Chromium Chemistry (2): K2Cr2O7

Potassium dichromate(K2Cr2O7) is a typical example of Cr(VI) compounds, which are all strongly oxidizing, extremely toxic and carcinogenic. Thus they are controlled.

However such compounds are very useful to bypass the Cr(III) exchange inertness, so we decided to make some.

2Cr3+ + 3[CO3]2- + 3H2O = 2Cr(OH)3 + 3CO2

2Cr(OH)3 + 3ClO- + 4OH- = 2[CrO4]2- + 3Cl- + 5H2O

2[CrO4]2- + 2H+ = [Cr2O7]2- + H2O

Attempt 1

20.00g(0.04mol) KCr(SO4)2.12H2O was dissolved in 200mL water to form a purple solution. 8.28g(0.06mol) K2CO3 was dissolved in water and added to that solution. Reaction quickly happened upon stirring, gas evolved and green precipitation appeared. This is Cr(OH)3, and is collected by filtration and washed. Filtration must be done after evolution of CO2 has ceased. This is normally very painful but this time filtration was very fast.

Then, the Cr(OH)3 was thrown into some water, 4.48g(0.08mol) KOH was added and 110g bleach(~4% NaClO, ~0.06mol) was added. Base can neutralize the H+ formed in this reaction, thus eliminating the possibility of toxic Cl2 formation.

Due to unknown reasons the reaction seems to be very slow, while it was quite quick when I tried it during my early days, probably because I added KOH this time. And reaction was not very complete, so I added a little more to dissolve all, but it seemed to be not that case, so I gave up and simply decanted to collect the solution.

Then sulfuric acid is added until the solution became very acidic. Chlorine was released due to excess of NaClO. To be honest testing pH was hard as the paper was quickly bleached. Be very careful and avoid breathing too much! I am now in the hospital and I don't know how did I come here. I was nearly dead at home, suffering from coughing, sneezing and difficulties breathing.

OK, we survived it! Then solution was boiled to minimal volume, but during this the solution turned nearly black and chlorine was evolved. Obviously we accidentally added way too much acid and greatly enhanced the oxidizing power!

By the way, to dispose the extremely toxic Cr(VI) solution, it was found that EtOH or H2C2O4 does not work well. FeSO4 is the best as it can reduce and precipitate Cr(VI) at any pH. The Fe(OH)3 stain caused by this can be easily removed with solution of H2C2O4.

Attempt 2

So, we have to repeat this experiment that nearly killed me. Same procedure was followed, but we calculated the theoretical amount of sulfuric acid to be 5.24g 37.4%(0.02mol). However since some sulfuric acid is consumed by ClO- we have to use more. We decided to first filter(this is in fact possible as basic [CrO4]2- is not very oxidizing), boil off some volume, add such amount first, and after chlorine evolution ceased carefully add more, repeat this until pH is adjusted to around 1.

Then solution was concentrated to minimal volume(~10mL), cooled, solution decanted, crystals washed with minimal water and finally dried. Solution was in fact a little opaque but there is already no chance of filtration. In fact far before this crystals already produced in a large quantity, maybe the data on Wikipedia is wrong. Cooling the solution to ice cold in a PTFE beaker before decanting can give more, but if you accidentally freeze it you need to wait for it to melt.

However, the product was found to be dirty and the residual acid and NaCl even caused reduction!

Very unluckily we did not even leave a photo during this experiment.

Clearly, decanting is not a good way to remove the solution hiding in the crystals, we accidentally precipitated other salts out and we surely cooled too much so solution was somewhat frozen. A sand core funnel is a good choice to filter. Also it is said that bleach contains huge amount of NaCl which may be very hard to remove from product.

In fact, long long ago, during a failed preparation of Cr2+, I observed that Cr3+ is oxidized by anode very easily, forming a yellow to brown stuff and is soluble, so it should not be CrO2 but soluble chromate.

Attempt 3

Anode: 2Cr3+ - 6e- + 7H2O = [Cr2O7]2- + 14H+

Cathode: 2H2O + 2e- = H2 + 2OH-

Net: 2Cr3+ + 7H2O = 3H2 + 8H+ + [Cr2O7]2-

Due to the low solubility of K2SO4 we decided to replace sulfate with acetate first, and this way the generated AcOH can be directly released as a free acid. This is possible as both H2SO4 and CrO3.xH2O are surely much stronger than AcOH.

20.00g(0.04mol) KCr(SO4)2.12H2O was dissolved, and a solution of Ba(OAc)2 from 15.80g(0.08mol) BaCO3 and 12.00g(0.20mol) AcOH was added. I used an excess of acetic acid to dissolve as much as possible but there is obviously some insoluble stuff. Doesn't matter. This solution was filtered, which is really very painful and I filtered a few times through the same paper to make it clear. If I used sand-core funnel it might be permanently blocked as almost nothing can dissolve BaSO4.

Then the dark purple solution was electrolyzed with a partitioned cell, and the anode is in the outer cell. Such apparatus is introduced before. Initially the current was very low even at a voltage of 12V, but only the inner cell generated gas(H2) which means oxidation is going. Also foam was generated in the inner cell, probably due to the local production of Cr(OH)3, and current dropped to nearly 0. This was easily solved by adding some dilute acetic acid into the inner tube. Also it seems that a white stuff of unknown composition and a yellow byproduct of PbCrO4 or BaCrO4 or something was generated in a small amount. Current was measured to be 0.5mA which is obviously way too low.

Later, we found that anode was seriously corroded though not completely. We consider it to be due to the damn electricity down during night, as PbO2 can react with Cr(III) and Pb can react with Cr(VI). But this is not the case, and in fact even when current is there the anode is still corroded. So we gave up.

In the final solution much yellow, white and green precipitation was observed. We decided to use FeSO4 to destroy Cr(VI) and to precipitate Pb2+(if exist).

Damn... but I have a great, very great idea now!

Attempt 4

Well, first question, how to get rid of NaCl and NaClO and every disgusting chlorine compound that can not only pollute your product but also kill you painfully? In fact, we can use calcium salts to precipitate CaCrO4. This is probably much more soluble than CaSO4, unlike what we thought at first, though different sources give very different data, but of course much less than K2CrO4 or CaCl2 or KCl so we can in principle get much, as long as we use very concentrated solutions. Then we can easily convert it back with K2CO3, as CaCO3 is very insoluble.

Well, maybe CaCrO4 is still too soluble so I decided to use SrCl2 to precipitate SrCrO4 instead. Ba won't work as BaCrO4 is very insoluble. SrCl2 is very cheap and not controlled, and might be useful in the future. Best properties of Sr is not only the appropriate solubility of SrCO3 and SrCrO4, but also the dramatically increased solubility of the latter when heated.

Second, how to get K2Cr2O7 without K2SO4? According to the Pourbaix diagram of Cr, [Cr2O7]2- is already the dominant species at pH of 7, so acetic acid should also be able to acidify the solution, and KOAc is very very soluble so it won't bother us.

To be honest, at this point I found CrO3 in some online shops, but at this point we are already unstoppable, and we amateur chemists should not depend too much on the mercy of the government.

After concentrating the yellow solution to ~100mL, 13.33g(0.05mol) SrCl2.6H2O in minimal water was added. This 25% excess is for the potential excess of KOH, as Sr(OH)2 is also only slightly soluble and may compete with SrCrO4 formation. I accidentally used too much water but in fact not a big deal.

Well, seems that agents from Tianjin is even worse than I thought, and when the solutions are mixed Cl2 was formed!

Even when hot a huge amount of precipitation appeared, and after full cooling down, we got... much yellow solid and a slightly orange solution. Obviously acid in SrCl2 converted some chromate to [Cr2O7]2- which cannot be precipitated, but we could only continue. Filter on a sand-core funnel to get the SrCrO4.

Since Sr(OH)2 is now unlikely, we decided to use no excess of K2CO3. So, 5.52g(0.04mol) K2CO3 was dissolved in 100mL water and SrCrO4 was added. Solution quickly gained yellow color, and to promote full conversion we boiled it for a while. Solution is yellow and solid is very slightly yellow. Filter to get the K2CrO4 solution without chloride.

Now its finally time to acidify. IDK how much AcOH should be added, but since AcOH is surely unable to turn it into CrO3 we can simply add excess. So, 6g(0.10mol) AcOH was added, and after CO2 evolution(from K2CO3) ceased the solution is nice orange and pH is 4.

The last thing is to boil it to minimal volume(you can allow crystals to form), and after cooling much crystals would appear. You need to stir and use glass rod to scratch wall during this to prevent violent boiling. Filter on a sand-core funnel and use minimal ice-cold water to wash it. Something white, probably KOAc or even K[H(OAc)2] can be observed but washing can remove it. The volume is in fact not very small, but AcOH in it may reduce solubility of both salts.

Yield: 1.43g(24.3%). Loss, in my opinion, is largely during the SrCrO4 precipitation, and we may need to manually adjust pH next time. Such product is very pure but contains a little bit unknown white solid.

This is less than 0.005mol so we cannot prepare our complex now, but already very good.

Attempt 5

Modifications:

1. K2CO3 in the first step is increased to ~9g to ensure full precipitation.

2. During precipitation of SrCrO4, SrCl2.6H2O was increased to 16g(0.06mol, 50% excess) and KOH(in the form of solution and ABSOLUTELY NOT SOLID) was added carefully into the boiling mixture until pH is adjusted to ~12. Boiling is to increase the solubility of Sr(OH)2. This time, upon cooling the solution became a normal pale yellow instead of intense orange.

3. 6.90g(0.05mol, 25% excess) of K2CO3 is used to extract chromate.

Yield: 2.63g(45%). Nearly doubled, but still too low and we need to think about reasons. From color we can in fact conclude that most steps are almost quantitative, for example, KCr(SO4)2->Cr(OH)3, Cr(OH)3->K2CrO4, SrCrO4->K2CrO4 and K2CrO4->K2Cr2O7. So loss could only be from the precipitation of SrCrO4 and the crystallization/washing of K2Cr2O7. If we need to do it once more we would explore ways to reduce loss. But, let's call it a day...

All in all, DO NOT use reagents with abnormally low price!!!

Attempt 6

Well, we are back to test a really amazing way...

Same way was used to produce the chromate solution.

It was filtered and treated with a solution of 0.05mol Ba(OAc)2.

A cream-colored stuff separated immediately but it cannot settle so we boiled it for a while and it now settle quickly, but it can still cross paper almost completely so we used G6 funnel and it can now be stopped, only a nearly colorless solution crossed.

This solid is BaCrO4, and may contain a little amount of BaCO3 but doesn't matter.

2.76g(0.02mol) K2CO3 and 2.64g(0.02mol) (NH4)2SO4 was mixed in some water and boiled until no ammonia was released(being endothermic it can take a long time). K2CO3 can be in a small excess to remove any ammonia, since ammonium can reduce Cr(VI). Well I accidentally made twice the amount but this can be fixed simply by dividing the solution into half.

The BaCrO4 was thrown into it, almost nothing happened and 5.24g(0.02mol 37.4%) H2SO4 was added. Solution immediately gained color and became orange when boiled and stirred. It is impossible to check progress so I just heated until I got tired. This was filtered and solution is not as orange as I expected, solid was still somewhat yellow. We can conclude that this reaction is very incomplete, probably due to the same structure of BaSO4 and BaCrO4.