Physics Help Forum Voltage Drop Across Electrodes

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 Feb 16th 2016, 08:50 PM #1 Junior Member   Join Date: Feb 2016 Posts: 2 Voltage Drop Across Electrodes Hello, I am a chemist running an electrochemical cell, and I am hoping for some pointers on how to calculate the voltage drop across electrodes as a function of position. The simplest embodiment of the cell consists of a rectangular sheet electrode surrounded by a second wire electrode surrounding the frame of the rectangular electrode. The two electrodes are immersed in a water/salt solution with a conductivity I can measure. I also know the sheet resistance of the electrodes used. Let's say I apply 0.5 V between the two electrodes. I know the voltage drop at the middle of the rectangular electrode should be less than that closer to the perimeter due to the IR drop of the solution, but I am unclear as to how to quantify this given known dimensions and conductivities of the electrodes and solution involved. Do I need to apply Gauss's law? Would this be an easy problem to solve analytically or is this something that would get crazy quickly? Any starting points for me would be much appreciated. I haven't taken physics since freshman year of college so it's been a while for me. Even terms for me to look up for further reading would be of use. Thanks in advance. Sincerely, Cesium
 Feb 17th 2016, 09:36 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 The answer to your question as posed is really quite simple: the voltage drop across the electrodes is equal to the voltage you apply to those electrodes. However, I'm guessing that what you really meant to ask is this: how to calculate the voltage potential in the solution as a function of position - is that right? Yes, you can apply Gauss's Law. Let's set up a coordinate system centered on the center if the inner electrode, let's let 'a' = the distance from the origin to the outside of the inner electrode, and 'b' = the distance from the origin to the inside of the outer electrode. The distance 'x' is measured from the origin, and can range between a and b. From Gauss: E(8Lx) = Q/epsilon, where Q is the total charge on the inner electrode. V(x) = integral from a to x of E(x) dx, which is: V(x) = Q/(8L epsilon) ln(x/a). The total voltage drop is V_T = Q/(8L epsilon) ln(b/a). This is equal to the voltage you apply to the electrodes. So the voltage at any point a <=x<=b, as a function of the voltage applied to the electrodes is: V(x) = V_T ln(x/a)/ln(b/a) Last edited by ChipB; Feb 17th 2016 at 09:43 AM.
 Feb 17th 2016, 10:20 AM #3 Junior Member   Join Date: Feb 2016 Posts: 2 Hi Chip, Your reply is much appreciated. I am interested in the voltage drop in the solution near different points along the electrode surface. So, I'm actually interested in the area from the middle of the inner electrode out to its edge (0 <= x <= a). I'm thinking the middle of the inner electrode should experience a smaller "effective" voltage drop since some of that voltage drop will drop over the IR of the solution. I'm a little confused by your solution too, in that when x = a, Vx = 0. Sincerely, Cesium Edit: Is this as simple as the following? Eapplied = Eactual + I*Rsln and Rsln = l/(c*A) where R is the solution resistance, l is the distance between the inner electrode point of interest and the outer electrode, c is the conductivity of the solution, and A is the area of the electrode (the area up to the point of interest) Last edited by Cesium; Feb 17th 2016 at 10:29 AM.
 Feb 17th 2016, 10:47 AM #4 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 Have you heard of 'overvoltage' in relation to electrochemical cells? This is posted in high school physics. Are you seeking a high school answer? Electrochemistrycan be quite complicated. Last edited by studiot; Feb 17th 2016 at 11:59 AM.
Feb 17th 2016, 11:21 AM   #5
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Join Date: Jun 2010
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 Originally Posted by Cesium I am interested in the voltage drop in the solution near different points along the electrode surface. So, I'm actually interested in the area from the middle of the inner electrode out to its edge (0 <= x <= a). I'm thinking the middle of the inner electrode should experience a smaller "effective" voltage drop since some of that voltage drop will drop over the IR of the solution.
The voltage at every point on the inner electrode is the same, assuming we can ignore bulk resistance of the material that the electrode is made from. The fact that solution is in contact with it is immaterial. I have attached a figure that may help you see why this is - a voltage V_T is applied to electrode B and electrode A is hooked to ground. The resistance of the solution is simulated using resistors - the figure shows 5, but it could be simulated using thousands if you want. The voltage at points 1, 2, 3, 4, and 5 are all the same, and is equal to V_T, regardless of the fact that there is current flow through those nearby resistors.

 Originally Posted by Cesium I'm a little confused by your solution too, in that when x = a, Vx = 0.
That's correct. I didn't state it explicitly, but the potential at each point is measured relative to the inner electrode, so it ranges from 0 at x = a to V_t at x=b..

 Originally Posted by Cesium Edit: Is this as simple as the following? Eapplied = Eactual + I*Rsln and Rsln = l/(c*A) where R is the solution resistance, l is the distance between the inner electrode point of interest and the outer electrode, c is the conductivity of the solution, and A is the area of the electrode (the area up to the point of interest)
I'm not following what you are attempting here. What do you mean by "the area up to the point of interest?"
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Last edited by ChipB; Feb 17th 2016 at 12:43 PM.

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