Chapter 17: Q28P (page 430)

For a rotating disk electrode operating at sufficiently great potential, the redox reaction rate is governed by the rate at which analyte diffuses through the diffusion layer to the electrode (Figure 17-15b). The thickness of the diffusion layer is
$δ = 1.61 D^{1 / 3} V^{1 / 6} ω^{- 1 / 2}$
whereis the diffusion coefficient of reactant $(m^{2} / s), v$ is the kinematic viscosity of the liquid
$(= v i s c o s i t y / d e b s i t y = (m^{2} / s), v$ and $ω$ is the rotation rate (radians/s) of the electrode. There are $2 π$ radians in a circle. The current densityis localid="1655441451764" $(A / m^{2})$ is
localid="1655441445229" $C u r r e n t d e b s i t y = 0.62 n F D^{2 / 3} v^{- 1 / 6} ω^{1 / 2} C_{0}$
where nis the number of electrons in the half-reaction, Fis the Faraday constant, and localid="1655441459070" $C_{0}$ is the concentration of the electroactive species in bulk solution localid="1655441466748" $(m o l / m^{3}, not m o l / L)$ Consider the oxidation oflocalid="1655441474339" $F e {(C N)}_{6}^{4 -}$ in a solution of localid="1655441479067" $10.0 m M K_{3} F e {(C N)}_{6} + 50.0 m M K_{4} F e {(C N)}_{6}$ at +0.90V(versus S.C.E.) at a rotation speed oflocalid="1655441490849" role="math" $2.00 \times 10^{3}$ revolutions per minute. 27The diffusion coefficient oflocalid="1655441497131" $F e {(C N)}_{6}^{4 -} is 2.5 \times 10^{- 9} m^{2} / s$ and the kinematic viscosity islocalid="1655441503345" $1.1 \times 10^{- 5} m^{2} / s$ Calculate the thickness of the diffusion layer and the current density. If you are careful, the current density should look like the value in Figure 17-16b.

Short Answer

Expert verified

The current density and thickness of diffusion layer of rotating disc electrode is $7.84 \times 10^{2} A / m^{2}$ The thickness of diffusion layer of rotating disc electrode is $1.53 \times 10^{- 5} m$

Step by step solution

Define Current density

The thickness of the diffusion layer is

$δ = 1.61 D^{1 / 3} v^{1 / 6} ω^{- 1 / 2}$

where is the diffusion coefficient of reactant $(m^{2} / s), v$ is the kinematic viscosity of the liquid $(= viscosity / density = (m^{2 / s}), v$ and $ω$ is the rotation rate (radians/s) of the electrode. There are $2 π$ radians in a circle. The current density $(A / m^{2})$ is

$Current density = 0.62 {nFD}^{2 / 3} v^{- 1 / 6} ω^{1 / 2} C_{0}$

where is the number of electrons in the half-reaction, is the Faraday constant, and $C_{0}$ is the concentration of the electroactive species in bulk solution $(mol / m^{2} not mol / L)$

Calculate the current density and thickness of diffusion layer of rotating disc electrode.

$ω$ is the rate of rotation.

Conversion of revolutions per minute to radians per second.

$(2.00 \times 10^{3} revolutions / \min) (\begin{matrix} \frac{1 \min}{60 s} \end{matrix}) (\frac{2 πradians}{revolution}) = 209 rad / s δ = 1.61 D^{1 / 3} v^{1 / 6} ω^{- 0.5} = 1.61 (2^{5} (1^{11} (209_{rad / s)} 1.53 \times 10^{- 5} m$