Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: fine on April 17, 2018, 12:44:23 PM
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Hi. I'm working on supercapacitors and i need to use CV to find it's areal capacitance. I'm using a software to find area under the curve. I don't know which area to use. Should it be the area under the curve during voltage increase (image 1) or the area of positive current even when voltage starts to decrease (image 2). I have also attached an image showing the formula for calculation of areal capacitance.
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(https://www.chemicalforums.com/index.php?action=dlattach;topic=94990.0;attach=9898)
What do each of the symbols mean?
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v is the scan rate during cyclic voltammetry
m is the area of the capacitor
Delta V is the voltage range applied to the capacitor during cyclic voltammetry (Vmax - Vmin)
Integral I dV is the charge stored on the capacitor during oxidation or reduction cycle.
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I hoped to understand the idea behind, but units don't work. Capacitance is measured in Farads, [itex]F=\frac C V[/itex] (Coulomb/Volt), which is not what this formula yields. Unless I am making some mistake scan rate is V/s, area is m2, ΔV is V, so we get
[tex]\frac C {\frac V s m^2 V} = \frac {C\times s} {V^2 m^2} = F \frac s {Vm^2}[/tex]
Doesn't make much sense.
Are you sure this integral shoudl yield a charge? Why dν and not dt?
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The formula calculates areal capacitance which is measured in F/m^2. The integral is done over x-axis which in cyclic voltammetry is Voltage hence the dV and not dt. If you look at the image the integral yields result as mA.V. The scan rate in my case is in mV/s and area is in cm^2.
(mA.V)/(cm^2. mV/s . V)=F/cm^2