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Help with Nerst Plank Equations
Posted Aug 3, 2015, 8:19 p.m. EDT 2 Replies
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Hi,
I have modelled a simple nerst plank equation with copper electrodes and copper sulphate.
The equations on both electrodes is copper dissociating and plating (modelled as flux).
The problem is that I get zero voltage, or almost zero.
Is there something in my model that is completely removing/ nullifying the electric field.
Thanks
I have modelled a simple nerst plank equation with copper electrodes and copper sulphate.
The equations on both electrodes is copper dissociating and plating (modelled as flux).
The problem is that I get zero voltage, or almost zero.
Is there something in my model that is completely removing/ nullifying the electric field.
Thanks
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2 Replies Last Post Aug 4, 2015, 9:58 a.m. EDT
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Posted:
9 years ago
Hi
There are several issues that you should take into account:
1. Your system is binary which means that it can be solved in closed form.
2. Potential is formed by two factors: ohmic drop and diffusion potential. Since you had the same diffusion coefficient for both Cu and SO4 there is no diffusion potential. I changed that and I got very nice potential profiles.
3. Your current was over the limiting current, making the surface concentration negative on the deposit side. That, in turn, makes the calculation of the potential crazy because you have to take logarithm of the concentration. I found that the limiting current is around 175 A/m².
4. The charge numbers in your simulation were 100,000 for Cu and -100,000 for SO4; I changed them to ±2.
5. You cannot impose more than one BC. Since you have the flux condition on both boundaries, nothing can be said about the potential. The potential you see in the plot is the Galvanic potential, electrode potentials are defined through the Nernst equations (or via kinetic equations).
Attached my solution, Comsol version 5.1.
There are several issues that you should take into account:
1. Your system is binary which means that it can be solved in closed form.
2. Potential is formed by two factors: ohmic drop and diffusion potential. Since you had the same diffusion coefficient for both Cu and SO4 there is no diffusion potential. I changed that and I got very nice potential profiles.
3. Your current was over the limiting current, making the surface concentration negative on the deposit side. That, in turn, makes the calculation of the potential crazy because you have to take logarithm of the concentration. I found that the limiting current is around 175 A/m².
4. The charge numbers in your simulation were 100,000 for Cu and -100,000 for SO4; I changed them to ±2.
5. You cannot impose more than one BC. Since you have the flux condition on both boundaries, nothing can be said about the potential. The potential you see in the plot is the Galvanic potential, electrode potentials are defined through the Nernst equations (or via kinetic equations).
Attached my solution, Comsol version 5.1.
Attachments:
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Posted:
9 years ago
Lasse,
Thanks a lot for the instant help.
Thanks, I thought that I had to define potential on both sides, maybe I was overdefining it.
Also actually I had changed copper and sulfate concentration back (to z_cu), but I guess I could not save it.
Thanks a lot for the instant help.
Thanks, I thought that I had to define potential on both sides, maybe I was overdefining it.
Also actually I had changed copper and sulfate concentration back (to z_cu), but I guess I could not save it.