While calculating charge in each peak obtained from Cyclic Voltammetry, Can I calculate the area of the peak and divide it by scan rate to get the charge?
Based on Q = It, you can use area of peak divide by scan rate. By the way, you have to carefully check the unit of them.
You can (and Bapi Bera gave a good methodology for doing that), but unless you know the specifics of the system, you really shouldn't. This is because the CV response consists both of faradaic and capacitive (i.e. double electric layer reorganisation) currents. When evaluating electrochemical processes we tend to want to compare faradaic currents (e.g. when determining the electrochemical reversibility of a process) rather than the capacitive currents. This is particularly important when redox reactions of your system cause changes in its relevant properties (e.g. surface area, conductivity, morphology, etc.) and in the case of some systems leads to absurd results.
Dear Pradeep Subedi
Charge is equal to peak area. First determine the peak area from the voltammogram then the peak area you determined is equal to the charge of that voltammogram . If you are using origin 8.0, to calculate the area under curve you want to calculate , I suggest you to follow this step: Analysis ---> Mathematics ---> Integrate
Hi Pradeep,
If I undrstand , the charge is Q=⌡Idt whuch is proportional to the surface under the peak.
The charge of the blank (without studed species) should be deducted from Total charge carried out with activated species.
Yosef Nikodimos Sir, well, will we use the value of integrated area as it or it is in cm2 and we have to change into m2 by dividing 100*100 and one thing more will we also have to convert mV/s scan rate to V/s by dividing with 1000 or not ? plz guide.
Dear Dr Faisal Ali
Use it as it is. Its unit is Coulomb. When we say the integrated area is the same with charge, it is in magnitude but the unit is Coulomb.
Then this unit will be canceled out by the unit of Faraday, when we calculate surface concentration of the electroactive species (C) using the formula
C = Q/nFA
Q = is the charge consumed in coulomb obtained by integrating the peak area.
A = is electrochemical active electrode surface area in cm2
F = is Faraday constant (9,6485 C mol−1) and
n = number of electrons participated in the electrode reaction
For more details about this formula, you can check it from this paper at: Article Electrochemical Determination of Metronidazole in Tablet Sam...
you can check equations (1-3)
I hope it helps.
The background currents should be subtracted using a baseline created in Origin. The baselines are created individually for each half of the voltammograms using 300 or more points automatically placed on the curve through the potential range, points are then deleted from the potential range of the peak.The remaining points are interpolated using B‐spline function to produce the baseline....
It is quite easy. This can be easily figured through the tutorial link given below - https://www.youtube.com/watch?v=FS3JxZmZnYY&t=29s
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Let me simplify a few things.
a) You can be misled if you think that the area of a 'standard Cyclic voltammogram' is charge. No it is not. The y-axis is current (in ampere let's say) and X axis is applied potential (in volt). Naturally the integration will give you Ampere-volt (or Watt). Charge is not Watt.
b) to get charge you need to either:
i) Get a plot of Current with X axis as time (check your exported data, one of the columns usually is time) and then integrate it, or,
ii) yield values in watt from standard cyclic voltammogram as I said above and the divide it by the scan rate.
Also if you want to be sure, calculate charge by both procedure and compare :)
Cheers and good luck!
To determine the area under the peak, when you do the integration of CV peak you will always get the unit of Ampere.volt. to simplify ampere is equal to charge per sec. so the final unit after integration is always (C/s).V or C.V/s then you need to divide it with scan rate (C.V/s)/( V/s) so in the end you will get coulomb or charge.
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