The different masses between two electrodes has me stuck on the calculation of capacitance (in F/g) and setting current (in A/g).
If you can measure the cell capacitance, C-cell in F, and also know the total mass of the active materials on both the positive and negative electrodes, M-total (= M-positive + M-negative), the specific capacitance, C-s in F/g, of the cell is simply the ratio of Ccell / M-total.
Note that you must measure the C-cell using the supercapacitor cell, instead of trying to derive the C-cell value, or C-s value from the specific capacitance of each of the positive and negative electrodes.
Thank for your suggestion!
However, when my mate set current to run the test, the problem emerge── the unit of current density is A/g, and the mass are quite different between two electrodes(one is metal oxide and one contains carbon only).
Could you give us some suggestion?
It's incorrect to take means of two electrodes weight as the mass to calculate F/g and A/g, isn't it?
If you are to report the specific capacitance of the supercapacitor, i.e. the cell, you must include the total mass of both electrodes to normalise the capacitance of the cell. By the way, the current from any power supply or potentiostat can only be A in unit, not A/g. Also, A/g should never be called "current density" which refers to units such as A/cm^2.
If we meet some problems in the future, may I ask you for suggestion?
(I am undergraduate student.)
Prof Chen, to test complete cell, for example paper by T.Brousse in JES 2004, it mention that
"During charging between 0 and 2.2
V, the potential of the MnO2 electrode increased from 20.1 to 0.92
V vs. Ag/AgCl. At the same time, the potential of the negative elec-
trode decreased from 20.1 to 21.28 V vs. Ag/AgCl." Initially in CV scan, MnO2 was scanned in range of 0-1V and activated carbon from 0-(-1.2)V..What are the reasoning behind this behaviour.
Thanks.
Would you please check your cited text from Brousse's paper? It is unlikely that the potential had changed from 20.1 V to 0.92 V.
Prof Chen if we want to calculate asymmetric capacitance then as u said we first need to calculate C-cell of two electrode and then to divide it on total mass. my question is the formula of C-cell will be Idt/dv? i.e. excluding the total mass at this stage and further which formula actually will be followed to calculate current? what will be the mass value?
Mr. George, will you please help me to under stand the fact that in case of asymmetric supercapacitor some authors multiplied the specific capacitance value with 4. Is it necessary to multiply for all asymmetric supercapacitor? please help.
- Badges
- Science topic
- Similar topics
- Chemistry
- Electrochemistry