# Synthesis of sodium ferricyanide

I need to synthesize sodium ferricyanide, $$\ce{Na3Fe(CN)6}$$, for some benchtop experiments where I don't want any potassium ions in solution. The way I am doing that right now is oxidizing $$\ce{Na4Fe(CN)6}$$ using sodium permanganate, $$\ce{NaMnO4}$$, and filtering out the $$\ce{MnO2}$$ that forms. However, upon reviewing the SDS for sodium ferrocyanide, I found that strong oxidizing agents are an incompatible material (I am not sure why). Is there a better way to go about doing this synthesis?

• How do you define a "better" way? What aspect of your present method is making you unsatisfied? Jan 21 at 18:05
• Usually this operation is made by bubbling gaseous chlorine $\ce{Cl2}$ into a sodium ferrocyanide solution. You will obtain sodium ferricyanide plus sodium chloride, which usually does not disturb further reactions. Jan 21 at 21:14
• @orthocresol I edited the original post. The SDS mentions strong oxidizing agents are incompatible with sodium ferrocyanide.
– LGS
Jan 22 at 20:37
• @OscarLanzi you can buy $\ce{NaMnO4}$ commercially.
– LGS
Jan 22 at 20:39

As @Maurice said, it can be made by bubbling gaseous chlorine into a sodium ferrocyanide solution. You will obtain sodium ferricyanide plus sodium chloride, which will not disturb further reactions.

$$\ce{2Na4[Fe(CN)6] + Cl2 -> 2Na3[Fe(CN)6] + 2NaCl}$$

There is also a patent which mentioned electrolytic oxidation of sodium ferrocyanide:

$$\ce{2Na4[Fe(CN)6] + 2H2O -> 2Na3[Fe(CN)6] + 2NaOH + H2}$$

You can find a detailed description of the setup in the patent.

• Your method can work but you have to find a way to separate the insoluble MnO2. Jan 22 at 5:36
• This seems like a good way to do it and I see that it's listed as a common reaction done with potassium ferrocyanide to form potassium ferricyanide. However $\ce{Cl_2}$ is a strong oxidizing agent, and those are supposed to be incompatible with $\ce{Na_3[Fe(CN)_6}]$ according to the SDS.
– LGS
Jan 22 at 20:44
• Is there a way to do this and know the exact concentration of NaCl formed? Because I suppose I won't know how much of $\ce{Cl_2}$ I passed reacted with the system?
– LGS
Jan 22 at 21:34