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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

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

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

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Although the exact proof needs complicated, empirically-derived equations, the essential idea is simple.

Cobalt(III) is a highly charged cation of a moderately electronegative(1.8) atom, so it wants electrons badly.

When bare-naked cobalt(III) is near water, it gets so mad at the highly electronegative, "greedy" oxygen atoms of water's not sharing its lone pairs that it smashes the oxygen-hydrogen bonds of water and "takes away" the bonding pair(s) of electrons of water. Cobalt(III) is reduced to cobalt(II), and the electron-taken-away, zero-oxidation-state oxygens end up pairing with each other to form dioxygen gas- just like how the exceedingly greedy King Louis XVI ended up on the Guillotine and "cut apart".

However, things change when cobalt(III) is near ammonia. Nitrogen is less electronegative than oxygen (meaning that it likes sharing its lone pair to cobalt more than oxygen), and although nitrogen is still much more electronegative than cobalt, the ammonia molecule doesn't mind forming a dative bond to cobalt, even if that would mean that it would gain a formal positive charge. This is obvious, since the atmosphere around the positively charged nitrogen in the ammonium cation is already less "greedy" than that of the oxygen atom in the oxonium cation- when one replaces one hydrogen in the ammonium cation with the even-less-electronegative-than-hydrogen cobalt, the positively charged nitrogen becomes even less "greedy". Sure, it does want more electrons under extraordinary situations, but it doesn't want more electrons at ordinary circumstances and hence doesn't need to (and doesn't) "take away" electrons from the oxygen-hydrogen bonds of water. The nitrogens have 8 electrons each near themselves, the cobalt has 18 electrons near itself, and the hydrogens have 2 electrons each near themselves. Everybody is happy- just like socialist democratic Scandinavia.

Hence, hexaamminecobalt(III) is stable in water while hexaaquocobalt(III) isn't.

Remember that the oxidation state theory assumes all-ionic bonds, and that ionic bonds are "taken-away electrons" while covalent bonds are "shared electrons".

Omitting numerical redox reaction coefficients here since they're easily calculated and off-topic

Considering that ammonium is much less acidic, and hence much less "greedy", than oxonium

Ammonium salts do decompose to ammonia gas and the parent acid when heated; a similar reaction also occurs with hexaamminecobalt(III) as well.

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