YT2095
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IDEs to ATEs or PERxATEs
I`m curious about the actualy process that takes place during this, largely because of the excess Halogen that seems to form.
this occures monst natably in a mChloride cell, and quite Visibly in in an mBromide or mIodide cell.
for example we use the Potassium salt of each, to My thinking it would be a good idea yo add a little Excess of the Hydroxide during this to absorb
the excess halogen.
I`m told this would be incorrect.
we have KCl + H2O, and wish to make KClO3.
hydrogen is given off at the Cathode, the Halogen at the anode (and often wasted into the air).
now the only difference between the 2 salts is 3 Oxygens, and so we need to get those from the H2O, thus Hydrogen is liberated at the cathode allowing
this.
there should be NO Chlorine lost!?
but since it IS, then surely adding a little of the metal hydroxide should help "Lock up" the free halogen and re-pressent it back into the equasion
again as unreacted material.
I just Know I`m missing something simple here, but I just can`t figure it
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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tumadre
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1 KCL+ H2O+elec.=== KOH +CL2 +H2
2 CL2 + KOH +KCL + H2O == kCLO3
I don't understand the second part
The problem is the CL2 is a gas
adding KOH... I don't know.. try it
what if you kept the pressure really high,
say 2 bar BELOW the pressure required to explode the H2 and CL2
would the extra pressure help disolve the CL2 in the water?
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woelen
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I can imagine that adding a very small amount of KOH helps trap the chlorine in solution. The Cl2 reacts with hydroxide to form chloride and
hypochlorite, the latter being converted to chlorate at somewhat elevated temperature (and also chloride).
If too much KOH is added, then I expect the cell efficiency to drop dramatically. The following side reaction then will occur at the anode:
4OH(-) - 4e --> 2H2O + O2
What you want at the anode is:
2Cl(-) - 2e --> Cl2
And subsequently: Cl2 + 2OH(-) --> Cl(-) + ClO(-) + H2O
So, I think that adding a minute quantity of KOH indeed may help reduce loss of chlorine gas.
With bromide and iodide I think that adding KOH is less useful, because with these, no gas is formed. With bromide, there may be some bromine vapor,
but it is only a VERY limited quantity and I do not think that adding KOH is beneficial. To my opinion, good mixing of the electrolyte is a better
option. At the cathode also hydroxide is formed and it is better to use that hydroxide, instead of extra added hydroxide.
===================================================
In an ideally constructed cell, no excess halogen is formed and no halogen is lost. Look at the following equations:
At the cathode:
6H2O + 6e --> 3H2 + 6OH(-)
At the anode we have the following reaction:
6Cl(-) - 6e --> 3Cl2
With perfect mixing (no loss of chlorine by bubbling away):
3Cl2 + 6OH(-) ---> 3Cl(-) + 3ClO(-) + 3H2O
You see that half the chlorine is converted back to chloride, the other half is converted to hypochlorite.
At elevated temperature the hypochlorite disproportionates to chlorate and chloride:
3ClO(-) ---> ClO3(-) + 2Cl(-)
So, the net reaction is that for each 6 electrons produced at the cathode (and consequently absorbed at the anode) one chloride ion is converted to
chlorate:
Just sum up all equations I have given above (that is why I started off with the multiplier 6).
6H2O + 6e --> 3H2 + 6OH(-)
6Cl(-) - 6e --> 3Cl2
3Cl2 + 6OH(-) ---> 3Cl(-) + 3ClO(-) + 3H2O
3ClO(-) ---> ClO3(-) + 2Cl(-) +
---------------------------------------------------------------------
6H2O + 6Cl(-) + 3Cl2 + 6OH(-) + 3ClO(-) --> 3H2 + 6OH(-) + 3Cl2 + 3Cl(-) + 3ClO(-) + 3H2O + ClO3(-) + 2Cl(-)
You see that there are many things which cancel out on left and right sides and what remains is the following:
Cl(-) + 3H2O --> 3H2 + ClO3(-)
You see? No excess chlorine, no additional hydroxide needed, it perfectly balances. In practice, what happens is that some chlorine is lost and also
some oxygen is formed at the anode. This results in a lower yield (less than 1 ClO3(-) ion per 6 electrons) and also in formation of excess hydroxide.
But with a well constructed cell, with good mixing, this loss can be kept low. With bromine or iodine instead of chlorine, this problem of loss is
even lower.
[Edited on 25-7-06 by woelen]
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12AX7
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The ideal pH for chlorate electrolysis is slightly acid anyway (~6), go figure.
Tim
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YT2095
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well I`ve been running my sodium Iodate cell now for 2 days almost, the H2 liberated at the cathode is nearly non now at least at the same voltage
I`ve been running at (5.2v). but the soln is a Very deep brown and Iodine has condensed around the vent also.
"Instinct" says to me, add some more NaOH to clear the iodine in the soln and let it carry on some more, maybe another day.
if it were a Chlorate cell I`de simply stop when there was no more Cl2 gas to smell at the set voltage, but being Iodine it`s very Visual and so
clearly pressent in soln, not only that, but after filtering, it`s going to be a Real Sh!t to get nice clean white NaIO3 crystals from it ![:(](./images/smilies/sad.gif)
and I don`t relish the idea of heating this soln as is often done with chlorate cells to convert the last bits of OCl.
the smell I can live with, the color everywhere? Naaah!
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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woelen
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Did you add some dichromate to the iodate cell? That should not be done with iodide, because dichromate is a sufficiently strong oxidizer to oxidize
iodide and then you'll have excess iodine.
Could you please describe in detail how the cell was setup. Did you start with NaI, or any other chemicals? It looks as if your cell indeed is too
acidic, but I don't understand at all, where the hydroxide, formed at the cathode, would have gone, unless some other chemicals were present in your
initial cell, besides NaI.
There is no need to fear heating a solution of an iodate. This stuff is not very reactive. On the other hand, there also is no real need to heat,
because IO(-) does not exist in that solution anyway. ClO(-) is much more stable than IO(-) and when iodine is added to an alkaline solution, then it
reacts to iodate and iodide at once, the IO(-) ions only existing as transients for VERY short time.
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YT2095
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sure, basicly a few months ago I added elemental Iodine to a strong soln of NaOH, I did this until there was a slight brown tinge to the soln then I
put in prill at a time of NaOH until it Just turned clear (White paper test).
there were crystals present in 100ml of soln, so thats a rough idea of it`s saturation, I then added another 100ml of water making 200ml of this soln
(still clear at this point).
I put this into my glass Halate cell.
it has 2 carbon electrodes, and nothing else (incl dichromate was added).
as the liquid level dropped it was topped up with plain tap water.
the power was turned on and then turned up until H2 was liberated at the cathode and I2 was falling into soln (along the bottom of the cell) I watch
this for about an hour on and off to make sure the carbon anode isn`t being pulverised by overdriving the cell, it wasn`t.
I just let it run for 2 days, stirring occasionaly with a glass rod.
there is Some carbon pulverisation at the anode after this time and it`s about 1 1/4 less it`s original diameter (nothing a filter won`t remove).
but the liquid itself is like the darkest coffee brown (clear though) that you can imagine.
I`ve turned the power off now and done nothing else beyond stir it twice, awaiting advice from here before I make another move.
it`s also not the danger that worries me with the heating of it, I just don`t want brown stains all over the walls (no fume cuboard yet).
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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woelen
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Quote: | sure, basicly a few months ago [....]there were crystals present in 100ml of soln[...] |
Was this container tightly closed for these two months, or was air allowed to come in?
If air is allowed to come in, then in the course of a week or so, all NaOH is converted to Na2CO3, and if air is allowed to come in longer, then
NaHCO3 is formed. NaHCO3 is only sparingly soluble, so these crystals you mentioned could well be NaHCO3.
With all this NaHCO3 around, I perfectly can imagine that the liquid becomes so brown. It acts as a buffer, taking away much (if not all) of the
hydroxide, formed at the cathode, and hence, your iodine, formed at the anode, is not reacting with hydroxide.
If your container was perfectly closed for these two months, then this is not an explanation of your observations, but if it was not, then I'm quite
sure that this is the cause of the problems. Try adding some of the liquid to some concentrated HCl. If it bubbles, then it contains
carbonate/bicarbonate.
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YT2095
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nope no fizzing at all, the soln went a white ish color at first like a ppt formed, there was also some dark material that formed but only lasted a
second or 2, and then the soln was a turbid yellow/brown, no observable gas evolved.
and yes the flask was left open after the crystals had formedand been dry, but over the month+ it took on water from the air and became a sludge with
a brown Iodine like tinge. this is what I made up to 200ml for my cell.
\"In a world full of wonders mankind has managed to invent boredom\" - Death
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woelen
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Although you don't have any fizzing I still think that you have had a lot of (bi)carbonate in your cell, together with some free iodine as a starting
point. If you keep an alkaline liquid or wet solid in contact with air, then it will absorb moist and CO2 and be turned into (bi)carbonate fairly
quickly. Especially if bicarbonate is present in the solution, then any hydroxide, formed at the cathode, is quickly taken up by the bicarbonate,
leaving the iodine at the anode unreacted in the liquid. The reason that you don't have any fizzing may be that the solution is too dilute. CO2
dissolves in water quite well.
The white precipitate you had with conc. HCl is most likely NaCl (which is insoluble in conc. HCl). On dilution of the acid, the NaCl also dissolves
again. The dark material is iodine, which quickly reacts with further iodide to form triiodide ion.
I would suggest the following. You don't want to loose the (expensive) iodine. Try all over again, but first save your iodine.
Add acid to your liquid, e.g. dilute H2SO4. Then take a small sample of your liquid and add a small sample of bisulfite. If this causes darkening of
the liquid and formation of black precipitate, then continue adding this, until the liquid does not become darker. If this does not cause darkening,
then do nothing at this step. In this step, any excess iodate is reduced to iodine and iodide and the bisulfite is oxidized to sulfate.
Then add H2O2 (10 ml of 3% H2O2 will do, I expect), until all iodine precipitates out of solution. This forms a fairly easy to separate dark
grey/black slurry at the bottom. The remaining liquid is just light brown. Now, you'll have more than 95% of all your iodine as a solid. Decant the
liquid and keep the precipitate. Add a lot of water and shake, let settle again and decant the liquid. The loss of iodine due to this is very low.
Iodine is almost insoluble in water. Repeat this rinsing step another time and then decant most of the water. Now, to the remaining water with iodine,
add some fresh NaOH or KOH, until the liquid becomes just pale yellow. At this point I think it will be slightly alkaline. Any traces of H2SO4 left,
from the acidification are not a problem at all, these simply form Na2SO4 in the solution.
With this solution, you can try the electrolysis again. Now you start with a solution, which is not spoiled with (bi)carbonate.
[Edited on 26-7-06 by woelen]
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YT2095
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can`t I just add some HI and destroy the bicarb that way?
then do the NaOH thing and re-electolise?
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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woelen
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If you have HI then that could do the job, but I would not make a mix of an acid and KI or NaI to do this. If you do that you get too much foreign
ions in solution and your cell will suffer from major side reactions at the anode.
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YT2095
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I have a small 15ml bottle of HI(aq) already, anyway, I`ve done that and put in the NaOH after stiring well, the soln became clear again and I`ve
started the cell back up again.
This time there isn`t the strong smell of Iodine coming from it and nothing condensing at the vents.
so far so good it would seem ![:)](./images/smilies/smile.gif)
Thnx Woelen!
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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YT2095
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de nada ![:(](./images/smilies/sad.gif)
the cell is finished now (the anode has gone) and so I decided to filter out the carbon that was making the soln look black.
it`s not JUST the carbon, the soln is the deepest brown almost black alot like iodine tincture with a Very strong iodine smell also.
I don`t want to give up, but this is very disheartening, something`s going really wrong here, I can`t see me getting nice white crystals from this
junk easily
\"In a world full of wonders mankind has managed to invent boredom\" - Death
Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine)
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