Chapter 17: Q38P (page 431)

Peak current $(I_{P})$ 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 $I_{P}$ versus $\sqrt{v}$ 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 $0.0201 {cm}^{2}$ , and the concentration of reactant is 1.00mM.

Short Answer

Expert verified

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

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

is the number of electrons

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

C is the concentration (mol/L)

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

v is sweep rate ( V/ s )

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

$The formula for I_{P} is : I_{P} = (2.69 . 10^{8}) . n^{\frac{3}{2}} . A . C . D .^{\frac{1}{2}} . V^{\frac{1}{2}} So, the slope is equal to I_{P} / v^{\frac{1}{2}} . The diffusion coefficient is : D = \frac{{slope}^{2}}{{(2.69 . 10^{8})}^{2} . 1^{3} . {(0.0201 . 10^{- 4} m^{2})}^{2} . (1 . 10^{- 3} M)^{2}} D = 7.83 . 10^{- 10} m^{2} / s$