That´s a problem when you work with very low resistance solutions. There are several methods to measure the resistance solution. I don´t know what is your purpose to measure the resistance, but one way to determine it is to run voltammograms with ferrocene at different scan rates, and then try to mach the experimental voltammograms with simulated by adjusting the resistance value. Other way is to run LSV with eletrolyte solution (just like you are describing, I think) imposing a voltaje with a very short time duration, andtry to adjust the settings experiment to get the greatest number of data values, and from the responses you could fit the tendency of the first part of the LSV with the equation i = Cv(1-exp(-t/RC)) which C= capacitance, v = scan rate, and R = resistance.

That´s a problem when you work with very low resistance solutions. There are several methods to measure the resistance solution. I don´t know what is your purpose to measure the resistance, but one way to determine it is to run voltammograms with ferrocene at different scan rates, and then try to mach the experimental voltammograms with simulated by adjusting the resistance value. Other way is to run LSV with eletrolyte solution (just like you are describing, I think) imposing a voltaje with a very short time duration, andtry to adjust the settings experiment to get the greatest number of data values, and from the responses you could fit the tendency of the first part of the LSV with the equation i = Cv(1-exp(-t/RC)) which C= capacitance, v = scan rate, and R = resistance.

@Pablo Daniel Astudillo-Sanchez

Also, why is the I independent of the scan rate in EDLC but not is pseudocapacitor? I understand that Q=CV, so I=C(dv/dt)------------- for this reason I should be flat in both EDLC and psuedocapacitor case.... But practically they are not?! WHY?

I

In both EDLC and pseudocapacitor, the CV current at every potential is always proportional to the scan rate. If you observe that the current does not vary with the scan rate, something is incorrectly arranged in your measurement.

On the other hand, the CV should always be a rectangle for either the EDLC or pseudocapacitor. Peak-shaped CVs should NOT be used for capacitance analysis, even though in such cases the current at every potential can also be proportional to the scan rate.

@George Zheng Chen Dear Dr. Chen, Sorry

Dear Dr. Chen, Sorry

Sorry, my question was wrong. I apologise for my mistake. My question is, why current does not change with Voltage (at a given scan rate)?  This is the reason  why  we get rectangular curve(with no inclination). I am looking for an equation to understand this.  Similarly if we have some series resistance than the CV plot will be inclined. So where does this ESR fit into the Q=CV equation? 

dQ=CdV-----------------------.>Idt=CdV---------------------------->I=C(dV/dt)------------------------------>I=C*scan rate....................... So how can I prove that Current should be constant with Voltage? And also where can I introduce the series resistance into this equation to account the slope of the CV?

Thanks

@George Zheng Chen

Dr. Chen, Way back I remember reading one of your paper, where you had mentioned the reason for the round edges of the CV curve. I forgot the justification?! Could you please mention the name of your paper if you remember? Also, how can we analyse the increase in current (after the EDLC potential window) using capacitance equation?

Regards

The equations in your message are correct without considering the effect of the ESR. To include the effect of the ESR into the equation, you will need to bring the so called iR drop into the equation. In other words, the actual voltage (for a two-electrode cell) working on the cell, Vactual is difference between the applied voltage, V, and the so called iR drop, ViR = I*ESR, i.e.

Vactual = V - I*ESR.

I = C(dVactual/dt) = C(d(V-I*ESR)/dt) = C(dV/dt) - C*ESR(dI/dt)

You then need to solve this differential equation to obtain the I~V relation.