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Normally gold can't be oxidizes by oxygen, but in the presence of $\ce{CN}^-$ it works.

In general, I have understood that this is due to the fact that complexation $\ce{[Au(CN)_2]^-}$ decreases the concentration of gold and thus also the single potential (Nernst's equation).

But I don't quite understand how the complex formation can occur beforehand? For the complex formation, the gold must be oxidized beforehand, so I'm not really getting anywhere with this explanation... Without oxidation no complexation, without complexation no reduction of concentration, without reduction of concentration no reduction of oxidation potential, without reduction no oxidation of gold by oxygen... Somehow this is a circular argument.

Or does the dissolving process start with a small complex formation (because in chemistry it is always the case that a small part of something is formed, although the energy for this is actually not sufficient) which then slowly favors the oxidation and this then favors the further complex formation so that it becomes more and more? So a slow-starting self-reinforcing process?

To summarize: Oxygen alone cannot oxidize gold. Why then does it work in the formation of the cyanide complex? What happens before? The oxidation or the complex formation, whereby during the complex formation Au+ must already be present, but how is that possible?

My intuition tells me it is about the relation between redox process and thermal + kinetic stability.

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  • $\begingroup$ It's not like you can ever have bare Au+ in solution - it's always coordinated already during oxidation. Oxygen and cyanide are both coordinated to the surface. $\endgroup$
    – Mithoron
    Commented Jul 5, 2023 at 0:21
  • $\begingroup$ @Mithoron yes, but Oxygen alone can't oxidize Au. So how would it work that it forms a complex "during oxidation" if the oxidation potential is too high beforehand? $\endgroup$
    – iwab
    Commented Jul 5, 2023 at 0:32
  • $\begingroup$ I could give you some explanation how in equilibrium there's, like, always a bit of product, but who says it's gonna oxidize it alone? Did you heard about concerted mechanisms? $\endgroup$
    – Mithoron
    Commented Jul 5, 2023 at 1:15
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    $\begingroup$ Then it's about time too look it up ;) $\endgroup$
    – Mithoron
    Commented Jul 5, 2023 at 1:45
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    $\begingroup$ Here is a good discussion, although unanswered: chemistry.stackexchange.com/questions/164487/… $\endgroup$
    – Nilay Ghosh
    Commented Jul 5, 2023 at 4:32

1 Answer 1

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Water cannot spontaneously get over the walls of a vessel. But it can do it via a tube to another place below the liquid level. That is commonly used for many liquids in vessels with an upper access.

The mechanism of gold dissolution in oxygenated cyanide solution is not based on direct oxidation by oxygen.

It is based on the big difference of the respective standard reduction potentials of the following first two reactions, leading to the summary third rection:

\begin{align} \ce{Au(s) + 2 CN-(aq) &-> [Au(CN)2]-(aq) + e-} &E^{\circ}=\pu{−0.6 V}\tag{1}\label{1}\\ \ce{4 e- + 2 H2O(l) + O2(aq) &-> 4 OH-(aq)} &E^{\circ}=\pu{+0.401 V}\tag{2}\label{2}\\ \ce{4 Au(s) + 8 CN-(aq) + 2 H2O(l) + O2(aq) &-> 4 [Au(CN)2]-(aq) + 4 OH-(aq)} \tag{3}\label{3} \end{align}

It means that

  • The reaction $\ce{Au(s) + 2 CN-(aq) <=> [Au(CN)2]-(aq) + e-}$ is at the standard conditions at equilibrium if the electrode potential (wrt SHE) is $\pu{−0.6 V}$
  • The reaction $\ce{4 e- + 2 H2O(l) + O2(aq) <=> 4 OH-(aq)}$ is at the standard conditions at equilibrium if the electrode potential is $\pu{+0.401 V}$.

It is obvious the electrode cannot have simultaneous both potentials, therefore some net reactions must be happening.

It is a kind of a single place galvanic cell, where the golden metal is at the same time the cathode and the anode, spending simultaneously both cathodic and anodic materials and using electricity for itself.

Gold in alkaline cyanide solution is a strong reducer, considering the redox potential value. The Au+ ions release from the metal is made much easier by reaction with cyanide. This is pushing the metal potential (wrt SHE) toward $\pu{−0.6 V}$ , at which the rate of the backward reaction (gold deposition) at standard conditions would be the same.

OTOH, oxygen as oxidant is being reduced to water by such a low potential, pushing the gold potential toward $\pu{+0.4 V}$, at which the rate of the backward reaction (oxygen formation) at standard conditions would be the same.

The gold metal potential will reach such a potential where is established the steady state of the same rate of both listed reactions $\eqref{1}$ and $\eqref{2}$ wrt the the zero net rate of electron release/capture.

For the same reaction with silver, it even reduces water to hydrogen to keep the potential high enough to dissolve silver, so oxygen is not needed.

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