Calculate Diffusion Coefficient K4Fe(CN)6 Cyclic Voltammetry

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Discussion Overview

The discussion revolves around calculating the diffusion coefficient of ferricyanide from cyclic voltammetry data. Participants are examining the relationship between peak current and scan rate, as well as addressing potential errors in the calculations and units used.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a calculation attempt for the diffusion coefficient using the equation ipc=(269000)n3/2AD1/2C v1/2, noting a discrepancy between their result and the theoretical value.
  • Another participant questions the appropriate units for peak current (ipc), suggesting it should be in amperes (A) rather than milliamperes (mA).
  • Further discussion on whether the choice of units affects the slope calculation, with some participants asserting that it does not impact the numerical value of the slope.
  • There is a challenge to the understanding of how units interact in the context of the calculations, indicating a need for clarification on the relationships between the variables involved.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct units for peak current or the implications of unit choice on the calculations. The discussion remains unresolved regarding the source of the discrepancy in the diffusion coefficient calculation.

Contextual Notes

Participants express uncertainty about the concentration used in the calculations and its impact on the results. There are also unresolved questions about the proper treatment of units in the mathematical relationships presented.

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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



ipc=(269000)n3/2AD1/2C v1/2

A=0.56 cm2
n=1
C= 0.000005 mol/cm3

The Attempt at a Solution



I ploted the ipc vs. v1/2 and found the slope of the line which was approx. 76.747

ipc/v1/2 = 76.747

Isolating for D =[76.747/((269000)n3/2AC)]2

=[76.747/((269000)13/20.56x0.000005)]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...
 
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What units ipc should be in? mA or A?
 
Borek said:
What units ipc should be in? mA or A?

it should be A but does that make a difference when the slope is calculated? since v1/2 is in mV/s and the slope would pretty much amount to the same value of 76.747?
 
zeromaxxx said:
it should be A but does that make a difference when the slope is calculated? since v1/2 is in mV/s and the slope would pretty much amount to the same value of 76.747?

That's not how it works.
 
zeromaxxx said:
it should be A but does that make a difference when the slope is calculated? since v1/2 is in mV/s and the slope would pretty much amount to the same value of 76.747?

That's not how it works.