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electrolsys basics

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Dr_Doggy

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so I got some salt water, when i apply it to a supply i get current based on volts and resistance, first for simplicity lets assume that the salt to water ratio won't change throughout the process.

As i understand there is a minimum of 2v(or so) required to start the electrolysis process, so when at the ideal voltage i can calculate the current based on Vapplied /Resistance of load = Icurrent at most effective output.
but when i go under that voltage the load resistance doesnt change so I can still calculate amps.
however what if i go over that voltage, naturally current increases, output increases, but efficiency drops due to heat losses(i think)

so my questions are:
when im under ideal voltage is the current i measure on ammeter the heat loss?
when im at ideal can i measure: current on ammeter = ((heat lost @(ideal -0.01degree) ) + output electrolysis)?
will the reaction reflect in the current of the circuit?

I guess I am wondering if there is a way to identify the ideal voltage or efficiency of my reaction (since it may change with shape/volume of container) without measuring the output gasses, and instead only measuring using the current and maybe temperature (or similar)?

Maybe im confused: when i measure current am i measuring the energy that got converted to heat and electrolysis output?
or
is there a way to control a lm317 or similar to ensure that Im always at ideal reaction voltage regardless of my volume & surface area?
 

Dr_Doggy

Well-Known Member
thanks! I am aware of that, more over im trying to understand whats happening behind the reaction, I even went as far as the chemistry forums where I learned about enthalpy, entropy and "Gibbs FREE(!) Energy". but there is still something somewhere i'm not getting... and i think it lies in these 3 questions:

when im under ideal voltage is the current i measure on ammeter the heat loss?
when im at ideal can i measure: current on ammeter = ((heat lost when lower than ideal ) + output electrolysis)?
will the reaction reflect in the current of the circuit?
 

Dr_Doggy

Well-Known Member
also 1.23v is not much output, what happens when i want to turn it up, is there a way to maintain the ideal efficiency, maybe by pulsing, ?

again I may be getting wires crossed with everything but all this talk of "various activation barriers and energy shells" makes me think of the tea laser and how the electronics are used to control/adjust for lots of corona(light output) and less thick full arcs(dielectric breakdown & other losses)
 

Dr_Doggy

Well-Known Member
another interesting thing is that NaCl min volts is 2.7, so i think that means if I stay under 2.7v there should be no chlorine gas produced(?)
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
I'll bet the electrode type and area have to matter. When "played" with electrolysis of water, I did use platinum screens. The inverted test tube in water to make H2 and O2. You can look at the reaction product volume as its being produced.
 
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