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Old Sep2-07, 12:27 AM                  #1
chanller

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Exclamation Electrolysis of water 2

Talking about electrolysis, what happens to the excess power input when gas production rate remains constant even though power is increased? Droplets of water settled at the top of my gas storage cylinder, does this mean excess power is converted into heat which vaporises water in the electrolyte?
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Old Sep2-07, 05:58 PM                  #2
Cesium

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Power = current X voltage

There is a minimum amount of voltage that you'll need to perform electrolysis. Any extra voltage will decrease efficiency and this extra energy will escape as heat which can dramatically increase the temperature of the solution.

Also, water evaporates.

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Old Sep3-07, 12:21 AM                  #3
JGM_14

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Talking

How is this all set up?
When the gas bubbles pop some of the water evaporates from the travel through the gas, not only that but water can vaporize well below freezing, e.g. ice sublimes.
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Old Sep3-07, 12:54 AM       Last edited by chanller; Sep3-07 at 12:56 AM..            #4
chanller

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Exclamation

Cesium:
So you mean that I cannot just connect a battery to the electrolyser, I need to know the 'maximum' (well I think it can take any voltage lower than its maximum but beyond this maximum it starts producing heat. Just like your mobile phone which needs a charger and cannot be connected directly to a supply of 240V and also a battery with emf lower than that of the battery of the phone should, I guess, be able to run the phone though it discharges more rapidly!)

But does anyone know how I can find the maximum voltage of my home-made electrolyser, well assuming that everything written above is true!

JGM_14 : I hope that what i have written above answers your question!
Evaporation takes place slowly but what I found is that at room temperature droplets of water fills the inner surface of the gas storage cylinder at the cathode! And I mean only at the cathode.
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Old Sep3-07, 02:24 AM                  #5
Gokul43201
 
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I believe the answer to your electrolysis questions is that before you actually go about performing an experiment, you should make yourself thouroughly familiar with the theory - that means completing the chapters on electrochemistry in a text like Atkins.

You can't just go about hooking up an arbitrarily chosen battery to an electrolytic cell and hope for the desired results. You need to at least know the electrode potentials and the overvoltage for the given conditions.
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Old Sep3-07, 05:23 AM                  #6
chanller

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If i use 1 molar sulphuric acid at rtp what voltage do you think i should use? 1.5V will be good?

Gokul43201 : I have not yet reached that chapter at school, and I desperately need to finish a project before a given time, that's why i need help!!!!!!
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Old Sep3-07, 12:30 PM                  #7
Cesium

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Perhaps someone else can it explain it better than I can:

It doesn't take a particularly high voltage to split water. There is a
threshhold voltage necessary to split water at all. Above that voltage,
the splitting will go faster. But voltage isn't all of the story. If you
have a sufficient voltage, the current determines the amount of water that
can be split. The amount of power (energy per time) that goes into the
splitting of water is the product of the votage and current. Efficiency
will be the amount of water split per amount of energy used; for maximum
efficiency, you will want to use as low a voltage as you can, and a high
current. If you use a higher voltage, the extra energy escapes as heat.
taken from http://www.newton.dep.anl.gov/askasc.../chem99440.htm

1.5V will be enough to split water; what is your goal with this project? I'd probably use a bit more, say maybe 5V, to overcome the resistance of the cell. Remember it is the amount of current that determines how much water will be electrolyzed. So if you want more hydrogen/oxygen gas, the more current the better. Computer power supplies are a good and cheap (free) source of low voltage & high current electricity.

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Old Sep3-07, 02:40 PM                  #8
chanller

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Cesium:
If the maximum number of gas produced per second depends on the size of the electrodes then I assume the larger the electrode size, the lower the resistance and hence with a battery of emf equal to the voltage you adviced, the current should be maximum and hence the amount of gas produced depends on the size of the electrodes!

Well i am trying to find the efficiency of a fuel cell! Power input to produce gas over power output.
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Old Sep3-07, 03:27 PM                  #9
mrjeffy321

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The rate at which water is electrolysized into Hydrogen and Oxygen gas depends on the rate at which charge flows into the electrolytic cell; in other words, the current controls the rate of gas production.
The current through the cell depends on the voltage applied across the cell and the resistance of the cell. The effective resistance of the cell depends on several factors including the electrode size and position, electrolyte concentration, and others.

There is a minimum voltage which one must use, or exceed, in order to electrolytically break water into H2 and O2 due to the nature of the electrochemical reactions occurring in the cell. One can calculate this minimum voltage if one was so inclined by working it out from the electrochemistry going on, but it is not very high. If you want large gas production rates you will probably want to use a higher voltage than the minimum so as to increase the current, but with this ‘over-voltage’ there will be increased heat loss and efficiency will drop.

What type of fuel cell are you referring to? There are many different types, each with their own efficiencies. I think for a Proton Exchange Membrane (PEM) fuel cell, 50% efficiency is not an unreasonable estimate.
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Old Sep3-07, 07:26 PM       Last edited by chanller; Sep4-07 at 01:34 AM..            #10
chanller

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It's a PEM fuel cell, 30% platinum carbon electrode, nafion as membrane. and i have been given a voltage current density curve for gas flows of oxygen 0.06SLPM 1.2 bar pressure and hydrogen 0.04 SLPM, 1.0 bar pressure which shows max. current at voltage 0.93V (which is the max. voltage), 0.5A. From the voltage current density curve I find that as voltage is increasing from zero, current decreases and power increases until the maximum is reached. Hence if i vary the gas flows with constant ratio of o2 : h2 and gas pressures and i use max. voltage, gas flows should be directly proportional to current, and hence power, at that voltage. Thus an efficiency can be found, which depends on the constant gas pressures and gas ratio used.

Without increasing voltage i can vary the electrode size to vary power, current or gas production (flow rates of hydrogen and oxygen with same ratio) in the electrolyser!
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Old Sep5-07, 11:22 AM                  #11
AbedeuS

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Just for me to clarify too...as many people say differant things on the subject, I'm right to assume that voltage in an electrolyser just determines the enegy an electron has between two points, so when your electrons have sufficient energy to break the bonds in water your main bottleneck is getting enough electrons with said energy through the solution to produce the gas you want, this is the current, and any excess energy the electrons has is expended as thermal energy yeah?
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Old Sep5-07, 02:24 PM                  #12
chanller

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Yeah.
Well, can anyone tell me if it is right to say that maximum power of a hydrogen oxygen fuel cell varies directly (with constant efficiency) with gas flow rate at constant pressure at room temperature
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Old Sep5-07, 03:12 PM                  #13
AbedeuS

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The maximum power or the maximum voltage? The only limit on power is how much water there is to electrolyse and the rate at which the resultant gases can be pumped off to ensure there isnt some crazy high pressure zone of elemental gases. As for maximum voltage, that should remain roughly constant at constant temperature? Although I'm not 100% sure on that.
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Old Sep6-07, 10:37 PM                  #14
JGM_14

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Originally Posted by chanller View Post
Cesium:
So you mean that I cannot just connect a battery to the electrolyser, I need to know the 'maximum' (well I think it can take any voltage lower than its maximum but beyond this maximum it starts producing heat. Just like your mobile phone which needs a charger and cannot be connected directly to a supply of 240V and also a battery with emf lower than that of the battery of the phone should, I guess, be able to run the phone though it discharges more rapidly!)

But does anyone know how I can find the maximum voltage of my home-made electrolyser, well assuming that everything written above is true!

JGM_14 : I hope that what i have written above answers your question!
Evaporation takes place slowly but what I found is that at room temperature droplets of water fills the inner surface of the gas storage cylinder at the cathode! And I mean only at the cathode.
You can just connect it to the battery if the voltage is higher than that of the minimum required to spilt water, but if you have low amperage there will not be much gas produced.
I made an electrolytic cell once that the resistance of the wire itself generated enough heat to melt the insulation, and it was copper wire, but i was electrolysing a saturated salt (sodium chloride) solution. Have you considered running the gases through a dessicator?

Some basic info.
http://www.crscientific.com/electrolysis.html
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