Comparison between different ways to determine diffusion coefficient and by solving Fick’s equation for spherical coordinates

Abstract

The following document covers the determination of the diffusion coefficient of two powder materials: LiFePO₄ and LiMn₂O₄ by using potentiostatic intermittent titration technique (PITT) and impedance spectroscopy methodology and compares relevant results with the following relation: $D = \frac{{R^2 I_0 }}{{3\alpha Q}}$, which is obtained by solving Fick’s spherical coordinate equation (where I ₀ is the initial step current in the PITT experiment, R is the particle radius, Q is the charge that intercalated during the step, and α is the percentage of the theoretical intercalated charge). This procedure allowed the verification of the validity of the spherical model for the powder materials, the accuracy of the expression proposed for the diffusion coefficient determination, and the correctness of the measures that had been taken.

Appendix

The solution has this form: c(r,t) = X(t)Y(r) + K ₁.

X(t) is only t dependent, Y(r) is only r dependent, and K ₁ is a constant:

$$\frac{{\partial X}}{{\partial t}}Y = D\frac{1}{{r^2 }}\frac{\partial }{{\partial r}}\left( {r^2 \frac{{\partial Y}}{{\partial r}}} \right)X$$

(19)

$$\frac{{\partial X}}{{\partial t}}\frac{1}{X} = D\frac{1}{{r^2 }}\frac{\partial }{{\partial r}}\left( {r^2 \frac{{\partial Y}}{{\partial r}}} \right)\frac{1}{Y}$$

(20)

the left side of the Eq. 20 is only a function of the time, while the right side is only a function of the radius. Therefore, both parts have to be constant, and the solution is:

$$c - K_1 = C{}_1\exp \left( { - Kt} \right)\frac{{A\exp \left( {i\sqrt {\frac{K}{D}r} } \right) + B\exp \left( {i\sqrt {\frac{K}{D}r} } \right)}}{r}$$

(21)

Because this equation must be also valid when r = 0, we impose A = −B, and then assuming C ₁ A = C ₂, we have:

$$c - K_1 = C_2 \exp \left( { - Kt} \right)\frac{{\exp \left( {i\sqrt {\frac{K}{D}r} } \right) + \exp \left( {i\sqrt {\frac{K}{D}r} } \right)}}{r}$$

(22)

assuming $C_2 = \frac{{K_2 }}{{2i}}$ to have a real solution and knowing that

$$\exp \left( {x + iy} \right) = \exp \left( x \right)\left[ {\cos \left( y \right) + i\sin \left( y \right)} \right]$$

(23)

we can write:

$$c - K_1 = K_2 \exp \left( { - Kt} \right)\frac{{\sin \left( {\sqrt {\frac{K}{D}} r} \right)}}{r}$$

(24)

$\sqrt {\frac{K}{D}} R = \lambda \pi$ for λ = 1, $K = \frac{{\pi ^2 D}}{{R^2 }}$ and therefore:

$$c - K_1 = K{}_2\exp \left( { - \frac{{\pi ^2 }}{{R^2 }}Dt} \right)\frac{{\sin \left( {\frac{\pi }{R}r} \right)}}{r}$$

(25)

the boundary conditions are:

$$t \to 0;\;r = R;\;c = c_s \Rightarrow \quad K_1 = c_{\text{s}} $$

(26)

$$t \to 0;\;r = 0;\;c = c_0 \Rightarrow K_2 = \frac{{\left( {c_0 - c_{\text{s}} } \right)R}}{\pi }$$

(27)

so:

$$c_{\text{s}} - c = \frac{{\left( {c_{\text{s}} - c_0 } \right)R}}{\pi }\exp \left( { - \frac{{\pi ^2 }}{{R^2 }}Dt} \right)\frac{{\sin \left( {\frac{\pi }{R}r} \right)}}{r}$$

(28)