Dear Physics Forum: Calculus problem (Physical Chemistry)

In summary, the conversation was about the Randles-Sevcik equation used in cyclic voltammetry, which relates the peak current to the scan rate. The gradient of the graph of I against v was also discussed, and it was expressed as dI/dV = k/2√v, where k is a constant. The user also requested a step-by-step procedure for finding this gradient.
  • #1
Luke Attigan
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0

Homework Statement



In the electrochemists technique of cyclic voltammetry, the peak current, I, is a simple function of the scan rate, v, according to the Randles-Sevcik equation:

Homework Equations



I = 0.4463nFAc(nF / RT)^1/2 D^1/2 V^1/2

Where n is the number of electrons transferred, F is the Faraday constant, D is the diffusion coefficient, and C the concentration of analyte. Write an expression to describe the gradient of a graph of I (as "y") against v (as "x") ie. dI/dV

The Attempt at a Solution



As a chemist, my maths isn't particularly strong, but I tried to answer the problem as best as possible. I treated the equation posted above as I = k*v^1/2, where k is treated as a constant and the answer I got was dI/dV = 1 ÷ 2*√v (assuming that constants when differentiated become zero).

I would be delighted if a member could show me a thorough step by step procedure on how the answer should actually look if I'm incorrect in my logic.

I appreciate your time and concern,

Kindest Regards,

Luke.
 
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  • #2
You started out with ##I = k v^{\frac 1 2}## so$$
\frac{dI}{dv} =\frac k {2v^{\frac 1 2}}$$which is what I think you did. The constant just stays there as a multiple.
 
  • #3
Thank you very much mate. I thought that was correct and as usual I've over complicated it more than required.

Appreciated fully.

L.
 

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