As stated in articles, electrochemical surface area (ECSA) is calculated by measuring electricity passed in H adsorption/desorption region by CV, but these peaks are adsorptive in nature so they are directly related to scan rate.In formula presented for this calculation, Qads is normalized by 210uC/cm2 (The value corresponding to fully-covered polycrystalline Pt ) but Qads is scan rate-dependent and will be varied if a different scan rate is chosen.How is it possible to account for scan rate when we want to accurately estimate ECSA .
If you check your units there you will realize that you should be integrating the current with respect to time (not potential) in order to get a charge in Coulomb, C.
If you check your units there you will realize that you should be integrating the current with respect to time (not potential) in order to get a charge in Coulomb, C.
It is important to explain how do you make the integration of your CV,
And what is the way you use to do the calculation for your charge.
because normally what you get from potentiostat is in volt in both x and y axis, but if you use computerized setup, the program convert the unite of Y axis to current (A).
in first case, you have to divided the integration number you get by both scan rate and the resistance of the potntiostat you use to convert it to charge. while in second case you have to divide it only on scan rate. so you will have the charge you need for ECSA without potential dependent as mention by Dr. Wickman
For the adsorption or capacitive behaviours, the charge is constant, but the current is scan rate dependent. In order to calculate the charge, you should use this equation (q = i*E/scan rate). i = current, E = potential, and i*E = the area under the Hupd peaks in the i-E curve or in a CV. For accuracy, you have to subtract the double layer charge by using the double layer current as your baseline.
let use do the units together i=q/sec (A=C/sec or mA=mC/sec), E (V or mV), and scan rate (V/sec or mV/sec). So let us replace the units in the equation (q = i*V/scan rate) you will get [(C/sec)*V)/V/sec)], which equal to C.
In order to get the real surface area, you have to divide the charge in mC by 0.21 mC/cm2 and you will get the area in cm2.
I hope that could help
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