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DerAlte
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Chromium Revisited
A stroll through the oxidation states and colors of Chromium. Pt1.
About 25 yrs. ago I did a series of simple experiments to show my teenaged son how varied this fascinating element’s chemistry was. Recently I felt
a yen to repeat these and did so. I thought it might be worth while to suggest some ideas, and chromium seems to be rarely mentioned, so here goes.
Note 1: All chromium compounds are poisonous, in the order Cr[II] < Cr[III] << CrO4 - - and Cr2O7- - .
Figures for LD50 rat oral are approx. 1900,1800,180, 25 mg/kg. NACN is around 6mg/kg. Hence the chromates are very poisonous, about 1/4 as lethal as
cyanide. They can be absorbed though the skin, so wear gloves.
Note 2: The colors quoted here are as I perceive them, in daylight under partly cloudy skies. In general of many of the Cr compounds show quite
different colors in artificial light, different grades of fluorescent and especially incandescent, due to illuminating spectrum. Even eyes differ –
my wife sees blues where I see greens. Photo sensors also vary – the only way is to see them for yourself and decide. Do the experiments!
We will run through the oxidation states of Cr, starting at 0, the metal itself (and, due to the definition used, that of a few organic Cr compounds
such as carbonyl, Cr(CO)6).
(1) Metallic Cr, Cr[0], can be made by the amateur quite easily by thermite reduction of the oxide Cr2O3 (lots of data on this site, UTSE). Another
method is by electrolysis of acidified Cr2(SO4)3 on to a copper cathode. Good hard shiny coatings are hard to do and usually the coat can be peeled
off. A mercury cathode can be used to make an amalgam. An iron cathode can be used (if inserted with current on) and will produce a black powered Cr
metal easily scraped off.
(2) For the following experiments, prepare a saturated solution of about 50-100cc of a chromic Cr[III] salt, usually sulphate or chloride. Chrome Alum
for growing crystals is often available and this is what I use, KCr(SO4)2.12H2O. This and the sulphate are both dark reddish-violet crystals.
DO NOT HEAT to assist solution for reasons below, dissolve at RT, first grinding the alum or sulphate to a fine powder. {If you use
the chloride, CrCl3.6H2O it will probably come as dark green crystals rather than violet}. In solution the alum acts the same as the sulphate, the K+
are merely spectator ions.
The Cr[I] oxidation state does not exist as far as I know (AFAIK), nor do distinct hydrides, although the metal can occlude hydrogen like all
transition metals.
(3) Aqua Complexes:
(3a) Heat a small amount of the violet (purple by transmitted light) solution of Cr[III] sulphate to boiling for a few minutes. The solution turns
from violet to deep green. On cooling it preserves the green color. It will stay green for several days to weeks but finally return to the original
violet. The green form is chemically different from the violet form as can be shown by precipitation of the sulphate ion with a barium salt,
and also by electric conductivity or freezing point measurement. The mystery was solved in the early 1900s; the violet ion is the complex Cr(H2O)6+++
; the dark green one maybe [Cr(H2O)4(SO4)]+; or perhaps [Cr2(H2O)l0(SO4)]++++ or the covalent [Cr2(H2O)6(SO4)3], or a mixture of all three.
Only the violet form will crystallize. If the green form is heated to dryness, only green scales result. Further cautious heating produces an
orange-yellow product which is the anhydrous form with perhaps some decomposition. Strong heating (bright red heat) causes decomposition and SO3
evolution: Cr2(SO4)3 --> Cr2O3 + 3SO3
For the chloride of Cr[III], Wiki has:
“Chromium(III) chloride (also called chromic chloride) is a violet coloured solid with the formula CrCl3. The most common form of CrCl3 sold
commercially is a dark green hexahydrate with the formula [CrCl2(H2O)4]Cl.2H2O. Two other hydrates are known, pale green [CrCl(H2O)5]Cl2.H2O and
violet [Cr(H2O)6]Cl3. This unusual feature of chromium(III) chlorides, having a series of [CrCl3−n(H2O)n]z+, each of which is isolable, is also
found with other chromium(III) compounds.”
This indicates the facility with which Cr forms colored complexes, in this case with water. It is also interesting to note that both Cr[III] sulphate
and chloride, if anhydrous, are virtually insoluble in water. The crystalline form of sulphate is Cr2(SO4)3.12H2O or, according to the above,
[Cr(H2O)6]2(SO4)3.6H2O, and is very soluble.
(4) Cr[III] Compounds
{The temptation to insert a Pourbaix or Eo/pH diagram here was great – if you understand them they tell a lot at a glance. But many do not and it
might merely confuse.}
The salts of Cr+++ with strong acids all give an acid reaction in solution – in other words Cr(OH)3 is a weak base; being a bit more technical, the
ion Cr+++ has a pKa of about 4 whereas a strong base ion like Na+ has pKa~14.
(4a) Take a test-tube of the violet sulphate solution and drip in some fairly concentrated solution of NaOH (or KOH). Until the hydroxide neutralizes
some of the acidity nothing happens, then a flocculent precipitate of a light gray-green color begins to settle out. If too much hydroxide is added,
this precipitate re-dissolves.
The precipitate is hydrated Cr[III] hydroxide, Cr(OH)3.xH2O. If instead the green form of sulphate is used, the formation is slower but the same
precipitate obtained. If done carefully, the liquid becomes colorless and depleted of Cr[III] ions. If the precipitate is left long enough (or
centrifuged) the color becomes a very distinct green color and compacts, probably due to loss of water.
The reaction is simply Cr+++ + 3(OH)- + xH2O --> Cr(OH)3. xH2O (s); x is approx. 3 when dried carefully. In alkaline solutions Cr[III] always
exists as the hydroxide, except when the pH is high.
(4b)Addition of excess hydroxide causes the precipitate to redissolve giving a light green solution. This is usually stated to be due to chromite ion
Cr(O2)-, or a hydrated form of this. This is still a Cr[III] ion, probably O=Cr—O-. Reducing the pH with acid re-precipitates the hydroxide.
(4c) Using NH4OH a similar precipitate with a grayish color is first formed. As further hydroxide is added carefully, again the liquid will clear. If
excess ammonia is added, and the solution left for an hour or so, this gelatinous ppt. will turn a light purple, quite different in color from the
sodium hydroxide case. On standing for a few days, or on prolonged boiling, the purple color disappears as ammonia is lost, leaving a gray-blue
hydroxide. My guess is that the purple is a pentamminechromium complex; [Cr(NH3)5.(OH)x](3-x)(OH). yH2O.
{The colors of the Cr[III] ammine chloride complexes are; tri+++, yellow; tetraCl2+, green; pentaCl++, purple; non ionic tri Cr(NH3)3Cl3, violet and
the ion Cr(NH3)2Cl4- , orange-red (Pauling, General Chem.)}.
See Brauer, p1345ff, for more on hydroxides and the organic complexes of Cr[III].
(4d) The hydroxide has a very low solubility product in water, about 6x10^-31. It can be purified by washing with large amounts of water and can be
dried as a dull green powder, hydrated, and is a useful source of Cr[III] salts with common mineral acids. On heating to red heat it loses water to
become a bright green oxide, 2Cr(OH)3 - -> Cr2O3 + 3 H2O. This oxide, the most stable form of Cr[III] under mild conditions, is usually insoluble
in acids but can be reacted with fused alkali hydroxides to form chromates – see later on. Structure is O=Cr-O-Cr=O.
Most of the Cr[III] salts can exist as blue-violet or green types when hydrated. As large crystals the violet forms look black; as powder, reddish.
Chromium and potassium sulphate are isomorphous and crystallize together in all proportions, giving crystals that vary from lightest pink to almost
black. Chrome alum is the equimolecular mixture. It can form massive octahedral crystals with patience – I have grown them up to 7cms. long looking
as black as coal.
Insoluble Cr[III] salts are the normal phosphate {blue} CrPO4, can be hydrated; sodium carbonate solution produces a similar precipitate to hydroxide
that may be a basic carbonate or Cr2(CO3)2.xH2O. The color is purple – Cr[III] also produces CO2 complexes.
Part II to follow – Cr[II], Cr[IV] and Cr[VI] compounds.
Der Alte
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woelen
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Very nice read. Please let us indeed know your personal experiences with the other chromium oxidation states.
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DerAlte
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Thanks, woelen.
Although some may think this whimsy, Mea Maxima Culpa! I meant to post in General Chemistry but after posting something else here, and under the
effect of a couple of beers, it landed up here. It might get a better audience if moved but I don't know how to do that.
Regards
Der Alte
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Polverone
Now celebrating 21 years of madness
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Thread Moved
11-12-2011 at 09:47 |
DerAlte
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Chromium Revisited – Part II
A bit more about Cr2O3 and Cr(OH)3:
In the first part of this series attention was paid to Cr[III] compounds and a few of their complexes. The most stable form of Chromium under normal
conditions is in fact Cr2O3 which to my surprise has a higher heat of formation (lower enthalpy) than even the fluoride (that entropy term at work, no
doubt!). Again I am severely tempted to add a Pourbaix diagram at this point, but again I resist for two reasons; one, it may confuse and two, I
cannot find a suitable simple one at the 1M Cr+++ concentration indicative of typical experimental conditions. For those familiar with the Eo/pH
diagrams, the one in Wiki under Chromium can be consulted, but it is for 10^-5M Cr+++ concentration. Change of concentration changes the position of
vertical boundaries especially but moves all boundaries except, of course, the (dashed) water lines.
You will, for instance, never see Cr(OH)3 on Pourbaix diagrams (OK, I wouldn’t bet on it) because they show the most thermodynamically stable form
under the Eo and pH condition of the axes at STP. That is Cr2O3; it covers a wide area.
2Cr(OH)3 -- > Cr2O3 + 3H2O + 2.13E05 j/mole
(see www.argentumsolutions.com/publications/CorrVol39p4881983.pdf for a discussion of issues re Eo/pH diags for Cr/water: it has several but they
cannot be copied – secured pdf)
In other words, CR(OH)3 is metastable at RT and the equilibrium lies well to the right in the above equation. You can show this experimentally as
follows. {There, I knew I ought never to have mentioned Pourbaix diagrams….}
(4e) This takes a bit of time to carry out. One must first produce a working quantity of Cr2(OH)3.xH2O from a Cr+++ compound as outlined previously
– aim to get 5-10 g reckoned as unhydrated. Wash copiously with water several times. The let settle and decant, finally filtering through a fine
mesh filter paper. Dry carefully in an oven at ~ 120-140C for several hours, until the mud becomes a powder of a dull green color.
Try reacting some of this with dilute acid – it should react fairly rapidly and give the violet form of the Cr[III] salt.
Scrape the powder from the filter paper and place in an evaporating dish and heat in the oven to around 250C. This still does not totally dehydrate
the hydroxide but also does not seriously convert it to oxide. This can be proven by reaction with acid or strong alkali. Now take the powder and heat
slowly on an iron spoon over a propane flame. Somewhere around a dull red heat (500-600C, I would guess) if you are lucky you may see it caloresce –
suddenly brighten and then dim. {This did not work the first time I tried it – it needs a fair quantity (at least 5g) to work, IMHO}. Now heat
strongly. On cooling the power still has a dull green color but it almost totally non-reactive to strong acid or alkali solutions, showing that Cr2O3
is very unreactive.
The experiment demonstrates that Cr(OH)3 is changed to the oxide by heating strongly enough, but is not exactly unstable. In fact it can be kept
indefinitely at RT in the dried state without losing reactivity. The calorescence is probably due to a change of state (entropy change) as the
amorphous structure left on ignition of the hydroxide changes to crystalline Cr2O3 at some transition temperature..
Other ways to make chromic oxide are to heat ammonium dichromate (NH4)2Cr2O7 or K2Cr2O7 with sulphur. Both ways tend to produce a bright green
product, especially the second (which is fun to perform!).
(NH4)2Cr2O7 -- > N2 + 2H2O + Cr2O3
K2Cr2O7 + S - -> K2SO4 + Cr2O3
(4d)Heat some solid potassium hydroxide (or NaOH) with about the same quantity of potassium nitrate (or NaNO3; chlorates can be used) in a nickel
(or SS) crucible or dish until melted together. Add some Cr2O3 made by one of the above methods or otherwise available. It reacts to produce Chromate,
K2Cr2O4, a yellow substance. The Cr[III] state is oxidized to the Cr[VI] state easily under the extreme alkaline conditions – even air will do
this, slowly. More of this later.
It is worth noting that Chromium is quite a reactive metal, slightly less so than Zinc but more so than iron, yet chromium is plated on to iron (over
Ni or Cu) and is very resistant to corrosion. This is believed to be due to the stability of Cr2O3 formation. The metal does dissolve in fairly
dilute HCl but nitric acid passivates it, and sulphuric also has little effect, and even when concentrated has to be heated to react with any speed.
Enough for now; Pt. III to follow, Cr[II] and Cr[VI} oxidation states.
Der Alte
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Sedit
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Quote: Originally posted by DerAlte  | Thanks, woelen.
Although some may think this whimsy, Mea Maxima Culpa! I meant to post in General Chemistry but after posting something else here, and under the
effect of a couple of beers, it landed up here. It might get a better audience if moved but I don't know how to do that.
Regards
Der Alte |
I think its a good read and is right where it belongs, in the fundamentals. All to often the little things in chemistry are over looked and this is
fundamental Chromium chemistry so my hats off to you. It literally takes me weeks to write something of this nature if not longer due to lack of focus
and understanding so since I can read faster then I can write I am always happy to see someone do the dirty work for me especially when its Cr
compounds 
Knowledge is useless to useless people...
"I see a lot of patterns in our behavior as a nation that parallel a lot of other historical processes. The fall of Rome, the fall of Germany — the
fall of the ruling country, the people who think they can do whatever they want without anybody else's consent. I've seen this story
before."~Maynard James Keenan
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blogfast25
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Interesting point about Cr(OH)3, Der Alte. I didn't know it was so stable. Cr(III)'s great gusto for coordination complexes may explain something
here.
[Edited on 13-12-2011 by blogfast25]
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DerAlte
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@ blogfast
I don't know at what temperature Cr(OH)3 begins to change to the oxide. Moderately heated samples still dissolve in acids, etc.; after red heat esp.
after calorescence, it does not. But Cr2O3 almost looks like an oxidising agent in thermite, due to the reductive powers of Al! You are our expert
guru on thermites, it's too long ago sice I did that one. Is Cr2O3 noticably less impressive than, say, Fe2O3?
Regards,
Der Alte
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blogfast25
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| Quote: Originally posted by DerAlte | @ blogfast
I don't know at what temperature Cr(OH)3 begins to change to the oxide. Moderately heated samples still dissolve in acids, etc.; after red heat esp.
after calorescence, it does not. But Cr2O3 almost looks like an oxidising agent in thermite, due to the reductive powers of Al! You are our expert
guru on thermites, it's too long ago sice I did that one. Is Cr2O3 noticably less impressive than, say, Fe2O3?
Regards,
Der Alte |
The chromium thermite runs a little tamer, as reflected by the HoF of both oxides: Cr2O3 = - 1140 kJ/mole, Fe2O3 = - 826 kJ/mole.
The Cr2O3 needs a little heat boosting: KClO3 + Al or K2Cr2O7 + Al (I don't recall the precise figures but have logged them) to obtain lump metal.
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DerAlte
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Correction to Part 1 Above.
Der Alte wrote:
| Quote: | | sodium carbonate solution produces a similar precipitate to hydroxide that may be a basic carbonate or Cr2(CO3)2.xH2O. The color is purple –
Cr[III] also produces CO2 complexes. |
I re-ran this one today and now find that the color quoted is nearer lavender or light blue gray. Nor does it dissolve in excess Na2CO3. A short
boiling does not convert the color, except to maybe reduce the bluish tint to slightly more gray. Try it and see!
But it is not green as several texts have it. (But I have not yet tried the green form of the sulphate) All texts also say this is
not a carbonate but a hydroxide and on further effort I tend to agree. If the precipitate is washed many times with water, to eliminate carbonate
contamination, and treated with acid, very little CO2 is evolved. It dissolves to give the violet form of Cr[III]+++ aqua complex.
I think the difference in color is probably due to crystal size, a frequent cause of such phenomena. As for CO2 complexes, this too is ruled out. I
think this is a Der Alte aberration; I must have been thinking of the carbonyl, Cr(CO)6.
Remembering that the (violet) ion is [Cr(H2O)6]+++ and is acidic, we have in water
[Cr(H2O)6]+++(aq) + H2O(liq) < -- > [Cr(H2O)5(OH)]++(aq) + H3O+(aq);
So one would expect Na2CO3 + Cr+++ salt to produce CO2 provided the pH is low enough. It does (best seen in conc. Cr salt solution, else the CO2
remains dissolved). As the salt is gradually neutralized by OH- ions from the dissolved Na2CO3 or NaOH a sequence of exchanges between the H2O ligands
and the OH- ligands can be assumed to take place:
[Cr(H2O)6]+++(aq) => [Cr(OH)(H2O)5]++(aq) => [Cr(OH)2(H2O)4]+(aq)
=> [Cr(OH)3 (H2O)3] (Solid precipitate);
And if the pH exceeds the ~11 afforded by Na2CO3 in solution, as in the case of NaOH solution, this precipitate dissolves to form negative aqua
complex ions:
=> [Cr(OH)4(H2O)2] - => [Cr(OH)5(H2O)] - - => [Cr(OH)6] - - -
as the pH goes from 11+ to 14+.
Whether the so-called chromites like NaCrO2 really exist I do not know. Mellor has:
“...when hydra ted chromic oxide is treated with sodium
hydroxide, the formation of chromite precedes the formation of the hydroxide.
With sodium hydroxide below 102V, primary sodium chromite is formed, whilst
above 102V the soln. contains also tertiary sodium chromite. Potassium chromite
is similarly produced ; below 82V-alkali only the primary chromite is formed, whilst
above 82V the soln. contains also secondary chromite. From soln. of potassium
chromite which have stood for a long time, needle-shaped crystals of the formulaCr2O3.3K2O.8H2O have been obtained.”
(V stands for the inverse of concentration, liters/mol, IIRC). Best of luck if you want to try it!
The precipitated hydroxide aqua complex does dissolve in strong NH3 solution, in spite of its weakness as an alkali. It switches water for NH3 in the
hexa coordinated complex,
[Cr(H2O)6]+++ + 6NH3 => [Cr(NH3)6]+++ + 6H2O etc.
For such beauties as Hexaamminechromium (III) Chloride, etc., see Brauer – you are on your own.
The overall message is that Cr[III] preferentially forms hexa-coordinated compounds, ionic and covalent, unless anhydrous. And anhydrous CrCl3, eg, is
insoluble if pure – and covalent.
I performed the original experiments over a few months last year. These notes come from that record.
Der Alte
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blogfast25
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| Quote: Originally posted by DerAlte | So one would expect Na2CO3 + Cr+++ salt to produce CO2 provided the pH is low enough. It does (best seen in conc. Cr salt solution, else the CO2
remains dissolved). As the salt is gradually neutralized by OH- ions from the dissolved Na2CO3 or NaOH a sequence of exchanges between the H2O ligands
and the OH- ligands can be assumed to take place:
[Cr(H2O)6]+++(aq) => [Cr(OH)(H2O)5]++(aq) => [Cr(OH)2(H2O)4]+(aq)
=> [Cr(OH)3 (H2O)3] (Solid precipitate);
And if the pH exceeds the ~11 afforded by Na2CO3 in solution, as in the case of NaOH solution, this precipitate dissolves to form negative aqua
complex ions:
=> [Cr(OH)4(H2O)2] - => [Cr(OH)5(H2O)] - - => [Cr(OH)6] - - -
as the pH goes from 11+ to 14+.
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The solutions of trivalent cations like Fe3+ and Al3+ don’t form carbonates with Na2CO3 because of the acidity of the hexa-aqua cations
(M(H<sub>2</sub>O)<sub>6</sub><sup>3+</sup>, as you stated. Whether this is also the case with chromium (III) I
don’t know but it sounds likely.
In the case of bivalent cations of D-block elements, some form actual carbonates (FeCO3, NiCO3, CoCO3, MnCO3 e.g.), others form basic carbonates
(copper for instance) and some both (zinc for instance).
If Cr (III) really does form a basic carbonate, it is more likely to form from concentrated solutions of Cr and soda than from diluted ones and CO2
will inevitable be released:
(x+2y) Cr<sup>3+</sup>(aq) + 3(x/2 + y) Na2CO3(aq) + (3x/2+n) H2O(l) ==== > xCr(OH)3.yCr2(CO3)3.nH2O(s) + 3(x + 2y) Na+(aq) + 3x/2
CO2(g)
I doubt whether such a basic chromium carbonate exists but there’s one way to find out: isolate and wash the precipitate carefully to remove any
free carbonate and check whether CO2 is released on dissolving the filter cake in strong acid…
[Edited on 16-12-2011 by blogfast25]
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DerAlte
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Carbonate complexes do exist - eg Brauer cites Carbonatotetraamminecobalt(III) Sulfate [Co(NH3)4CO3]2SO4 • 3 H2O, but a quick web search
revealed none for Cr.
Der Alte
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blogfast25
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An interesting factoid about chromium is that its amphoterism cannot be used to separate it from iron, at least not in 'realistic' separations.
Some time ago, I co-precipitated Fe (II) and Cr (III) with NaOH, using a low Fe/Cr ratio, then added excess NaOH and sure enough the chromium went
into solution as chromite (Cr(OH)<sub>4</sub><sup>-</sup> .
"Problem solved", I thought until someone here claimed I didn’t know what I was talking about.
Now I’m all for proving myself wrong so I dissolved quite a bit of stainless steel in 30 % HCl with the goal of recovering the chromium. I
precipitated the metals as hydroxides, then added extra NaOH but the chromium did not dissolve! Either it had completely co-precipitated with the
Fe(OH)2 or possible something akin to ‘hydrated Chromite’ (the mineral) had formed.
Only by ‘fusing’ the wet precipitate with an excess of solid KOH did the Cr solvate as potassium chromite. Quite strange…
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DerAlte
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@blogfast
I would have treated the NaOH neutralized solution precipitates (or the redissolved solution) (assumedly from something like 18/8 SS)with NaOCl 5%-6%
; to convert to chromate. Ni and Fe would remain as hydrate oxides or hydroxides - it is damn difficult to produce ferrates and both Cr(OH)3 and
Cr(OH)4 - (dissolved "chromite") are a facile oxidation to CrO4-- ion at higher pH. H2O2 is a much weaker oxidant at high pH than NaOCl but still
works.
{More later in the Cr ptIII post to come.}
Regards,
Der Alte
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blogfast25
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| Quote: Originally posted by DerAlte | @blogfast
I would have treated the NaOH neutralized solution precipitates (or the redissolved solution) (assumedly from something like 18/8 SS)with NaOCl 5%-6%
; to convert to chromate. Ni and Fe would remain as hydrate oxides or hydroxides - it is damn difficult to produce ferrates and both Cr(OH)3 and
Cr(OH)4 - (dissolved "chromite") are a facile oxidation to CrO4-- ion at higher pH. H2O2 is a much weaker oxidant at high pH than NaOCl but still
works.
{More later in the Cr ptIII post to come.}
Regards,
Der Alte |
Ooops. I forgot to mention that with the KOH was mixed in an an amount of KClO3: the leachate of the fusion contained chromate, not chromite.
My bad.
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DerAlte
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Chromium Revisited – Part IIIa
IUPAC Gold Book defines a transition element as one whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete
d sub-shell. The electronic structure of Cr in the ground (unionized gaseous) state is [Ar].3d5.4s1; it has six valence electrons outside the
argon electron structure.
For technical reasons (as the physicists are wont to say whenever the explanation is too involved in a cloud of Quantum Mechanics) the 4s level has a
lower energy level than 3d in the atoms K, atomic number (Z) =19 and Ca Z=20. That is, 4s is bound more strongly to the nucleus. And hence the d
levels are not occupied first. The simplified explanation is that electrons in the s orbitals penetrate the Ar shells and hence see a higher part of
the nuclear charge, while the d orbitals do not. As Z increases the Ar core decreases in size due to the increased nuclear charge and the 4s levels
begin to approach the 3d levels.
The first appearance of the 3d orbitals is at Sc, [Ar].3d1.4s2; past Zn, [Ar].3d10.4s2, the 3d and 4s shells are filled and the 4p subshell starts at
Ga. At vanadium Z=23 the 4s and 3d levels approach equal energy, and at Cr Z=24 the 4s electron actually has a higher energy by 0.96 eV
(Pauling). This means the 4s electron is lost first in compounds of Cr.
The 3d block elements have no chemically similar elements of lower atomic weight and represent a new series with new chemical properties. Neither zinc
nor scandium ions have any colored compounds AFAIK (unless with transition metal anions, of course!) although they are part of the d block elements.
They can form complexes, of course.
I don’t consider them true transition elements, only those from Ti to Cu. These elements form at least one chemically stable ion with a
partly filled d sub-shell. The d orbitals can, due to the Pauli exclusion principle, each have two electrons with opposed spin, but the d shell first
fills up each with only one in each, due to the fact this is the lowest energy state and these are parallel rather than opposed.
Ti has configuration [Ar].3d2.4s2, Cu has [Ar].3d10.4s1. But Cr does not follow the usual Building Rule (in German Aufbau) for the electronic
configuration. Vanadium has [Ar].3d3.4s2 as expected but Cr has [Ar].3d5.4s1 instead of [Ar].3d4.4s2; the latter corresponds to an excited state of
neutral Cr.
Past this point only MS or PhD students and researchers steeped in QM fog fear to tread; I certainly dare not. Things get complicated in a hurry with
hybrid orbitals, splitting fields etc. that make the more familiar sp hybrids of the carbon atom look like child’s play.
The bottom line is that Cr is essentially hexavalent due to its five 3d’s and one 4s electron and the lone 4s is lost first on compound formation,
but no Cr(I) oxidation state exists AFAIK. The states labelled Cr[0] (AFAIK) (except the metal, by definition) are all coordinated covalent compounds
with organics like Cr(CO)6 and the structure is then 3d6. 4s0; Cr(II)++ is 3d4.4s0; Cr[III]+++ is 3d3.4s0 and so on.
The complex formation so prevalent in Cr and other transition metals is explained by the Cr atom ion having vacant bonding orbitals that act as a
Lewis acid. Suitable ligands that act as Lewis bases then donate electron pairs to form the bond eg :NH3 or H2O: or EDTA with its six sites.
The colors of Cr and true transitional compounds are due to transitions between levels of the d electrons which fall within the visual range 380nm to
750nm, or photons with energy 3.28 to 1.59eV.
Enough Preamble and on to the Chemistry. Metallic Cr was mentioned above in Pt.1. We pass on to Cr[II] compounds.
Chromium[II] compounds
These are all strong reducing agents and hence can only be made by even stronger reducing agents. The fundamental equation is thus, using Cr[III]
compounds as the Cr source, and ignoring aqua complexing on both ions,
Cr+++(aq) + e- -- > Cr++(aq) Eo = -0.42 V
This can be achieved electrolytically by cathodic reduction or chemically by dissolving metallic Cr in dilute acids or using the Zn/acid method.
Electrolysis requires a divided cell. The acid solution needs Cr metal. The Zn/acid method is easier and quicker but introduces Zn salts into the
solution.
Zn(s) -- > Zn++(aq) + 2 e- Eo = 0.782 V
So 2Cr+++(aq) + Zn(s) < --> 2Cr++(aq) + Zn++(aq) dE 0.386 V …overall reaction
However, if we look at the overall reaction as (sulphate case with H2SO4, Zn)
Cr2(SO4)3 +Zn -- > 2CrSO4 + ZnSO4, it looks as if adding acid is not necessary. But it is;
so also
2H+(aq) + 2e- - - > H2(g) Eo=0.000V
Zn(s) + 2H+(aq) -- > Zn++(aq) + H2(g) Eo=0.782V
Hydrogen is formed as a by-product of the zinc-acid reaction but the reduction of Cr+++ occurs on the surface of the Zn. Does this hydrogen also
reduce Cr[III]?
{ Herein lies the 200 year old mystery of the “nascent” hydrogen so popular in the ancient texts – still not properly solved AFAIK. If,
instead of zinc you use iron, hydrogen is still produced but the Cr[III] is not reduced.
For Fe(s) -- > Fe++(aq) + 2e- Eo=0.44V: This is 0.02V higher than the Cr[III] to Cr[II] reduction potential potential so any effect would
have a very low equilibrium constant K. Using electrochemical reasoning one does not expect reduction; K is too low. And so it turns out. Yet H2 is
produced as usual and this H2 – in its “nascent” form – also cannot effect the reduction.
What is special about the “nascent’ form produced by Zn, compared with that produced by Fe, if it is in fact effecting the reduction?
It is almost certainly not produced as H + H -- >H2; the H-H bond strength is 436 kJ/mol or 436/94.5 eV = 4.6 eV and there is nothing energetic
around powerful enough to achieve this energy. If the H2 does not enter the reaction why is the acid (source of H+ ions) needed? If it does, in what
respect is it an excited form of hydrogen?
The texts merely produce hand-waving or sweep this one under the rug. Few will dare to even mention the stoichimetry of the Zn/acid reduction. This
subject is well worth a thread on its own. If any of you theoreticians knows of interesting recent papers on this horny old problem I would be
fascinated to read them! }
As usual I ramble. One more point before I turn to the practical experiments. All the texts mutter about ‘instant oxidation by oxygen in the air’
and ‘Cr[II] is capable of reducing water to hydrogen’ and quote very elaborate precautions to avoid these calamities. Do not be put off! Yes, they
occur:
4Cr2+ + O2 + 4H+ → 4Cr3+ + 2H2O dE = + 1.64 V … oxidation of Cr++
2H+ + 2 e– → H2 Eo = 0.000 V
Cr3+ + e− → Cr2+ E0 = - 0.410 V
2 Cr2+ + 2H+ → H2 + 2Cr3+ d E = + 0.410 V …reduction of H+ in H2O
These are energetic, especially the oxidation, so air must be somehow excluded.
To Be Continued in Part IIIb…
Regards,
Der Alte
[Edited on 21-12-2011 by DerAlte]
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blogfast25
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| Quote: Originally posted by DerAlte | These are all strong reducing agents and hence can only be made by even stronger reducing agents. The fundamental equation is thus, using Cr[III]
compounds as the Cr source, and ignoring aqua complexing on both ions,
Cr+++(aq) + e- -- > Cr++(aq) Eo = -0.42 V
This can be achieved electrolytically by cathodic reduction or chemically by dissolving metallic Cr in dilute acids or using the Zn/acid method.
Electrolysis requires a divided cell. The acid solution needs Cr metal. The Zn/acid method is easier and quicker but introduces Zn salts into the
solution.
Zn(s) -- > Zn++(aq) + 2 e- Eo = 0.782 V
So 2Cr+++(aq) + Zn(s) < --> 2Cr++(aq) + Zn++(aq) dE 0.386 V …overall reaction
However, if we look at the overall reaction as (sulphate case with H2SO4, Zn)
Cr2(SO4)3 +Zn -- > 2CrSO4 + ZnSO4, it looks as if adding acid is not necessary. But it is;
so also
2H+(aq) + 2e- - - > H2(g) Eo=0.000V
Zn(s) + 2H+(aq) -- > Zn++(aq) + H2(g) Eo=0.782V
Hydrogen is formed as a by-product of the zinc-acid reaction but the reduction of Cr+++ occurs on the surface of the Zn. Does this hydrogen also
reduce Cr[III]?
{ Herein lies the 200 year old mystery of the “nascent” hydrogen so popular in the ancient texts – still not properly solved AFAIK. If,
instead of zinc you use iron, hydrogen is still produced but the Cr[III] is not reduced.
For Fe(s) -- > Fe++(aq) + 2e- Eo=0.44V: This is 0.02V higher than the Cr[III] to Cr[II] reduction potential potential so any effect would
have a very low equilibrium constant K. Using electrochemical reasoning one does not expect reduction; K is too low. And so it turns out. Yet H2 is
produced as usual and this H2 – in its “nascent” form – also cannot effect the reduction.
What is special about the “nascent’ form produced by Zn, compared with that produced by Fe, if it is in fact effecting the reduction?
It is almost certainly not produced as H + H -- >H2; the H-H bond strength is 436 kJ/mol or 436/94.5 eV = 4.6 eV and there is nothing energetic
around powerful enough to achieve this energy. If the H2 does not enter the reaction why is the acid (source of H+ ions) needed? If it does, in what
respect is it an excited form of hydrogen?
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I don’t believe the hydrogen plays ANY role, nascent or otherwise, although I used to believe that it did.
Not produced as H + H -- > H2? Then as what? The hydrogen produced is, erm, hydrogen… H2. Clearly H3O+(aq) + M(s) === > H(g) + M+(aq)
+ H2O(l) and 2 H(g) === > H2(g), no? What other mechanism could there be?
In:
Zn(s) + 2H+(aq) -- > Zn++(aq) + H2(g) Eo=0.782V
For ΔG you have to take into consideration also the ionisation energy of Zn, the heat of formation of H3O+ and the solvation energy of Zn++…
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Bezaleel
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| Quote: Originally posted by DerAlte | { Herein lies the 200 year old mystery of the “nascent” hydrogen so popular in the ancient texts – still not properly solved AFAIK. If,
instead of zinc you use iron, hydrogen is still produced but the Cr[III] is not reduced.
For Fe(s) -- > Fe++(aq) + 2e- Eo=0.44V: This is 0.02V higher than the Cr[III] to Cr[II] reduction potential potential so any effect would
have a very low equilibrium constant K. Using electrochemical reasoning one does not expect reduction; K is too low. And so it turns out. Yet H2 is
produced as usual and this H2 – in its “nascent” form – also cannot effect the reduction.
What is special about the “nascent’ form produced by Zn, compared with that produced by Fe, if it is in fact effecting the reduction?}
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A thought: could it be that zinc hydride is involved as a catalyst? Zinc hydride is a reasonably stable compound, though sources disagree upon its
stability in aqueous environments. If ZnH2 is indeed involved in the reaction, magnesium or scandium might work too.
Another thought: has it been tested whether zinc in solution works when bubbling H2 through solution? Here again, maybe zinc might catalyse the
reduction of Cr3+ to Cr2+ by (diatomic) hydrogen.
Edit:
| Quote: | | I don’t believe the hydrogen plays ANY role, nascent or otherwise, although I used to believe that it did. |
We all were taught to believe that.
Anyway, if the absence of Zn (which reacts with H3O+) cause the reaction not to run, then what is the mechanism?
[Edited on 21-12-2011 by Bezaleel]
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blogfast25
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| Quote: Originally posted by Bezaleel |
A thought: could it be that zinc hydride is involved as a catalyst? Zinc hydride is a reasonably stable compound, though sources disagree upon its
stability in aqueous environments. If ZnH2 is indeed involved in the reaction, magnesium or scandium might work too.
Another thought: has it been tested whether zinc in solution works when bubbling H2 through solution? Here again, maybe zinc might catalyse the
reduction of Cr3+ to Cr2+ by (diatomic) hydrogen.
Edit:
| Quote: | | I don’t believe the hydrogen plays ANY role, nascent or otherwise, although I used to believe that it did. |
We all were taught to believe that.
Anyway, if the absence of Zn (which reacts with H3O+) cause the reaction not to run, then what is the mechanism?
[Edited on 21-12-2011 by Bezaleel] |
Catalysts do NOT affect the ΔG or ΔE of reactions, both of which are state functions. They only affect reaction speed.
I don’t understand your last point: Zn reduces Cr (III) to Cr (II), obviously in the absence of Zn the reaction cannot proceed.
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ScienceSquirrel
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I made Chromium II acetate using this procedure, it is reasonably easy.
http://alpha.chem.umb.edu/chemistry/ch371/documents/Synthesi...
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blogfast25
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So Cr (II) acetate is poorly soluble in cold water, dimeric and deep red in colour? Interesting...
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ScienceSquirrel
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It is pretty good.
The solution starts off yellow, goes green then blue and then a red precipitate forms when you add the sodium acetate.
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Bezaleel
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| Quote: Originally posted by blogfast25 |
Catalysts do NOT affect the ΔG or ΔE of reactions, both of which are state functions. They only affect reaction speed.
I don’t understand your last point: Zn reduces Cr (III) to Cr (II), obviously in the absence of Zn the reaction cannot proceed.
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I assume both of us have ruled out that hydrogen radicals or hydrogen in any other form reduces Cr(III) to Cr(II). Then, if Zn alone were responsible
for the reduction, why does the solution need to be acidic?
Der Alte said: | Quote: | However, if we look at the overall reaction as (sulphate case with H2SO4, Zn)
Cr2(SO4)3 +Zn -- > 2CrSO4 + ZnSO4, it looks as if adding acid is not necessary. But it is; |
[Edited on 21-12-2011 by Bezaleel]
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blogfast25
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| Quote: | | Quote: Originally posted by Bezaleel | I assume both of us have ruled out that hydrogen radicals or hydrogen in any other form reduces Cr(III) to Cr(II). Then, if Zn alone were responsible
for the reduction, why does the solution need to be acidic?
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[Edited on 21-12-2011 by Bezaleel] |
Who says that's correct? Cr<sup>3+</sup> solutions are always acidic anyway, as he correctly points out higher up: hydrolysis of the
hexa-aqua trivalent chromium cation means that there's always quite a bit of H3O+ around in such solutions. So testing whether the reduction would
take place in neutral conditions isn't possible... I'd be surprised if the reduction of, say Cr2(SO4)3, with fine zinc powder would require acid...
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DerAlte
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I knew if I mentioned the word ‘Nascent’ all hell would break loose..
Blogfast wrote:
| Quote: | | Who says that's correct? Cr3+ solutions are always acidic anyway, as he correctly points out higher up: hydrolysis of the hexa-aqua trivalent chromium
cation means that there's always quite a bit of H3O+ around in such solutions. So testing whether the reduction would take place in neutral conditions
isn't possible... I'd be surprised if the reduction of, say Cr2(SO4)3, with fine zinc powder would require acid... |
The acidity of hexaaqua-Cr[III] is around 4-5, I believe; ie about the level of 10^-5 N acid. Take your 10N plus HCl and dilute a million times and
pour over zinc (dust if you have it). Watch for bubbles of H2; but I don’t think we can wait that long. Or if you want to neutralize, add a drip of
NaOH severely diluted. That’s the order of acidity you are talking about. I am sure a minute amount of Cr+++ will go to Cr++ but I defy anyone to
detect it! Also Cr[II] is scarcely acidic, but I’m not sure where I read that. Goggle it.
If the reaction occurs without acid on the zinc it’s very slow, in spite of the dEo or dGo prediction. And therein lies the rub…
@ScienceSquirrel
Stole my thunder re the acetate! See how I did it in the next part.
@Bezaleel | Quote: | | A thought: could it be that zinc hydride is involved as a catalyst? Zinc hydride is a reasonably stable compound, though sources disagree upon its
stability in aqueous environments. If ZnH2 is indeed involved in the reaction, magnesium or scandium might work too.Another thought: has it been
tested whether zinc in solution works when bubbling H2 through solution? Here again, maybe zinc might catalyse the reduction of Cr3+ to Cr2+ by
(diatomic) hydrogen…. |
I believe the hydride theory has been put forth in the literature, among scads of others.
As for diatomic hydrogen, if it worked, why would all those organic reductions not use it? – you could avoid all that zinc and acid contamination.
No, it’s always zinc/ acid – or the hydrides.
Good reducing agents (High dEo or dGo) are hard to find while oxidants are 10 cents a dozen.
| Quote: | | Anyway, if the absence of Zn (which reacts with H3O+) cause the reaction not to run, then what is the mechanism? |
That’s the $64,000 question…
Regards,
Der Alte
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blogfast25
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Der Alte:
2Cr+++(aq) + Zn(s) < --> 2Cr++(aq) + Zn++(aq) ΔE 0.386 V …overall reaction
With Nernst that gives an equilibrium constant K (@ 298 K) of about 30, so not much, but I’d still like to see how a ‘neutral’ Cr (II) solution
reacts with fine zinc powder: the hugely increased surface area must have a positive effect on reaction rate. Unfortunately I’ve sold my last 100 g
of it…
My point about Cr (III) salts being acidic was that since as on paper the reduction should proceed, maybe the presence of H3O+just catalyses it? In
which case even low concentrations would have some effect.
I’m still not convinced the hydrogen plays a part but it’s impossible to separate the variables here: no increased acidity w/o hydrogen
(‘nascent’ or diatomic, LOL).
Conundrum 101.
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