It is often said that "while cobalt(III) per se is strongly oxidising, hexamminecobalt(III) chloride is stable, even in concentrated hydrochloric acid, due to the strong donor properties of the ammine ligands". However, according to LibreTexts, the cation is actually only kinetically stable in the presence of hydronium ions, eventually turning into hexaaquocobalt(III) (which is strongly oxidising) and ammonium ions in the thermodynamic limit.

In pure water, the hydronium ion only exists in minute amounts, due to the high equilibrium constant of the generalised Comproportionation of hydronium and hydroxide to water. This might or might not counterbalance the high thermodynamic instability of the hexamminecobalt(III) ion against ligand exchange in acidic water.

My question now follows- is the hexamminecobalt(III) cation thermodynamically stable in pure water?

UPDATE: All reaction conditions are assumed to be STP.

  • $\begingroup$ I'm afraid in many senses this question is not that useful. Firstly it would be useful to tie down the conditions more tightly, temperature, pressure, concentrations etc. But even with that for chemistry you really can't just ignore kinetics, because "slow" really can mean "not observed" and in that case if you do ignore kinetics the answer to your question is whatever is the lowest energy state for system of interest, probably a mixture of cobalt oxides, ammonia and maybe nitrogen and water (or arguably a lump of iron). I suspect this is not what you want. $\endgroup$
    – Ian Bush
    Commented Sep 22, 2022 at 7:35
  • $\begingroup$ @IanBush thank you for the comment. I've re-worded the question to make things more clear. $\endgroup$ Commented Sep 22, 2022 at 13:34
  • $\begingroup$ Have you made any attempts at computing the stability of the complex or reaction energy with water or hydronium? It may also help if you think about the possible elementary steps involved in transforming hexamminecobalt(III) into hexaaquocobalt(III). You can start to prove this if you can identify a pH-dependent step and identify rates of elementary steps. $\endgroup$
    – Stephen
    Commented Sep 22, 2022 at 13:36
  • $\begingroup$ Just deleted the "what are its decomposition products" part altogether as, CTTOI, that part of the question mixes kinetics and thermodynamics. $\endgroup$ Commented Sep 22, 2022 at 13:38

1 Answer 1


I think it is tricky to do the analysis in "pure water", since you cannot have a charged solution. Maybe what you mean is an aqueous solution at neutral pH?

Based on what I saw in the LibreTexts link you shared, it looks like you can do some back of the envelope type analysis. They give us the following ligand exchange reaction $$ \ce{[Co(NH_3)_6]^{3+}} + 6\ce{H_3O^+} \to \ce{[Co(H_2O)_6]^{3+}} + 6\ce{NH_4^+}, \tag{1} $$ which has an equilibrium constant of $K_{eq} = 10^{64}$. The standard state reaction free energy at $T = 298.15$ K can be computed as $$ Δ G^\circ = -k_B T \ln K_{eq} = -3.79~\text{eV}. \tag{2} $$ Under more general conditions, you can write the thermodynamic driving force as $$ Δ G = Δ G^\circ + k_B T \ln\left(\frac{a_{\ce{[Co(H_2O)_6]^{3+}}} ~ a_{\ce{NH_4^+}}^6 }{ a_{\ce{[Co(NH_3)_6]^{3+}}} ~ a_{\ce{H_3O^+}}^6 } \right), \tag{3} $$ which can be rewritten as $$ Δ G = Δ G^\circ + k_B T\ln\left(\frac{a_{\ce{[Co(H_2O)_6]^{3+}}} ~ a_{\ce{NH_4^+}}^6 }{ a_{\ce{[Co(NH_3)_6]^{3+}}}} \right) + 6\ln(10)k_B T \times \text{pH}. \tag{4} $$ Now unfortunately this analysis is still going to be a little limited since you would need to define the ionic strength of the solution to compute mean ion activity coefficients (using something like the extended Debye-Hückel equation). You also need to take into account other possible reactions that could occur, such as conversion of ammonium into ammonia gas, maybe buffering effects from dissolved CO$_2$ gas, etc. But what you can do from here is at least check the conditions under which the hexamminecobalt(III) complex will be stable (i.e., for what set of ion activities will $Δ G > 0$).

Setting $Q$ to be the activity ratio in the Eq. 4, you can make a plot like the one below, which plots the logarithm of the activity ratio against pH. The dashed line denotes where $Δ G = 0$. From this, you can get an idea of the stability of the complex under a variety of conditions (barring any additional reactions taking place). You can see that you'd need to push the reaction pretty far to the right in order for the reverse reaction to become thermodynamically favorable again. That is a long winded way to say that hexamminecobalt(III) is pretty unstable under most conditions within the scope of this analysis.

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  • $\begingroup$ +10. Beautiful answer! I've just edited the first chemical formula to use ChemJax. Perhaps you can do the rest yourself? $\endgroup$ Commented Sep 23, 2022 at 0:34
  • $\begingroup$ Thanks, Nike. It's been updated. $\endgroup$
    – Stephen
    Commented Sep 23, 2022 at 1:30
  • $\begingroup$ Such a beautiful answer! $\endgroup$ Commented Sep 23, 2022 at 3:02

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