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Topic: Electrochemistry: Cyclic Voltammetry  (Read 2239 times)

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Offline fine

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Electrochemistry: Cyclic Voltammetry
« on: April 17, 2018, 12:44:23 PM »
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.

Offline Borek

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Re: Electrochemistry: Cyclic Voltammetry
« Reply #1 on: April 18, 2018, 02:49:39 AM »


What do each of the symbols mean?
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Re: Electrochemistry: Cyclic Voltammetry
« Reply #2 on: April 18, 2018, 02:53:16 AM »
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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Re: Electrochemistry: Cyclic Voltammetry
« Reply #3 on: April 18, 2018, 10:17:11 AM »
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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Re: Electrochemistry: Cyclic Voltammetry
« Reply #4 on: April 18, 2018, 10:32:15 AM »
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

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