# Decomposition of Gold Hydroxide

On heating will gold hydroxide $\ce{Au(OH)3}$ decompose to form pure gold or will a gold oxide (e.g. $\ce{Au2O3}$) be stable? Does anyone know the temperature of these reactions?

• en.wikipedia.org/wiki/Gold(III)_hydroxide Feb 7, 2017 at 21:42
• I wasn't sure how much to trust wikipedia... so simply $\ce{2Au(OH)3}$ > $\ce{Au2O3}$ + $\ce{3H2O}$ at 160 degrees? Will $\ce{Au2O3}$ just form Gold and oxygen gas at higher temperatures?
– BJH
Feb 7, 2017 at 21:48
• books.google.pl/… It looks like, but hydration seems to complicate this. Feb 7, 2017 at 22:03

There is yet no consensus on what occurs when gold(III) hydroxide is heated. Part of the problem is that the exact structure of $\ce{Au(OH)3}$ is unknown, or dubious at best. It has been inconsistently formulated as "truly" $\ce{Au^{III}-(OH)3}$$^{\ \ \ [\text{a}]}, oxide-hydroxide hydrate \ce{AuOOH*H2O}$$^{}$, oxide hydrate $\ce{Au2O3*xH2O}$$^{\ }, even an acid-like hydrate \ce{H[AuO2]*H2O}$$^{\ }$. I am going to use $\ce{[AuOOH*H2O]^{*}}$ with an asterisk in equations to keep this controversy in mind.

At temperatures below $130\ ^\circ \mathrm{C}$, the decomposition is probably best described by$^{\ }$

$$\ce{[AuOOH*H2O]^{*} ->[below 130\ ^\circ \mathrm{C}] AuOOH + H2O}.$$

Around $140\ ^\circ \mathrm{C}$ the oxide-hydroxide decomposes further.$^{\ }$

$$\ce{2AuOOH ->[140-150\ ^\circ \mathrm{C}] Au2O3 + H2O}$$

Gold(III) oxide is already considerably unstable from $155\ ^\circ \mathrm{C}$ .$^{\ }$ It is not definitively shown what reaction follows. Some posit $\ce{AuO}$$^{}, which might not exist or actually be better described as \ce{Au[AuO2]}$$^{}$, others propose a combined product of elemental gold $\ce{Au}$ and $\ce{Au2O}$$^{}. There are studies that rule out these products, at least as isolable intermediates, and instead propose \ce{-> Au + O2}$$^{\ \ \ }$.

$$\ce{2Au2O3 ->[150-165\ ^\circ \mathrm{C}] 4AuO + O2}$$

$$\ce{4Au2O3 ->[150-165\ ^\circ \mathrm{C}] 4Au + 2Au2O + 5O2}$$

$$\ce{2Au2O3 ->[150-165;\ 300\ ^\circ \mathrm{C}] 4Au + 3O2}$$

The proponents of the bottom reaction, a more direct $\ce{-> Au + O2}$ claim that the oxides were mistaken for a complex mixture of $\ce{Au}$, $\ce{Au2O3}$, and interstitial or dissolved $\ce{O2}$.$^{}$

The authors$^{}$, nevertheless, still note

It is not yet clear how this “dissolved” oxygen should be described. Is it really dissolved in interstitial metal lattice positions, are there trapped $\ce{Au2O3}$ oxide clusters in a densely sintered metal matrix, or encapsulated oxygen gas in lattice imperfections of the gold?$^{}$

If the intermediates $\ce{AuO}$ and $\ce{Au2O}$ exist, then not for too long. The decomposition of these oxides would most likely consist of multiple interconversions between $\ce{AuO}$ and $\ce{Au2O}$. For example$^{\ \ [\text{b}]}$,

$$\ce{2Au2O + O2 <=> 4AuO},$$ $$\ce{Au + AuO <=> Au2O},$$ $$\ce{Au2O -> Au + O2}.$$

Over $300\ ^\circ \mathrm{C}$ only gold metal $\ce{Au}$ and dioxygen $\ce{O2}$ remain.$^{}$ Another experiment$^{}$ supports this conclusion but differentiates between surface and subsurface oxides. The latter are found to be stable well over $300\ ^\circ \mathrm{C}$$^{}. More research is required to shed light into what is actually the main decomposition pathway. An additional way of expressing the process at lower temperatures is^{}$$\ce{6AuOOH -> 4Au + Au2O3 + 3O2 + 3H2O.}$$[\text{a}] \ce{Au(OH)3} is presumed to assume a square planar coordination with four \ce{-OH}-groups.^{} A most recent study provides evidence for a four-coordinated linear polymer.^{} [\text{b}] Equations are written on the analogy with copper.^{\ }  Daisuke Kawamoto, Hiroaki Ando, Hironori Ohashi, Yasuhiro Kobayashi, Tetsuo Honma, Tamao Ishida, Makoto Tokunaga, Yoshihiro Okaue, Satoshi Utsunomiya, Takushi Yokoyama. 'Structure of a Gold(III) Hydroxide and Determination of Its Solubility'. Bulletin of the Chemical Society of Japan, (2016), 89(11), 1385-1390. DOI: 10.1246/bcsj.20160228.  H. Karik, K. Truus. (2003). Elementide keemia. (p 313)  C. Klanner, N. Weiher, E. A. Willneff, S. L. M. Schroeder. 'In Situ Quick XAFS Studies of the Thermal Decomposition of Gold(III) Oxide'. Hasyweb Annual Report 2000, (2000), part 1, 46, 2986.  R. Ripan, I. Ceteanu. (1969). Chimia metalelor. II volume. (p 768-769)  I. V. Mironov. 'Properties of Gold(III) Hydroxide and Aquahydroxogold(III) Complexes in Aqueous Solution'. Russian Journal of Inorganic Chemistry, (2005), 50(7), 1115  Κ. V. Astakhov, A. G. Elitsur, K. M. Nickoluev. Zhurnal obshcheĭ khimii, (1951), 21, 1753.  A. G. Massey, N. R. Thompson, B. F. G. Johnson, R. Davis. (1973). The Chemistry of Copper, Silver, Gold. (1975 reprint)  O. Diaz-Morales, F. Calle-Vallejo, C. de Munck, M. T. M. Koper. 'Electrochemical water splitting by gold: evidence for an oxide decomposition mechanism'. Chemical Science, (2013), 4, 2334-2343. DOI: 10.1039/C3SC50301A.  C. Klanner, N. Weiher, E. A. Willneff, S. L. M. Schroeder, C. Figulla-Kroschel, M. Jansen. 'Thermal Stability and Decomposition Kinetics of Crystalline Gold(III) Oxide and Disordered Gold(III) Oxides/Hydroxides'. Hasyweb Annual Report 2001, (2001), part 1, 46, 5440.  M. Peuckert, F. P. Coenen, H. P. Bonzel. 'On the Surface Oxidation of a Gold Electrode in 1\ \mathrm{N} \ce{H2SO4} Electrolyte'. Surface Science, (1984), 141, 2-3, 515-532.  L. K. Ono, B. R. Cuenya. 'Formation and Thermal Stability of \ce{Au2O3} on Gold Nanoparticles: Size and Support Effects'. Journal of Physical Chemistry C, (2008), 112(12), 4676–4686. DOI: 10.1021/jp711277u. Wikipedia quotes on the decomposition of gold(III) hydroxide: It is easily dehydrated above 140 °C to gold(III) oxide. Moreover, the equation of the decomposition is given here:$$\ce{2Au(OH)3 -> Au2O3 + H2O}$$If we heat up further, the auric oxide would decompose:$$\ce{2Au2O3 ->[160-290 C] 4Au + 3O2}$$Mind you, you will not become a millionaire by extracting gold in this way. The gold obtained is very impure. :) • Thanks for the chemiday.com link, that's really useful. The same site also lists the decomposition of$\ce{Au2O3}$at 160 - 290 °C by$\ce{2Au2O3 -> 4Au + 3O2}\$ here
– BJH
Feb 8, 2017 at 19:34
• @BJH Thank you, i could include that in the answer. Feb 9, 2017 at 12:34