[lviorel]

Hi,
I did electrolysis of potassium ferrocyanide solution, I used iron electrodes. I know that at the cathode was this reaction: 2H+ = H2, and on anode developed Prussian blue. I need to desribe reaction at anode.

Thanks.

[Borek]

Look at the Prussian blue formula. How does it differ from the ferrocyanide present before electrolysis?

[lviorel]

It differs with Fe(3+) ion. But where do these ions came from? From Fe (0) electrode or can they come from [Fe(CN)6](4-) ions?

[Borek]

Think about the amount of cyanide present.

[lviorel]

I don´t know, but I suppose that it is from Fe electrode. But can be also this reaction: [Fe(CN)6](4-) = Fe(3+) + 6(CN)(-) + e(-) ?

[Borek]

Have you liberated any cyanide, or did it all end in Prussian blue precipitate?

[lviorel]

it end in Prussian blue precipitate.

[Borek]

All cyanide

end in Prussian blue precipitate.

That rules out this reaction:

[Fe(CN)6](4-) = Fe(3+) + 6(CN)(-) + e(-)

But in fact this reaction points out the correct answer. Assuming for a moment that you have Fe³⁺ and stoichiometric amount of CN^- in the solution - will they react?

[lviorel]

I think they will react to [Fe(CN)6](3-).

[lviorel]

So these complexs are stable and they won´t react to (CN)- and Fe3+ is from electrode.

[Borek]

I think they will react to [Fe(CN)6](3-).

Correct. Now, let's assume you have not posted your conclusions yet

Fe³⁺ reacts with CN^-. Summing it up with the reaction you have posted earlier:

[Fe(CN)₆]^4- = Fe³⁺ + 6CN^- + e^- = Fe(CN)₆^3- + e^-

or, without nonexisting "intermediate"

[Fe(CN)₆]^4- = Fe(CN)₆^3- + e^-

You see now what is going on?

[lviorel]

Yes. Thak you very much.