Giddaerappa Kuntoji the charge and discharge time depends on current. If you are talking about GCD, then there is losses in the supercapacitor (self discharge, etc) and hence the amount of charge on supercapacitor is less. I = Q/t for a constant current Q decreases then t also decreases. Now, depending on the type of material used, the construction etc the mechanism changes. E.g. Migration of the active electrolyte between two electrodes of the device in case of AEESC Article Mechanism investigation and suppression of self-discharge in...
The main point of departure (for q-losses) in the supercapacitors is the overestimation in potential window(s) of electrolytes' (analytical) stability, for a certain case of useful/practical WE-CE. There are other, minor, self discharge issues (electrolytes' impurities, aging, etc.), also.
First of all, there are also many capacitors that discharge time longer than charging time, which may be due to hydrogen-evolution or other side effects.Second, the phenomenon that the discharge time is shorter than the charging time is common in many solid-state flexible supercapacitors. It is very likely that the internal resistance of the device is relatively large, resulting in a large voltage drop. In addition, the supercapacitor still has a serious self-discharge phenomenon, and even some other micro-short-circuit phenomena will lead to the loss of capacity and shorten the discharge time.
it would be, please, impressive, if you would take a shine to name/refer to a case(s)[1] (about: Qdisch. >[?] Qcharge).
1. Partial quot. from Yunlong Yang : "First of all, there are also many capacitors that discharge time longer than charging time, which may be due to hydrogen-evolution or other side effects. ..."
If you have a purely double-layer system, your charge-discharge currents are the same, and you are charging and discharging within the same window (i.e. V0 -> Vmax -> V0) you would not expect your discharge time to be longer than your charge time due to the first law of thermodynamics.
Imagine you are charging the supercapacitor by supplying energy, and by discharging it you are depleting that energy. If you are discharging for longer under the same rate of depletion to that of applying, you are somehow getting more energy out of your supercapacitor than you put in.