Dear Physics Forum: Calculus problem (Physical Chemistry)

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SUMMARY

The discussion focuses on the application of the Randles-Sevcik equation in cyclic voltammetry, specifically deriving the gradient of the peak current (I) with respect to the scan rate (v). The equation is given as I = 0.4463nFAc(nF / RT)^(1/2) D^(1/2) v^(1/2). The user, Luke, initially simplified the equation to I = k*v^(1/2) and derived dI/dV = 1 ÷ (2*√v). A fellow forum member confirmed that Luke's approach was correct, clarifying that the constant remains as a multiplier during differentiation.

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  • Understanding of the Randles-Sevcik equation in electrochemistry
  • Basic calculus, specifically differentiation techniques
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  • Knowledge of constants such as the Faraday constant (F) and diffusion coefficient (D)
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Luke Attigan
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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.
 
Last edited:
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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.
 
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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