gold and copper destined for electric applications in which any oxygen impurities greatly decrease conductivity.

Another great thing about electrolysis is that it can be used to easily apply a thin layer of the metal over a piece, if the piece is conductive itself. This is how gold coated audio jacks are produced, by using the audio jack adapter as the cathode.

| improve this answer | |
$\endgroup$
4
$\begingroup$

You say:

I searched everywhere, but I didn't get the answer.

When I type youtube gold recovery from chloroauric acid into Google, the first link is to a Youtube video with a visible demonstration of converting chloroauric acid to gold.

The video shows that metabisulfite can reduce chloroaurate to form tiny tiny grains of gold powder, as a reddish dust. A chemical reaction for this might be:

$$\ce{4 HAuCl4(aq) + 3 Na2S2O5(aq) +9H2O -> 4 Au(s) + 6 NaHSO4 + 16 HCl(aq)}$$

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.