Suppose a supercapacitor electrode is being tested within the potential range of 0-0.4 V in both three electrode cell, two electrode cell and in a symmetric supercapacitor assembly.We know that a three electrode system always overestimates the value. Is there any relation between the two measurement, provided the mass of a single electrode and the working potential is same.
Again suppose two material A and B. A can be operated within the potential range of -0.5V to +0.2V, whereas, B can be operated within the potential range of 0V- 1 V. If we fabricate an asymmetric supercapacitor,would their asymmetric assembly be able to give a cell voltage of 1.5V? and if A is symmetrically assembled, what should be the working potential and cell voltage?
In your case, assuming the equipotential is zero, your negative electrode will have a working potential window of 0.5 V (from 0 to -0.5V) and your positive electrode a window of 1.0 V (from 0 to 1.0 V). This means that your negative electrode should have a capacitance value twice of that of the positive electrode because the amount of charge stored is equal in both electrodes, i.e. Q = C+*E+ = C-*E- where C+ and C- are the capacitance values, and E+ and E- the potential windows of the positive and negative electrodes, respectively.
First of all, the measurements in three electrode (this is also some time known as single electrode measurement) and two electrode cells should give you very comparable results, if not identical. I am not sure where your impression "a three electrode cell always overestimates the value" comes from, but it is wrong.
In terms of expressing the results from different measurements in a comparable manner, the specific capacitance of the electrode material is often derived and reported. Using the three electrode cell, you can obtain the specific capacitance directly. In the symmetrical design, the two electrode cell capacitance as measured can be converted to the specific capacitance of the electrode material by multiply the cell capacitance either (1) by 2 and then divided by the total mass of the materials on the two electrode, or (2) by 4 and then divided by the mass of the single electrode.
If you obtain very different values of specific capacitance from the two methods, it means one or both of your cells were not properly assembled. In such a case, you need to read and learn more about supercapacitors.
Please note that you CANNOT use an asymmetrical two electrode cell to measure the specific capacitance of either of the two electrode materials. In this case, the specific capacitance of each electrode material has to be measured by the three electrode cell.
Regarding your question on potential window, if your positive electrode has a window from 0 to +1 V and your negative electrode from -0.5 to +0.2 V, you can indeed have a cell voltage up to 1.5 V. However, if you may or may not achieve this maximum cell voltage depends strongly on the capacitance ratio of the two electrodes, and also the so called equipotential which is the potential of both the positive and negative electrodes when the cell voltage reached zero during discharging. One suggestion from me is that you should never simply assume that the positive and negative electrodes must have the same capacitance value. You can read my following papers for more details.
Electrochim. Acta, 86 (2012) 248-254.
http://dx.doi.org/10.1016/j.electacta.2012.07.033
Energy Environ. Sci., 3(10) (2010) 1499 - 1502.
http://dx.doi.org/10.1039/c0EE00228c
In your case, assuming the equipotential is zero, your negative electrode will have a working potential window of 0.5 V (from 0 to -0.5V) and your positive electrode a window of 1.0 V (from 0 to 1.0 V). This means that your negative electrode should have a capacitance value twice of that of the positive electrode because the amount of charge stored is equal in both electrodes, i.e. Q = C+*E+ = C-*E- where C+ and C- are the capacitance values, and E+ and E- the potential windows of the positive and negative electrodes, respectively.
Prof. Chen, your information is very much useful to me.
Further I want to ask:
'Suppose an electrode can be operated within the potential range of -0.5V to +0.2V, when tested in a three electrode cell. I want to fabricate a symmetric supercapacitor using the same electrode'.
1. What should be the working potential and what would be the cell voltage?
2. During specific capacitance calculation which potential value (V) to be considered?
3. What would be the case when the working potential of the electrode material is within -0.5V to 0V (tested in a three electrode cell)?
1. For the symmetrical cell, the voltage window is 0.7 V at maximum assuming the mid potential (-0.15 V) between -0.5 and 0.2 V is the equipotential.
2. In an ideal case (e.g. very rectangular CVs), any potential or potential range between -0.5 and 0.2 V should give you the same capacitance. In practice, you should use the full potential range explored (instead of individual potentials) to derive the capacitance
3. As mentioned above, the potential range does not matter if the material behaves closely to that of a capacitor (e.g. a very rectangular CV). However, if your CV shows larger (or smaller) current between -0.5 and 0 V, then the capacitance derived would be also larger (or smaller).
Suppose the working potential of an electrode is -0.5 V to 0 V.
If your potential ranges for both the positive and negative electrodes are referenced to the same reference electrode (or potential), my answer is yes, i.e. these two types supercapacitors belong to asymmetrical supercapacitor, and the maximum cell voltage will be 2 V, as long as you have balanced the electrode capacitance accordingly.
The following is the link to our recent paper on this topic and the paper is Open Access.
Cell voltage versus electrode potential range in aqueous supercapacitors,
Sci. Rep., 5 (2015) Article No. 9854.
http://dx.doi.org/10.1038/srep09854
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