For any type of supercapacitors (or a two-electrode cell), the charging-discharging should always be from 0 V to the designated cell voltage for charging, and from the designated cell voltage to 0 V for discharging. The designated cell voltage is the maximum cell voltage by which the cell still behaves like a capacitor, e.g. the CV still retains the rectangular shape.

According to your CVs, the maximum cell voltage in your case seems to be 0.5 V which is in agreement with your GCD plot on which the voltage profile starts to approaches a plateau when the voltage reaches beyond 0.5 V. The voltage plateau indicates the occurrence of a non-capacitive process that is likely the decomposition of the electrode material or the electrolyte.    

For a symmetrical cell, discharging the cell to the negative voltage range is simply recharging the cell in the opposite direction, but it does not generate very useful information (unless the purposes are to confirm the symmetry of the cell). For an asymmetrical cell, discharging to the negative voltage range may damage the cell.  

To me it looks fine, ideal capacitors show a triangular charge discharge profile, in your case you might be getting some pseudo-capacitance while charging to 0.4V or degradation of your material or electrolyte.  But I think you can calculate the capacitance during discharge using the method you mentioned. Just a suggestion, try fixing the voltage rather than fixing the time.  

Thank you for your feedback, Umair. 

I understand that the shape of the curve is ok for a real capacitor. But my concern is about the negative values. Shouldn't the discharge stop at 0V ?

I am not sure about your system, but here is my opinion. If you look at your CD profile, the positive potential regime is exactly the opposite of the negative potential regime. So when you apply positive charging current, you are building up the charges on each electrodes. Lets say positive charges on electrode A while negative charges on electrode B. At +0.4 volts, you electrodes have been charged up. While applying the negative discharging current you are depleting the charges from your electrode upto 0V. This gives you the amount of charge you extracted back which can be used to calculate capacitance. However, below 0V, electrode A is storing negative charges while electrode B is storing positive charges ( You have reversed the polarity), Everything is same, accept the polarity of the electrode. So calculating capacitance from +0.4 to -0.4 will give you twice the value of your actual capacitance. So my suggestion is to use discharge curve till 0V to find the real capacitance. 

It appears to be as "OK", for a symmetrical supercapacitor SC.  For a symmetrical SC you can cross "0V", (Shouldn't the discharge stop at 0V ?).   Also,

1.   What are your electrodes ?

2.   Explain more, please; What do you mean by the statement  : "... my counter electrode in short-circuit with the reference"  ?

For any type of supercapacitors (or a two-electrode cell), the charging-discharging should always be from 0 V to the designated cell voltage for charging, and from the designated cell voltage to 0 V for discharging. The designated cell voltage is the maximum cell voltage by which the cell still behaves like a capacitor, e.g. the CV still retains the rectangular shape.

According to your CVs, the maximum cell voltage in your case seems to be 0.5 V which is in agreement with your GCD plot on which the voltage profile starts to approaches a plateau when the voltage reaches beyond 0.5 V. The voltage plateau indicates the occurrence of a non-capacitive process that is likely the decomposition of the electrode material or the electrolyte.    

For a symmetrical cell, discharging the cell to the negative voltage range is simply recharging the cell in the opposite direction, but it does not generate very useful information (unless the purposes are to confirm the symmetry of the cell). For an asymmetrical cell, discharging to the negative voltage range may damage the cell.  

The charge-discharge tests you did looks like a half-cell test. Usually, we don't say that is the voltage of a supercapatior. Instead, it is a potential of your working electrode versus the reference. When you are testing the performance of a real supercapacior,  the cell voltage should start from 0 V then go to a designated cell voltage during charging. For the aqueous supercapacitors, this voltage is limited by the water splitting reaction occurring in one electrode.

Hey Camila not sure if you will still need a feedback on this but here it goes. I also got similar kind of result when I was manually sweeping between the voltages. In your instrument, there should be an option to stop the discharging when the voltage reaches zero. At that point the potential will automatically switch to charging once it reaches 0V.

First of all, I think the problem is technical. I agree with Anishkumar Manoharan. May be you need to change the experiment technique from chronopotentiometry to galvanostatic charge/discharge or cyclic chage/discharge technique.

Camila Bramorski It is correct to use +2 mA for charging and then -2 mA for discharging, and this is indeed called constant current charging and discharging. The word constant refers to the fact that current remains constant during charging or discharging, instead of being the same for both charging and discharging which is impossible.

The important operation in the experiment is to make sure that the cell voltage does not go beyond the maximum cell voltage when charging with +2 mA, and not below 0 V when discharging with -2 mA. The charging time may differ from the discharging time, depending on how reversible the process is.

Hello, im conducting a similar test for graphne based supercapacitor and using 3 electrode system to measure the specific capacitance using CV & GCD.

During GCD, The discharge goes to 0 very quickly and then in negative potential. How do I stop it at 0V, since the discharge should stop at 0V and how do I improve the dircharge time in graphene electrode? Any advice would be appreciated.