Question

Measuring the size of a microelectrode by cyclic voltammetry.

(a) Redox chemistry for ferrocyanide in Figure 17-32 was given at the beginning of Section 17-5. Write the analyte half-reaction that occurs at the upper plateau near 0.4 Vand at the lower plateau near 0 V(versus S.C.E.).

(b) The limiting current ${\mathrm{I}}_{limit}$, which is the difference between the upper and lower plateaus, is related to the radius of the disk-shaped electrode (r)and the diffusion coefficient (D)and bulk concentration (C)of the analyte:

${\mathrm{I}}_{limit}4nFDCr$

Where nis the number of electrons in the half-reaction and F is the Faraday constant. In this equation, the units of concentration should be $\mathrm{mol}/{\mathrm{m}}^3$to be consistent with the other quantities in SI units. The diffusion coefficient for ferrocyanide cited in the reference for Figure 17 - 32is $9.2\times {10}^{-10}\mathrm{m}/\mathrm{s}$in water at${25}^{°}\mathrm{C}$. Calculate the radius of the microelectrode.

Short answer

a. The half-reaction for the upper plateau and lower plateau of the given graph has to be written.

b. The radius of a microelectrode in cyclic voltammetry for ferrocyanide has to be calculated.

Step-by-step solution

Define Karl Fischer titration

Half reaction is one of oxidation or reduction in redox reaction. Half reaction is given by oxidation state change in reactants or products in redox reaction.

Limiting current is given as

${\mathrm{I}}_{\mathrm{limit}}=4\mathrm{nFDCr}$

r is radius of electrode.

D is diffusion coefficient.

C is analyte's concentration

F is faraday constant.

n is number of electrons transported in half reaction.

a) Determine the half-reaction for the upper plateau and lower plateau of the given graph using Karl Fischer titration

$$\begin{gathered} {\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]}^{4-}+4\mathrm{NaF}\to {\mathrm{Na}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]+4{\mathrm{F}}^{-} \\ \mathrm{the}\mathrm{half}\mathrm{reaction}\mathrm{is}\mathrm{given}\mathrm{as} \\ [{\mathrm{Fe}}_{]}^{{\left(\mathrm{CN}\right)}_6}+4\mathrm{NaF}\to {\mathrm{Na}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]+4{\mathrm{F}}^{-} \\ [\mathrm{F}{\mathrm{e}}_{]}^{{\left(\mathrm{CN}\right)}_6}\stackrel{\mathrm{oxidation}}{\to }[\mathrm{Fe}{\left(\mathrm{CN}\right)}_{6]}+4{\mathrm{e}}^{-} \\ 4\mathrm{NaF}+4{\mathrm{e}}^{-}\stackrel{\mathrm{Reduction}}{\to }+4{\mathrm{F}}^{-} \end{gathered}$$

b) Determine the radius of the microelectrode

We have equation$={\mathrm{I}}_{\mathrm{limit}}=4.\mathrm{n}.\mathrm{F}.\mathrm{D}>\mathrm{r}.\mathrm{C}$

So, the radius is:

localid="1654945789785" $$\begin{gathered} \mathrm{r}=\frac{{\mathrm{I}}_{\mathrm{limit}}}{4.\mathrm{n}.\mathrm{F}.\mathrm{D}.\mathrm{C}} \\ \mathrm{r}=\frac{18.5.{10}^{-9}\mathrm{A}}{4.1.96485\mathrm{C}/\mathrm{mol}.9.2.{10}^{-10}{\mathrm{m}}^2/\mathrm{s}.10\mathrm{mol}/{\mathrm{m}}^3} \\ \mathrm{r}=5.21.{10}^{-6}\mathrm{m} \\ \mathrm{r}=5.21\mathrm{\mu m} \end{gathered}$$