If i got it well, you have to charge the double layer. i.e. you have a positive current corresponds to the double layer charge, and by reverse the scan, you have to have also double layer charge,but with negative sign.

negative currents usually correspond to reduction peaks in the usual accepted convention for cathodic reactions (i.e. from maximum potential to minimum potential) (negative currents) and anodic reactions (positive currents)

For an ideal EDLC, it is notable that a shape of CV is rectangular without deformation and also symmetric with respect to the zero current line(axis). But, for practical EDLC,  the shape is inclined due to an electrode resistance including ion transfer resistance and electrode material resistance. Hence, at the initial time of charging( or discharging), the current is increased( or decreased) with a slop from a minimum to a maximum( or from a maximum to minimum).      

Hatem, If I understand what you are saying, the current is positive during charging and negative during discharge. I think I understand this concept, but experimental scans and published CV scans almost always have the current sign traversing a current "zero" during both charging and discharging. Is there any physical significance to this "negative" current component at the onset of charging or to the "positive" current component at the onset of discharging?

Jong-Huy, I do not understand how your answer relates to current "sign" or the apparent change in sign during both charging and discharging in CV scans.

the negative current at the onset of charging curve could be any slow side process still going on (e.g. electrolyte decomposition) which has a negative current. It is similar to the ORR in the positive going scan.

the positive current at the onset of discharging is still any oxidative reaction or side process.

Exactly saying, the current  from CV is 'a response current' on the weep(scan) rate in a given potential window. The reason that the sign of response current is not instantly reversed according to the changed sign of scan rate is due to a nonideal polarization of porous electrode caused by internal resistance within  pore structure, as an iR drop effect. Formore inforamtion, please refer to "Electrochemical Capacitor and Hybrid Power Sources", Proceeding Volume 2002-7(ISBN 1-56677-325-3) published by ECS, 2002, pp 5-8.      

According to my knowledge, as you increase the potential the current increases due to the polarization of charges at the electrode/electrolyte interface. After a while complete polarization occurs hence the rectangular shape. When reversing the potential, you are actually depolarizing the charges (which will again take some time for complete depolarization) i.e the double layer formed is destroyed. Basically alignment and realignment of charges causes the current reversing sign in the CV curve of EDLC material

Your question needs to be more accurately define, and the answer depends on if you are dealing with cyclic voltammograms recorded in a two electrode capacitor (either laboratory or commercial types) or a single electrode in a three-electrode cell.

For the former (a two-electrode capacitor), the current is always positive for charging and negative for discharging (as pointed out by one of contributors above). For symmetrical capacitors, it does not matter how you connect the cell because the positive and negative electrodes in such a capacitor are identical. However, for an asymmetrical capacitor, the polarity of each electrode is defined, and you must follow the polarity of the capacitor for charging and discharging because reversing the polarity may damage the capacitor.

To derive the specific capacitance (which is usually applicable to an electrode material), it would be simpler and more meaningful if the capacitor is the symmetrical type, and the specific capacitance is 4 times of the cell capacitance against the mass of each electrode, or 2 times of the cell capacitance against the total mass of the two electrodes.

For asymmetrical capacitors, the calculation requires the knowledge of individual electrode materials, and at least the specific capacitance of one of the two electrodes. Otherwise, it is impossible to derive specific capacitance for any one of the two electrode materials in the asymmetrical capacitor.   

As for the sign of the current, it is actually due to the switch of the direction of the  voltage scan (for the cell, or if you use the term potential scan, it should refer to a electrode) at the end of charging or discharging. When the scan increases the voltage (or potential), dU/dt  (U is voltage and t is time) is positive, and the current is also positive. If the scan decreases the voltage, dU/dt is negative, and the current becomes negative. This is one of the  main reasons why cyclic voltammograms of capacitors (and also supercapacitors) are always rectangular.

For the latter (a single or working electrode in the three-electrode cell), it will then depend on if you use the material for the positive or negative electrode in the supercapacitor. However, because potential is always measured against the potential of a given reference electrode in the three-electrode cell, the current is always positive for the positive potential scan, and it becomes negative for the negative potential scan.

George, Thanks for trying to help. I don't think electrochemists confuse cyclic voltammetry, a fundamentally three electrode experiment, of EDLC electrodes with analysis of two electrode capacitors. Your statement that "because potential is always measured against the potential of a given reference electrode in the three-electrode cell, the current is always positive for the positive potential scan..." is simply incorrect. I refer you to the attached file provided by Gamry, on CV of EDLCs. In addition, your own recent paper [Akinwolemiwa, Peng, & Chen, JES, 162, A5054-A5059, 2015, attached] in Fig. 1(b) illustrates the CV of an ideal EDLC, which has approximately equal portions of positive and negative current at the potential scan extremes, following scan reversal. I want to thank Hatem and Jong-Huy for their helpful answers. 

John, I am not very sure what makes you think my statement "because potential is always measured against the potential of a given reference electrode in the three-electrode cell, the current is always positive for the positive potential scan..." as simply incorrect.

The Gamry Note shows the CV current in the same way as I described, i.e. positive current during positive potential scan and vice versa. However, the effect of dU/dt on the current sign is not discussed in Garmy's Note.

Nevertheless, as I have noticed, Figs 4 and 5 in Garmry's Note are actually incorrectly presented. The x-axis is presented as Vf vs. ref, but the CVs could have only been recorded on or derived from a two electrode cell, which means that there could not be any reference used. Also, recording the CV of a capacitor into the negative voltage range is theoretically incorrect. The voltage change should always go from 0 V to the maximum voltage and reversed to 0 V.

Of course, in my reply to your question, the context is the CVs of capacitors or supercapacitors. In some other cases, e.g. electro-catalysis, current signs do not always follow the potential scan directions.

Hope you are not confused by the difference between the "potential of a single electrode" and the "voltage of a cell" which is the difference between the potentials of the positive and negative electrodes in the cell. I do notice terms such as "cell potential" and rarely "electrode voltage" in the literature, but these are incorrect strictly speaking.