I performed chronoamperometry analysis at -0.4V for the period of one hour. However, I don't how to calculate the total charge.
The area under the curve (i.e. the integral) is the charge. Plot your current versus time data in a program like OriginLab or Excel, then ask the program to perform a numerical integration from t = 0 to t = 1 hr. If it is not already like this, you may want to plot the time in terms of seconds, rather than hours. This will mean that you do not have to convert the units after you do the integration. Remember, an Amp, (A) is equal to Coulombs/Second (C/s).
side note: determining the total charge by numerical integration is not a very accurate methods for short times (i.e. t
Dear Anatole Borisov
I have calculated the integrated area by plotting current versus time data in OriginLab. However, as per the definition Charge (Q) = I*t (i.e. Ampere.sec=Columb). let's say the integrated area is -2.3. Therefore my concern is whether the integrated area value corresponds to charge (Q) ?
If your units of your current versus time plot are Amps and seconds, respectively, then the integration result will be in Coulombs. If that's what you get -2.3 from, then the amount of charge passed is equal to -2.3 C.
Thank you Anatole Borisov for your valuable suggestions.
Yeah, units of current vs time plot are Amps and seconds.
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