Although more details about the experiment would be useful, this behaviour appears reasonable - essentially, if the same charge is to be passed through the system (assuming the same concentration of electroactive species), a larger current density will achieve this in a shorter time.

To give more details: your solution has a given amount of electroactive species, which can accept a limited charge. You can calculate how much charge, based on the concentration of those species and the electrode reactions you predict / know to take place in your system.

Knowing this value and dividing it by the current passing through the system (product of current density and electrode area) you will get the electrolysis time. The more you know about your system, the more accurate will this estimation be; do not expect a perfect fit, however, as various non-ideal properties of the system and constraints of your epxeriment might affect how much of the electroactive species are electrolysed.

thank you sir for your nice reply @ Tomasz Jarosz

Dear Sajjad,

@ Tomasz Jarosz give you a comprehensive answer. But from your figures for 1 A/g (graph33) and 5 A/g (presentation1) somethings also happens. Namely, idealu, charging times should be equal to discharging times, which is not the case with your electrode. Hence Coulombic efficiency is far from one (Qd/Qc). So probably during charging electrolyte decomposition occurred as well (probably oxygen evolution).

This behavior is the electrode voltage with time during charging and discharging process of an electrochemical electrode. It reflects the capacitive behavior of the electrode. As the charging current increases the the time required to build up the same charge or the same voltage will decrease proportionally. The same happens for the discharging process. So when the current increases the charging discharging time cycle decreases proportional to the current.

It is so that the charging time T= Q/I = CV/I

Best wishes