Sciencemadness Discussion Board - Electrolytic Hydrogenation/Reduction questions

Electrolytic Hydrogenation/Reduction questions

Electra - 11-12-2013 at 10:24

I recently got assigned a project for my chemistry class to research how Electrolytic reductions are done in industrial settings. The only thing we have done on the subject is very simple calculations about electrochemistry. Things such as calculating the time to reduce or oxidize 24g of material if the input is xx-amps @ 200mA per cathode surface area. A lot of the calculations come out to be 5-6 hours for such a small amount of material. The equations we worked with never dealt with watts, and I know watts = joules/sec = amps * voltage. I know Joules would be important if we are taking electrical energy and making some chemical modification.

Based off the little I've learned in this class on this subject, I am not sure how these industrial reactions would be done. I'm supposed to create a diagram, a presentation, calculations for them.

Industrial settings are going to work with a lot more than 24g at a time, and 24/g per 6 hours isn't realistic. Assuming proper cooling, what variables would a company need to increase in their system to pull off 250, 500g, 1000g, per hour?

I would think since 200mA per square cm of cathode is supposed to be maintained then the only option would be to increase the voltage? More joules/sec entering the system = more chemical reactions = more heat = more needed cooling?

Why hasn't any of the electrochemistry problems we've done accounted for both voltage & amps, but only one or the other? If anyone could provide me with some equations that you think would be useful for my project it would be very appreciate!

-Jenn

WGTR - 11-12-2013 at 11:15

If you are limited to a certain current density, then the only option is to increase the surface area of the electrode,
so that you can increase the overall current. Electrochemical reactions are dependent on current, since they are
reduction/oxidation reactions. Electrochemical systems can be very complex. If you increase the current by raising
the voltage (increasing the current density), then an entirely different reaction may take place on your electrode.

In my short and rather non-illustrious lifespan, I've realized that most chemists are not good in electronics, and most
electronic people know little about chemistry. It's a shame, because the two fields go hand-in-hand. An electrical
engineer can sound like a retard talking about chemistry to a chemist. Someone well-versed in electronics can
become frustrated reading through chemical journals, trying to decipher the obtuse ways in which chemists record the
electrical parameters of their experiments. In many ways things have gotten better with time, and if you're good
in both fields, then that is the best of both worlds.

To answer your question more specifically, though, current is important from the standpoint of determining faradaic
efficiency. Voltage is important from the standpoint of determining what reaction is actually occurring (cyclic
voltammetry) in the solution. Power efficiency doesn't really become an issue until you get into industrial situations,
where energy becomes a major cost factor.

If someone disagrees, please correct me on this. I'm not exactly an authority on the topic.

[Edited on 11-12-2013 by WGTR]

Electra - 11-12-2013 at 11:37

I have understood that to maintain the correct mA/surface the surface would need to increase in order to increase the current. I am not an expert on electricity, but I've heard wall outlets are limited at 15-20amps, and cannot go over this.

If voltage is unable to change, and amps can't go over 20, then that would still only result in a very limited reaction rate. Unless of course the voltage were really high? I feel like this is wrong. Are amps or voltage limited in this manner? Even from a non industrial standpoint, the reactions we've talked about in class of reducing/oxidizing 25/g of material taking 5-6 hours seems incredibly way too long.

I proposed a chemical scenario to someone well versed in electronics and they said if you have the amps at 5, with the corresponding surface area ratio correct, then increasing the voltage will indeed increase the rate of the chemical reaction.

You said increasing the voltage increases the current density, but I was told this is not the case? Was that a mistype? Or?

WGTR - 11-12-2013 at 11:47

An electrochemical cell approximates a resistor (that is, a very non-linear one). Think Ohm's Law for this. An increase
in voltage causes an increase in current for a given resistance. All other things being equal, it's impossible to increase the
voltage across a given cell without increasing its current.

Electra - 11-12-2013 at 12:20

So ideally for industries to upscale these reactions they would minimize the voltage while maximizing surface area and keeping the current high?

With hypothetical numbers: If there is 5000cm^2 of surface area, @ 200mA per cm^2, the current would be set at 1000, and the voltage be calculated based on the resistance? Would this cause a reaction to occur at a faster rate? ie, Larger surface area, larger current = faster reaction?

In the book it says for instance 1-mol of [some molecule] is completely reduced when 5 faradays of electricity have passed. From what I understand 1 faraday = 96,485 Joules. 5 faradays = 482,425 Joules then. If you wanted that to occur in an hour and Voltage & Current are in play then Voltage has to be whatever 5 faradays divided 3600 seconds, divided by the current is? This is assuming minimal resistance. Do I have the right idea here? If this is the case I think I know what to do with my project, but if not then I am totally lost.

Electra - 11-12-2013 at 12:54

I just got finished talking with an Electrical Engineer friend and they said if the resistance of the system is minimal then a person would use a resistor, determined by calculating it based off the voltage, to set the needed current for the system? Then with the current and voltage known the amount of Joules passing through the system, and thus Faradays passing through can be determined?

WGTR - 11-12-2013 at 13:13

Quote: Originally posted by Electra

With hypothetical numbers: If there is 5000cm^2 of surface area, @ 200mA per cm^2, the current would be set at 1000, and the voltage be calculated based on the resistance?

You could, but you'd have to figure out a way to calculate the effective resistance of the cell up front. An electrochemical cell
is essentially a variable resistor, that varies somewhat in value depending on voltage (in other words, it's non-linear over a
large range). Unless you're doing more advanced chemistry (which I can't do), this is usually done the other way around.
You set the voltage to whatever gives you the current that you want under the prevailing conditions in the cell. This
usually requires some empirical lab work. You can, however, approximate the voltage needed across the cell by looking at
the difference between the two half-cell reaction potentials. Maybe someone else could give a better answer, as I'm
reaching the limits of what I can intelligently answer.

Quote: Originally posted by Electra

Would this cause a reaction to occur at a faster rate? ie, Larger surface area, larger current = faster reaction?

Quote: Originally posted by Electra

So ideally for industries to upscale these reactions they would minimize the voltage while maximizing surface area and keeping the current high?

I'll let you figure these out

Quote: Originally posted by Electra

In the book it says for instance 1-mol of [some molecule] is completely reduced when 5 faradays of electricity have passed. From what I understand 1 faraday = 96,485 Joules. 5 faradays = 482,425 Joules then. If you wanted that to occur in an hour and Voltage & Current are in play then Voltage has to be whatever 5 faradays divided 3600 seconds, divided by the current is? This is assuming minimal resistance. Do I have the right idea here? If this is the case I think I know what to do with my project, but if not then I am totally lost.

I think you're confusing joules with coulombs. You might want to rethink that one and try again. Farads are also not normally
used to measure the passage of current. You could use coulombs or ampere/hours for that.

[Edited on 11-12-2013 by WGTR]

Electra - 11-12-2013 at 13:40

When speaking of Faradays I was not meaning Farads. According to Wiki 1 Faraday Unit = 96,485 (96.485 kJ) per volt gram equivalent. Am I misinterpreting this?

If my above interpretation is wrong, then how does one calculate the time it takes for 5 Faradays to pass if the voltage and current is known?

With the above I was thinking 5 Faradays could be passed in an hour for instance if the Joules/Sec was at 134.02. If the resistance of the system changes, does that also mean the wattage changes, thus, the voltage must be readjusted to correct the wattage if a specific current is to be maintained?

WGTR - 11-12-2013 at 14:57

Ah, see...I learned something today. In the electronics world, beginners sometimes use the term "faraday" when they're searching
for a sophisticated way of saying "farad", so that confused me a bit. Anyway, I think you have the right idea. One joule
equals one watt/second. Joules per hour divided by 3600 will also give watt/seconds. Then watts divided by current will give
voltage.

To answer your last question, the answer is "yes".

[Edited on 11-12-2013 by WGTR]

Electra - 11-12-2013 at 15:33

For a more simple question, am I correct to interpret 1 Farad as 96,485 Joules of electrical work as per wikipedia?

WGTR - 11-12-2013 at 16:05

No, you had it right before. It was my mistake. 1 faraday = 96,485 J per volt gram equivalent.

1 farad = 2 J per volt^2 = 1 coulomb per volt.

[Edited on 12-12-2013 by WGTR]

Metacelsus - 11-12-2013 at 16:12

A farad is a unit of capacitance. You are thinking of Faraday's Constant, the total charge on a mole of electrons (96485 Coulombs/mole).

Electra - 11-12-2013 at 16:43

These units are so confusing. Chemistry is so much easier to understand.

Cheddite Cheese,

So does 1 Coulombs/mole = 1 joule? If not then I do not see any definable way to measure Faradays passed in a given time period based off wattage(Joules/Sec). I haven't had much luck finding the answer on google either.

Metacelsus - 11-12-2013 at 18:45

No. One joule, in this case, is one coulomb times one volt.

To calculate coulombs passed, multiply current (in amperes) by time (in seconds). If you need to calculate current from power and voltage, divide power by voltage.

Luckily, I have interests in both chemistry and electronics.

Electra - 11-12-2013 at 19:18

So the amount of Faradays passed in a given time period has nothing to do with voltage at all, but is entirely dependent upon current and nothing else? If 1 Faraday is just 96,485 coulombs then how does this relate to Joules/Sec. Are you implying that more coulumbs cannot be passed in a given time period by increasing the voltage, but only by increasing the current? This is all assuming the system is drawing the amount of watts that are being outputted.

Metacelsus - 11-12-2013 at 19:30

Increasing the voltage will also increase the current, and thus the power. However, different voltages may cause different reactions to take place. If you want to increase the current without increasing the voltage, then you must use a different electrode configuration, such as having electrodes closer together or using ones with a higher surface area.

Quote: Originally posted by WGTR

An electrochemical cell approximates a resistor (that is, a very non-linear one). Think Ohm's Law for this. An increase
in voltage causes an increase in current for a given resistance. All other things being equal, it's impossible to increase the
voltage across a given cell without increasing its current.

Ohm's Law: Voltage = Resistance times Current

The "resistance" of an electrochemical cell varies with voltage.

Magpie - 11-12-2013 at 19:34

Quote: Originally posted by Electra

Industrial settings are going to work with a lot more than 24g at a time, and 24/g per 6 hours isn't realistic. Assuming proper cooling, what variables would a company need to increase in their system to pull off 250, 500g, 1000g, per hour?

I think you are all making this way more complicated than it needs to be.

Why stop at 1000g/hr. Aluminum reduction plants no doubt put out tons/hr. It's just a matter of making a larger facility, ie, a facility with not only larger cells, but of many, many cells. So an aluminum plant may only have a voltage requirement of a few volts DC but have a huge power requirement and huge current requirement.

As I understand it there is an optimum voltage for each given electrolytic reaction. This is usually a small direct current (DC) voltage like around 5 volts, but varies with the particular circumstances. Electrode current density (amps/square inch of electrode surface) may also have optimum values. Everything else: power (watts), current (amps), size of the equipment, etc will just scale up with the desired production rate.

Relevant equations: P=IV. V=IR. P=V^2/R. R would be the cell resistance in ohms, P = power in watts, I = current in amps, and V = voltage in volts.

A Faraday is just a convenient quantity of electrons as it will be the number needed to reduce a mole equivalent of metal cations of the form M+. Reducing a mole of M++ cations will take 2 Faradays, etc.

example:

Al+++ + 3e- -----> Al

Three Faradays of e- required per mole of aluminum.

[Edited on 12-12-2013 by Magpie]

[Edited on 12-12-2013 by Magpie]

Artemus Gordon - 13-12-2013 at 18:12

Quote:

I'm still a noob at chemistry, but I have been an electronics engineer for over 30 years. I'm pretty sure an e-cell would behave like a battery in reverse.

A battery has an inherent voltage, and it will try it's best to always deliver that voltage. So a 1.5v battery will increase or decrease the current based on the demands of the circuit it is connected to in order to maintain that 1.5 volts.

An e-cell to electrolyze water, for example, needs 1.23v to drive the reaction (if I remember correctly). That doesn't change, no matter how many moles per second are being reacted. If you had a 10v power supply connected to this e-cell and a resistor in series, the resistor would drop the remaining 8.77 volts. Ohm's Law tells you that the current through a resistor = V/R, so you can control how much current flows by changing the resistance. An 8.77 ohm resistor would set the current at 1 amp (1 coulomb/second) and a 4.385 ohm resistor would set it to 2 coulombs/sec.

All this power (8.77 volts times x amps) flowing through the resistor gets turned into heat, so that is a big source of inefficiency. One way to improve this is to have several e-cells in series. Using my example, that 10 volt supply could drive up to 8 cells in series and only 0.16 volts needs to be dropped by the resistor. (I am assuming zero resistance in the rest of the circuit, which of course is not realistic.)

Also Electra, you should know that there are things called transformers that can "trade" voltage for current or vice-versa. If you used USA-standard electricity at about 120 volts which can typically handle up to 15 amps, a transformer is capable of dividing down that 120 volts to, for example, 12 volts with an equivalent multiplication of the current. In this case, if you drew 50 amps out of the transformer, it would only load your wall socket by 5 amps.

Artemus Gordon - 13-12-2013 at 19:15

Regarding my previous post, I was assuming that the e-cells have been designed to work with a particular current level. One would need to know how many moles/sec each cell was expected to deliver and choose electrode areas and spacing between the electrodes to work at the current level needed for that. So the things said by WGTR and the others are not wrong. But assuming that the cells were properly designed and constructed to handle, for example 10 amps, I expect that the power supply would then be chosen to deliver maybe 70% of that to the cells. And yes, there will be inefficiencies in the cells that would require a voltage somewhat higher than the theoretical value one would find in a chemistry manual, but in a well-designed system, I would think that the variance should be within a small fraction of a volt of that theoretical in each cell.

AJKOER - 14-12-2013 at 05:57

Artemus G.:

If one had a battery cell situation with say a copper electrode and replaced it with a colloidal suspension of Cu (that is, an exponential increase in surface area). What would happen?

My guess: an increase in heat with a much shorter battery life.

[Edited on 14-12-2013 by AJKOER]

Artemus Gordon - 15-12-2013 at 16:26

AJKOER
I'm not quite sure what you are asking here. If it is the case that the SA of the copper electrode is the limiting factor of whatever reactions are taking place, then yes, the reaction would proceed at a higher rate with a bigger SA. I believe that in commercial batteries it is usually the rate that ions can diffuse through the membrane that separates the two half-cells that is the limiting factor in how much current the battery can deliver. So if the battery was short-circuited, the current would rise until it hit the limit, whether that limit was due to electrode SA, or ion diffusion, or something else, and yes, a shorted battery can get pretty hot sometimes.

But my point in my second post is that the circuit that is designed to use this battery must not demand that much current. The current drawn would be ensured to be at a level the battery could easily deliver (or else the designer would call for a higher capacity battery). So in this case, by limiting the amount of current that flows out of one electrode and into the other (I'm just going to avoid the whole positive current flow vs. negative current flow conventions to minimize confusion), the load circuit becomes the factor limiting the reaction rate inside the battery.

Connecting an electrolysis cell directly to a constant-voltage supply (which is what a battery is, and what most power supplies are) is equivalent to shorting out a battery. In my example, there is an 8.77v difference between what the supply is designed to provide and what the reaction requires to drive it. That voltage WILL be dropped somewhere. If there isn't a load resistor or other current-limiting circuitry, then the voltage will be dropped across all the (very small) parasitic resistances of the power supply, the e-cell, and the wires connecting them, and the current will rise to some unpredictable level. At that point, you probably have no idea what will happen. If most of that resistance is in the wires or the supply, you might burn out the circuit. If most of the resistance is in the electrodes, they will get hot but the solution they are in will sink the heat and you might be OK, but it would be a waste of power which would be unacceptable in an industrial-scale plant.

Basically what I am saying is that there needs to be a mechanism to limit the current that is sent to the e-cell, and provided the e-cell is designed to work at that current level, you won't have unexpected voltages, heat, or explosions. You could then quintuple the size of the electrode and it would make no difference because the electrode is not the factor limiting the reaction rate.

[Edited on 16-12-2013 by Artemus Gordon]

Magpie - 15-12-2013 at 20:28

If I was running an aluminum plant I would absolutely minimize the use of any resistor in the cell circuit. That just produces waste heat. I would tune the voltage to the cell requirement using a transformer.

What you might do on a lab scale for convenience is totally different compared to what an aluminum plant would do where the object is to make money.

Artemus Gordon - 15-12-2013 at 21:15

Quote: Originally posted by Magpie

If I was running an aluminum plant I would absolutely minimize the use of any resistor in the cell circuit. That just produces waste heat. I would tune the voltage to the cell requirement using a transformer.

What you might do on a lab scale for convenience is totally different compared to what an aluminum plant would do where the object is to make money.

Yes, of course. I was trying to keep things simple for Electra, although I did point out that several e-cells could be run in series to mop up excess voltage, and I did mention transformers.