Dear Abdul,

Can you tell what kind of problem? if you are worry about specific capacitance value of the composite, then check specific capacitance from charge-discharge curve.

Plot your graph in origin. Go to analysis . there you will find mathematics.Under mathematics select integration and choose absolute area. then you will get the area under the curve. Now you have to divide it by 2. As your area should consist of either oxidation or reduction.(Though as the two halves are not identical it will  not give exact result)

Then just divide that value by "active mass" and scan rate.

And you will gate the specific capacitance.But this is just the pseudo capacitance.

Dear Abdul,

Agree with Kheribot that value of specific capacitance of composite check with charge discharge curve. Based on my knowledge, the shape of the CV curve indicating the type of capacitance e.g. pseudo capacitance or ideal capacitance. The formula for measuring capacitance from CV is same. Check with others.

 I am trying to find the Area under the curve using origin but this is the result. what am I doing wrong?

Dear Gourav Bhattacharya, you need to be careful when using origin to calculate the area, as there're two options: mathematical and absolute, in which absolute is not accurate as when we sum up those areas, we need the negative values as summation will then  "minus off" the area which is the area enclosed in the CV. We need to choose "mathematical"

We know that an area is simply the area under the curve (from the curve to the x-axis) which we need to take the area under the anodic scan minus the cathodic scan

One way to check is to separately calculate the area under the anodic and cathodic scan and then manually minus them. You get the same result as the "mathematical"

Dear John,

Mathematical area of the whole CV would give 0 for a perfectly symmetric CV for CHARGE DISCHARGE systems! Absolute areas should be used----because -values under F(x) baseline and the negative current region value will add up. So you will end up getting the area of the entire curve!   Absolute area to be used if the area under the wcurveureve is calculated

Dear Deepak,

I understand what you meant, this attached picture on the left shows a perfect CV for a supercapacitor. When calculating the area, the green trapezoid give a positive value and the red trapezoid also give a positive value because V final - V initial = -ve value multiply with the current value which is also -ve. Hence either mathematical or absolute will give a +ve value. You won't get a 0 for the area calculated.

Problem arises when our CV is not always a perfect square that has the x-axis cut across the middle perfectly. If we look at the CV on the right, the green trapezoid's area in the charging curve is already larger than the graph itself and we know that the area has to be minus off with the red trapezoid. Therefore we need the red trapezoid's area to be -ve. Using origin. The change in V is -ve because V final < V initial, however the current is in the positive region hence the red trapezoid's area has a negative value.

Using mathematical method from the origin pro will add up the +ve green and the -ve red which is the right way. If we take the absolute area of both green and red then the area will be way larger than it's suppose to be which will give an overestimation of the capacitance of your sample. 

One of the potentiostat software that we were using also display the area but the area calculated by the software is also giving the absolute area of the graph (which is also wrong) and we have contacted the company and discussed with them thoroughly. After having their software developer to look deeper into this matter they also agreed that their calculation in the software was a mistake. So we need to be careful when reporting the specific capacitance value and don't simply depend on the value displayed on the software as well. Deviation between the mathematical and absolute area will become more significant if the CV curve obtained has high overpotentials or have many prominent redox peaks especially for NiO and PANI.

Another to add on is that after you get the accurate area of the graph, the area need to be divided by 2 before dividing it with the scan rate, mass and potential window. Or another way is to only calculate the area under the negative sweep current. This has also been a hot debate over many researchers many published journals also didn't divide with a factor of 2 as well. 

Hi John,

Cool explanation and illustration- Thanks!

Of course the entire charge under the curve will have Anodic and cathodic, so one should divide this by 2. But the most accurate way is either calculate anodic charge or cathodic(they have to be same if they are in reversible region) and the do the capacitance calculation.

Regarding the area explanation, I checked with one of my plots. I see that absolute area works perfectly fine for me. But the way I go about this is I calculate the area under the anodic or cathodic alone and then take the absolute. We also have Nova software, they use similar technique and we end up getting the same results.

But please note: I am not a mathematician, so my views might be wrong. But I am kind of confident of my charge values when I see my results as an electrochemistry student

regards