Dear Hadar, in general, oxygen evolution as well as oxide formation is pH-dependent, I attached a number of papers which illustrate this. From the theory, one might calculate the potentials by means of the thermodynamic equilibrium conditions which will result in a potential for both reactions.

This means that you have to solve the Nernst equation for both reactions as a function of pH. For the oxygen evolution this means:

2H2O -> 4H+ + O2 + 4e-

And the Nernst equation would be E= E0 + 0.059/4 * log (conc H+)^4

0.059/4 because 4 electrons are involved. And the exponent 4 to the H+ concentration, because 4 H+ are involved.

Results e.g. for pH 3, i.e. conc H+=10^-3 mol/l, a potential of about 1.05 V (E0=1.23 V), and for pH 10, it will be around +0.64 V.

Now you have to calculate the same for the oxidation of gold, which will show a different behaviour, because the E0 is different, and the number of transferred electrons per reaction is different!

The equation is Au2O3 + 6H+ + 6e- -> 2 Au + 3H2O (sorry, the reaction above is written in the oxidation direction).

The Nernst potential would be E= E0 + 0.059/6 * log (conc H+)^6

E0=1.5 V, and then e.g. for pH 3: The activity of all solid materials (i.e. Au-oxide and Au metal) is equal to "1", thus only the pH is important.

Results would be (from a thermodynamical point of view)

+1.32 V for pH3, and ca. +0.9 V for pH 10.

It should be mentioned however, that gold hydroxide formation is more likely for alkaline pH. And the real structure, i.e. more loosely bound atoms have to be considered, practically in a changed value for E0.

gold generally undergo reduction on applying 1.5V.

but gold is exist as nano particle, things will be different.

see

http://pubs.acs.org/doi/abs/10.1021/jp304157a

I agree with Jacob, small particles with their large surface/volume ratio may be oxidized easier (i.e. at lower potentials) compared to the bulk materials - e.g. because the surface atoms are bound weaker compared to the bulk, or to a regular surface.

Thanks Jacob and Dirk. In that case, a more general question, at what standard reduction potential gold oxidation happens and in what potential oxygen evolution? Is gold oxidation happens at lower or higher potentials than oxygen evolution?

Dear Hadar, in general, oxygen evolution as well as oxide formation is pH-dependent, I attached a number of papers which illustrate this. From the theory, one might calculate the potentials by means of the thermodynamic equilibrium conditions which will result in a potential for both reactions.

This means that you have to solve the Nernst equation for both reactions as a function of pH. For the oxygen evolution this means:

2H2O -> 4H+ + O2 + 4e-

And the Nernst equation would be E= E0 + 0.059/4 * log (conc H+)^4

0.059/4 because 4 electrons are involved. And the exponent 4 to the H+ concentration, because 4 H+ are involved.

Results e.g. for pH 3, i.e. conc H+=10^-3 mol/l, a potential of about 1.05 V (E0=1.23 V), and for pH 10, it will be around +0.64 V.

Now you have to calculate the same for the oxidation of gold, which will show a different behaviour, because the E0 is different, and the number of transferred electrons per reaction is different!

The equation is Au2O3 + 6H+ + 6e- -> 2 Au + 3H2O (sorry, the reaction above is written in the oxidation direction).

The Nernst potential would be E= E0 + 0.059/6 * log (conc H+)^6

E0=1.5 V, and then e.g. for pH 3: The activity of all solid materials (i.e. Au-oxide and Au metal) is equal to "1", thus only the pH is important.

Results would be (from a thermodynamical point of view)

+1.32 V for pH3, and ca. +0.9 V for pH 10.

It should be mentioned however, that gold hydroxide formation is more likely for alkaline pH. And the real structure, i.e. more loosely bound atoms have to be considered, practically in a changed value for E0.

Besides the excellent description provided by Dirk Luetzenkirchen-Hecht, I would add that you have to keep an eye on the solution composition. For example, if the NP are coated with a layer of surfactants (typically used during the synthesis), you will see a change in the potential.

Oxygen evolution always proceeds at oxidized surface...its hardly that you can generate oxygen on gold surface which is not already covered with oxide

You have to add fluoride ions in solution and gold would be oxidised easyly

Using a proper ionic liquid as a solvent will help u to oxidize gold to a desired oxidation state.

I am sorry, i want to say cyanide ions instad floride ions.

at all cases oxygen or hydrogen evolution occur with pH dependency. At higher acidic media the oxygen evolution took place at high anodic potential. to deplace metals oxidation potential you can ad a complex reagent.

Dear Hadar, one of possible ways to reduce oxidation potential of gold below oxygen evolution potential is addition of complexing ions like chloride: Au + 4Cl- => AuCl4- +3e. Gold is oxidized and dissolved as tetrachloroaurate