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Posted

this is the third re-electrolysis or my barium chlorate, eventualy it becomes Perchlorate.

if we have a group 1 metal, like Sodium for example, typicaly it goes; mCl, mOCl, mClO3, mClO4.

 

but what happens if electrolysis of a perchlorate is allowed to continue?

 

does the water then just split as H and O?

does that Chlorine get ripped off entirely leaving the m-hyroxide?

 

anyone know or tried this?

Posted

We need to use molten compound instead of an aqueous solution, right?

I know very little about it.

Below's my idea.

The fact is just like that of electrolysis of sulphate, nitrate.

Posted

Electrolysis of chlorides gives the following reaction:

 

anode: 2Cl(-) --> Cl2 + 2e

cathode: 2H2O + 2e--> H2 + 2OH(-)

 

When the liquid is mixed, you get the following secondary reaction:

 

Cl2 + 2OH(-) --> Cl(-) + ClO(-) + H2O

 

On heating and long standing you get the following reaction:

 

3ClO(-) --> 2Cl(-) + ClO3(-)

 

So, the chlorate does not form at the anode, it forms indirectly through the intermediate hypochlorite.

 

As you see, Cl(-) is produced from all these reactions again and that Cl(-) also is used up again at the anode. Finally, only ClO3(-) is left. Before that happens, however, a new net reaction comes into play at the anode:

 

H2O + ClO3(-) --> H(+) + HClO4 + 2e

 

Probably the intermediate reaction is as follows:

 

2ClO3(-) --> Cl2O6 + 2e

Cl2O6 + H2O --> HClO3 + HClO4 --> 2H(+) + ClO3(-) + ClO4(-)

 

The H(+) is neatralized by OH(-) from the cathode.

 

There also is a side reaction at the anode during the entire electrolysis process:

 

2H2O + 4e --> 4H(+) + O2 + 4e

 

As long as chloride, Cl(-) is present, the formation of Cl2 is strongly favoured. When Cl(-) is depleted, then the conversion of ClO3(-) to ClO4(-) via the Cl2O6 intermediate is favoured, but with increasing relative concentration of ClO4(-), nothing but the water is left and finally only water is oxidized at the anode, giving oxygen and acid.

 

But the oxygen and acid-forming reaction also occurs already when quite some chloride is present and also when quite some chlorate is present. This causes the yellow color in chlorate electrolysis cells. When ClO3(-), Cl(-) and H(+) are present at the same time, then a complicated reaction occurs, in which ClO2 (deep yellow) is formed.

Posted

fascinating!

I`ve noted on several occasions a distinct yellowing of the soln, I passed it off as some form of contaminant in the carbon electrodes, the odd part was that when crystalised anyway there was some yellow "edging" around the clear crystals, and all throughout the whole lot stunk to high heaven of Bleach (the OCl), and gave off an extreme gas that can only be described as a cross between NO2 and Chlorine for smell but with a white vapor (very nasty stuff!).

 

so you`re saying at the ClO4 stage, any further will give the ClO2 complex then?

I`m really curious to find out what happens if this is just kept running ad infinitum... :)

Posted
fascinating!

I`ve noted on several occasions a distinct yellowing of the soln' date=' I passed it off as some form of contaminant in the carbon electrodes, the odd part was that when crystalised anyway there was some yellow "edging" around the clear crystals, and all throughout the whole lot stunk to high heaven of Bleach (the OCl), and gave off an extreme gas that can only be described as a cross between NO2 and Chlorine for smell but with a white vapor (very nasty stuff!).

 

so you`re saying at the ClO4 stage, any further will give the ClO2 complex then?

I`m really curious to find out what happens if this is just kept running ad infinitum... :)[/quote']

The ClO2 gas is formed from chlorate and chloride at acidic conditions. So, when the electrolysis is done ad infinitum, then at a certain point all chlorate is converted to perchlorate and then only hydrogen and oxygen are produced (until all water is used up, but that far you will not come, before that the matter solidifies and your process comes to an end).

 

The yellow stuff mostly is formed when there still is quite some chloride in the solution and when there is chlorate. At this point still Cl2 is formed from chloride ions, but also acid is formed (from the ClO3(-) and from H2O, see previous post) and at that point most of the yellow ClO2 is formed. When only perchlorate is left in solution, then hardly any ClO2 will be formed anymore.

 

I did an experiment, making ClO2 at high concentration. At the end of the page you see the reaction equations on how it is formed from chlorate and chloride under acidic conditions. This is exactly what happens in the electrolysis cell. You do not see gaseous yellow ClO2, because in the electrolysis cell it remains dissolved in water.

 

http://woelen.scheikunde.net/science/chem/exps/clo2/index.html

 

 

Another bad effect which can occur during electrolysis is back-reduction of ClO(-), ClO3(-) and ClO4(-) at the cathode. An example reaction equation is:

 

ClO(-) + H2O + 2e --> Cl(-) + 2OH(-)

ClO3(-) + 3H2O + 6e --> Cl(-) + 6OH(-)

ClO4(-) + 4H2O + 8e --> Cl(-) + 8OH(-)

 

These latter reactions are REALLY bad, they destroy your nice chlorate and so on.

Posted

Hmmm... I can see your point, and for Chlorate production this Would be somewhat "less than usefull" :)

 

ok, so 2 extra questions, When is the best time to know when to stop (ie/ your ideal chlorate/perchlorate mix with NO ChlorIDE)?

 

and secondly, if we DO permit this reaction to go to the extreme and get all the chlorine out and all that lovely added O2, do we end up with the m-Hydroxide?

ignoring the fact it would go dry ultimately (we just keep adding the water to preclude this event).

Posted
ok, so 2 extra questions, When is the best time to know when to stop (ie/ your ideal chlorate/perchlorate mix with NO ChlorIDE)?

Here is the point where a current source instead of a voltage source comes into play. If you use a current source for your cell, then you'll see that the voltage, needed to maintain the same current slowly rises. The lowest voltage is needed for conversion of your Cl(-) to Cl2, a higher voltage is needed for conversion of ClO3(-) to Cl2O6 and yet another voltage is needed for electrolysis of the water, forming O2. Unfortunately, the bumps are not sharp, the observed voltage is increasing slowly.

 

Determining when to stop best is not easy and depends on what you want. The problem is that at no point in time there is only chlorate and nothing else. You see a decrease in chloride concentration, an increase in chlorate concentration and even, when still a lot of chloride is present, you also will have some perchlorate in solution already. So, an answer to this question is not easy.

 

I would work as follows:

 

Take a solution of well-known concentration.

Start electrolysis and measure the voltage with a given current from a current source.

Every now and then sample the amount of chlorate in the solution and keep track of the voltage at that point. From these measurements, you can keep a record and use that in future electrolysis systems. You can even automate it with a current source, which is switched off, when a certain threshold voltage (with some hysteresis built in) is exceeded.

 

and secondly, if we DO permit this reaction to go to the extreme and get all the chlorine out and all that lovely added O2, do we end up with the m-Hydroxide?

ignoring the fact it would go dry ultimately (we just keep adding the water to preclude this event).

No, you'll not end up with the metal hydroxide. Ideally, for each 2OH(-) produced on the cathode either 2H(+) is produced at the anode, or a Cl2 molecule is produced. In practice, due to losses of chlorine gas (bubbling out of solution) and due to losses of OH(-), due to absorbtion of CO2 and due to losses of small bubbles around the cathode, things are more complicated, but when no chlorine gas is lost into the air and no OH(-) is lost due to droplets being lost into the air, the mix finally will contain mostly ClO4(-) with some ClO3(-), ClO(-), Cl(-) and an equilibrium is reached. At the cathode, H2, OH(-) and back-reduced Cl(-) are formed and at the anode, O2, H(+), Cl2, ClO4(-) are formed keeping the system in equilibrium. In the meantime there also always will be some ClO2 in solution, which with OH(-) forms ClO3(-) and ClO2(-). ClO2(-) in turn is converted to ClO3(-) and Cl(-) again:

 

3ClO2(-) --> Cl(-) + 2ClO3(-), very similar to reaction of ClO(-)

 

So, the final outcome will be (assuming no gasses are lost to the air, except H2 and O2, and water is replenished):

 

- moderate amounts of Cl(-), due to back-reduction at anode

- moderate amounts of ClO(-), due to reaction of Cl2 and OH(-)

- small amounts of OH(-), due to small excess of OH(-), formation of ClO2 takes H(+)

- small amounts of ClO2, due to presence of Cl(-) and ClO3(-)

- small amounts of ClO2(-), due to reaction between ClO2 and OH(-)

- moderate amounts of ClO3(-), due to disproportionation of ClO(-) and ClO2(-)

- large amount of ClO4(-), due to final oxidation of ClO3(-)

Posted

Excellent answer! thank you :)

 

as for the 1`st part, I`ll be perfectly happy with a Chlorate/perchlorate mixture with NO Chloride, hence this it the 3`rd time around of electrolysis, then filter and evaparate/crystalise, re-hydrate, electrolise and so forth....

 

actualy the product the 1`st time around was good, the second time was even better, and this will be the last time now, with hopefully insignificant ChlorIDE levels.

 

it`s a darn pity that electrolysis of plain HCl(aq) didn`t give a Chloric acid, that would make life a little easier :)

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