Bard discusses the diffusion layer in the first chapter, section 1.4 (introduction to mass transfer controlled reactions). This 1D diffusion layer encompasses the linear concentration gradient. Beyond this distance, the concentration is effectively the same as the bulk solution.
\[\delta(t)=\text{ diffusion layer over time}\]
The moles of a molecule electrolyzed (M) at an electrode is approximately the difference in concentration between the bulk solution and electrode multiplied by the volume encompassed by the area of the electrode and half the diffusion layer. \[M \approx [C_{bulk}-C_{electrode}] \frac{A\delta(t)}{2}\]
For an electrode potential where a molecule is completely reduced/oxidized the \(C_{electrode}\) is effectively zero. \[\text{Where }C_{electrode}=0 \]
\[M = C_{bulk} \frac{A\delta(t)}{2}\]
This equation should essentially be: \[ Mass = Concentration * Volume \]
Therefore: \[ Volume = A*\frac{\delta(t)}{2}\]
Since volume can be described by an area times height (z): \[ Volume = A*z \]
\[ z = \frac{\delta(t_s)}{2}\]
The diffusion layer is defined in terms of \(D_{ap}\) and the scan time \(t_s\): \[\delta(t) = 2 \sqrt{D_{ap}t_s}\]
So finally z can be defined purely in terms of \(D_{ap}\) and the scan time \(t_s\): \[ z = \sqrt{D_{ap}t_s}\]
Therefore the volume A*z interrogated by the electrode depends on Dap (and scan time)!
How can we effectively figure out this scan time, \(t_s\)?? I suppose this could be estimated empirically by measuring on ferrocene with a known \(D_{ap}\)?
Because of the dependency of z on \(D_{ap}\), I don’t think we should try to define Concentration / \(D_{ap}\) using potential step, because we would have to guess a z value…which itself depends on the value we are trying to estimate. Intead, I propose that we estimate \(D_{ap}\) using the titration approach, which should be independent of the Volume/area/z (assuming that \(D_{ap}\) and C are the same for SWV and GC on the same sample).
Then we can use that \(D_{ap}\) in our \(D_{phys}\) estimate, which necessarily depends on a Volume/z value.
Recall that for \(D_{phys}\) We start with the dilution of an initial mass with 1D diffusion next to a no flux boundary: \[C(t) = \frac{2M_0}{A_{xy}\sqrt{4\pi D_{phys}t}} \tag{eq. 1}\] Peak Square Wave current is given by: \[I_{swv} = \frac{nFA_{xy}C \sqrt{D_{ap}}}{\sqrt{\pi t_p}} \psi \tag{eq. 2}\] Concentration can be broken into mass / volume \[C = \frac{M}{V} = \frac{I_{swv}\sqrt{\pi t_p}}{nFA_{xy}\sqrt{D_{ap}}\psi} \tag{eq. 3}\] Therefore initial mass would be: \[M_0 = \frac{I_0\sqrt{\pi t_p}V}{nFA_{xy}\sqrt{D_{ap}}\psi} \tag{eq. 4}\] Substituting into eq. 1 gives: \[\frac{I_{swv}\sqrt{\pi t_p}}{nFA_{xy}\sqrt{D_{ap}}\psi} = \frac{2 \frac{I_0\sqrt{\pi t_p}V}{nFA_{xy}\sqrt{D_{ap}}\psi}}{A_{xy}\sqrt{4\pi D_{phys}t}} \tag{eq. 5}\] Simplified: \[I_{swv} = \frac{2I_0V}{A_{xy} \sqrt{4 \pi D_{phys}t}} \tag{eq. 6}\] The relevant volume should be the surface area of the electrode * the height of the biofilm? \[\frac{V}{A_{xy}}=z \tag{eq. 7}\] Below I plot this normalized decay curve: \[\frac{I_{swv}}{I_0} = \frac{z}{\sqrt{ \pi D_{phys}}}t^{-1/2} \tag{eq. 8}\] Fitting the normalized decay curve againt t^-(1/2) yields this slope: \[slope = m = \frac{z}{\sqrt{ \pi D_{phys}}} \tag{eq. 9}\] Rearranging gives Dphys: \[D_{phys} = \frac{z^2}{\pi m^2} \tag{eq. 10}\]
Slope of the linear fit of GC vs. SWV titration plot: \[slope = m = \frac{\Delta I_{gc}}{\Delta I_{swv}} \tag{eq. 1}\] Generator collector peak current: \[I_{gc} = nFSCD_{ap} \tag{eq. 2}\]
Square Wave peak current: \[I_{swv} = \frac{nFACD_{ap}^{1/2}}{\pi^{1/2}t_p^{1/2}} \psi \tag{eq. 3}\] Substituting into eq. 1: \[m = \frac{I_{gc}}{I_{swv}} = \frac{nFSCD_{ap}}{\frac{nFACD_{ap}^{1/2}}{\pi^{1/2}t_p^{1/2}} \psi} \tag{eq. 4}\] Simplifying: \[m = \frac{S\pi^{1/2}t_p^{1/2}D_{ap}^{1/2}}{A\psi} \tag{eq. 5}\]
\[m^2 = ({\frac{S\pi^{1/2}t_p^{1/2}D_{ap}^{1/2}}{A\psi}})^2 \tag{eq. 6}\]
\[m^2 = \frac{S^2\pi t_p D_{ap}}{A^2\psi^2} \tag{eq. 7}\] Rearranged for Dap: \[D_{ap} = \frac{m^2 A^2 \psi^2}{S^2\pi t_p} \tag{eq. 8}\]