Model for the impedance of a disk electrode according to Newman's models, published in the papers with the following DOIs

• 10.1149/1.2423795
• 10.1149/1.2407464
• 10.1149/1.2424003

Laplace equation and boundary conditions

The first boundary condition describes that the capacitive current equals the current due to electric field:

$$ i=C \frac{\partial\left(V-\Phi_0\right)}{\partial t}=-\left.\kappa \frac{\partial \Phi}{\partial z}\right|_{z=0, r \leq r_0} $$

(later: can add faradaic current too). Potential on the electrode: $V=V_0 e^{j \omega t}$. Potential outside double laver: $\Phi_0$. Newman only considers the diffusion layer (net 0 free charge). Newman considers the double layer as an ideal capacitor.

$$ \begin{aligned} & \text { BC2: }\left.\kappa \frac{\partial \Phi}{\partial z}\right|_{z=0, r>r_0}=0 \\ & \text { BC3: } \Phi\left(z^2+r^2 \rightarrow \infty\right)=0 \end{aligned} $$

Coordinate transformation

New coordinates: ellipsoidal coordinates; $\xi$ : distance from the disk; $\eta$ : angular coordinate.

$$ \begin{aligned} & z=r_0 \xi \eta \\ & r=r_0 \sqrt{\left(1+\xi^2\right)\left(1-\eta^2\right)} \end{aligned} $$

The Laplace equation (Poisson equation for zero net charge in the electrolyte) can then be written as

$$ \frac{\partial}{\partial \xi}\left[\left(1+\xi^2\right) \frac{\partial \Phi}{\partial \xi}\right]+\frac{\partial}{\partial \eta}\left[\left(1-\eta^2\right) \frac{\partial \Phi}{\partial \eta}\right]=0 $$

Newman uses the method of separation of variables; he defines

$$ \Phi=P(\eta) Q(\xi) $$

so that the Laplace partial differential equation can be rewritten into a system of two ordinary differential equations:

$$ \begin{aligned} \frac{d}{d \eta}\left[\left(1-\eta^2\right) \frac{d P}{d \eta}\right]+n P & =0 \\ & \frac{d}{d \xi}\left[\left(1+\xi^2\right) \frac{d Q}{d \xi}\right]-n Q=0 \end{aligned} $$

The solutions to these equations are Legendre polynomials, in Newman's notation $P_{2n}(\eta)$ and $M_{2n}(\xi)$.

To describe the behavior at different frequencies, we define the potential in the solution as $\Phi=V_0 e^{j \omega t} U(r, z)$. If $\Phi$ satisfies Laplace's equation, then $U$ also satisfies it, because all $r$ and $z$-dependence is in $U$. Thus we obtain

$$ U=\sum_{n=0}^{\infty} B_n P_{2 n}(\eta) M_{2 n}(\xi) $$

Solving for the coefficients $B_n$

Using the expression for $\Phi$ in $\mathrm{BC}_1$ we get

$$ j \Omega\left(1-\sum_{n=0}^{\infty} B_n P_{2 n}(\eta)\right)=-\frac{1}{\eta} \sum_{n=0}^{\infty} B_n P_{2 n}(\eta) M_{2 n}^{\prime}(0) $$

$M_{2 n}^{\prime}(0)$ was given in doi: 10.1149/1.2424003. Multiply the above equation by $\eta P_{2 n}(\eta)$ and integrate from 0 to 1 :

$$ B_m=-\frac{4 m+1}{M_{2 m}^{\prime}(0)} j \Omega\left(\int_0^1 \eta P_{2 m}(\eta) d \eta-\sum_{n=0}^{\infty} B_n \int_0^1 \eta P_{2 m}(\eta) P_{2 n}(\eta) d \eta\right) $$

Thus we obtain an infinite set of equations:

$$\left(\frac{1}{j \Omega \mathbf{D}}+\mathbf{M}\right) \vec{B}=\vec{a}$$

Here,

$$D_{mm} = -\frac{4 m+1}{M_{2 m}^{\prime}(0)}$$

(a diagonal matrix),

$$a_m = \int_0^1 \eta P_{2 m}(\eta) d \eta$$

and

$$M_{mn} = \int_0^1 \eta P_{2 m}(\eta) P_{2 n}(\eta) d \eta$$

The infinite-dimensional linear system can be approximated by taking the first few dimensions, because $B_n$ and the integrals become smaller and smaller for increasing $n$. Then we can solve this system with numpy.

Calculating the impedance

The impedance only depends on $B_0$. Following Newman the total current is given by

$$ I=\int_0^{r_0} i 2 \pi r d r=-2 \pi r_0{ }^2 \kappa \left.\int_0^1 \eta \frac{\partial \Phi}{\partial z}\right|_{z=0} d \eta $$

so

$$ I =-2 \pi r_0 \kappa V_o e^{j \omega t} B_0 M_{o^{\prime}}(0) $$

which, with $M_0{ }^{\prime}(0)=-2 / \pi$, leads to

$$ Z=\frac{V}{I}= \frac{1}{4 r_0 \kappa B_0}. $$