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dann2
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Stannous Chloride
Hello,
I'm going around in circles.
Will Tin Chloride (Stannous Chloride, SnCl2) decompose with heat, in the presence of O, to Tin Oxide. I know Tin Tetrachloride (Stannic Chloride)
will. Can Stannous Chloride be easily made. I can obtain Tin Oxide, Tin metal and HCl easily.
Also looking for Antimony Trichloride. I believe this is easy to make with Antimony Trioxide and HCl? Is it possible to start with Antimony Metal.
TIA,
DANN2
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gsd
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It is very easy to make SnO.
Gently boil tin chips in HCl till it dissolves completely. Filter the solution of suspended impurities. What you get is solution of SnCl2 in dilute
HCl.
Now precipitate SnO by adding stoiciometric quantity of NaOH dissolved in water. In some patent literature it is mentioned that adding abut 5-10 %
Na2CO3 to caustic lye, and carring the precipitation reaction at about 80 Deg C, gives good yields.
wash the precipitate 3/4 times with water to get black / grey SnO.
Please note dry SnO is pyrophoric!
gsd
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Eclectic
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Be aware that SbH3 is highly toxic...
Sb is supposed to dissolve in concentrated hydrochloric acid to produce SbCl3 solution, recoverable as anhydrous SbCl3 by distillation.
I wouldn't try this without using very good ventilation...
Don't use alloys as your Sb source, especially Sb/Zn 
[Edited on 4-23-2007 by Eclectic]
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dann2
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| Quote: | Originally posted by gsd
It is very easy to make SnO.
Gently boil tin chips in HCl till it dissolves completely. Filter the solution of suspended impurities. What you get is solution of SnCl2 in dilute
HCl.
Now precipitate SnO by adding stoiciometric quantity of NaOH dissolved in water. In some patent literature it is mentioned that adding abut 5-10 %
Na2CO3 to caustic lye, and carring the precipitation reaction at about 80 Deg C, gives good yields.
wash the precipitate 3/4 times with water to get black / grey SnO.
Please note dry SnO is pyrophoric!
gsd |
Hello,
Thanks for replys.
I need to make the Tin (IV) Oxide (SnO2) on Titanium Metal using a Tin compound. The Patent askes for Stannic Chloride to be decomposed with heat to
Tin oxide. I am wondering if Stannous Chloride will do instead of the Stannic Chloride as I can obtain Stannous Chloride easily.
I can purchase lots of Tin Oxide at the ceramics store but that is no good to me.
The Tin Oxide on the Titanium must be doped with Antimony Oxide so I need a compound of Antimony that will decompose using heat to Antimony Oxide.
Both the decomposable by heat Antimony and Stannic (or hopefully Stannous) Chloride are applied in a alcohol solution to the Titanium and the Ti
heated.
Quote from US Pat. 4,040,939:
A solution for the semi-conductive intermediate coating was prepared by mixing 30 milliliters of butyl alcohol, 5 milliliters of hydrochloric acid
(HCl), 3.2 grams of antimony trichloride SbCl.sub.3), and 15.1 grams of stannic chloride pentahydrate (SnCl.sub.4.5H.sub.2 O).
A strip of clean titanium plate was immersed in hot HCl for 1/2 hours to etch the surface. It was then washed with water and dried. The titanium was
then coated twice by brushing with the solution described above.
The surface of the plate was dried for ten minutes in an oven at 140.degree. C. after applying each coating. The titanium was then baked at
500.degree. C. for 7 .+-. 1 minutes.
The theoretical composition of the semi-conductive intermediate coating thus prepared was 81.7 percent SnO.sub.2 and 18.3 percent antimony oxides
(calculated as Sb.sub.2 O.sub.3).
End quote:
Wondering if Stannous would do instead of the Stannic Chloride??
If yes, I guess I would need to use less of the Stannous to keep the molar ratio of Sn to Sb the same as the patent.
(excluding the water in the Stannic/Stannous)
As you explained above I can make SnCl2 easily enough but how do I extract the SnCl2 from the HCl?
I had not thought of looking up patents for to make chemicals! (embarrased face here)
Thank You,
Dann2
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not_important
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You need to keep tin chlorides in fairly acid solutions. The hydrates can be made by evaporating the solutions (NOT to dryness) until crystals form,
then cooling and collecting the tin salt on a filter. In the case of SnCl2 you will need to evaporate it almost to dryness, as it is very soluble in
water, although strong HCl may reduce the solubility. If you do take the solution all the way to dryness, especially without enough HCl around, you
likely will get some hydrolysis to basic chlorides.
SnCl2 is a fairly strong reducing agent. If you expose a solution of it to air, more quickly be bubbling filter air through the solution, the
solution will grow cloudy as hydrated oxide of Sn(IV) is formed. Adding concentrated HCl will dissolve that forming a solution of SnCl4 in HCl.
Evaporate solution to get crystals of SnCl4.5H2O
SnCl2 will go to SnO unless oxidised by air. As you are making a semiconducting film, not getting too much deviation from the actual stoichiometries
would seem to be a good idea. Try Sn(II), but if things don't work quite right then switch to Sn(IV).
There will be a lot of HCl coming off those solutions during evaporation, you'll want to trap it to keep all the metals around from corroding. Same
goes for SbCl3.
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Eclectic
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Thanks for that patent number. I'd been looking for it.
Some of the earlier Diamond Shamrock patents:
US3865703, US3917518
Maybe with some additional useful experimental results using Sn/Sb mixed metal oxides to depassivate titanium electrodes.
As long as the Sn:Sb ratio is near 90:10, it may not be that critical.
Dissolve your metals or oxides in concentrated HCL, mix, maybe add some HNO3 or H2O2 to bring up the oxidation state, evaporate to syrup, add alcohol,
paint, bake, and repeat.
(That's what I plan to try. Let me know if you get it working first.)
[Edited on 4-24-2007 by Eclectic]
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Rosco Bodine
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You probably shouldn't try to directly isolate these chlorides in pure form because they require the presence of some excess HCl to prevent hydrolysis
. You don't actually need to isolate the chloride salts you have made .
For convenience , the equivalent mixture of the patent
would be made simply by working with strong solutions of the metal chlorides still residing in the solutions in which they were made , and having HCl
in excess for their stability . You know how much metal or metal oxide was dissolved to make the chloride , so you can calculate the amount of the
chloride contained per unit volume of the solution , and just measure out the volume of liquid which contains your required quantity of chloride .
Do that measurement for both the antimony and the tin
chloride strong solutions , and just add those quantities
to your butanol , letting the HCl component simply be that
HCl which is in excess as a stabilizer for the chloride solutions .
I didn't see any specific data for the lifetime of this electrode on a perchlorate cell , which would seem to be the real " acid test " for the
performance , so I remain a bit dubious about the value this particular patents method . It would seem that if their testing did reveal
some suitability for use in a perchlorate cell , then that
data would be presented . It's not likely they forgot to test that capability or that it performed superbly in
perchlorate production and they neglected to disclose
that value .
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Eclectic
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One of the older patents uses 90:10:6 Sn:Sb:Ru for a hypochlorite electrode with a lifetime before passivation of 600+ hours at 1A/in^2 if I read it
correctly. That's without PbO2 overplate. The Sn:Sb oxide seems to do a good job of preventing the formation of insulating TiO2 under strongly
oxidizing conditions.
I suspect the Diamond Shamrock folks are more interested in hypochlorite and chlorate bleaching agents, and chlor/alkali process electrodes than
perchlorate cells.
[Edited on 4-24-2007 by Eclectic]
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Twospoons
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You might also try doping the SnO2 with Indium, as in ITO glass. Source: indium/tin lead-free solder. Indium content is usually about 2%.
Helicopter: "helico" -> spiral, "pter" -> with wings
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dann2
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Hello,
Thanks for replys. US Pat. No3940323 uses Molybdenium as the dopant (instead of Sb, Indium etc).
It also says that there are lots of salts that are ok for thermal decompositon to SbO2 + Dopant (Mo).
_______________________________
Stannic and Stannous Chlorides, Oxychlorides, Alkoxides, alkoxy halides,
resinates, amines and the like (to quote exactly)
_______________________________
Are any of these's compounds used anywhere that access to them is straightforward for the 'non chemist'.
Cheers,
Dann2
Edit = wrong Pat. No.
[Edited on 25-4-2007 by dann2]
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Eclectic
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There is a lead free solder 95/5 that is 5% antimony, balance tin.
Dissolve it in concentrated HCl (outside, good ventilation) and you are half way there.
Fluorine doped tin oxide is also conductive. Here is an abstract that seems to indicate that tin oxide doped with both fluorine and antimony may be
more conductive than either alone:
Sb, F doped SnO2 (This paper indicates highest conductivity at ~5mol% Sb. F content is reduced by baking @500C on a glass substrate)
Fluorine concentations in F doped tin oxide films seem to run 3-4% and are obtained by adding a bit of NH5F2 (ammonium bifluoride) to the Sn chloride
solution before spraying it on the substrate and baking.
You may be able to get everything you need for this OTC via 95/5 solder, brick cleaner, and aluminum cleaner 
[Edited on 4-26-2007 by Eclectic]
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Eclectic
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(from "Industrial and Engineering Chemistry" Vol. 16, No. 4)
The following procedure is recommended for the treatment
of an alloy containing tin or antimony or both:
Treat 1 gram of the borings with 20 cc. of concentrated
nitric acid and 10 cc. of water. Evaporate to dryness over a
flame or a steam bath. Add 10 cc. of dilute nitric acid and
warm. Filter and wash with dilute nitric acid. The filtrate
will contain all the metals except the tin and antimony, which
are retained as oxides on the filter paper. Up to this point
this is a regular procedure for alloys. Wash the oxides into a
beaker with 50 cc. of a strong solution of sulfur dioxide and
digest from 3 to 5 minutes at 60" to 70" C. Heat to boiling,
add about 10 cc. of concentrated hydrochloric acid, and boil
until all sulfur dioxide is removed. If necessary to remove
traces of the oxides from the filter paper, this can be treated in
a similar manner. Tin and antimony will now be in solution
as chlorides of the higher valence in the presence of traces of
sulfate, which is a convenient form for further treatment.
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dedalus
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SnCl2 always makes a lot of white "metastannic acid" sludge even if you dissolve it in acid. Incredibly easy to oxidize.
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alancj
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Can stannous chloride (SnCl2) be made by disolving SnO2 in HCL acid? Or is the oxide inert to the acid?
Thanks
Alan
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12AX7
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SnO2 has Sn(4+), how exactly does it get to Sn(2+) by dissolving in HCl?
Tim
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alancj
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would I just get a solution of stannic chloride then?
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UnintentionalChaos
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Stannic Chloride reacts quite violently with water releasing clouds of HCl, though it does form hydrates. I imagine that dissolving SnO2 in HCl would
require a large excess of concentrated acid and would make nice plumes of Cl2 (analagous to MnO2 + HCl, which makes MnCl2 and used to be a means of
producing Cl2 before the chlor-alkali process). Don't gas yourself. If your oxide is calcined, it will be hellishly difficult to dissolve it in the
acid. I think the addition of small amounts of H2O2 helped the dissolution along in a discussion about MnO2 IIRC.
[Edited on 5-14-07 by UnintentionalChaos]
Department of Redundancy Department - Now with paperwork!
'In organic synthesis, we call decomposition products "crap", however this is not a IUPAC approved nomenclature.' -Nicodem
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alancj
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| Quote: | Originally posted by UnintentionalChaos
Stannic Chloride reacts quite violently with water releasing clouds of HCl, though it does form hydrates. I imagine that dissolving SnO2 in HCl would
require a large excess of concentrated acid and would make nice plumes of Cl2 (analagous to MnO2 + HCl, which makes MnCl2 and used to be a means of
producing Cl2 before the chlor-alkali process). Don't gas yourself. If your oxide is calcined, it will be hellishly difficult to dissolve it in the
acid. I think the addition of small amounts of H2O2 helped the dissolution along in a discussion about MnO2 IIRC.
[Edited on 5-14-07 by UnintentionalChaos] |
Well, I know that PbO2 will dissolve in hot HCL, I never noticed any large amount of Cl. but then again I did it in a test tube in a small amnt. I
thought SnO2 might do the same thing. I was wanting to make SnCl4 .5H2O. Following what not_important said:
| Quote: | SnCl2 is a fairly strong reducing agent. If you expose a solution of it to air, more quickly be bubbling filter air through the solution, the solution
will grow cloudy as hydrated oxide of Sn(IV) is formed. Adding concentrated HCl will dissolve that forming a solution of SnCl4 in HCl. Evaporate
solution to get crystals of SnCl4.5H2O
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Maybe hot/concentrated HCL in excess plus bubbling air into the solution would work to dissolve the SnO2. I haven’t bought any yet. Would I be
better off just buying metallic tin? What do ya'll think?
Btw, why would calcined Sn02 be any different than something that was (say) precipitated? I have had trouble with getting calcined litharge to
dissolve in acid (I ground it to a powder- with dust mask on) but precipitated litharge dissolves in nitric acid just fine.
Thanks for the help,
Alan
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not_important
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SnO2 isn't a strong enough oxidiser to form Cl2 from hydrochloric acid.
As SnO2 is already fully oxidised, there's nothing to be gained by bubbling air through SnO2 + HCl(aq), and you'll lose some HCl that way.
If you're after SnCl2, then buying tin or making it from SnO2, if you already have that, is the way to go. Reacting with hydrochloric acid and
evaporation with the exclusion of oxygen will crystallise out SnCl2.2H2O.
Calcining something tends to increase the crystal size, fuse very find particles into larger ones, cause solids in metastable forms to release energy
and change to the more stable form, and so on. All of these tend to reduce the reactivity of the solid, by reducing surface area and/or forming more
inert forms.
In the case of tin, precipitated SnO2 is a hydrated form of the oxide, which is somewhat more reactive. Usually the hydrated form is calcined to get a
product of fixed composition.
Heating SnO2 with strong alkali, or by fusing the oxide with an alkali, will given a stannate such as Na2SnO3 (+3H2O as a solid). Solutions of these
precipitate 'metastannic acid' on treatment with CO2 (or exposure to air), one of the hydrated oxides. This dissolves in acids fairly easily.
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alancj
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So SnO2 will form a solution of SnCl2 with HCL, then?
If that's true, and I wanted to make SnCl4.5H20 by the above quoted procedure, then I would first dissolve my Sn02 in an excess of conc. HCl, and once
dissolved, bubble air through the solution to oxidize it to SnCl4. Correct me if I'm wrong.
The metastannic acid info you mentioned is interesting, as well as that on calcined oxides. Thanks.
Alan
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alancj
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| Quote: | Originally posted by Eclectic
Fluorine doped tin oxide is also conductive. Here is an abstract that seems to indicate that tin oxide doped with both fluorine and antimony may be
more conductive than either alone:
Sb, F doped SnO2 (This paper indicates highest conductivity at ~5mol% Sb. F content is reduced by baking @500C on a glass substrate)
[Edited on 4-26-2007 by Eclectic] |
Anyone have access to this paper?
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woelen
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No, SnO2 does not make a solution of SnCl2. Oxidation state does not change from +4 to +2! I think not_important made a typo by telling that SnO2 is
needed for making SnCl2. What actually is needed is SnO, which also is available commercialy. Sometimes it is offered on eBay.
The commercial SnO2 does not dissolve at all in any acid, no matter what concentration and what temperature. It simply is inert. The only reasonable
solvent for SnO2 is molten NaOH or molten KOH, not something you want to play with if you don't have the experience and the right equipment (remember,
molten NaOH/KOH dissolves glass and also attacks many metals).
So, go for SnO instead of SnO2.
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not_important
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Ahh - meant make tin metal from SnO2, as opposed to buying the metal. Do not try to speak in one language while typing in another.
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alancj
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| Quote: | Originally posted by not_important
Ahh - meant make tin metal from SnO2, as opposed to buying the metal. Do not try to speak in one language while typing in another.
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I understand… I guess I misunderstood what the "it" was referring to. Would I just heat SnO2 with carbon in air to reduce it to metallic tin? Are
there different ways to do it?
| Quote: | Originally posted by woelen
No, SnO2 does not make a solution of SnCl2. Oxidation state does not change from +4 to +2! I think not_important made a typo by telling that SnO2 is
needed for making SnCl2. What actually is needed is SnO, which also is available commercialy. Sometimes it is offered on eBay.
The commercial SnO2 does not dissolve at all in any acid, no matter what concentration and what temperature. It simply is inert. The only reasonable
solvent for SnO2 is molten NaOH or molten KOH, not something you want to play with if you don't have the experience and the right equipment (remember,
molten NaOH/KOH dissolves glass and also attacks many metals).
So, go for SnO instead of SnO2. |
You say commercial SnO2, is that to say that a non-calcined form of SnO2 would dissolve in acid? Just out of curiosity…
From my reading of the sodium metal thread I was under the impression that NaOH doesn’t really attack steel, or stainless steel. Could a porcelain
dish or crucible be used as well? I have both porcelain and a very nice 18/8 stainless steel cup I found (it’s wider than it is tall and probably is
about 400ml). I would think it would work quite well. But what do I know, I’m just “harmless.” 
Thanks for the clarification guys,
Alan
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not_important
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porcelain gets chewed up by strong base just about as much as glass, it's the SiO2 content that does it.
Non-calcined SnO2 is generally one of the hydrated form, which dissolve in acids to a greater or lesser extent. That precipitated by bases from SnCl4
solutions, or by or acids or CO2 from stannate solutions, tends to me more soluble in acids than that made from the action of strong HNO3 on tin
metal.
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