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blogfast25
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Oxidation of SnCl2 to SnO2
I had quite a large batch of ‘high grade’ SnCl2.2H2O which turned out to be largely insoluble, presumably due to hydrolysis, perhaps also part
oxidation to Sn(IV). In order not to waste the product completely I decided to convert it to SnO2.
The batch was mixed with about the same weight of water, stirred so that some of it dissolved and then strong ammonia was added to about neutral pH.
The resulting slurry was filtered (it filtered easily) and washed with hot tap water, then hot DIW. (the filtrate still smelled faintly of NH3).
The filter cake was a light yellow sandy substance, presumably mostly Sn(OH)2.nH2O. It was dripped ‘dry’, then put in a defunct but clean, shallow
aluminium pan and heated on an electrical plate. Right away the product started discolouring to grey. I thought this might be Al pick up and
transferred the material to a Pyrex oven dish but on heating the discolouration continued: by that time the stuff was almost dark grey!
But low and behold, on further heating the colour reverted back to a much lighter colour, with patches of light yellow here and there. After a couple
of hours, occasionally stirring and turning, the now dry product looked like the synthetic Cassiterite I bought from eBay some years back: off white
to light grey.
I presume that during air oxidation of SnO an intermediate oxide of dark (black?) colour is formed, something like:
SnO + ½ O2 → SnO2
SnO + SnO2 → Sn2O3 (SnO.SnO2)
Sn2O3 + ½ O2 → 2 SnO2
Wiki mentions the formation of something like that on heating of SnO in inert atmosphere through disproportionation to Sn and SnO2 but here we are in
oxidising conditions.
[Edited on 23-8-2012 by blogfast25]
[Edited on 23-8-2012 by blogfast25]
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AJKOER
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Interestingly, I was looking at an old and more recent sources and the reaction of SnCl2 with aqueous NH3 may result in compound salts, (NH4)2SnCl4 or
even (NH4)2SnCl6, mixed in with the Sn(OH)2.
One possible undesirable reaction sequence:
SnCl2 + 2 NH3 + 2 H2O --> Sn(OH)2 + 2 NH4Cl
SnCl2 + 2 NH4Cl --> (NH4)2SnCl4
Or on net:
2 SnCl2 + 2 NH3 + 2 H2O --> Sn(OH)2 + (NH4)2SnCl4
Not knowing to what extent this or other reaction sequences could take place, I would not be surprised if the yield of SnO2 was less than expected.
Recent reference: "Competition Science Vision", Aug 2000, page 789.
Link: http://books.google.com/books?id=jegDAAAAMBAJ&pg=PA789&a...
Old Source: "Hand book of chemistry", Volume 5, by Leopold Gmelin, page 94.
Link: http://books.google.com/books?id=OgI5AAAAMAAJ&pg=PA94&am...
[Edited on 23-8-2012 by AJKOER]
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unionised
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I think it's disproportionation to Sn and SnO2.
At the surface the conditions are oxidising, but throughout the bulk material they are reducing.
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blogfast25
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AJ:
I doubt if the NH3 played up. After washing only a trace remained. If there is such a complex (SnCl2.NH3) it could have slowed things down a bit, but
not much more. On adding the strong NH3 to the slurry, no colour change was observed at all. That only happened on heating in air.
Quote: Originally posted by unionised  | I think it's disproportionation to Sn and SnO2.
At the surface the conditions are oxidising, but throughout the bulk material they are reducing. |
Funnily enough, when it reverted back to 'normal' colour this happened FIRST at the bottom of the material. Of course that was also the hottest layer.
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AJKOER
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Blogfast25:
Upon further research, before seeing your comment, I removed the possible formations of the SnCl2.NH3 and SnCl2.2NH3 abducts as these may be created
only by the anhydrous SnCl2 reacting with ammonia gas. Although I would not be surprised if these abducts, upon being created, were capable of
forming hydrates (in a similar fashion to CaCl2).
However, per the reactions, for example:
6 SnCl2 + 2 H2O + O2 --> 2 SnCl4 + 4 Sn(OH)Cl (s)
SnCl2 + 2 NH3 + 2 H2O --> Sn(OH)2 + 2 NH4Cl
SnCl4 + 2 NH4Cl --> (NH4)2SnCl6 (Ammonium chlorostannate or pink salt)
any resulting formation of (NH4)2SnCl4 or even (NH4)2SnCl6, could reduce your yield of SnO2.
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woelen
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Isn't the dark color simply due to formation of SnO from Sn(OH)2? Subsequent change of color to white again is due to oxidation of SnO to SnO2.
SnO is a dark substance, nearly black. SnO2 is white. Sn(OH)2 also is white. So, these changes of color you observed are totally understandable.
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blogfast25
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| Quote: Originally posted by AJKOER | Blogfast25:
Upon further research, before seeing your comment, I removed the possible formations of the SnCl2.NH3 and SnCl2.2NH3 abducts as these may be created
only by the anhydrous SnCl2 reacting with ammonia gas. Although I would not be surprised if these abducts, upon being created, were capable of
forming hydrates (in a similar fashion to CaCl2).
However, per the reactions, for example:
6 SnCl2 + 2 H2O + O2 --> 2 SnCl4 + 4 Sn(OH)Cl (s)
SnCl2 + 2 NH3 + 2 H2O --> Sn(OH)2 + 2 NH4Cl
SnCl4 + 2 NH4Cl --> (NH4)2SnCl6 (Ammonium chlorostannate or pink salt)
any resulting formation of (NH4)2SnCl4 or even (NH4)2SnCl6, could reduce your yield of SnO2.
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Yep, I did think SnCl2 ammoniacates would only be formed in anhydrous conditions.
As regards pink salt, I made significant quantities of it a while back (it’s reported on this forum somewhere). It’s a white, crystalline water
soluble compound that on strong heating completely sublimes (presumably to NH4Cl and SnCl4) leaving no trace. The trouble with your hypothesis though
is that there’s precious little NH3 to begin with. The oxidation of Sn (II) to Sn (IV) in wet conditions in air is clearly slow. I didn’t measure
yield quantitatively (impossible to do, as I didn’t really know what I started from) but the amount I obtained seemed reasonable, allowing also for
the usual synthesis losses.
| Quote: Originally posted by woelen | Isn't the dark color simply due to formation of SnO from Sn(OH)2? Subsequent change of color to white again is due to oxidation of SnO to SnO2.
SnO is a dark substance, nearly black. SnO2 is white. Sn(OH)2 also is white. So, these changes of color you observed are totally understandable.
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Woelen, I didn’t know SnO was black: the hydrated product is a light yellow, so I assumed SnO would be light coloured like SnO2. The discolouration
started when the slurry was really quite wet still (it kind of ‘bumped’ badly at first). But it’s possible. Cu(OH)2 is blue and dehydrates to
black CuO even when wet.
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AJKOER
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| Quote: Originally posted by blogfast25 | Woelen, I didn’t know SnO was black: the hydrated product is a light yellow, so I assumed SnO would be light coloured like SnO2.
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OK, per the source "A Text Book of Inorganic Chemistry", by Anil Kumar De, page 376, to quote:
"When freshly precipitated, the oxide has the composition 2SnO.2H2O. It is usually an olive-green powder which gives greyish crystals in contact with
water."
Link:
http://books.google.com/books?id=PpTi_JAx7PgC&pg=PA379&lpg=PA379&dq=(NH4)2SnCl4+preparation&source=bl&ots=mp0VF3Nxrl&sig=pp8dW
ha5AkWpxijaJDbt2X9qldM&hl=en#v=onepage&q&f=false
______________________________________
The SnO disproportionation alluded to previously is noted to occur in the solid state only. Per a Russian source, to quote (translation):
"Monoxide is produced by heating a solution of tin SnCl2 and alkali. In the solid state, it has a tendency to dismutation by the scheme:
2 SnO = SnO2 + Sn
but in the liquid (mp. 1040 ° C) and gas (bp. 1425 ° C) is stable. "
Link:
http://translate.google.com/translate?hl=en&sl=ru&tl...
I found another source ("Concise Encyclopedia Chemistry" by Walter de Gruyter) which notes a partial disproportionation to Sn and SnO2 upon melting at
1,080 C.
[Edited on 25-8-2012 by AJKOER]
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blogfast25
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| Quote: Originally posted by AJKOER | | Quote: Originally posted by blogfast25 | Woelen, I didn’t know SnO was black: the hydrated product is a light yellow, so I assumed SnO would be light coloured like SnO2.
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OK, per the source "A Text Book of Inorganic Chemistry", by Anil Kumar De, page 376, to quote:
"When freshly precipitated, the oxide has the composition 2SnO.2H2O. It is usually an olive-green powder which gives greyish crystals in contact with
water."
Link:
http://books.google.com/books?id=PpTi_JAx7PgC&pg=PA379&lpg=PA379&dq=(NH4)2SnCl4+preparation&source=bl&ots=mp0VF3Nxrl&sig=pp8dW
ha5AkWpxijaJDbt2X9qldM&hl=en#v=onepage&q&f=false
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Hmmm... I've precipitated "Sn(OH)2.nH2O" many times, also from really pure SnCl2 solutions, and have always obtained something off-white to light
yellow. I might confirm this again with pix soon. Maybe even tonight.
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blogfast25
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Below: a part ingot of 99.9 % Sn. Nice and shiny. Yum!
A bit of hack saw fun and 50 ml of 30 % iron free HCl later (the chunk of Sn reacting):
It took about 2 h to dissolve. The solution of SnCl2 was crystal clear and colourless.
Left tube: after adding strong NH3 to a few ml of the SnCl2 solution, Right tube, same treatment but with added 5 M NaOH (demonstrating amphoteric
behaviour of Sn (II) - sodium stannite formed):
Now it gets really interesting. First up: the left hand tube emptied in a shallow pyrex dish (kitchen towel background): note how white the
precipitate of ‘tin (II) hydroxide’ really is:
Then a time series, taken 2 – 3 mins apart with that dish on high heat on the hot plate:
Clearly the Sn (II) oxide hydrate dehydrates from white to brown/black before oxidation reverses the colour back to whitish tin (IV) oxide. During the
run I could see crystalline structure being formed and the end-product shows crystalline areas too.
[Edited on 24-8-2012 by blogfast25]
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AJKOER
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Blogfast25:
Nice pictures!
Does the last picture show a touch of green and grey in places (hard to say)?
The next time you do this, I think it would be interesting to react SnCl2 with aqueous Na2CO3:
SnCl2 + Na2CO3 --> SnO (s) + 2 NaCl + CO2 (g)
and note whether the SnO looks the same. If it does not, I would not be too surprised as I would not normally describe a salt, as did my source (which
I also saw described precisely the same way elsewhere) by its 'usual' color. Note, in the case of Al(OH)3 created via ammonia (actually this
modification is called 'Aluminum oxide hydrate') bears no resemblance to Aluminum hydroxide via H2O2.
[Edited on 25-8-2012 by AJKOER]
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blogfast25
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| Quote: Originally posted by AJKOER | Blogfast25:
Nice pictures!
Does the last picture show a touch of green and grey in places (hard to say)?
The next time you do this, I think it would be interesting to react SnCl2 with aqueous Na2CO3:
SnCl2 + Na2CO3 --> SnO (s) + 2 NaCl + CO2 (g)
and note whether the SnO looks the same. If it does not, I would not be too surprised as I would not normally describe a salt, as did my source (which
I also saw described precisely the same way elsewhere) by its 'usual' color. Note, in the case of Al(OH)3 created via ammonia (actually this
modification is called 'Aluminum oxide hydrate') bears no resemblance to Aluminum hydroxide via H2O2.
[Edited on 25-8-2012 by AJKOER] |
No, that is a camera effect: the real thing is remarkably white, with the odd darker patch where oxidation hadn't completed yet. Lighting in my lab is
far from optimal: 2 TL strip lights only leads to poor white balance.
I'll try the Na2CO3 route later on.
Not sure what you mean by 'Aluminum hydroxide via H2O2'...
[Edited on 25-8-2012 by blogfast25]
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AJKOER
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OK, I believe I did this experiment once. Slowly dissolve Al in Na2CO3.H2O2 (I used a generic Oxyclean brand). This resulted in a white amorphous and
insoluble form of Al(OH)3.
If one tries this experiment with H2O2 and a strong base, the Al(OH)3 dissolves, but weak and the Al(OH)3 precipitates.
I will retry to verify, but unfortunately, this is not a quick reaction.
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blogfast25
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This is the Sn (II) oxide hydrate from adding Na2CO3 saturated solution to yesterday’s SnCl2 solution:
Indistinguishable from the one made with NaOH with the naked eye. And white…Almost snow white…
[Edited on 25-8-2012 by blogfast25]
[Edited on 25-8-2012 by blogfast25]
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Heuteufel
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I once made SnO according to the procedure described in Brauer (precipitating hydrated tin(II)-oxide with Na2CO3 from
SnCl2 and heating the hydrate on a NaCl-Water bath (110 °C). If the experiment is done right, beautiful blue-black cristals with metallic
sheen are formed!
According to Brauer in air, SnO oxidized to SnO2 above 220 °C, a temperature easily achieved by a hotplate.
If you want to make SnO2 it is simplest to react Sn with nitric acid (attention:formation of copious amounts of NO2!).
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blogfast25
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| Quote: Originally posted by Heuteufel | I once made SnO according to the procedure described in Brauer (precipitating hydrated tin(II)-oxide with Na2CO3 from
SnCl2 and heating the hydrate on a NaCl-Water bath (110 °C). If the experiment is done right, beautiful blue-black cristals with metallic
sheen are formed!
According to Brauer in air, SnO oxidized to SnO2 above 220 °C, a temperature easily achieved by a hotplate.
If you want to make SnO2 it is simplest to react Sn with nitric acid (attention:formation of copious amounts of NO2!).
|
Very interesting, Heuteufel. I will probably try that with the remaining SnCl2 solution.
Here the purpose was simply to salvage some SnCl2.2H2O 'gone bad'...
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blogfast25
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Well, I tried to make anhydrous SnO but didn’t succeed, following Brauer’s method (drying hydrated SnO at 110 C, essentially).
I’d run out of Na2CO3, so for the first run I used NaHCO3 to neutralise the SnCl2 solution. Darkening started immediately but stopped at about olive
green, after about 1 h on the sat. brine bath. 1 h later and no further darkening had occurred.
Then a test using strong NH3 but that SnO hydrate precipitate hardly darkened at all.
Perhaps it really is dependent on sodium carbonate as a neutralising agent to obtain a form of SnO hydrate that dehydrates slowly to anhydrous SnO at
110 C?
[Edited on 27-8-2012 by blogfast25]
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S.C. Wack
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Maybe it would be good to read what everyone was citing.
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AJKOER
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| Quote: Originally posted by AJKOER | OK, I believe I did this experiment once. Slowly dissolve Al in Na2CO3.H2O2 (I used a generic Oxyclean brand). This resulted in a white amorphous and
insoluble form of Al(OH)3.
If one tries this experiment with H2O2 and a strong base, the Al(OH)3 dissolves, but weak and the Al(OH)3 precipitates.
I will retry to verify, but unfortunately, this is not a quick reaction. |
Yes, in a few hours one has snow flakes in water ( Al(OH)3 ). A tip would be to quickly manually separate the snow flakes from the still visible
strips of Al foil. In 2 days, the Aluminum strips have nearly completely broken down into black silicon specks and cannot be easily separated from the
Al(OH)3.
For those interested in the chemistry, the Na2CO3 allows the penetration of the Al2O3 passive layer permitting a hydrolysis reaction:
2 Al + 6 H2O --> 2 Al(OH)3 (s) + 3 H2 (g)
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blogfast25
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Huh? 'Splain, please SC?
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S.C. Wack
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Sometimes, even articles in German from 1922 have a word or three, somewhere, on why certain guidelines have to be followed?
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blogfast25
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| Quote: Originally posted by S.C. Wack |
Sometimes, even articles in German from 1922 have a word or three, somewhere, on why certain guidelines have to be followed? |
Ok SC, you can quit speaking in tongues now.
Here’s the German abstract in your link:
”1. Es wurde das Gleichgewicht der Reaktion SnO2 + 2CO ⇋ Sn + 2CO3 [sic] studiert und gefunden, daß dieses Gleichgewicht von beiden
Seiten her erreichbar und von Menge und Zusammensetzung des Bondenkörpers unabhängig ist.”
Says nothing relevant to synth of SnO. I’ll be trying that again this WE, this time using Na2CO3 as a base.
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AJKOER
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Here is some interesting material per Wiki:
"Blue-black SnO can be prepared by heating the tin(II) oxide hydrate, SnO.xH2O (x<1) precipitated when a tin(II) salt is reacted with an alkali
hydroxide such as NaOH.[3] Metastable, red SnO can be prepared by gentle heating of the precipitate produced by the action of aqueous ammonia on a
tin(II) salt.[3] SnO may be prepared as a pure substance in the laboratory, by controlled heating of tin(II) oxalate (stannous oxalate) in the absence
of air.[4]
SnC2O4 → SnO + CO2 + CO
Tin(II) oxide burns in air with a dim green flame to form SnO2.[3]
2 SnO + O2 → 2 SnO2
When heated in an inert atmosphere initially disproportionation occurs giving Sn metal and Sn3O4 which further reacts to give SnO2 and Sn metal.[3]
4SnO → Sn3O4 + Sn
Sn3O4 → 2SnO2 + Sn
SnO is amphoteric, dissolving in strong acid to give tin(II) salts and in strong base to give stannites containing Sn(OH)3−.[3] It also
dissolves in strong acid solutions to give the ionic complexes Sn(OH2)32+ and Sn(OH)(OH2)2+, and in less acid solutions to give Sn3(OH)42+.[3]"
Link: http://en.wikipedia.org/wiki/Tin(II)_oxide
Interestingly, a key phrase above on the preparation of red SnO via ammonia may be 'gentle heating'. Also, the statement that SnO is amphoteric may be
significant.
____________________________________________
On SnCl2, again per Wiki:
"Tin(II) chloride can dissolve in less than its own mass of water without apparent decomposition, but as the solution is diluted hydrolysis occurs to
form an insoluble basic salt:
SnCl2 (aq) + H2O (l) <==> Sn(OH)Cl (s) + HCl (aq)
Therefore if clear solutions of tin(II) chloride are to be used, it must be dissolved in hydrochloric acid (typically of the same or greater molarity
as the stannous chloride) to maintain the equilibrium towards the left-hand side (using Le Chatelier's principle). Solutions of SnCl2 are also
unstable towards oxidation by the air:
6 SnCl2 (aq) + O2 (g) + 2 H2O (l) → 2 SnCl4 (aq) + 4 Sn(OH)Cl (s)
This can be prevented by storing the solution over lumps of tin metal.[2]"
Also, the following comment "Tin(II) chloride also behaves as a Lewis acid, forming complexes with ligands such as chloride ion, for example:
SnCl2 (aq) + CsCl (aq) → CsSnCl3 (aq)
Most of these complexes are pyramidal, and since complexes such as SnCl3 have a full octet, there is little tendency to add more than one ligand. The
lone pair of electrons in such complexes is available for bonding, however, and therefore the complex itself can act as a Lewis base or ligand."
Link: http://en.wikipedia.org/wiki/SnCl2
My take from the above is that SnCl2 with a dilute ammonia solution could form Sn(OH)Cl and further complex.
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blogfast25
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| Quote: Originally posted by AJKOER |
My take from the above is that SnCl2 with a dilute ammonia solution could form Sn(OH)Cl and further complex.
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It's possible but in this run the ammonia was actually 25 - 30 % and an excess of it was used. Alkalinoty should really be high enough to hydrolyse
all the way down to 'Sn(OH)2.nH2O', IMHO.
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S.C. Wack
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I don't care what the abstract says? I'm obviously talking about the article itself cited by your reference, and also mine (which also mentions Ann.
chim. phys.[5] 27, 145 (1882), attached hot and unread from gallica's press) which I don't have, and linked to? I'll download the large narod archive
that it's in (at some point) but there is other downloading on my near-Luddite internet.
Attachment: N0034867_PDF_143_180DM.pdf (779kB)
This file has been downloaded 715 times
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