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Author: Subject: Multilayer Metal Oxide / Titanium Anodes for Chlorate/Perchlorate
tentacles
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[*] posted on 21-3-2008 at 20:11


I have household ammonia, I think it weighs out to about 11%, my nitric is at least 97%..

I already have Co(NO3)2 and Mn(NO3)2 prepared, and dried. I have Bi2O3 (pyro grade) to make Bi(NO3)3 and I was planning on using the SnCl4 to make the alpha etc and neutralized with some diluted nitric acid.

SnCl4 7.03g - I will start with the appropriate amount, neutralize with NH4OH, wash and HNO3
Co(NO3)2 1.126g
Mn(NO3)2 0.59g
Bi2O3 .60g

I made a spreadsheet if you want a copy to check the math.. I'll just label some shit and post it up...

edit: I have some glycerin for my beer brewing (to preserve yeast samples)

[Edited on 21-3-2008 by tentacles]

Attachment: BMC-TO anode calcs.xls (15kB)
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[*] posted on 21-3-2008 at 21:31


dann, I made a writeup (word doc), a sort of how-to on making a Ti/Co/Mn chlorate anode. I'm not sure I want to spoon feed anyone, but if you want it for your site, you are welcome to it.

One of these days I will try to compile a list, or spreadsheet, of all the various anodes that have been reported on here. At least, the tested ones.
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[*] posted on 21-3-2008 at 21:37


I have the pdf Merck tables for densities / molarities on common acids and bases including NH4OH , I can post if needed . The tetrahydrate of Mn nitrate should be mole weight of 251.01 .

I think the easiest way of calculating the precursor solution
would be to use "precursor equivalents" which provide
the correct percentage ratios of oxides after pyrolysis ,
and then account for the dilution to whatever matches the
~8% SnO2 . I have a good idea how to dilute to the required
concentration , based on weight , using an adjusted reciprocal of 8% as a mulitplier to determine the end weight for the mixed composition . I'm working it out and I'll post the details along with the weights .

IIRC the 8% SnO2 loading was about maximum before the smooth SnO2 coatings transitioned to cracked mud sort of coatings on sintering . It's exactly like the scenario with coatings of lacquer , there's an optimum solvent to "resin"
ratio which forms the best continuous film residue as the solvent leaves .

This whole thing will need to be done on weights to get it
accurate , the first time , not knowing the solution densities
for such a mixture .

I think the easiest way to do this is simply to look at the
percentages as being grams in a total of 100 grams of
total oxides after pyrolysis . So 82% SnO2 becomes 82 grams of SnO2 , 8% MnO2 becomes 8 grams MnO2 , ect .
Then just calculate the molar equivalents of precursors required for the x number of grams needed for that oxide .

The dilution weight will then have to be derived from the outcome of those calculations .

From what I can see already .....
we do arithmetic differently :P
Here's a quick look based on the first calculations I have done on parts totaling 100 of the mixed oxides as related
to precursors for producing 82% SnO2 , 8% MnO2 , 5% Co3O4 , 5% Bi2O3 .......precursor quantities as follows
should produce 100 grams total of the above mixed oxides
represented in their designated percentages .

82 g SnO2 from 190.77 g SnCl4 - 5 H2O (nitrate precursor)

190.77 g SnCl4 - 5 H2O ------> 199.6 g Sn(NO3)4

8 g MnO2 from 23.1 g Mn(NO3)2 - 4 H2O

5 g Co3O4 from 18.13 g Co(NO3)2 - 6 H2O

5 g Bi2O3 from 5 g Bi2O3 (nitrate precursor)

with 4.06 g HNO3 -----> 10.41 g of Bi(NO3)3 - 5 H2O

A solution weighing 1025 grams containing the above
precursors would have an 8% SnO2 loading .
That dilution involves ~784 ml H2O , less whatever
small additional weight of acid , glycerin , ect. is used .

A reasonable sized experimental batch would probably
be one tenth to two tenths the above quantities .

My inclination would be to not dilute it that much and
try it at the syrupy stage first to see how it goes .
It may be that the "glassiness" expected of this mixture
is not as susceptible to the cracked mud kind of problem
as appears at 8% loading on ordinary SnO2 sols and
it may stand a much higher loading which would be good
because it will build thickness faster if it does work out
that way .

It might also be worthwhile to use a mixture of chlorides
and nitrates to possibly utilize the oxygen surplus of the nitrates in aiding the conversion of the chlorides to the oxides , as something towards an "oxygen balanced"
mixture . I would still keep it on the oxygen rich side
however by perhaps 150% of theory for O2 balance if that modification is tried , and still the higher temps are going to be needed to assure completion . This is another totally unreferenced idea , so I can't point to any justification .
It could aid the stability of the precursor solutions , and spread the pyrolysis and diffusion process across a range of temperature , improving the coating by gradualizing the sintering . Or it may not help at all .


[Edited on 22-3-2008 by Rosco Bodine]
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[*] posted on 22-3-2008 at 20:21
9.75% Bi2O3 doped SnO2 optical quality film US6777477


The question was asked earlier about possibly using the alpha stannic oxyhydroxide directly as a coating precursor .
It can't be used directly , but after peptization which is very easy :D , it definitely can be used . This is the exception I should have mentioned before .

The patent US6777477 has been brought up several times before as being of possible ( probable) interest , and this seems like a good point to bring it up once again . This patent was mentioned most recently on page 7 of this same thread
http://www.sciencemadness.org/talk/viewthread.php?goto=lastp...
where I was observing that all of these coatings schemes
involve a sol transition , and indeed there are a few schemes
where the peptized , relatively pure hydrosol gotten from
the alpha stannic oxydroxide and its dopants can be used
as a baked coating precursor . This is the exception for
the matter where the directly precipitated alpha oxyhydroxide cannot be used as a coating , as its particle size is too large and it does not sinter to an adherent film
but dusts instead . Ammonia or organic amines change the
surface property and reduce the size of the colloidal particles , acting something like a detergent or emulsifier ,
so that the precipitated larger particles resuspend , and
"redissolve" , something like a controlled reversal of the
neutralization which precipitated the oxyhydroxide , to
form a hydrosol , which is something like an induced supersaturated solution , a colloidal dispersion of the hydrated oxides . Upon evaporation of the water , along with
loss of the volatile amine which is acting as the dispersant , this system will dehydrate and sinter to an adherent film .

See Example #2 for the 9.75% Bi2O3 doped SnO2
http://www.sciencemadness.org/talk/viewthread.php?action=att...
Attachment: US6777477 Sb2O3 doped SnO2 via ammonia soluble derivative.pdf (67.31 KiB)

This patent method avoids the nitrates of tin and avoids alcoholates , yet reports a way of making use of chloride precursors to produce optical quality films via dip coating , which are highly doped with either antimony or bismuth ....
a result which fits our purpose nicely .

When the oxidative soak deposition proved unworkable
using SnCl2 because of it reducing the cobalt spinel interface
coating , this patent method was suggested as an alternative for applying a "sealing layer" over the spinel ,
and it still seems like this should work fine .

This 9.75% Bi2O3 doped SnO2 could serve as a working anode coating , it may even work as an interface coating
but my guess is it would do best to use the Co3O4 spinel
as the interface and then seal it and overlayer it with this
patent composition . The Bi doping is in the range of those
Bi doped SnO2 described in US4272354 as useful , at about
the lowest Bi2O3 in the 9:1 to 4:1 SnO2 to Bi2O3 preferred ratios . The Bi doping percentage could likely be increased
above the 9.75% which provides the optical quality film of
the US677477 patent , to achieve the higher percentage doping described in the US4272354 perchlorate anode patent .

Earlier on page 6 of this thread Xenoid pointed to this
US4272354 patent as a principal interest , and indeed there is something of a nexus between the US4272354 patent
and the US677477 patent Example #2 , for Bi2O3 doping of SnO2 . Sorry for the aggravation if this seems circular ,
once again , but these two patents are very significant
if indeed the Bi2O3 doping should prove out as valuable
for the working coating dopant which indeed is the required
"special ingredient" catalytic for perchlorate .

Simply combining the cobalt spinel interface method which
is already proven , with this Bi2O3 doped SnO2 sealing and working coating , may be the simplest baked coating scheme
which is workable for a perchlorate anode .
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[*] posted on 22-3-2008 at 22:15


Rosco: the only information I could find said that Mn(NO3) was a hexahydrate - hence the higher mass. Even if it is a tetrahydrate, my crystals are still a bit damp.

I tried to get another damn replacement heatgun today and got the "you need a receipt" runaround, even though I have the box and it's clearly not a year old. So I picked up a B&D unit at HD marked to hit 540C. It's also a bit higher amperage.
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[*] posted on 22-3-2008 at 23:07


Yeah I knew the heatgun will be pressed harder on the temp needed for the SnO2 compositions . I still need a few parts for my tube furnace .

CRC gives for Mn(NO3)2 - 4 H2O colorless to pink d 1.82
mol.wt. 251.01 mp 25.8C bp 129.4C
sol 100ml H2O @0C 426.4g , v sol al

from a manufacturers product data

Mn(NO3)2 anhydr. mol.wt. 178.95
d @25C 1.54 50% sol H2O

http://www.sigmaaldrich.com/catalog/search/ProductDetail/RIE...
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[*] posted on 23-3-2008 at 20:57


Pink? All of the Mn(NO3)2 that I've made is definitely brown. The crystals are also brown.

My CRC (CD version) lists a hexahydrate as well as a tetra - it says the tetra is pink and the hexa is rose. So why are my Mn(NO3)2 solutions, and crystals, brown?

[Edited on 23-3-2008 by tentacles]
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[*] posted on 23-3-2008 at 21:17


My 35 year old CRC says colorless to rose monoclinic
d 1.82 . IIRC there is a hexahydrate which exists in solution at ordinary temperature . But if you have the solid at room temperature it is probably the tetrahydrate .
Using a prepared reagent which is so hygroscopic ,
what I would do is go by solution density entirely for
an already existing nitrate solution , or boiling point
indicating the tetrahydrate . But the way to be sure
of molarity is probably to make the nitrate from the carbonate with nitric acid .

The brown color may be some decomposition , or impurity .
A drop of nitric acid may clear it up , it is probably pH sensitive . Seems like I have seen the commercial 50%
packed with as much as 5% HNO3 .

I have looked into this further and found conflicting references , as well as some acknowledgement in the literature of conflicting references and some attempts at explanation of why . The composition of the hydrates of manganese nitrates will usually be a mixture of different hydrates which can have very close melting points which
confounds identification , and it is also possible for a
manganic as well as a manganous oxidation state to be coexistant , complicating the matter further .

This is why it is difficult to be sure precisely what quantity of
manganese is actually present , absent analysis of a sample ,
or direct preparation from a more easily quantifiable intermediate like the carbonate . Know you got to love that :D


[Edited on 24-3-2008 by Rosco Bodine]

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[*] posted on 24-3-2008 at 08:05


Hell, If I'm already going to be dissolving crap in nitric acid, no reason I can't measure out mn carbonate too. I might as well just start with Co carbonate as well. I have changed my spreadsheet to reflect this, plus I fixed a bunch of the crappy math I had developed. I guess I shouldn't work on spreadsheets after a few beers.

SnCl4-5H2O 8.16g
CoCO3 .68g
MnCO3 .35g
Bi2O3 .80g

And it looks like it will consume ~6.47mL of HNO3

I do have something to say about that roasted anode - when I went to etch it, I sanded it very briefly with an old piece of sandpaper. (I use 1000 grit to remove anode coatings) I didn't remove all the coating, apparently, because after etching it in hot HCl for 20 minutes, I pulled it out and the coating was still there. This seems like a pretty good sign to me, the previous coatings such as Co, Mn, would dissolve in the HCl.

[Edited on 24-3-2008 by tentacles]
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[*] posted on 24-3-2008 at 09:53


You are right , that is a very good sign .

The coating which did sinter is tough enough to endure an etching bath and not just rinse away . It reveals the vitrification to an adherent film of ceramic material did occur . So *if* the conductivity and catalytic activity remain present for that vitrified coating .....bingo! ,
one baked coating perchlorate anode
is what it (very probably:P) will be .

The dopant levels may bear significant increasing of
their percentages for increased activity of a working coating , but for the interface and near interface sealing coatings it is probably best to keep the dopant levels
conservative , aiming for conductivity and a good sealing effect .

From the higher percentages allowable for Bi2O3 doping
of SnO2 , even to the point where it would seem the
SnO2 is the dopant for the Bi2O3 ....the physical compatability is much better for bismuth and SnO2 than
is the case with antimony . The Bi2O3 is a sinterable
film former all by itself , as is SnO2 but of course there
isn't enough conductivity for the pure films to be useful ,
as they are quite good dielectrics . But with added
conductivity from the cobalt and manganese doping ,
then a ceramic semiconductor material is produced .

I wonder if silver oxide or lead oxide might even be useful as an included dopant in a composition like this . If it is
well entrapped in a glass like coating it may be perfectly
stable and could have catalytic activity .
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[*] posted on 25-3-2008 at 08:59


Making the stannic oxohydroxide - it takes rather more ammonia than I expected. Also, it comes out almost like a jelly for a few minutes. It does settle out at least somewhat. I rinsed and it's settling better now. I probably won't have time to make the stannic nitrate and do a decent job of testing it, so I will wait until tommorow to add the nitric acid and carbonates/oxide.

[Edited on 25-3-2008 by tentacles]
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[*] posted on 25-3-2008 at 12:29


Dumping the gell into a larger volume of very faintly alkaline cool ammonia water ( pH 7.5 - 8 ) may facilitate breakup of the gell and settling as a manageable , rinseable precipitate .

You have to be careful with the excess of ammonia because it will only to a point help settle the precipitate , but then more will begin to peptize that hydrated oxide to
a hydrosol , with some conversion to ammonium stannate .

From what I have gotten from a couple of patents the
alkalinity should be 7 to 8 pH , higher pH will peptize ,
lower pH will gell .....so it is quite sensitive .

Also it looks like solutions in the range of 5-10% SnCl4
will gell , so it may be that lower concentrations would be
better . I have never done this precipitation so this would
need to be checked , but what I would suggest is using
relatively dilute SnCl4 in water , say a cold 2-3% solution ,
mixed with cold ammonium hydroxide also of 2-3% concentration with good agitation , the quantities being
calculated at the neutralization equivalents . For convenience what I would do is simply match the two prepared equivalent solution volumes by dilution of the lesser to match the greater volume , and then do a
simultaneous rapid pouring of the two cold solutions together into a beaker with a large stirbar already spinning , and check the pH immediately on the still cold solutions , adjusting it as needed with dilute NH4OH or HCl
until about 7.5 to 8 pH is achieved , then gradually warm
the stirred mixture to perhaps 60C - 70C .

*If* my guess is correct .....
( yeah another one of those maybe huge "ifs" :D )
this sort of manipulation should facilitate a controlled precipitation of a settling and decantation rinseable ,
filterable precipitate , as opposed to having a beaker full of white jello :D .

I have looked for a published preparation but so far
haven't found one .

Edit: deleted remarks about suggesting possible usefulness
of NH4NO3 in part substituted for some of the NH4NO3 as per Lowenthal , as it appears likely that the hydrated metastannic rather than the hydrated alpha stannic oxide
desired would be the product of the Lowenthal method .

[Edited on 25-3-2008 by Rosco Bodine]
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[*] posted on 25-3-2008 at 14:42


I must have done something right for once, I just kept adding NH4OH solution until the SnCl4 turned and stayed turbid. I had it in a beaker on my stirplate while I added the ammonia from a 20ml syringe. I initially dissolved the ~8g of SnCl4 in 40ml of cold water. By the time I was done I must have had 100ml in the beaker. It definitely got a little soupy there, then I added 30ml of 35% H2O2 and heated. Turned the beaker into skim milk, at about 60C I'd say. I wasn't monitoring the temp. I gave it a few minutes and the temp went up to 80C or so, then I turned off the heat and stirrer and let it cool. The presumed stannic OHO settled down, I pulled off the clear part and washed with ~400ml. Pulled off as much liquid as possible and then put it in the fridge for tommorow.

Also mixed up 18ml ~50% HNO3. A bit strong but I'll be adding it to however much water ends up with the stannic OHO, probably 30-50ml. The HNO3 is in the freezer.
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[*] posted on 25-3-2008 at 15:49


That's a coincidence you should mention the H2O2 , as I remembered H2O2 being mentioned in a couple of references as producing precipitation similarly as does NH4NO3 , even in acidic but dilute solutions . I have been looking for a couple of hours to try to find that reference which was basically two sentences with no elaboration .
I think I even wrote something about it in one of these threads but I can't remember where . There's so much data I have been accumulating I can't keep track of it anymore :D

Those proportions that I put down above which were
derived from the parts per hundred , totaling one hundred
are correct . But you would have to adjust the carbonate weights if you use the carbonates instead of the nitrates
of cobalt and manganese . Instead of what I showed above for the cobalt nitrate hexahydrate , you would
use 7.4 grams of cobalt carbonate and 10.6 grams of
manganese carbonate , 5 grams Bi2O3 , and the stannic nitrate derived from conversion of 190.8 grams of stannic chloride pentahydrate . The additional HNO3
for the two carbonates will be 19.5 grams HNO3 pure basis plus 4.1 grams HNO3 for the Bi2O3 , 25 grams HNO3
pure basis would probably be required . And water about 750 ml for these quantitites would put the SnO2 loading at about 8% , but I would try it more concentrated .
An additional 137.2 grams HNO3 pure basis , will be needed for conversion of the stannic oxyhydroxide to stannic nitrate.
For the neutralization of the 190.8 grams of SnCl4 - 5 H2O
would be needed 141.35 ml of 29.4% 26 degree Baume
NH4OH .

You can derive a multiplier for any desired quantity
and multiply with these figures to make a different sized batch .

BTW I found a published English translation of the Lowenthal article today , which is the same article as woelen translated
a few weeks ago . And please note that the SnCl2 shown
in the article is actually in todays corrected formula SnCl4 ,
as the numerical formulas of the time had the number of Chlorines wrong by one half , even though the designation
of stannous compounds and stannic compounds with respect
to oxygen content of oxides is correct . The "protochloride"
designation is the stannous salt , while the "perchloride" designation is the stannic salt .

Edit: Some references say that the precipitate of hydrated
stannic oxide gotten by the Lowenthal method may be rinsed with 10% nitric acid , which indicates that is the metastannic hydrated oxide , rather than the alpha stannic hydrated oxide which is required for dissolution in HNO3 to form the nitrate . So the Lowenthal method is likely useless
if the intention is producing an intermediate for conversion
to stannic nitrate .

[Edited on 25-3-2008 by Rosco Bodine]

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[*] posted on 25-3-2008 at 17:21


Well, as far as the H2O2 causing the oxohydroxide to precipitate while still acid, I'm not sure how acid yesterday's solution was, but it did not even begin to precipitate. Could have been too concentrated. It worked fine as it was today, though.

I was just using the procedure you've mentioned a few times on the board in various places (here, and in the tin nitrate thread).
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[*] posted on 25-3-2008 at 17:52


From what I have read it seems that the alpha stannic oxyhydroxide is unstable , so it is a delicate matter to
precipitate it in manageable form where its density and level of hydration are just right . The concentration of the precursor solutions should be dilute and the neutralization
and precipitation condition controlled . There's other
materials which show similar difficulty .

I know in that US6777477 patent they describe pH 8
as precipitating , but then pH 10.5 is peptizing and that
isn't a very wide transition . That's where the NH4NO3
and heating could help . But then heating tends to polymerize the alpha to the meta form .

You may have the H2O2 use mixed up with another purpose . I can't find the reference on H2O2 and I'm thinking it may have applied to the stannous salt for another purpose and product , like for raising the SnCl2 to SnCl4 .....so you may not have what you think ,
but a metastannic oxyhydroxide instead . It's the hydration state and level of polymerization which
determines which isomer you have . The only other place I recall H2O2 being used was with the peroxyorganic acid
substituted metastannic acid sols .

Wait a minute ....I found that obscure reference on H2O2
See page 6 of the attached excerpt , (page 270 item #14) .
This may be a variation on the Lowenthal synthesis because
a careful neutralization is used in the Lowenthal method
also whose purpose is in fact to produce an easily filtered
and clean and pure precipitate to facilitate analysis . However
the precipitate is not really the hydroxide but a basic hydroxide or "oxyhydroxide" for the Lowenthal method ,
and I suspect these may be the same products . Very interesting , I meant to look into this before but something else came up and I forgot about it .

*Stannous* salts are used with this H2O2 here though , I
was right about that :D . I suppose what probably happens
here is something similar to the oxidative soak deposition ,
but here the process is a bulk precipitation so rapid that
the hydration level is maintained . This could be a shortcut
method from the metal or SnCl2 , however to a precipitate
which might be filterable . Somehow I don't think this
would go as well as working from the already oxidized
stannic salt , as per Lowenthal or ordinary neutralization
with NH4OH . Later references have indicated the Lowenthal method likely produces the metastannic hydrated oxide which is unreactive with nitric acid .

However ammonium carbonate , and likely ammonium bicarbonate , should produce the desired alpha stannic oxyhydroxide in the same manner as does ammonium hydroxide , and use of these carbonate of ammonia makes
the neutralization pH less critical , as the precipitated alpha
stannic oxyhydroxide is insoluble in excess of ammonium carbonate , not forming ammonium stannate and peptizing
as would be the case with an excess of ammonium hydroxide .

[Edited on 25-3-2008 by Rosco Bodine]

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[*] posted on 25-3-2008 at 18:23


Ah, reading back I realize it may/was not needed since I started with the stannic chloride. No big loss, certainly. What I seem to be getting from all this discussion and whatnot, is that I really need to get my hands on a pH meter. Better yet, a pH controller which could serve as both meter and a cell regulator.

Rosco, any idea how a pH probe would hold up in a chlorate/perchlorate cell environment? I'm thinking maybe not so good, if for no other reason than the plastic won't hold up to that kind of abuse. Let us know what you think.

I guess I'll find out if it's the metastannic oxyhydroxide tommorow when I try to dissolve it with the HNO3.
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[*] posted on 25-3-2008 at 18:31


I'm not sure about the pH monitoring . What would
be great is some sort of color indicator , nothing to break or corrode:D

See my edit above , I found that H2O2 reference .

BTW , using only 8 grams of SnCl4 , is that just a test run ,
because you are making not much , like a large test tube amount .

After doing some more reading , it appears that the most desirable reagent for neutralization of the stannic chloride
to produce alpha stannic oxyhydroxide with least difficulty ,
is ammonium carbonate . The neutralization pH is not
critical , as the precipitate is not soluble in an excess of
the ammonium carbonate . Now that I have rediscovered
this , I am nearly certain this was declared in one of the patents which has been posted , or perhaps one that I have
already but have not posted .....this bicarbonate and/or
carbonate neutralizer seems extremely familiar and I know I have seen it somewhere before .

[Edited on 26-3-2008 by Rosco Bodine]
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[*] posted on 25-3-2008 at 22:05


I don't really see any reason to make more than a test batch, after all, how many anodes can ~80ml of solution make? A crapload, anyways.

How would I make ammonium carbonate? Ammonia + carbonic acid? I suppose it's a bit of a moot question, if the current method works without too much trouble at least.

No reason I can't hook up my carbonation chamber and crank out some fizzy water if that's what it takes.
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[*] posted on 25-3-2008 at 22:27


If you do careful work , the carbonate or bicarbonate isn't really necessary .

It's really only useful for reducing the need for pH accuracy.
By using the bicarbonate of ammonia , there is no need for
simultaneous addition sort of mixing of cold solutions kept
slightly acidic and then titrated after mixing to a barely alkaline pH for precipitation on warming .

Using the bicarbonate allows simply running in the stannic chloride into a large volume of the bicarbonate , where the
instantly precipitated stannic hydroxide is not affected by
the excess bicarbonate . See US4775412 which uses
this method .

This is the same result as using the method described in preparing the intermediate of US6777477 . Actually
I have rediscovered this too , and searching I found
the exact same parallel had been noted before in another thread , about a year ago .
I *knew* I had seen the bicarbonate used before , but forgot what was the advantage for it . You just have to have more accurate measuring of your reactants when alternately using the ammonium hydroxide and watch the pH so it doesn't go basic enough to peptize - redissolve the precipitate you want to filter and keep for conversion to the nitrate . Keep the endpoint pH 7.5 to 8 and you're fine .
Let it go to pH 10 - 10.5 and you will have a peptizing system underway . How reversible the peptization is I'm not sure , maybe not reversible at all . It could very well be that
the peptization product involves a conversion to a metastannic acid hydrosol , and that involves a change
not reversible by something trivial like a change in pH .
Fortunately the peptization requires some hours to proceed , so if you overshoot the target pH you have time to adjust if you don't delay .
http://www.sciencemadness.org/talk/viewthread.php?goto=lastp...



[Edited on 26-3-2008 by Rosco Bodine]
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tentacles
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[*] posted on 26-3-2008 at 09:22


It would seem I don't do careful work, because the nitric didn't dissolve the precipitate in the least.

How much ammonium carbonate would you expect I'd need? It looks like I'd have to make it myself to get any reasonable quantity. I've tried to find baker's ammonia (NH4HCO3) here but no place seems to carry it. Going to try a pharmacy next.
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[*] posted on 26-3-2008 at 11:41


The problem likely was the H2O2 .

The H2O2 evidently polymerized the alpha stannic oxyhydroxide to the metastannic oxyhydroxide ,
which is something I had a worry may occur .

Try the dilutions and procedure I first suggested and
it will probably work okay . It may precipitate fine from those dilute solutions even slightly incompletely neutralized and slightly acidic . In other words it may be better to err on the side of incomplete neutralization by
a percent or two , to prevent peptization by going to far
with the ammonium hydroxide .

The neutralization of one mole stannic chloride requires
4 NH3 , so that would be 2 moles of the normal carbonate , or 4 moles of the bicarbonate .

If all of this proves awkward then just use one of the
Ordway described methods of dissolving the tin metal in
an aqua regia that is predominately nitric acid , and bear
the lesser percentage of SnCl4 that is mixed product .
The bulk of the tin will be present as the nitrate and
will still likely be at an oxygen surplus . And there will
probably be enough excess acid there for simply adding
your dopant precursors to neutralize that excess acid .
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[*] posted on 26-3-2008 at 14:03


I will try again with the ammonium hydroxide, without the H2O2. The carbonate is proving difficult to find, up here at least. And it just seems like a bitch to make if I go the ammonia + carbonic method, plus that would be a very dilute solution by that time.
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[*] posted on 26-3-2008 at 21:12


As sort of a side experiment ....

If you haven't tossed out the metastannic acid , try
adding about a quarter to three tenths gram of iron filings
and some nitric acid , boil it down to a residue after the iron dissolves but don't overheat it . See if a little water
then dissolves everything .
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[*] posted on 26-3-2008 at 22:25


It's already long gone.. but who knows, tommorow's may turn out just as badly.
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