Vinicy - 25-11-2005 at 09:14
The Electrolysis of KI
the % yield of product, Iodine, of my experiment is very low. How can I increase the % yield of Iodine.
The electrode is Graphite
At standard condition
The concentration of KI solution is 1 M
DC current 3.0 volt
Anode and Cathode are in the same beaker
[Edited on 26-11-2005 by Vinicy]
[Edited on 26-11-2005 by Vinicy]
BromicAcid - 25-11-2005 at 10:27
What were you conditions? Strong KI solution? Did you have an separation of your electrodes, not just in distace but in some sort of membrane?
Possibly a stupid question but you were using DC correct? How much current, did it cause the cell to heat up too much, could you give us more
conditions for the electrolysis you preformed?
Marvin - 25-11-2005 at 10:48
Keeping the produced alkaline fluid away from the liberated iodine is somewhat important. Something that should also be born in mind is that
liberated iodine will be soluable in the potassium iodide solution. (I choose to assume its in solution and not molten).
garage chemist - 25-11-2005 at 14:04
Just acidify the KI solution, otherwise the iodine yield will always be very low, due to reaction with the alkali produced at the cathode.
Without acid, you'll just get potassium iodate (assuming your electrodes hold up- what were you using?).
If you just want to make iodine, mix KI with HCl and H2O2. If the amounts are stochiometric, an almost quantitative yield results, with the
supernatant liquid being only very slightly brown due to abscence of iodide.
IrC - 25-11-2005 at 14:08
Would it be possible to do this as a molten salt in a stream of O2 instead of a water solution, where iodine vapor was collected, and allowed to cool
and crystallize? Just a thought off the top of my brain without actually researching it.
Fleaker - 25-11-2005 at 18:41
If you are after elemental iodine, there are better ways than aqueous electrolysis of the salt. Could always use HCl or MnO2 with a cold finger for
the iodine to deposit on, or Caro's acid (I've used this for 100mL runs for producing elemental bromine).
Vinicy - 26-11-2005 at 10:19
PLEASE ~
BromicAcid - 26-11-2005 at 10:44
I just noticed you edited your inital post.
Your electrolysis will give you iodine and hydroxide anion in your solution. The problem being that iodine will react with hydroxide anion. The
solution is to either divide the electrodes with some sort of membrane to limit mixing, or to acidify the mixture first like Garage Chemist mentioned.
This will help eliminate the hydroxide as it is produced to increase your yield of iodine. Also noteworthy is that elemental iodine reacts with the
iodide ion to give the triiodide ion (I<sub>3</sub><sup>-</sup>
which is soluble so as you are forming iodine it will be soluble in the solution much more so then if no iodide was present. So your
solution will turn brown rather then just start producing solid iodine. As a result your electrolysis may not seem productive until it runs for some
time and your yield may seem crummy until most of this complex is oxidized.
Vinicy - 26-11-2005 at 11:31
Can this be possible ?
When the process is occuring. Hydroxide ion is produced. So the solution is basic.
2H2O + 2e- --> H2 + 2OH- E0 = -0.83 V (Basic Solution)
While
I2 + 2e- --> 2I- E0 = 0.615
H20 will be produced at anode instead of Iodine.
neutrino - 26-11-2005 at 11:49
The first reaction is undoubtedly occuring at your cathode. The hydroxide ions formed there and the iodide ions at your annode are reacting with your
iodine before it can come out of solution:
I<sub>2(s/aq)</sub> + I<sup>-</sup><sub>(aq)</sub> -->
I<sub>3</sub><sup>-</sup><sub>(aq)</sub>
I'm not sure, but the hydroxide + iodine will probably do this:
OH<sup>-</sup> + I<sub>2</sub> --> IO<sub>3</sub><sup>-</sup> + I<sup>-</sup>