The reduction potential for $\ce{Au^3+ + 3e- -> Au}$ is $1.503 \text{ V}$. The reduction potential of $\ce{Cl2 + 2e- -> 2Cl-}$ is $1.36 \text{ V}$. So, the $E^o$ for $\ce{2Au^3+ + 6Cl- -> 2Au + 3Cl2}$ is $0.143 \text{ V}$.

If the reaction $\ce{2Au^3+ + 6Cl- -> 2Au + 3Cl2}$ has a positive E standard ($E^o$ = $0.143 \text{ V}$), then how is it possible for $\ce{AuCl3}$ to exist in aqueous solution without $\ce{Au^3+}$ oxidizing $\ce{Cl-}$ to $\ce{Cl2}$?

  • 2
    $\begingroup$ There is not considered $\ce{Au^3+ 4 Cl- <-> [AuCl4]-}$. $\endgroup$
    – Poutnik
    Jan 10, 2020 at 5:12
  • 5
    $\begingroup$ $\ce{AuCl3}$ does not exists. It is an empirical formula for $\ce{Au2Cl6}$ (doi.org/10.1107/S0365110X58000694). $\endgroup$ Jan 10, 2020 at 5:45

1 Answer 1


Gold(III) chloride does not exist as $\ce{AuCl3}$. According to Wikipedia, the name gold trichloride is a simplification of the name, which is referring to the empirical formula, $\ce{AuCl3}$. The X-ray defraction studies revealed that the chemical exists as a chloride-bridged dimer, $\ce{Au2Cl6}$ (Ref.1):

Structure of Au2Cl6

In $\ce{Au2Cl6}$, each gold center is square planar, and the bonding in $\ce{Au2Cl6}$ is considered somewhat covalent (Wikipedia). Yet, your suggestion is somewhat true, since a recent research has shown that $\ce{Au2Cl6}$ is susceptible to decompose to metalic $\ce{Au}$ and $\ce{Cl2}$ gas at high temperatures or expose to $\ce{e-}$-beam irradiation (Ref.2).

Also, another research group (Ref.3) has shown that:

Gold chloride, supplied as $\ce{HAuCl4}$ solution, is rapidly adsorbed on the silicate surfaces, the Au(III) is reduced to metallic gold, and gold particles grow on the surface. SEM images show agglomerates of gold unevenly distributed on the surface of the silicates, including in some areas forming agglomerates, especially on quartz and feldspar.



  1. E. S. Clark, D. H. Templeton, C. H. MacGillavry, "The crystal structure of gold(III) chloride," Acta Cryst. 1958, 11, 284-288 (https://doi.org/10.1107/S0365110X58000694).
  2. Keita Kobayashi, Ryo Kitaura, Katsuhiro Sasaki, Kotaro Kuroda, Takeshi Saito, Hisanori Shinohara, "In Situ Observation of Gold Chloride Decomposition in a Confined Nanospace by Transmission Electron Microscopy," Materials Transactions 2014, 55(3), 461-465 (https://doi.org/10.2320/matertrans.MD201308).
  3. Sima Mohammadnejad, John L. Provis, Jannie S. J. van Deventer, "Reduction of gold(III) chloride to gold(0) on silicate surfaces," Journal of Colloid and Interface Science 2012, 389(1), 252-259 (DOI: 10.1016/j.jcis.2012.08.053).
  • 2
    $\begingroup$ There are other III chlorides that form the same dimeric structure, such as molten aluminium (III) chloride. $\endgroup$
    – J.G.
    Jan 10, 2020 at 16:05
  • 3
    $\begingroup$ Not really the same dimer. Aluminum (III) chloride has tetrahedral coordination not square planar. Blame it on a different electronic structure. $\endgroup$ Jan 22, 2020 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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