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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?

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    $\begingroup$ How do you define a "better" way? What aspect of your present method is making you unsatisfied? $\endgroup$ Jan 21 at 18:05
  • $\begingroup$ 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. $\endgroup$
    – Maurice
    Jan 21 at 21:14
  • $\begingroup$ @orthocresol I edited the original post. The SDS mentions strong oxidizing agents are incompatible with sodium ferrocyanide. $\endgroup$
    – LGS
    Jan 22 at 20:37
  • $\begingroup$ @OscarLanzi you can buy $\ce{NaMnO4}$ commercially. $\endgroup$
    – LGS
    Jan 22 at 20:39

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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.

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  • $\begingroup$ Your method can work but you have to find a way to separate the insoluble MnO2. $\endgroup$ Jan 22 at 5:36
  • $\begingroup$ 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. $\endgroup$
    – LGS
    Jan 22 at 20:44
  • $\begingroup$ 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? $\endgroup$
    – LGS
    Jan 22 at 21:34

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