You can also do the following:

The capacitance of electrode can be calculated on the basis of the following equation:

C = ∫ i(E )dE / (2v (E2 − E1 ))

where C is capacitance (in farad (F)), i(E) is the instantaneous current (in A), ∫ i(E)dE is the total voltammetric charge obtained by integration of positive and negative sweep in CV (0.1 M KNO3, 1.0 mM K3[Fe(CN)6], 20 mV/s) except faradic peaks portion (in A.V). (E2-E1) is the potential window width (in V).

Q/2*V*m .Here Q is the average integral area of CV curve (I).V is working voltage potential range(v). and m is mass (g)of the electrode material.so you can get gravimetric capacitance in F/g.For clear information go through the attached file.

Annamalai gave a good suggestion. Use good S/W to measure integral area of CV.

you can do a CV of your back ground in a narrow potential range and calculated from the intensities measured at the same potential

You can also do the following:

The capacitance of electrode can be calculated on the basis of the following equation:

C = ∫ i(E )dE / (2v (E2 − E1 ))

where C is capacitance (in farad (F)), i(E) is the instantaneous current (in A), ∫ i(E)dE is the total voltammetric charge obtained by integration of positive and negative sweep in CV (0.1 M KNO3, 1.0 mM K3[Fe(CN)6], 20 mV/s) except faradic peaks portion (in A.V). (E2-E1) is the potential window width (in V).

If i understand you after polymerisation by cyclicvoltammetry the adsorbed layer on a surface of electrode you can separate it and make the weight or any analysis

Thank you so much Annamalai, Araceli González and Fuad Asswadi . It is really good suggestion for me. Yes I polymerize polyaniline and tried to use as supercapacitor.

I think that Araceli's answer is the easier way i.e. "you can do a CV of your back ground in a narrow potential range and calculated from the intensities measured at the same potential". You only have to run a linear o cyclic sweep voltammogram at a potential window where faradaic processes do not take place. Then you can use the equation: C = i/v (where C (F), I (A) and v is scan rate (V/s)). You could find more information in Electrochemical Methods (Bard & Faulkner) Chapter 1.2. or Analytica Chimica Acta 615 (2008) 30-38.

In the absence of faradaic processes, the shape of cyclic voltammogram exactly follows that of differential capacitance curve, since j = C v (j = current density in A cm-2, C = capacitance in F cm-2, v = scan rate in V s-1).

If you give me your e-mail I can send you,as an example, our experimental curves of both Capacitance and CV on a nonadsorbed salt as KPF6.

I enclosed the curves as attachment

C = ∫ i(E )dE / (2v (E2 − E1 ))  why we use factor 2 in the denominator can any one answer me C = ∫ i(E )dE / (2mv (E2 − E1 )). in some paper the use factor 2 some did not use why?

the factor 2 in the denominator is due to the one CV cycle contains both charge (during forward scan charge stored on the system then  reverse scan charge delivered from the system). for real application point, we need to find how much charge can we get from the system so we divide the total charge in CV by 2

 In the reversible system the total charge must be devided by 2.

How to get E2 − E1  value in cV for capacitance measurement. Kindly provide with example

for E2 - E1 value just single scan to see where oxidation and reduction occur then select values between them. 

How we can calculate the i(E)  instantaneous current of a CV and what is the condition

Dear Neeraj I(E) the value of current as the function of potential, just integrate the curve the you will get the area.

Thank you for reply answer

Dear Firoz sir I am still confusing on integrating the i(E) its give 1/2 i2(E) 

Can you give any reference to solved  this 

Dear Neeraj, just you can do in origin easily, integration mean area under the curve. Suppose you have x and y column.

first add highest -ve value of current to y-column to get y-axis value positive (starting from zero)

Then just integrate the curve, you will get the are between two curves.

The area under the curve from E1 to E2 will  be positive and area from E2 to E1 will be negative. It will be atuo calculated.

If any problem, send me data.

Dear sir attached a CV file with normal condition which is followed by me

this graph is not plotted correctly, it should start from 0 to 1V current plotted in ampere. Just add in the y-column -15 then plot the graph and go to analysis and mathematics then integrate to get the charge. If potential will be in Volt, current in Ampere then charge will be in coulomb. This is not raw data so cannot plot

I can understand what KP and Araceli has said but I have a doubt. Now  for electrical double layer capacitive (EDLC) behavior the CV curve is symmetric within the experimental voltage window. May be due to this and subsequent integration within voltage limits, the 2 factor comes into the denominator. But should we divide by 2 even for asymmetric type capacitance where there is no redox reactions occuring

@soumya, i personally have not much knowledge about electrochemical characterization, but i feel for symmetric capacitors, where two electrodes have same capacitance, in that case, total capacitance is half of measured capacitance so we need to divide it with 2, whereas in asymmetric, electrodes are different materials, so we dont use 2( in the denominator). i guess asymmetric generally refers to 3 electrode systems, using Ag/AgCl or Hg/HgCl and Pt as reference and counter electrodes. Kindly correct me if i am wrong.

From the value of charge, current, potential and variable time data that's obtained automatically from cyclic voltmmetry diagrams of electrochemical electrode behaves as supercapacitor; How can calculate areal capacitance and specific capacitance?  

repeat the data and do as by professor Firoz Khan

You have to consider the mass of the active material only which you have used for electrode preparation. Exclude the mass of current collector.

how to determine the value of m in Q/2*V*m

The capacitance of electrode can be calculated on the basis of the following equation:

C = ∫ i(E )dE / (2v (E2 − E1 ))

this formula is for three electrode system or two electrode system ?

Agree with the response of Araceli González-Cortés

How can we calculate the areal specific capacitance of supercapacitor from CV and GCD ?

If we deposite active material 1cm2 on the Nickle form both sides.

The Cs is calculated from CV as follows:

Cs=Integration (current)/(scan rate*mass*potential window)

Firoz Khan

sir, if the result of the area is in coulomb(C), then how we suppose to get the result in F/g?