# [Co(H2O)6]3+ vs [Co(NH3)6]3+ oxidising water

Cobalt(III) complexes cannot be prepared by addition of the ligand to a solution of the metal cation because the aqua cation $$\ce{[Co(H2O)6]^3+}$$ has no stable existence in solution. It is strongly oxidising with an E° = +1.84 volts for $$\ce{[Co(H2O)6]3+ -> e— [Co(H2O)6]^2+}$$. It oxidises water according to the equation: $$\ce{2 [Co(H2O)6]^3+ -> H2O + 2 [Co(H2O)6]^2+ + 2H+ + 1/2O2} (2)$$

Complex formation, however, can modify the standard electrode potentials considerably (for example, E° = +0.10 volts for $$\ce{[Co(NH3)6]^3+ -> e— [Co(NH3)6]^2+}$$ and therefore, oxidation from Co(II) to Co(III) is easily accomplished in the presence of coordinating ligands".

Why is it that $$\ce{[Co(H2O)6]^3+}$$ oxidise water, but $$\ce{[Co(NH3)6]^3+}$$ does not? I understand how values of reduction energy potential effect the proceeding of a reaction, but what is it about $$\ce{[Co(H2O)6]^3+}$$ that makes it able to oxidise water more?

There are 2 competing principles.

$$\ce{[Co(H2O)6]^3+}$$ has strong oxidation effect, but...

$$\ce{[Co(NH3)6]^3+}$$ is much more stable than $$\ce{[Co(NH3)6]^2+}$$.

As the consequence, the former from the next 2 equations is significantly shifted to the right.

\begin{align} \ce{ [Co(H2O)6]^3+ + 6 NH3 &<=>> [Co(NH3)]^3+ + 6 H2O} \\ \ce{[Co(H2O)6]^2+ + 6 NH3 &<=> [Co(NH3)]^2+ + 6 H2O} \end{align}

Therefore

$$\frac{c_{\ce{[Co(H2O)]^3+}}}{ c_{\ce{[Co(H2O)]^2+}}}\lt\lt \frac{c_{\ce{[Co(NH3)]^3+}}}{ c_{\ce{[Co(NH3)]^2+}}}$$

As both complex pairs keep the same redox potential, their standard redox potentials are therefore very different.