Calculate Diffusion Coefficient K4Fe(CN)6 Cyclic Voltammetry

[zeromaxxx]

Homework Statement

Calculate the diffusion coefficient (cm2/s) of ferricyanide if cyclic voltammograms conducted on a solution of 1 mM KClO4 + 5 mM K4Fe(CN)6 at scan rates of 1, 2, 5, 10, 20, and 50 mV/s, resulted in peak currents of 76, 100, 175, 243, 348 and 552 mA. The electrode used for the experiment had been modified with a polymeric coating an effective area of 0.56 cm².

Homework Equations

i_pc=(269000)n^3/2AD^1/2C v^1/2

A=0.56 cm²
n=1
C= 0.000005 mol/cm³

The Attempt at a Solution

I ploted the i_pc vs. v^1/2 and found the slope of the line which was approx. 76.747

i_pc/v^1/2 = 76.747

Isolating for D =[76.747/((269000)n^3/2AC)]²

=[76.747/((269000)1^3/20.56x0.000005)]²

= 10 382.51

which is clearly wrong since the theoretical value of D is 7.6 x 10^-6

Can anyone point out my mistake? I have a feeling it's with the concentration though I'm not quite sure...

 

[Borek]

What units i_pc should be in? mA or A?

 

[zeromaxxx]

What units i_pc should be in? mA or A?

it should be A but does that make a difference when the slope is calculated? since v^1/2 is in mV/s and the slope would pretty much amount to the same value of 76.747?

 

[Borek]

it should be A but does that make a difference when the slope is calculated? since v^1/2 is in mV/s and the slope would pretty much amount to the same value of 76.747?

That's not how it works.

 

[Borek]

it should be A but does that make a difference when the slope is calculated? since v^1/2 is in mV/s and the slope would pretty much amount to the same value of 76.747?

That's not how it works.