We can know from the Randles–Sevcik equation that current is directly proportional to sqrt.of voltage , but what is the phenomenon occurs in the battery.
Dear Darbar,
Total current in a system, i = ifaradic +icharging and faradic current is fixed with scan rate.
Scan rate = dV/dt,
i = ifaradic + dQ/dt, i = ifaradic + dQ/dV*scan rate,
i = ifaradic + C*scan rate
when scan rate increases, the charging current will increase and actual capacitance is constant for a material. When scan rate increase you are allowing more current to flow.
Best Regards,
Bapi
Actually, not only the charging current increases with the scan rate, but also the faradaic current.
The faradaic current passed through the electrode is a direct proportional to the flux of the concentration at the surface of the electrode. The higher scan rate will make the slope higher since the diffusivity of active species is a constant. This will result in the higher current as described the Randles–Sevcik equation, which is a direct proportional to the square root of the scan rate. The charging current is a proportional to the scan rate. So the charging current becomes relatively more important at faster scan rates.
However, in the actual system, the charging current doesn't follow the capacitor relation. You should treat it as a CPE.
The details about it can be found here
Article Ohmic resistance and constant phase element effects on cycli...
Dear Darbar,
We have to consider the following factors
1) Fick's Laws
2) Randles-sevicik equation
3) disturbance in "Nerstian equilibrium"
4) concentration of active species at electrode surface.
Please refer 6th chapter in the following book "Instrumental methods in electrochemistry" by Southampton Electrochemistry Group.
https://www.elsevier.com/books/instrumental-methods-in-electrochemistry/pletcher/978-1-898563-80-8
May be helpful to you..
If the rate of change of potential difference (Voltage) increases in a given closed circuit. It lead to increasing current in that circuit. Because more number of charges (or ions) carriers were triggered by the potential difference.
Good question,
Charge (coulomb) = current (ampere) × time (second)
When the scan rate is increased the time will be decreased. So according to the above equation and to keep the same charge value , the current should be increased, that's way the peak current is getting bigger as long as the scan rate is increased.
Dear Devendrasinh Darbar ,
Because this question is a basic one.
When your device is an electrochemical electrode or any type of junction diode when you apply a voltage ramp dV/dt on them, the conduction current is independent of the the speed of the scan to great extent but the capacitive current increases with dV/dt. Therefore so in order to measure the static characteristics of the device you must reduce dV/dt such it becomes very smaller than the conduction current.
Best wishes
the volts linear sweep from initial to final potential runs , as the volt incleased per second say 20mv/second to 50mv/ second u will see current increased
in in general when u apply more volt in less time reaction is faster and will give rise to fast electron transfer from species to sensing electrodes also the v=IR
volts directly prop to current
Devendrasinh Darbar
well, as per my view of point,
- when you increase the scan rate the speed of the electron transfer/mass transfer were increasing because of scan rate is mv/s, by this type you increase the the rate of voltage applying in solution.
- As per above line the voltage rate increasing in time domain,the mass transform were increasing,the electrons attracted on working/sense electrode in higher rate,this way the current was increasing by increasing scan rate.
At higher scan rates the rate of diffusion is more than the rate of reaction. Hence, more electrolytic ions reach the electrode electrolyte interface whereas very few ions participate in the charge transfer reaction. Therefore, the current at higher scan rate increase.
Does the sentence of @Garima Chaturvedi means this?
The overall current at the electrode increases, indicating that more ions are reaching the electrode surface even though some of them are not contributing to the reaction.
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