I recently tried an experiment: I used a potentiostat to charge my electrochemical cell (copper electrodes in 10mM saline solution) by applying a constant bias voltage (100mV). I could clearly see a charging curve (see attachment). Before the cell was fully charged, I stopped the measurement and disconnected the cell from potentiostate. 1s later, I resumed the measurement, but found the discharging curve started with a burst and could match the previous one, indicating the double layer capacitor somehow self-discharged. My intuition is that a capacitor should maintain its charge when it is disconnected from circuit. Does any one have an explanation for my experiment?
Thanks,
Dan
The nanoscale[1] corrosion[2,3] behavior of Copper in a dilute NaCl solution is the main reason for this activity. So, your electrochemical cell is far from an ideal the double layer capacitor model, the way you might assume.
Also, by applying a constant bias voltage (100mV), you might make your own evaluation; you might measure a small (but a non zero steady state) current (in a nA range). If you could use a Cu-mesh[4], then the above current (at the same bias, 100mV) will heap, so, to be more easy to measure it, in (a sub-)uA range.
1. Nanoscale corrosion behavior of polycrystalline copper fine wires in dilute NaCl solution investigated by in-situ atomic force microscopy Article Nanoscale corrosion behavior of polycrystalline copper fine ...
2. Influence of chloride on the long-term interaction of copper with deoxygenated neutral aqueous solutions Article Influence of chloride on the long-term interaction of copper...
3. Corrosion performance of copper grounding material
Article Corrosion performance of copper grounding material
4. Copper Mesh https://www.americanelements.com/copper-mesh-7440-50-8
Dear Dan,
Thanks for asking this very good question. I have always thought about this and come to the following possible reasons.
1. The double layer formation is the totally dependant on the surface charge density. For the metallic surfaces (under no applied potential), when exposed to the some electrolytes, depending on the inner potential (energy required to bring an electron from infinity (arbitrary point) to the surface) and the ionic strength of the electrolyte these develop double layer in nanoseconds. The compromise potential is then called OCP (open circuit potential) include the potential drop in the inner, outer, diffuse and diffusion layer across the surface.
2. Under applied potential (+100 mV vs OCP), you are forcing the electrons to come out of the sample and in result expanding the space charge region at the surface hence attracting more anions to migrate toward the surface. Once you remove the electric field (particularly in metals) the surface will regain its original state. If you go beyond specific potential limits the applied electric field may overcome the surface energy and you will get the faradaic response.
3. In case of semiconductors or on carbonaceous materials there is permanent charge (n-type, p-type) or in the form of surface functional groups, respectively. Once you change the potential (either forward or reverse biased you will get the difference in current response indicating the faradaic activity. Similarly, adsorption/desorption of ionic species by interacting with the surface functional groups also influence the surface charge density.
Hi Kashif,
Thanks for you answer!
Could you clarify the second point: " Once you remove the electric field (particularly in metals) the surface will regain its original state " ?
I think I understand that by applying a bais potential, the electrons are forced to move from working electrode (WE) to counter electrode (CE) or accumulate on WE, creating net charges on WE and thus double layer. My intuition is that if we stop the bias potential somehow, say turning off the switch connecting WE to the potentiostat instantly fast (typical ~ ns), the charge may remain on electrode surface. Perhaps it is somehow related to circuitry of potentiostat
Dan
Dear Dan,
Thank you for the question for further elaboration of words. Here the 'original state' means the reversible behavior of surface during charging and discharging. In metals once you apply potential, the electrons move against the applied electric field, leading to increase in the potential gradient across the interface (electrode/electrolyte). If you don't cross the atomic binding energy limits (no non-capacitive, faradaic reaction), which is different for every metal and switch off the connection (stop applying potential), ideally the metal should reestablish its original state (macroscopically speaking, no change in OCP and exchange current). From the current transient profile you may get the surface capacity (charge). This capacity includes charge in the double layer and/or the charge involved in specific adsorption of ions at the surface. The specific adsorption of ions depends on the crystallographic structure and density of defects e.g, kinks, jogs, dislocations etc. on the surface.
You can clearly see the decrease in surface capacity in three successive cycles (from your attached figure) which presumably is associated with the surface coverage due to specific adsorption of ionic species.
The 'charge remain on the surface' means depending on the surface capacity (accounting the surface coverage) after potential excursion, the surface will try to regain its original state. But any difference (in terms of current and potential under no electric field before and after potential excursion) is totally a function of surface structure and its interaction with the ionic species.
The circuitry of the PS can also have minor influence, but i think this is very negligible because the very high internal resistance during open circuit mode would not let the charge to dissipate.
I hope i answered the question, if you still have any question please don't hesitate to contact me. You may also communicate through my email as well. I wish to learn more from the discussion. Take care
For you Au, look for the 'contribution of specific adsorption' on the surface as i mentioned in my last comment. Thanks
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