liftedresearch
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Electrolysis questions
Hello all,
Im having trouble figuring out what the follow sentence means
"After the electrolysis using a current of 1A at a temperature of 0°C (2.2 faradays per mole of benzyl chloride)
My first question is would you need to keep the current at 1 Amp, or could you use a higher current? So as to use a constant voltage.
Secondly, if the current has to stay at 1 Amp, how would I monitor the restistance of the solution to adjust the voltage accordingly?
If you could help me out it would be much obliged
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TheAlchemistPirate
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Why would you have to monitor the resistance of the solution? If you have to maintain 1 amp, why not just have an ammeter in the circuit, and simply
adust the voltage as the current raises or lowers? I don't know if you can you can use a higher current or not; to my knowledge all that limits how
much current you can use is heat. That heat may come from the resistance of your circuit, electrodes, or (most likely) your solution. If you have to
keep the cell at 0 degrees C, this might limit you. Some people will cool their solutions to get around this, others might increase their electrode
surface area (therefore decreasing their resistance in the solution and electrode).
Maybe I'm majorly oversimplifying things. Irc or someone else who knows what they're talking about might come around and give you a better answer.
Keep in mind I have never made a proper electrolysis cell so feel free to doubt me.
"Is this even science anymore?!"
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j_sum1
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You are right that the resistance of a cell changes as the concentration of ions in a solution changes. It also changes due to surface effects on the
electrodes and also due to the changing surface area of the electrodes as they erode or have material deposited on them. Additionally, the resistance
changes due to cell geometry.
There are two main variables. The first is the voltage. In any given application there is a minimum voltage required. There is often benefit in
keeping the voltage relatively low. Overpotential leads to side reactions and other issues that can be problematic. Obviously the voltage applied
has an effect on the current.
The second variable is the current. Obviously the lower the current the longer the reaction will take. High current leads to high current densities
which has an impact on your electrodes -- typically eroding your anodes.
So, you want to keep your voltage within certain parameters and your current below a particular value. Typically this is done through a regulated
power supply. Apparently they are not difficult to construct, but I bought mine. You set the maximum voltage and maximum current and away you go.
You can achieve the same effect by simply watching the current and adjusting the voltage as needed. A multimeter and a variable resistor in your
circuit will do the trick. Or, depending on your application you might get away with just using a battery or transformer either directly or with a
static resistor in series.
Finding a suitable current does depend on your application. If you are plating a metal and want a fine finish then you would control also the
temperature, the concentration, the presence of other ions in solution, the amount of stirring and probably a whole bunch of other things. And the
actual figure you arrive at depends again on cell geometry. There is no substitute for trial and error in my opinion.
Now to your question. I have never come across the unit, faradays per mole. I don't know what this means.
I always use a simple formula to calculate the time of a run.
t=nIF
t is time in seconds.
n is the number of moles of electrons needed to complete the reaction.
I is the current (obviously)
F is Faraday's constant which is roughly 96500 amp seconds per mole.
I usually allow 10% extra time to allow for inefficiencies, set a time of day that I will come back to collect my product and adjust the maximum
current accordingly.
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liftedresearch
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Solution volume
Thank you both for your great answers. My next question would be if by keeping your voltage within a certain allowed range, and your voltage constant
(or sufficiently close), how would the volume of the cell affect the reaction? By that i mean that a 10 liter electrolytic solution would have a
higher resistance (or am i incorrect in assuming that), than say a 1 liter solution. Would it be more efficient to increase the amount of electrolyte?
Or would the fact that both the 10L and 1L have the same concentration of electrolyte make the resistance of both solutions the same?
Thank you again
[Edited on 27-11-2015 by liftedresearch]
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j_sum1
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You mentioned voltage twice. I think that was a typo.
Your question is one of cell geometry. Like I said, trial and error.
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WGTR
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Quote: Originally posted by liftedresearch  | Hello all,
Im having trouble figuring out what the follow sentence means
"After the electrolysis using a current of 1A at a temperature of 0°C (2.2 faradays per mole of benzyl chloride)
My first question is would you need to keep the current at 1 Amp, or could you use a higher current? So as to use a constant voltage.
|
Faradaic reactions involve the transfer of charges across the double-layer interface at the electrodes. The measure of electric charge is
coulombs, which is equal to ampere-seconds. The rate of the reaction is directly tied to the current.
Additionally, the voltages present at both electrodes will often determine the products in an organic electrolytic reaction. A reference electrode is
required when measuring the voltages across each electrode interface. Also, the electrode material and surface finish will often affect efficiency,
especially if you're trying to isolate an intermediate oxidation/reduction product.
| Quote: Originally posted by liftedresearch | Secondly, if the current has to stay at 1 Amp, how would I monitor the restistance of the solution to adjust the voltage accordingly?
If you could help me out it would be much obliged |
For things like this a current-limited power supply can be used. I have several adjustable ones in my lab, and they are commonly available. You set
the current to 1A (in your example), and set the voltage limit high enough to allow 1A to flow under all operating conditions. Such a set-up doesn't
measure or account for electrode potentials. For something like this you'd need a more-expensive battery test cell or voltammetry set-up.
The value of current is meaningless, however, unless electrode dimensions are considered. A more useful value to know would be current
density, which would be the value of current per electrode unit area.
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