@J. Lavanya, the definite diffusion coefficient cannot be calculated without the area of electrode. The area can be quite easily determined by cyclic voltammetry experiment of 4 mM K3Fe(CN)6 in 1 M KCl, which diffusion coefficient is 6.3x 10^-10 m^2 s^-1.

Without the area, You can only calculate the quotient D/area. For certain applications, for example when changes of D, not definite values, are of interest, it would be enough. 

It can be calculated using the Randles–Sevcik equation if the relationship amongst both parameters is linear. The fact that it does not cross the origin means that the process is not only controlled by mass transfer. 

If you want, take a look on my paper where this fact happened while I was studying zinc electrodeposition: 

https://www.researchgate.net/publication/260636076_Electrochemical_Study_of_a_Simulated_Spent_Pickling_Solution

Article Electrochemical Study of a Simulated Spent Pickling Solution

As Jordi Carrillo told you, in this case you have to use the Randles-Sevcik equation for the obtaining of diffusion coefficient. You can use, for example, CalcTool - an online tool good for it (http://www.calctool.org/CALC/chem/electrochem/cv1).

If you know the concentration of ions diffusing, then you can get D from R-S analysis for the peaks if it is a reversible reaction of from the limiting current if it is an irreversible system. 

Dear Jakub,

There is the following reason of the situation "does not crosses the origin"  that mechanism depends on the scan rate. Try to check. Have you the different slopes of your dependence? If so, analyse each part separately. It can help. You will not see very fast processes at the slow scan rate and vice verse.

@Petr Jakubec - the point is that, Randles-Sevcik equation in its base form (as in the link) requires the current peak to be directly proportional to square root of sweep rate. And I wrote, that the dependence I obtained does not crosses through origin, so it is not direct proportion. Furhermore, it is better to calculate diffusion coefficient not from single point, but linear regression to obtain more reliable value. 

@Georgii Sokolsky - in some of the experiments, that the dependence does not exhibit reversibility, it is linear in all range of sweep rate (25-150mV/s), but does not crosses origin; in others, at certain sweep rate, change of slope occurs. 

It can be the result of complex mechanism or your process is not diffusion controlled. Try lower than 25 mV/s sweep rate and you will be sure about that...

As to reproducibility of results I have no ideas... It depends on conditions and may be purity of reagents, double distilled water, inert gas to bubble and etc. should be checked 

Did you correct the backgound currents? If not, you might try the measured CV substracted background CV at each scan rate to obtain the neat CVs.

Yes, the I have taken the background into account. Maybe I should have added before, that potential corresponding to the current peak changes with sweep rate. Increase of sweep rate causes movement of anodic peak to higher, and anodic peak to lower potentials, respectively. 

This fact may mean that your process is an irreversible one. I also observed this phenomenon in the paper attached previously.

Dear Jakub,

This sounds good. It is quite normal that the peak-to-peak separation increases with increasing the scan rate, for most quasi-reversible systems. You might make a series of plots for the peak currents against vx (x =0.5, 0.6, 0.7, 0.8, 0.9 and 1.0) to see which one has the best linear fit through the 0 point in the X and Y axes. For some systems, the rate-limiting step is a mixture process, i.e. no pure diffusion control. For example, the redox species studied is particially adsorbed on the electrode surface or confined in a film matrix. You might take a look at the Supporting Information (Figure S12, in particular) of our paper (attached).

I am not sure if the files were attached. If not, attached Again here. 

if i dont know the area of my modified electrode . how to use randles equation to calculate diffusion coefficient

@J. Lavanya, the definite diffusion coefficient cannot be calculated without the area of electrode. The area can be quite easily determined by cyclic voltammetry experiment of 4 mM K3Fe(CN)6 in 1 M KCl, which diffusion coefficient is 6.3x 10^-10 m^2 s^-1.

Without the area, You can only calculate the quotient D/area. For certain applications, for example when changes of D, not definite values, are of interest, it would be enough. 

J. Lavanya you should use any reference solution with a well-known D, as ferro cyanide, to determinate the area of your electrode, then you will be able to determine the D from your solution.

Another possible way is that you look for the charge needed to form monolayers from your modified electrode and determine, from this, the area obtained.

Thankyou sir

@Jakub Piekart, Could you show me the source of the diffusion coefficient of 4 mM K3Fe(CN)6 in 1 M KCl? Thank you.

@Lin Zeng: The value is quite widely cited, e.g. in Gao et al. "A cyclic voltammmetric technique for the detection of micro-regions of bmimPF6/Tween20/H2O microemulsions..." Green Chem. 2006, 8, 43.

@Jakub Piekart, Thank you very much. I found it.

Dear all,

Could you help me understand C and A in the Randles-Sevcik equation? I used 3 M KOH electrolyte, 3 is the value for C? Another question is: I used Ni foam (1 cm x 1 cm), so, 1 cm² is the value of A?

What are you determining? The concentration is of the chemical specie that is exchanging electrons with the working electrode.

Your surface is 1cm2 but I suggest you to use the same units for all the equation variables, that is, 3M (mol/liter = mol/dm3 = 1/1000 mol/cm3) if you use your area in cm2 .

Dear Jordi Carrillo Abad ,

I am working with electrode materials that exhibit battery-like behavior. For the cyclic voltammetry results, I did the graph of the current vs. square root of scan rate to talk about diffusion-limited processes. I would like to calculate the diffusion coefficient from Randles-Sevcik equation, but I'm not understanding some terms, such as C and A. The equation is:

ip = (2.687 x 105) n3/2 A D1/2 C v1/2

I used 3 M KOH electrolyte, 3 is the value for C? Another question is: I used Ni foam (1 cm x 1 cm), so, 1 cm² is the value of A?

What is the peak current that I should use? I used a scan rate range of 5-100 mV/s. Should I use ip for 100 mV/s?

In general, you should made a graphic with Ip vs v^(1/2). It should fit a straight line and from the slope you can obtain the D.

However, you should consider:

- The peak you choose is related to what electrochemical reaction? You have to know the number of electrons exchanged. Then, you have to use the C of the ionic specie involved in the reaction studied.

- If you use a foam as electrode you have to know that your geometrical area is lower than your active surface. Then you should first determine your active surface. You can do that by using a well-known electrolyte, as ferricyanide, with your Ni foam.