Is hexaamminecobalt(III) chloride thermodynamically stable in pure water?

Question

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.

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.