Author Topic: Calculating minimum cell voltage  (Read 1311 times)

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Offline kephra

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Calculating minimum cell voltage
« on: March 20, 2022, 10:20:34 AM »
When I started experimenting with production of colloidal silver, I tested various cell voltages, knowing that it must be greater than 3.5 which is the minimum voltage  required to reduce sodium and oxidize silver.  I did not see any increase in quality once I exceeded 10 volts, so I used 10 volts as a minimum voltage.

In a previous post, I stated that if you wanted to use a wider electrode spacing than 1.5 inches, you should increase the minimum cell voltage using this formula:
NewVoltage = 3.5 + 6.5 * NewSpacing/1.5
So for a 3 inch spacing, the minimum voltage would be 3.5+6.5*2 = 16.5 volts.
So why is it not 20 volts, when the spacing is now doubled?

The answer is that there are actually 4 electrodes in the cell, two physical, and two pseudo electrodes which are the bounday layers between the bulk water and the electrodes.  These pseudo electrodes are only a few nanometers ithick, and always produce the same voltage drop provided enough voltage is present.  This voltage drop at the silver anode through the boundary layer is 0.8 volts, and at the cathode is always 2.7 volts provided there is more than 2.7 + .8 volts available.  So changing the cell voltage only changes the voltage drop between the pseudo electrodes, and not the physical electrodes.  So the total cell voltage which you can measure with a voltmeter is 0.8 +6.5 +2.7 volts if you have 10 volts total.
The 6.5 volts in the bulk water is what moves the ions to and from the boundary layers where they interact with the physical electrodes. 

To make the subject even more complicated, the boundary layers are actually composed of 5 sublayers, but that is irrelevant to this discussion.

Spacing     Min Volts
----------------------------
1.5 inch 10 volts
2  inch  12.2 volts
2.5  inch  14.3 volts
3 inch  16.5 volts
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