Question

Peak current $\left({\mathrm{I}}_{\mathrm{P}}\right)$and scan rate (v) are listed for cyclic voltammetry of the reversible reaction Fe( II ) Fe(III ) of a water-soluble ferrocene derivative in 0.1MNaCLIf a graph of ${\mathrm{I}}_{\mathrm{P}}$versus$\sqrt{\mathrm{v}}$gives a straight line, then the reaction is diffusion controlled. Prepare such a graph and use it to find the diffusion coefficient of the reactant from Equation17 - 21for this one-electron oxidation. The area of the working electrode is$0.0201{\mathrm{cm}}^2$, and the concentration of reactant is 1.00mM.

Short answer

The graph between peak current and voltage for the given reaction has to be sketched.

Step-by-step solution

Define reversible reaction

In a reversible reaction, the peak current of first cycle is relative to strength of analyte and square root of the sweep rate

${\mathrm{I}}_{\mathrm{Pc}}=(2.69\times {10}^8){\mathrm{n}}^{\frac{3}{2}}{\mathrm{ACD}}^{0.5}{\mathrm{v}}^{0.5}$

is the number of electrons

A is the area of the electrode$({\mathrm{m}}^{2)}$

C is the concentration (mol/L)

D is diffusion coefficient of electroactive species $({\mathrm{m}}^2/\mathrm{s})$

v is sweep rate ( V/ s )

Determine the graph between peak current and voltage for the given reaction.

$$\begin{gathered} \mathrm{The}\mathrm{formula}\mathrm{for}{\mathrm{I}}_{\mathrm{P}}\mathrm{is}: \\ {\mathrm{I}}_{\mathrm{P}}=(2.69.{10}^8).{\mathrm{n}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.\mathrm{A}.\mathrm{C}.\mathrm{D}{.}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.{\mathrm{V}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \mathrm{So},\mathrm{the}\mathrm{slope}\mathrm{is}\mathrm{equal}\mathrm{to}{\mathrm{I}}_{\mathrm{P}}/{\mathrm{v}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}. \\ \mathrm{The}\mathrm{diffusion}\mathrm{coefficient}\mathrm{is}: \\ \mathrm{D}=\frac{{\mathrm{slope}}^2}{{(2.69.{10}^8)}^2.1^3.{(0.0201.{10}^{-4}{\mathrm{m}}^2)}^2.(1.{10}^{-3}\mathrm{M}{)}^2} \\ \mathrm{D}=7.83.{10}^{-10}{\mathrm{m}}^2/\mathrm{s} \end{gathered}$$