It is unlikely that you are dissolving Au in those conditions, unless you have a very acidic environment, those cyanide groups are not going to be removed from the complex! This system is used extensively in electrochemistry and the conditions you cite do not generate any cyanide nor do they produce corrosion of the Au electrodes.

The formation of Prussian blue on the other hand can very easily happen in these systems and the process has been studied extensively. It is very common, especially at high concentrations of ferrocyanide and/or ferricyanide, and you don't even need to initially have both in solution, as you turn one into the other as you run your CVs. Usually making sure that both of your electrodes are very clean at the begining and that you keep the concentration of K3(FeCN)4 low and that of the supporting electrolyte high (KCl should be fine, but if you are worried, you can also use other salts such as KNO3 or Na2SO4). If it only happens at your CE, you can also try using a carbon-based one instead of the gold, as metals make Prussian blue formation more likely (although Au is among the best in avoiding it, in terms of metals... Pt for example is much worse!)

Our comment is that the used K3[Fe(CN)6] complex in our experiment has impurities of ferric cations that caused Prussian blue formation as follow reactions:

K+ + Fe3+ + [FeII(CN)6]4−→ KFeIII[FeII(CN)6]

or

4Fe3+ + 3[FeII(CN)6]4−→ FeIII[FeIIIFeII(CN)6]3 

That is true?

What is your comment?

Thanks

You could be dissolving your gold electrode. Cyanidation of gold is a well known technique of gold extraction from ore. Cyanide groups in Fe(CN)6--- ion are less available to react than free cyanide, but the oxidating conditions of your cell could drive the formation of Au(CN)2-. If that's the case, the Fe+++ ions that free up from this reaction could then combine with Fe(CN)6--- to make Prussian Blue precipitate. This is just a hunch, though.

Additionally (or alternatively), chloride could be an issue. Gold chloride is water soluble, and while gold does not spontaneously react with chloride, it will do so with enough oxidative drive (e.g. gold electrowinning, or aqua regia).

Run your CV in FCN with a different supporting electrolyte, and in KCl without the FCN, and see what happens.

It is unlikely that you are dissolving Au in those conditions, unless you have a very acidic environment, those cyanide groups are not going to be removed from the complex! This system is used extensively in electrochemistry and the conditions you cite do not generate any cyanide nor do they produce corrosion of the Au electrodes.

The formation of Prussian blue on the other hand can very easily happen in these systems and the process has been studied extensively. It is very common, especially at high concentrations of ferrocyanide and/or ferricyanide, and you don't even need to initially have both in solution, as you turn one into the other as you run your CVs. Usually making sure that both of your electrodes are very clean at the begining and that you keep the concentration of K3(FeCN)4 low and that of the supporting electrolyte high (KCl should be fine, but if you are worried, you can also use other salts such as KNO3 or Na2SO4). If it only happens at your CE, you can also try using a carbon-based one instead of the gold, as metals make Prussian blue formation more likely (although Au is among the best in avoiding it, in terms of metals... Pt for example is much worse!)

1. Always there is a redox reaction at the anode that we usually ignore. In your case the reduction of [Fe(CN)6]3- is normal.

2. Try to reduce the potential window (700 mV is too large).

3. Try to reduce the concentration of K3[Fe(CN)6] at about 20...50 mM.