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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?

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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.

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