For my wood staining experiments I have purchased 1 kilo of ferrous sulphate as a source of iron(II). It turned to be quite oxidised (yellowish) and the its solution is muddy yellow instead of green. Does it make sense to try to recrystallise it in a hope that oxidation products are less soluble or insoluble in water?
5
-
2$\begingroup$ No, it probably doesn't. You could try to reduce it, though. $\endgroup$– MithoronCommented May 10, 2022 at 19:01
-
$\begingroup$ Ferrous sulfate solutions are unstable with respect to oxidation in water at neutral pH. No cure for the brown stuff. The oxidized products are already insoluble in water. $\endgroup$– ACRCommented May 10, 2022 at 19:10
-
$\begingroup$ BTW, podzol, the soil of pine forests, is often rich in ferrous iron, and when first exposed, beneath a layer of forest duff, it's first a pale green. Within fifteen minutes, it darkens to a reddish-tan as the iron is oxidized by the air. as @AChem states, most ferrous salts are not air-stable. $\endgroup$– DrMoishe PippikCommented May 11, 2022 at 0:13
-
$\begingroup$ How about filtering it? Ferric ions tend to aggregate; it might be easy, even if the rate is slow, to get what you need as you need it. $\endgroup$– James GaidisCommented May 12, 2022 at 19:26
-
$\begingroup$ If you were to dissolve it in dilute sulfuric acid then shake it with a solution of DEHPA (bis-2-ethylhexyl hydrogen phosphate) in an aliphatic kerosene then I am sure that the Fe(III) will extract while the Fe(II) will stay in the aqueous solution. You would have to work out a method of keeping the oxygen away from the purified product. You might want to read about the Fricke dosemeter. It so happens that radiation increases the rate at which ferrous sulfate reacts with oxygen $\endgroup$– Nuclear ChemistCommented May 26, 2022 at 17:06
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
You better make your turbid solution boil with some iron wires $\ce{Fe}$ and a bit of sulfuric acid. The turbidity observed in your solution is due to a mixture of two precipitates containing $\ce{Fe^{3+}}$ ions, produced by the oxidation reaction $$\ce{4 Fe^{2+} + O2 + H2O -> 4 [Fe(OH)]^{2+}}$$ then $$\ce{[Fe(OH)]^{2+} + 2 H2O -> Fe(OH)3(s) + 2H+}$$ or $$\ce{[Fe(OH)]^{2+} + SO4^{2-} -> Fe(OH)SO4}(s)$$ Whatever the chosen reaction, both are producing a turbid solution. And this turbid solution may be destroyed to produce $\ce{Fe^{2+}}$ ions if boiled with some iron wires or nails dipped into the acidified solution, which will reduce ferric substances into ferrous ions, according to : $$\ce{2 Fe(OH)3 + Fe + 6 H^+ -> 3 Fe^{2+} + 6 H2O}$$ or $$\ce{2 Fe(OH)SO4 + Fe + 2 H^+ -> 3 Fe^{2+} + 2 SO4^{2-} + 2 H2O}$$ In both cases, the precipitate has disappeared and all the iron in solution is under the form of ferrous ions $\ce{Fe^{2+}}$ and only these ions
