Simulation: electrode model
Variables & Constantes
There are 4 charge species to consider a priori in an electrochemical device:
a) Positive and Negative ions in the electrolyte in the pores and in the separator.
b) Electrons and holes in the electrode (solid electronic conductors).
Resistance is much smaller in carbon => the carrier path will be maximized in it.
At high frequency only the pores close to the electrode surface will be reach
In each electrode volume element there is a mixing of two homogeneous phases: the electrode matrix and the pores.
Φ1(x,t) is the electric potential in the matrix.
Φ2(x,t) is the electric potential in the electrolyte “far enough” from the pore surface. We aren’t studying the potential in the double layer….
Φ1(x,t) and Φ2(x,t) depend on the position in the depth of the electrode.
i(x) = i1(x) + i2(x) where i1 is the current in the electrode matrix and i2 is the current in the electrolyte
Current in the electrode matrix
Only electronic conduction in the electrode matrix (carbon),
Ohm law => i1:
Gradient definition => Φ
Liquid phase physical laws
Electrical field Coulombic force =>
Concentration gradient => Fick law
(multiplied by the charge) to get charges diffusion
Opposed to the migration
μ mobility [m2/Vs]Brownian μp = vd/ F. Electrical μe = vd/ E.
kB Boltzmann’s constant
n density of carriers
zj Nr of charge of specie j
c charge molar concentration
Faraday F = A e
c+ = c- = c
ρ = 0
Single charge carrier
In most of the case the electrolyte salt is made of single charge ions
z+ = -z- = 1
J+ = F N+
J- = -F N-
i2 = J+ + J- = F (N+ - N-)
Total current in the electrolyte
Charge storage at the interface
At the interface between the pores and the electrolyte there is a uniform capacitance C = C(x) per unit of pore surface.
The interface surface per unit of volume is given by a
The current involved per unit of volume is J(x,t)
The current of specie j Jj(x,t)
dq change of interface electrode charge
dq+ change of positive ion charge
dq+ /dq = dq- /dq = -1/2
Bring 4 electrons on the electrode surface => bring 2 negative ions and remove 2 positive ions in the electrolyte neighborhood.
Current of charge
J = (J+ + J-)
C capacity per pore surface unit
Consider an electrode volume unit,
Its capacitance is given by aC, where a is the pore surface per electrode volume unit and C is the capacitance per surface unit
The current which flow through the unit surface (cross section of the unit volume) is i2
div i2 = [i2(x+δx)-i2(x)] / δx
dρ/dt = charge trapped in the double layer = aC dΦ2/dt
Continuity equation at the interface
Equations in the electrode
Electronic conduction :
Equations in the separator
No electronic conduction (except selfdischarge through the separator)
i1(x,t) = 0
Field equations + Boundaries
G : charge separation (generation)
R : charge recombination