[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 $\ce{[Co(H2O)6]3+ -&gt; e— [Co(H2O)6]^2+}$. It oxidises water according to the equation:
  $$\ce{2 [Co(H2O)6]^3+ -&gt; H2O + 2 [Co(H2O)6]^2+ + 2H+ + 1/2O2} (2)$$
  
  Complex formation, however, can modify the standard electrode potentials considerably (for example, E° = +0.10
  volts for $\ce{[Co(NH3)6]^3+ -&gt; e— [Co(NH3)6]^2+}$ 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.

$$\begin{align}
\ce{  [Co(H2O)6]^3+ + 6 NH3 &amp;&lt;=&gt;&gt; [Co(NH3)]^3+ + 6 H2O} \\ \ce{[Co(H2O)6]^2+ + 6 NH3 &amp;&lt;=&gt; [Co(NH3)]^2+ + 6 H2O}
\end{align}$$

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.