For a three electrode system, I am getting a specific capacitance of 400 F/g at a potential difference of 1 V and scan rate of 5 mV/s. The same electrode has been used to make a symmetric supercapacitor and the obtain specific capacitance of the whole cell is 93 F/g (provided, all the experimental conditions are same as in case of three electrode system). So, the specific capacitance of the supercapacitor is approximately (1/4)th of the specific capacitance of the single electrode that has been obtained in three electrode system. In few literature, it has been mentioned that the specific capacitance of the cell should be half of the specific capacitance of the electrode obtained from three electrode system. Kindly, put some limelight over this.
Cell specific capacitance, Cs, cell, is not a widely used concept. My understanding is that it should be calculated as follows from the cell capacitance, Ccell, and the total mass of both the positrode, m+, and negatrode, m-.
Cs,cell = Ccell /(m+ + m-) .......................(1)
For a symetrical cell, m+ = m- = m, the above equation becomes
Cs,cell = Ccell /2m .......................(2)
For a symmetrical cell, the relation between cell specific capacitance and material specific capacitance, Cs, may be derived below.
1/Ccell = 1/(m+Cs) + 1/(m-Cs) = 2/mCs.......................(3)
Ccell = mCs/2.......................(4)
Bringing (4) into (2) leads to the equation below.
Cs,cell = Cs /4 .......................(5)
Deatied answers are given there:
https://www.researchgate.net/post/Do_three_and_two_electrdoe_system_in_supercapacitor_have_different_Specific_capacitance_F_g_even_with_the_same_electrode
Cell specific capacitance, Cs, cell, is not a widely used concept. My understanding is that it should be calculated as follows from the cell capacitance, Ccell, and the total mass of both the positrode, m+, and negatrode, m-.
Cs,cell = Ccell /(m+ + m-) .......................(1)
For a symetrical cell, m+ = m- = m, the above equation becomes
Cs,cell = Ccell /2m .......................(2)
For a symmetrical cell, the relation between cell specific capacitance and material specific capacitance, Cs, may be derived below.
1/Ccell = 1/(m+Cs) + 1/(m-Cs) = 2/mCs.......................(3)
Ccell = mCs/2.......................(4)
Bringing (4) into (2) leads to the equation below.
Cs,cell = Cs /4 .......................(5)
George Zheng Chen sir, I got whatever you have explained; still I am bit puzzled. I will be attaching a file along with this thread for a better explanation of my question.
1. Can we correlate the value of specific capacitance obtained from three electrode and two electrode systems?
2. If yes, then the value of specific capacitance of the supercapacitor (two electrode system) must be half of the specific capacitance of the single electrode (obtained from three electrode system).
3. But, if only two electrode system has been considered, then the specific capacitance of a single electrode is four times the specific capacitance of the supercapacitor.
(Talking about symmetrical two electrode system)
Your point 1 is correct, but point 2 is wrong, and point 3 is right.
The cell capacitance (not the cell specific capacitance) is a half of the electrode capacitance (not the electrode specific capacitance).
The two equations you gave should not be applied at the same time to both the CV of the electrode in a three electrode cell, and that of the two electrode cell. These two CVs are different. In other words, the current, I, in your first equation is not the same as that in your second equation. (If you do use the same electrodes to build the cell, the current on the CV of the two electrode cell would be a half of the current on the CV of the electrode in the three electrode cell, if all experimental conditions are the same in both cases.) Thus, in your second equation, you cannot equal the integration term to the specific capacitance of the electrode.
Tapas Das : The multiplier of 4 adjusts the capacitance of the cell and the
combined mass of two electrodes to the capacitance and mass of
a single electrode. - Meryl D. Stoller.
Please refer this paper-- "Best practice methods for determining an electrode material’s performance for ultracapacitors"
George Zheng Chen sir in the formula pic of Tapas Das we used mass in g or mg to calculate the specific capacitance of in unit F/g kindly tell.
Dear Tapan,
Here are some literatures for your reference in order to calculate the specific capacitance (both in 3-electrode, 2-electrode), energy density and power density (both 3 and 2-electrode).
Subramanya, B. and Bhat, D.K., 2015. Novel eco-friendly synthesis of graphene directly from graphite using 2, 2, 6, 6-tetramethylpiperidine 1-oxyl and study of its electrochemical properties. Journal of Power Sources, 275, pp.90-98.
Zhang, L. and Shi, G., 2011. Preparation of highly conductive graphene hydrogels for fabricating supercapacitors with high rate capability. The Journal of Physical Chemistry C, 115(34), pp.17206-17212.
Ghosh, K. and Yue, C.Y., 2018. Development of 3D MoO3/graphene aerogel and sandwich-type polyaniline decorated porous MnO2− graphene hybrid film based high performance all-solid-state asymmetric supercapacitors. Electrochimica Acta, 276, pp.47-63.
All the best.
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