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Since conductivity is the reciprocal of resistivity, it can be used to calculate resistance in the following equation:

$$R=\frac{\rho l}{A}$$

$$R=\frac{l}{\kappa A}$$

Could this, along with Ohm's Law be used to predict the current in an electrolytic cell at a given voltage? My initial calculations with conductivities of around $20000\mu S/cm$ and a voltage of 10V yielded me an abnormally high resistance and current less than 1 amp. My actual experiment with 10V had a current of about 2.5A, which means the resistance of the solution is about 4, which doesn't seem right?

So I wanted to ask if this just doesn't work for electrolytic cells or am I doing something wrong?

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    $\begingroup$ An electrolytic cell in DC scenario is in no means similar to a classical electronic resistor. While its conductivity does have influence of the passing current, it is far from a simple function. A model circuit simulating such a cell would be quite complicated. The most simple one is a power source in a serie with a resistance. (Both are current and current history dependent.) $\endgroup$
    – Poutnik
    Commented Feb 17, 2022 at 12:12
  • $\begingroup$ BTW, $20000\ \mu \pu{S}$ is not conductivity but conductance. $\endgroup$
    – Poutnik
    Commented Feb 17, 2022 at 12:38

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Sort of.

First it is important to mention that electrolysis uses up some voltage, regardless of current. So if you use 10 V and electrolysis of your reaction requires 2 V, then only 8 V will be available for the resistance.

Second is that your measuring device is affected by it too, and those usually use low voltage, where this effect is especially noticeable.

To measure the resistance of your electrolysis cell, you are better off using two measurments. One with a high voltage, low current, and a large value resistor. This way voltage drop in the electrolysis cell will be seen alone. And another one with high current, low voltage. This together with the previous result will help to get the resistance, when voltage drop from the previos result is subtracted from the available voltage in the second case.

And at higher current you may encounter gas bubbles that increase resistivity, and at higher still - plasma bubbles, that will decrease resistivity.

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  • $\begingroup$ Thank you for the answer, that cleared up a lot, but just another question on how electrolysis uses up current, I just wanted to make sure whether or not I also have the subtract the voltage required for unwanted reactions during electrolysis from the applied voltage? $\endgroup$
    – Alan Liu
    Commented Feb 18, 2022 at 3:14
  • $\begingroup$ @AlanLiu amount of electrolysis reaction is directly proportional to current * time. All of the added voltage above the treshold is wasted on speeding up the reaction. If you want to predict how current will change as you change voltage, yes, subtract reaction voltage from the supply voltage. $\endgroup$ Commented Feb 18, 2022 at 3:28

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