# 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}$$^{[0]\ [1]\ [2]\ [\text{a}]}, oxide-hydroxide hydrate \ce{AuOOH*H2O}$$^{[3]}$, oxide hydrate $\ce{Au2O3*xH2O}$$^{[0]\ [4]}, even an acid-like hydrate \ce{H[AuO2]*H2O}$$^{[5]\ [6]}$. 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$^{[3]\ [7]}$

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

Around $140\ ^\circ \mathrm{C}$ the oxide-hydroxide decomposes further.$^{[3]\ [7]}$

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

Gold(III) oxide is already considerably unstable from $155\ ^\circ \mathrm{C}$ .$^{[1]\ [3]}$ It is not definitively shown what reaction follows. Some posit $\ce{AuO}$$^{[3]}, which might not exist or actually be better described as \ce{Au[AuO2]}$$^{[3]}$, others propose a combined product of elemental gold $\ce{Au}$ and $\ce{Au2O}$$^{[1]}. There are studies that rule out these products, at least as isolable intermediates, and instead propose \ce{-> Au + O2}$$^{[0]\ [2]\ [7]\ [8]}$.

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

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

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

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 $\ce{O2}$.$^{[9]}$

The authors$^{[9]}$, 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?$^{[9]}$

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$^{[1]\ [3]\ [\text{b}]}$,

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

Over $300\ ^\circ \mathrm{C}$ only gold metal $\ce{Au}$ and dioxygen $\ce{O2}$ remain.$^{[9]}$ Another experiment$^{[10]}$ supports this conclusion but differentiates between surface and subsurface oxides. The latter are found to be stable well over $300\ ^\circ \mathrm{C}$$^{[10]}. More research is required to shed light into what is actually the main decomposition pathway. An additional way of expressing the process at lower temperatures is^{[9]}$$\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.^{[2]} A most recent study provides evidence for a four-coordinated linear polymer.^{[0]} [\text{b}] Equations are written on the analogy with copper.^{[1]\ [3]} [0] 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. [1] H. Karik, K. Truus. (2003). Elementide keemia. (p 313) [2] 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. [3] R. Ripan, I. Ceteanu. (1969). Chimia metalelor. II volume. (p 768-769) [4] I. V. Mironov. 'Properties of Gold(III) Hydroxide and Aquahydroxogold(III) Complexes in Aqueous Solution'. Russian Journal of Inorganic Chemistry, (2005), 50(7), 1115 [5] Κ. V. Astakhov, A. G. Elitsur, K. M. Nickoluev. Zhurnal obshcheĭ khimii, (1951), 21, 1753. [6] A. G. Massey, N. R. Thompson, B. F. G. Johnson, R. Davis. (1973). The Chemistry of Copper, Silver, Gold. (1975 reprint) [7] 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. [8] 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. [9] 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. [10] 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