semiconductive - 18-6-2021 at 10:13
I've been running several experiments about various nonaqeuous (organic) solvents that can be used in electrodeposition. Esters were a fairly useful
solvent.
One of the experiments led to a result that surprised me, and I'm curious about what's going on. So, I set up a different experiment to attempt to
synthesize esters with electrodes involved and see if any side reactions showed up.
Formic acid is supposed to be strong enough to cause Fisher/Spier esterification on it's own, although I expect the reaction to proceed slowly or to
be limited by the water produced. So, I decided to start the reaction with electricity running through via relatively inert graphite electrodes. The
electrodes should be able to lyse water with acid present. (1/2 O2 + H2 ) . If that's correct, the esterification reaction can continue nearly to
completion....
However, when I try the reaction ... I generally get far more hydrogen bubbles than oxygen bubbles. It's not a 2 or even 3 to one ratio, but more
like 10 to 1.
I'm producing far more hydrogen:oxygen in alcohol than in water.
When I do the experiment with formic acid and methanol, the solution slowly turns purple. I looked up Methyl formate, and it should be clear. So, I
think I have a side reaction going on.
It's possible that the graphite electrode polymer is causing discoloration of the solution (it's just a hi-polymer graphite pencil lead.); but when I
repeat the experiment with formic acid and isopropyl alcohol (which has 1% water), the solution does not turn purple and the amount of oxygen
produced at the anode has decreased over several days of running the experiment. So, I'm reasonably certain that the electrode plastic binder isn't
causing a significant color change.
I'm getting rid of a lot of hydrogen ... but the oxygen is not leaving the solution. I can't imagine much else would be produced as a gas.
So I assume that I have to be producing a peroxide of some kind.
The hydrogen bubbles appear immediately, and there's no diffusion delay going on from the beginning of a fresh solution; Theoretically, the ions
produced at the cathode should all be negatively charged and not 'want' to combine as long as both atoms involved are negatively charged ... since
they would repel each other.
This makes me wonder:
1) In chemical reactions such as electrolysis of water or peroxide formation ... where and under what conditions, exactly, does a bond form between
the hydrogen ions?
A positively charged hydrogen ion could be 'neutralized' at the negative electrode, or negatively charged....
Do negatively charged hydrogen atoms bump into neutral water molecules in the bulk liquid and lose one charge ... or do negative ions actually 'hit'
each other and then eject an electron?
What's the physical mechanism of charge neutralization and bond formation at an electrode ?
2) Since some kind of peroxide pretty much has to be forming, there's an explosion hazard issue I need to consider should I make a bunch of this
stuff. I know that Ethers form peroxides spontaneously, but I've never heard of esters doing so ... even though both of them have an oxygen bridge
between two organics, R-O-R'. (I'm a novice at chemistry, so explanations would be appreciated.)
I'm curious to know how the extra oxygen on a carboxylic acid can affect peroxide formation and potential for explosion. If this is not a perxode,
and I'm wrong ... I'd appreciate knowing what I am likely producing, and why.
In F/S Esterification, the carboxylic acid's oxygen is the FIRST thing to be changed by the presence of a positively charged hydrogen. So, it seems
to me that the double bonded oxygen might be even more reactive than hydroxide oxygen in carboxylic acids.
That makes me wonder:
If I'm forming a peroxide instead of an ester ... is it more likely that it will form with the carboxyl double bonded oxygen and temporarily leave the
-OH in-tact, or will it form with the at the hydroxide location ....?
As my electrochemical reaction proceeds, I expect that the point will come where all of the 'easier' site to peroxidise have been exhausted. How is
the reaction likely to proceed after that? (This is something I plan to test, but I'd like to understand the theory first... as it's probably going
to take a week to drive the first part of the reaction anywhere near completion.)
[Edited on 18-6-2021 by semiconductive]
semiconductive - 20-6-2021 at 08:34
See picture:
Gas bubbles in electrolyzed formic acid + 99% isopropyl.
40 milli amps of current through 0.5mm graphite electrodes, 3 or more volts applied.
This is after several days. There is heavy bubbling on the left electrode with gasses erupting off the electrode into the electrolyte horizontally
and then floating upward. (Hard to see, as my phone camera can't focus well on fast moving bubbles.)
BUT: There is very little forming on the anode, not even enough to fizz out into the electrolyte. by volume is 5% or less than is being produced at
the cathode.
There is a thin film, though, that has formed on the anode changing it's appearance very slightly. If I take it out of solution, I can't see anything
unusual under a microscope ... but under IPA it looks slightly different (lighter) than a graphite that has not been an anode.
Regarding the methanol version of the reaction, the mauve/purple color is slowly fading and the solution is becoming increasingly clear. The anode is
disintegrating and a film of some kind has built up on the cathode. Conductivity is dropping rapidly. I will replace the graphite electrodes,
decant off the solution, and see what happens.
[Edited on 20-6-2021 by semiconductive]
semiconductive - 25-6-2021 at 13:44
I Replaced the electrode and decanted. The reaction proceeded the same as before but the replacement graphite electrode did not (and is not) breaking
down.
After several more days, the anode gas began increasing steadily. It's now approximately 2:1 ratio by volume, so I think esterification is happening
and water is being broken down. It doesn't look like peroxides are going to build up in the vial beyond a certain amount before esterification
becomes the preferred reaction.
As a test:
I took straight formic acid and electrolyzed it with two graphite electrodes, and both electrodes bubbled and the anode disintegrated. So, I think
most of the electrode disintegration is caused by formic acid attacking the binder when it's very concentrated.
The gas release is very similar to water electrolysis, but I wouldn't expect oxygen to come free from a carbon bond ... so, I'm guessing the gas at
the anode is possibly CO2 + ethene ?
2 CH2OOH
2 CH2OO- 2 H+
2CH2 + 2 CO2 (loss of two electrons to anode, I'm not sure how to charge balance this)
H2C = CH2
So, this is also possible in the alcohol solutions and is a gas I didn't think of being possible before. DId some searches and came across kolbe
electrolysis. I still don't understand how the ions neutralize and dimmerization happens at the electrodes ... but it seems the idea isn't crazy and
it can happen.
[Edited on 25-6-2021 by semiconductive]
njl - 25-6-2021 at 16:59
The answers to many of these questions may be found in the chemistry of radicals. Reading up on radical electrochemistry would help you here. As to
peroxides, no you won't form any appreciable amount. First of all organic peroxides specify "perethers", the word you're looking for
is peresters. While peresters exist and are documented in the literature, you will have trouble making them unless you're trying.
They are far more unstable than peroxides, and decompose quickly enough at room temperature that they won't build up (if they form in the first
place). The most common I've read about are aromatically stabilized acids with bulky alkyl groups i.e. t-butyl perbenzoate. The mechanism for H2
production is also via radicals. As for the electrolysis of formic acid I would expect H2 and CO2.