Sciencemadness Discussion Board

Well-known reaction of KMnO4 and HCl: how well-known is it really?

woelen - 4-9-2020 at 11:17

Most of you will know the easy method of making chlorine gas of good purity by adding solid KMnO4 to concentrated HCl.

When this is done, then you hear a nice crackling/bubbling noise and chlorine is formed, even at room temperature. The liquid becomes very dark brown.

In literature this reaction mostly is described as follows:

2 KMnO4 + 16 HCl --> 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2

I obtained one paper (thanks macckone!), which goes a little more into detal:

By Max I. Bowman

This paper states that intermediate MnO2 is formed, which then is reduced further to MnCl2. I think that this still is not the whole story.

Some observations, with the use of very pure reagent grade chemicals:
- The liquid does not become totally colorless, not even after many days of standing. It remains brown/green.
- On heating (near boiling), the liquid also does not become completely colorless. Long after all chlorine is driven off and no smell of chlorine is present anymore, the liquid still remains green/brown.
- On addition of a small amount of H2O2, the liquid at once becomes perfectly colorless. The H2O2 under these strongly acidic conditons immediately reduces manganese in any oxidation state, larger than +2, to colorless manganese(II). A similar observation is made with SO2 as reductor (e.g. from sodium metabisulfite).

The above observations make clear that the brown/green color is due to some manganese compound in oxidation state, larger than +2.

Having these observations, I was inclined to think that maybe Cl2 can be in equilibrium with manganese in a higher oxidation state. In order to test this, I prepared a solution of a little MnCl2 in conc. HCl. This solution is colorless. I bubbled quite some chlorine through this liquid, such that a pale green liquid is obtained with a pale green atmosphere above it. This I gently heated and swirled, to keep the chlorine dissolved. When this is done, the liquid remains very pale green, just the color of dissolved Cl2. On stronger heating, the liquid becomes perfectly colorless again and the Cl2 simply is driven off. So, this makes clear that manganese(II) is not oxidized by Cl2 in the strong acid.

So, I get two seemingly incompatible observations:
1) A solution of KMnO4 in conc. HCl cannot be made completely colorless, not even after boiling for some time, or waiting for days. The same is true for a solution of MnO2 (which must be totally free of iron for this experiment!). Such a solution also cannot be made completely colorless, it remains green/brown.
2) On the other hand, a solution of MnCl2 in conc. HCl cannot be made dark brown/green with Cl2.

So, now I wondered what the quite stable brown/green material is, which is formed from MnO2 (or KMnO4) and conc. HCl. When a more careful observation is made, then different stages of the reaction can be observed:
1) Initially, there is formation of a very dark brown color, which in very thin layers looks a little orange/brown.
2) After some boiling, the color of the liquid shifts from reddish brown to a more greyish brown and on even longer boiling it shifts to green/brown (olive green).
3) After much longer boiling, or standing for days in a capped tube, the liquid becomes lighter olive green, like certain brands of olive oil.

I also treated a solution of MnCl2 in conc. HCl with ClO2/air mixtures. Just pour the intense yellow gas mix over a solution of MnCl2 in conc. HCl. When you do this, then a very dark brown compound is formed at the surface of the liquid. On shaking it is dark brown, very much like the color you get immediately after adding pure MnO2 or KMnO4 to conc. HCl. Is this due to oxidation of Mn(2+) by the ClO2? Or is this a complex? Could it be that the brown/green compound is a complex of Mn and ClO2, with the ClO2 being formed (in small amounts from the Cl(-) and the manganese in high oxidation state)?
The brown liquid, obtained from passing ClO2 over MnCl2, dissolved in conc. HCl, cannot be distinguished from the liquid, obtained by adding KMnO4 to HCl. It remains brown for days, on boiling it turns olive green and getting it colorless is not possible by boiling alone. Addition of a pinch of sodium bisulfite immediately makes it colorless.

As you can see, this "simple" and "well-known" reaction has much more to it than people usually think. It is a useful synthetic reaction for Cl2 (albeit a little expensive), but it is also very interesting to investigate in more detail. I think that there are many interesting species involved in this reaction as intermediate.

The latter experiment with ClO2 I did, because I observed that when I added a drop of bleach or some calcium hypochlorite to a solution of MnCl2 in conc. HCl, then I did get some brown coloration. This, however, can be explained, because hypochlorites always contain a little chlorate when they are somewhat older and that chlorate gives a mix of Cl2 and ClO2 on contact with HCl. I remembered that from experiments I did several years ago (in 2015). I now decided to revisit this experiment and having a somewhat more structured approach.

[Edited on 4-9-20 by woelen]

Bedlasky - 4-9-2020 at 13:25

This is really intersting Woelen! I would never have thought of that chlorine dioxide can be formed in this reaction.

S.C. Wack - 4-9-2020 at 13:47

Not so sure chemists of yesteryear (up to 1930 or so referenced by Mellor vol 12 pp 374-380) would agree with a couple of these statements...AFAIK they would not include oxygen and would point to Mn+3, such as can be obtained in the form of the K salt, if KCl is also added.

Perhaps the very dark brown is +4...this paragraph from Mellor sounds sensible if not correct:

[Edited on 5-9-2020 by S.C. Wack]

mncl.gif - 58kB

unionised - 5-9-2020 at 04:28

Cotton and Wilkinson also refer to the reaction as
MnO4- + 4 Cl- + 8 H+ --> Mn3+ + 2 Cl2 + 4 H2O
"The Mn3+ is only very slowly reduced to Mn2+ unless a catalyst such as Cu2+ or Ag+ is present"

Bedlasky - 5-9-2020 at 04:35

Remy also mentioned formation of [MnCl6]2- and [MnCl5]2- in reaction of MnO2 with conc. HCl.

woelen - 5-9-2020 at 11:15

Interesting remark about catalysts, such as Ag(+) or Cu(2+). I'll try the reaction with one of these and see whether the dark brown color disappears quickly in the presence of one of these catalysts.

unionised - 5-9-2020 at 11:25

i was hoping you would.
It's potentially important to anyone using the reaction to make chlorine.
It's more economical if the reaction goes to Mn(II).

Maurice VD 37 - 6-9-2020 at 12:02

Woelen's study is rather interesting. I would suggest to register the optical spectrum of the solution starting from the very beginning of the reaction up to the end. Permanganate spectrum is known to have three "fingers" in the green region. How do these fingers change with time ? Does one see new peaks appearing ?

woelen - 7-9-2020 at 02:11

I tried the experiment with copper(II) as a catalyst and indeed, this makes a difference!
I added a small pinch of copper sulfate to 4 ml of 30% HCl.
This gives a nice green solution, with a yellowish hue.
To this, I added a spatula of solid KMnO4. Initially, the reaction does not differ from one without adding the copper(II) sulfate. As long as there still is solid KMnO4, there is fast production of Cl2, and a crackling/bubbling sound. The liquid becomes very dark brown. Once all KMnO4 is used up, the liquid remains dark brown and production of Cl2 stops (or becomes slow).
Next, I gently heated the solution. This is where the difference occurs. Without copper(II), one needs to heat for a long time and quite strongly to get most of the manganese reduced to its +2 oxidation state. Even after long heating, the liquid remains fairly strongly colored. With the copper(II), the liquid quickly (within a minute or so) is green/yellow. Whether really all manganese is converted to the +2 oxidation state, or nearly all (e.g. 99+ %), is hard to see with the color of the chlorocuprate complex in the mix as well, but one thing is clear, from a practical point of view, all manganese is reduced completely.
From an academic point of view it may be interesting to determine whether a tiny fraction of the manganese still remains in a higher oxidation state or not ;)

unionised - 7-9-2020 at 08:46

I wonder if the legendary Cu(III) oxidation state is involved.
Ditto Ag(III) if that also acts as a catalyst.
Thanks for indulging my curiosity .

woelen - 7-9-2020 at 10:39

The use of copper(II) as catalyst indeed is very interesting. I still had a test tube standing, with a pale brown solution, which had been boiled for several minutes already and was standing there already for several days. I added just 2 drops of the spent solution of the experiment, given in my previous post (2 posts above this post). Next, I gently heated and just in a few tens of seconds the brown color was gone and a very pale lemon color remained (very dilute tetrachlorocuprate(II) complex).

It is hard for me to check whether the copper(III) oxidation state is involved. The mechanistic pathways for the redox reaction must be quite different in the presence of the copper(II), but I have no idea how it is affected.

kmno4 - 10-9-2020 at 15:50

The reaction is really complicated. It is because of chloride ligands, stabilizing higher oxidation states of Mn. More HCl gives more satable "browns". When Cl2 is prepared from KMnO4 (or MnO2), NaCl and H2SO4 (in excess), the mixture also becomes deeply brown, but it ends as pinkish solution after longer boiling.
There is similar old paper titled "THE ACTION OF HYDROCHLORIC ACID ON MANGANESE DIOXIDE" and the author made very laborious work to isolate Mn(III) and Mn(IV) compounds from the mixture (in organic solvens).