beeludwig
Harmless
Posts: 30
Registered: 7-5-2019
Member Is Offline
|
|
PbI2 and electrolysis questions
I have some lead iodide in solution (obviously, most is in precipitate). Would it be possible to electrolytically separate them in solution or would
I just split the water? If I wanted to split them this way, would I have to melt it down to the molten state?
Additionally, in a more generalizable sense, how can I predict a head of time in a salt solution, how could I predict ahead of time which ions will be
reduced and which will be oxidized?
Finally, could I chemically separate them by oxidizing the iodide to the iodine
Thanks for helping out guys. You're the best.
|
|
|
woelen
Super Administrator
Posts: 7977
Registered: 20-8-2005
Location: Netherlands
Member Is Offline
Mood: interested
|
|
Electrolysis of a solution of PbI2 will not work in practice. theoretically, you get I2 at the anode and Pb at the cathode. But because of the very
low concentration in solution, you will get a very low current and the electrolysis will be VERY slow.
A nice experiment could be to apply 5 volts to a solution of PbI2 with a graphite anode and any conductive cathode. You may be able to see formation
of some I2 at the anode, due to slight brown/yellow coloration of the solution near the anode. Given the low speed of the raction, I, however, doubt
you can observe anything at all.
--------------------------------------------------------------------------------------------------------------------------------------
At the anode, there are three materials which can be oxidized (which is drawing electrons from molecules or ions):
1) The anionic species, in this example, it is iodide ion.
2) Water
3) The anode material
The one, which is easiest oxidized will be oxidized.
With iodide and iron anode, the iron anode will be oxidized: Fe - 3e --> Fe(3+)
With iodide and graphite, the iodide will be oxidized: 2I(-) - 2e --> I2
With Fluoride and graphite, water will be oxidized: 2H2O - 4e --> 4H(+) + O2
Sometimes, when certain compounds are of comparable ease of oxidation, two competing reactions occur at the same time. E.g. with electrolysis of
dilute H2SO4 with platinum anode, at the cathode, there are two competing reactions:
2H2O - 4e --> 4H(+) + O2
2SO4(2-) - 2e --> S2O8(2-)
With chloride ions, the main product is chlorine, but also some water is oxidized, so you get a mix of mainly chlorine, containing a small amount of
oxygen.
At the cathode, things are the same, but now with reduction (pushing electrons upon ions or molecules). With e.g. a copper electrode, sodium chloride
in solution, there is
1) water
2) sodium ions
3) copper metal
towards which electrons can be pushed. In this case, water can easiest be reduced: 2H2O + 2e --> 2OH(-) + H2
With lead iodide and a copper cathode, the Pb(2+) ions easiest accept electrons: Pb(2+) + 2e --> Pb
With iodic acid, at the cathode, using a platinum cathode, iodine is formed at the cathode. Electrons are pushed upon HIO5 molecules, and via a number
of steps, this leads to formation of I2 and at the same time consumption of H(+).
|
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
