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Ted Mooney, P.E. RET
Pine Beach, NJ
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for Metal Finishing since 1989
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Boundary condition on bipolar electrode
2003
Hi all,
At first excuse me for my English. There is anode and cathode in galvanic bath, but between them bipolar electrode. The question, what voltage will be on bipolar electrode. Interests both mathematical model, and given your experiment. In details. There is anode, cathode. In simple case between them on centre is located bipolar electrode. The bipolar electrode is not connected neither to the voltage source, nor to the current source. The bipolar electrode is a plate. It is located perpendicular anode and cathode, something type:
| --- |. On anode voltage is 5 V. I interest, how boundary conditions can be given for bipolar electrode, current-potential curve of the bipolar electrode.
- Tambov, Tambovskaya Province, Russia
Nobody can give you quantitative answers to a problem posed without quantitative input, Vlad. But, in general, the conductivity of metal is very high and the conductivity of the electrolyte relatively low such that the potential of the bipolar electrode will be proportional to it's position between the anode and cathode. If it's midway, it's potential will be 2.5V
The side facing the cathode will act as an anode, the side facing the anode will act as a cathode. It will be pretty much as if you had two 2.5V cells sitting next to each other.
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey
2003
2003
Thanks for answer Ted.
You have written that on bipolar electrode, will be a voltage
2.5V. But I have a question. In simple case between anode and cathode on centre is located bipolar electrode. It is located perpendicular anode and cathode, i.e. on direction to motion ions. For instance, the bipolar electrode's length is a half of the distance from anode before cathode. I expect that potential will be 3.75V. I right? Or the potential 2.5V
- Tambov, Russia
2.5 V, approx., Vlad. Assuming the resistance of the metal in the bipolar anode is much less than the resistance in the solution, then virtually the entire 5 V drop is consumed by the resistance of the solution. The potential at any point on the bipolar anode will be approximately 2.5 V whether that point is nearest the anode or nearest the cathode. Of course if the bipolar anode was extremely thin or made of a very high resistivity metal, then the current might cause a voltage drop through it, but for practical purposes the entire 5 V drop occurs across the solution.
V=IR. Find the resistance of the whole system and you will know the current. Knowing the current, multiply by the resistance of the segment in question and you'll have the voltage drop across that segment. Assuming the resistance of the bipolar anode is very low, the voltage drop across it will be very low, so it's potential anywhere along it's length will be almost the same.
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey
2003
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