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