If you add a ligand to the metal centre, you will stabilize it, thus making it easier to form. So if you put, say, gold into a solution of cyanide ions, the gold will become easier to oxidize because the product [Au(CN)4]- is much more stable than simple hydrated Au(III) ions. If the ligand didn't make the ion more stable, it wouldn't form in the first place.

From thermodynamic point of view the reversible potential of metallic electrode (metal in water solution of his ions) is representing by Nernst equation:

EMe/Mez+ = Eo Me/Mez+ + RT/zF ln a Mez+

The activity of the metal ions in solution is calculated using complex formation reaction with ligands L

Mez+ + nL = [Me Ln]

The constant of complex formation is K compl. = a [Me Ln] / an L a Mez+

a Mez+ = a [Me Ln] / K compl. aL

Substitute a Mez+ in Nerst’s equation :

E = Eo Me/Mez+ - RT/zF ln K compl. + RT/zF ln a [Me Ln] = Eo compl. +RTzF ln a [Me Ln] - RT/zF ln aL

Finally we have new Nerst equation (if I didn’t any mistake).

Normally formation of the stable complexes decreases the thermodynamic potential of the metal (new Eo compl.). It means the metal anodic dissolution facilitates by complex formation like gold dissolution is taking place in king’s vodka.

In English, we use the Latin name aqua regia ("royal water").

You can apply the Nernst Eq'n to relate the oxidation potential to the decreased-by-the-complex-formation concentration of free metal ion. You could also add the Gibb's Free Energy for the formation for the complex ion to the Gibb's Free Energy of the oxidation, then reconvert that back to an oxidation potential. You should get exactly the same answer; it's just a different way of looking at the reaction.

Gert Van der Zwan: As far as I know, there are two types of electrochemical behaviour. One is that described by Darren George, in which electroactive is a complex ion as a whole (e.g. ferro-ferricyanide or ferrocene etc). In other words, the electrochemistry occurs without liberation of ligands from the metal centre. Here, the strength of the ligand field is reflected directly (some examples are e.g. in Advances in Inorganic Chemistry, Volume 61, p. 444, but also separate articles can be found, e.g. Labuda et al. Ligand-structure effect on electrochemical reduction of copper(II) complexes, Monatshefte für Chemie / Chemical Monthly 1992, Volume 123, Issue 8-9, pp 693-699).

The other behaviour is that one which describes in his answer A. Nazarov, for single complex described by Lingane and later developed to what is usually called DeFord and Hume formalism. Here, electroactive is free metal ion but not the complex(es). The shift in potential will not probably be reflected to ligand field. The analysis by polarography was published by D. DeFord and D. N. Hume, J. Am. Chem. Soc. 73 (1951) 5321–5322.

Gert, there is no simple relation between spin state (ligand field strength) and redox potential. Ligand field strength has an influence on complex stability but energetically much more important is the electrostatic interaction between metal ion and the ligands. When you look at the hydration energy of 3d metal ions M2+ as a function of the d-electron numbers you can see that the hydration energy increases linearly from d0 to d5 to d10 (decreasing metal ion size). The two well known humps in between are due to ligand field effects but the deviation from the base line d0-d5-d10 is less than 10%. Nearly all Fe3+ complexes have higher stability constants than the corresponding Fe2+ complexes - because electrostatics dominate nearly allways. The redox potentials of the Fe(CN)6 complexes tell us that [Fe(III)CN)6]3- is more stable than [Fe(II)CN)6]4- by 7 orders of magnitude. The fact that [Fe(II)CN)6]4- is regarded as non-toxic is due to its kinetic inertness, not its stability.

I'm not sure if I can add much to the great answers that have already been given, but this has been a topic of research interest for me in the past that has focused specifically on non-heme iron and manganese.

It seems that in addition to the electronic factors of the ligand, the geometry and sterics of the ligand can be related to the redox potential of the metal complex that's formed. So, in the case where one oxidation state of the metal favors a particular geometry over another (say, in cases with Jahn-Teller distortion), ligands that are more comfortably able to accomodate that geometry (due to their sterics) will shift the redox potential to favor redox activity more than a ligand that cannot accomodate that geometry. I think this phenomenon may be related to Gert's post above about the effect of the ligand field on the d-orbital splitting.

Not sure if this is a chicken/egg explanation, any thoughts from anyone else would be greatly appreciated because it's a concept that I've been trying to sort out in my mind for some time. Thanks!

Adding a ligand gives rise to a decrease of the standard potential of any redox couple. The new potential is a function of the dissociation constant of the complexes formed which constitute the new redox couple.