Cyclic Voltammetry App: Simulation walkthrough
Last updated: September 20, 2020
In this post, I’ll explain the cyclic voltammetry simulation I’ve created in greater detail. I’ve chosen to document this simulation with MATLAB code since its syntax is easy to follow. You can find the full MATLAB script file here.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% CVsim.m  Cyclic voltammetry simulation
% Peter Attia
% Based on Bard and Faulkner, Appendix B
% EC mechanism
% Updated September 20, 2020
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear, clc, close all
Again, this simulation is welldescribed in Bard and Faulkner, Appendix B. Highly recommended for all the details. This walkthrough is designed to serve as a “practical” guide to performing these simulations, as well as to make Bard and Faulkner’s occasionally dense writing style more accessible.
Set independent variables
%% INDEPENDENT VARIABLES %%
C = 1.0; % [=] mol/L, initial concentration of O. Default = 1.0
D = 1E5; % [=] cm^2/s, O & R diffusion coefficient. Default = 1E5
etai = +0.2; % [=] V, initial overpotential (relative to redox potential). Default = +0.2
etaf = 0.2; % [=] V, final overpotential (relative to redox potential). Default = 0.2
v = 1E3; % [=] V/s, sweep rate. Default = 1E3
n = 1.0; % [=] number of electrons transfered. Default = 1
alpha = 0.5; % [=] dimensionless chargetransfer coefficient. Default = 0.5
k0 = 1E2; % [=] cm/s, electrochemical rate constant. Default = 1E2
kc = 1E3; % [=] 1/s, chemical rate constant. Default = 1E3
T = 298.15; % [=] K, temperature. Default = 298.15
These variables are the “adjustable” parameters in this simulation. These parameters should be straightforward to someone interested in cyclic voltammetry simulations. I’ll note that $ k^0 $ is the electrochemical rate constant, with units of $ \text{cm/s} $ , and $ k_c $ is the chemical rate constant, with units of $ \text{1/s} $. You can read more about these terms on the fundamentals page.
Set physical constants
%% PHYSICAL CONSTANTS %%
F = 96485; % [=] C/mol, Faraday's constant
R = 8.3145; % [=] J/molK, ideal gas constant
f = F/(R*T); % [=] 1/V, normalized Faraday's constant at room temperature
These variables are fundamental physical constants. The Faraday constant is often normalized by $ RT $.
In my work, I don’t consider temperatures other than room temperature. If your work involves variable temperature, you can move $ T $ from “constants” to “independent variables”.
Set simulation variables
%% SIMULATION VARIABLES %%
L = 500; % [=] number of iterations per t_k (pg 790). Default = 500
DM = 0.45; % [=] model diffusion coefficient (pg 788). Default = 0.45
$ L $ and $ D_M $ are two intrinsic simulation variables.
$ L $ is the number of timesteps in our simulation. Our experiment will take some amount of real time ($ t_k $, as discussed below). For a trivial example, if we sweep from $ +1 \text{V} $ to $ 1 \text{V} $ and back at a rate of $ 1 \text{ V/s} $, we are simulating a total time of $ 4\text{ s} $. If $ L = 500 $, each timestep will be $ 4000 \text{ ms / } 500 = 8 \text{ ms} $.
Two notes on $ L $:

$ L $ controls the resolution of our experiment. A low value of $ L $ will lead to “choppiness” in the output; a high value of $ L $ is computationally expensive. Bard and Faulkner recommend $ L $ to be between 100 and 1000. However, “computationally expensive” in 2001 (the publication year of Bard and Faulkner) is different than “computationally expensive” today. My 2014 MacBook Air ran this simulation with $ L = 10000 $ in 0.4s. This simulation is an analytical form of a set of differential equations, so it’s already computationally inexpensive relative to its original differential form. In my experience, $ L = 500 $ produces smooth curves.

Changing $ L $ will change the magnitude of dimensionless current, so don’t change $ L $ when comparing two simulations  unless, of course, you’re interested in understanding the effect of $ L $ 😉.
$ D_M $ is the model diffusion coefficient. $ D_M $ is defined as $ D\Delta t / \Delta x^2 $. A “model diffusion coefficient” is used to ensure that one implicit assumption is satisfied: within a single timestep, material can only diffuse between nearestneighbor boxes. If $ \Delta x $ is too small for a given $ \Delta t $, the simulation becomes unphysical. For example, if $ D_M = 0.5 $, a masstransfer operation between a full box and an empty box will leave both boxes halffull, which violates Fick’s laws. Thus, we must satisfy the condition that $ D_M < 0.5 $. Bard and Faulkner recommend $ D_M = 0.45 $, which we retain here.
Since we’ve already defined $ D $ and $ \Delta t $ is controlled by $ L $, $ D_M $ specifies $ \Delta x $. The implication of this definition is that while $ x $ and $ t $ are physically independent, $ \Delta x $ and $ \Delta t $ are dependent in this analytical simulation. In other words, the spatial resolution is a function of the temporal resolution.
Calculate temporal and spatial step size
%% DERIVED CONSTANTS %%
tk = 2*(etaietaf)/v; % [=] s, characteristic exp. time (pg 790). In this case, total time of fwd and rev scans
Dt = tk/L; % [=] s, delta time (Eqn B.1.10, pg 790)
Dx = sqrt(D*Dt/DM); % [=] cm, delta x (Eqn B.1.13, pg 791)
j = ceil(4.2*L^0.5)+5; % number of boxes (pg 792793). If L~200, j=65
This section shows the derived constants. I’ll walk through each of them individually:

$ t_k $ (
tk
) is the total experiment time, or a “characteristic experimental duration”. $ t_k $ is calculated from multiplying the total voltage swept by a scan in one direction by two (to account for both sweeps) and dividing by the voltage sweep rate. 
$ \Delta t $ (
dt
) is the size of each timestep, given the total experiment time and the number of timesteps we desire, $ L $. 
$ \Delta x $ (
dx
) is the width of each box. It’s controlled by $ D_M $, as discussed above. 
$ j $ (
j
) is the number of ‘finite elements’, or ‘boxes’, in the length dimension. The size of each box, $ \Delta x $, is already set, but we can calculate the number of boxes required by estimating the diffusion layer length in units of boxes. 3D diffusion proceeds as $ 6(Dt)^{0.5} $. In units of ‘boxes’, we need $ j_{max} = 6(Dt)^{1/2}/\Delta x + 1 $. With some algebra, we find that $ j_{max} < 4.2 L^{1/2} $. I added the+5
since it can’t hurt.
Calculate reversibility parameters
%% REVERSIBILITY PARAMETERS %%
ktk = kc*tk % dimensionless kinetic parameter (Eqn B.3.7, pg 797)
km = ktk/L % normalized dimensionless kinetic parameter (see bottom of pg 797)
Lambda = k0/(D*f*v)^0.5 % dimensionless reversibility parameter (Eqn 6.4.4, pg. 236239)
These parameters control the extent of chemical and electrochemical reversibility. We can estimate the shape of the IV curve just by knowing the values of these parameters. I discuss chemical and electrochemical reversibility here.

$ k_c t_k $ (
ktk
) is the dimensionless kinetic parameter. Specifically, it is the dimensionless chemical kinetic parameter, capturing the effect of the $R \overset{k_c}{\rightarrow} Z $ reaction. $ k_c t_k $ controls the extent of chemical reversibility. 
$ k_m $ (
km
) is the normalized dimensionless chemical kinetic parameter. This value is convenient in future calculations. 
$ \Lambda $ (
Lambda
) is the dimensionless electrochemical reversibility parameter, defined by $ \frac{k^0}{(Dfv)^{0.5}} $. $ \Lambda $ is an indicator of the ratio between the rates of charge transfer and mass transfer.
Send warnings to user
%% CHEMICAL REVERSIBILITY WARNING %%
if km>0.1
warning(['k_c*t_k/l equals ' num2str(km) ...
', which exceeds the upper limit of 0.1 (see B&F, pg 797)'])
end
In my experience, this simulation only breaks under one condition, documented by Bard and Faulkner (pg 797). One limitation of the choice of a finiteelement model is that the approximation breaks down at extreme conditions. In this case, if $ k_c $ is too large, the rate of chemical reaction significantly exceeds the time resolution of the simulation, $ t_k/L = \Delta t $. For example, if $ k_c = 10 \text{ s}^{1} $ and $ \Delta t = 0.1 \text{ s} $, we turn over product faster than we can compute the changes in concentration. This condition leads to numerical instabilities.
According to the text, the limit is $ k_c t_k/L > 0.1 $. My code warns you about this, but allows you to proceed. If this condition is not satisfied, you may see infinite current “spikes” when you run the simulation. To study a system with a high value of $ k_c $, increase $ L $.
Preinitialize variables
Almost there. The next section is mostly preinitialization:
%% PREINITIALIZATION %%
C = C / 1000; % Convert C from mol/L to mol/cm3
k = 0:L; % time index vector
t = Dt * k; % time vector
eta1 = etai  v*t; % overpotential vector, negative scan
eta2 = etaf + v*t; % overpotential vector, positive scan
eta = [eta1(eta1>etaf) eta2(eta2<=etai)]'; % overpotential scan, both directions
Enorm = eta*f; % normalized overpotential
kf = k0.*exp( alpha *n*Enorm); % [=] cm/s, fwd rate constant (pg 799)
kb = k0.*exp((1alpha)*n*Enorm); % [=] cm/s, rev rate constant (pg 799)
O = C*ones(L+1,j); % [=] mol/cm^3, concentration of O
R = zeros(L+1,j); % [=] mol/cm^3, concentration of R
JO = zeros(1,L+1); % [=] mol/cm^2s, flux of O at the surface
Again, I’ll walk through the variables:

$ k $ (
k
) is the time index.k
is a vector. Sincek = 0:L
(inclusive), we require $ L + 1 $ boxes, instead of $ L $ boxes. 
$ t $ (
t
) is the simulation step time, in seconds.t
is a vector. 
$ \eta_1 $ (
eta1
) and $ \eta_2 $ (eta2
) combine to form $ \eta $ (eta
), which is a vector of the overpotential. We ploteta
at the end. 
$ E_{norm} $ (
Enorm
) is a vector of the dimensionless overpotential in volts. This form is convenient for ButlerVolmer expressions. 
$ k_f $ (
kf
) and $ k_b $ (kb
) are the forward and reverse electrochemical rate constants, given by the ButlerVolmer equation.kf
andkb
are vectors, with values that change withEnorm
. The rate constants have units of $ \text{cm/s} $. 
O
andR
are the concentrations of $ O $ and $ R $, respectively. The initial concentration of $ O $ is $ C $, and t he initial concentration of $ R $ is $ 0 $. TheO
andR
arrays have $ L + 1 $ rows (time steps) and $ j $ columns (length steps).O
andR
are indexed asO(k,j)
andR(k,j)
, or $ \text{(time, length)} $. Note that this convention switches the rows and columns in Bard & Faulkner’s convention. I find mine more intuitive to work with in MATLAB. 
J
is a vector representing the flux of species $ O $ ($ mol/cm^2s $). We need $ L + 1 $ columns (time steps).
Run main simulation
And finally, we run the simulation. We cycle through all possible times; within each timestep, we cycle through all lengths.
%% START SIMULATION %%
% i1 = time index. i2 = distance index
for i1 = 1:L
% Update bulk concentrations of O and R
for i2 = 2:j1
O(i1+1,i2) = O(i1,i2) + DM*(O(i1,i2+1)+O(i1,i21)2*O(i1,i2));
R(i1+1,i2) = R(i1,i2) + DM*(R(i1,i2+1)+R(i1,i21)2*R(i1,i2)) ...
 km * R(i1,i2);
end
% Update flux
JO(i1+1) = ( kf(i1+1).*O(i1+1,2)  kb(i1+1).*R(i1+1,2) ) ./ (1 + Dx/D*(kf(i1+1) + kb(i1+1)) );
% Update surface concentrations
O(i1+1,1) = O(i1+1,2)  JO(i1+1)*(Dx/D);
R(i1+1,1) = R(i1+1,2) + JO(i1+1)*(Dx/D)  km*R(i1+1,1);
end
% Calculate current density, Z, from flux of O
Z = n.*F.*JO * 1000; % [=] A/cm^2 > mA/cm^2, current density
I found Bard and Faulkner’s derivations opaque, since I was always mentally converting between dimenional and dimensionless variables. Additionally, the use of dimensionless current hides important subtleties in the simulation results. I much prefered the derivation published by Brown in this paper, which is written with dimensional variables. In this part of the walkthrough, I’ll primarily follow Brown’s derivations, and reference equations in Bard & Faulkner when appropriate.
Bulk concentrations
The bulk concentrations of $ O $ and $ R $ change due to diffusion and, in the case of $ R $, the subsequent chemical reaction. We start with Fick’s combined 1^{st} and 2^{nd} laws:
\[\frac{\partial C(x,t)}{\partial t} = D\frac{\partial C^2(x,t)}{\partial x^2}\]We apply a discrete approximation to arrive at Eqn B.1.6:
\[C(x,t+\Delta t) = C(x,t) + \frac{D\Delta t}{\Delta x^2}\big(C(x+\Delta x,t)  2C(x,t) + C(x\Delta x, t) \big)\]We substitute simulation variables to arrive at an expression for $ O $. The bulk concentration of $ O $ changes only by diffusion (Eqn B.1.9):
\[O(k+1,j) = O(k,j) + D_M\big[ O(k,j+1)  2O(k,j) + O(k,j1)\big]\]The dynamics of the bulk concentration of $ R $ are identical, except we also need to add the subsequent chemical reaction (Eqn B.3.5):
\[R(k+1,j) = R(k,j) + D_M\big[ R(k,j+1)  2R(k,j) + R(k,j1)\big]  \big(k_c \Delta t \big) R(k,j)\]Every timestep, $ k_c \Delta t $ molecules of $ R $ convert into the chemical product.
Current
Bard and Faulkner write an equation for current in Eqn B.4.9:
\[\frac{i(t)}{nFA} = k_fO(x=0,t)  k_bR(x=0,t)\]However, Bard and Faulkner use the American current convention,
but I prefer the IUPAC convention (negative currents for negative overpotential, or
potentials less than the redox potential).
This page
details the differences between the two conventions;
current conventions are one of the more confusing elements in electrochemistry,
as this paper documents.
I use the IUPAC current convention in the rest of this walkthrough.
If you prefer the American current convention used in Bard and Faulkner,
you can simply change plot(eta,Z)
to plot(eta,Z)
at the end of this script,
with no loss of fidelity.
Using the IUPAC current convention, the above equation becomes:
\[\frac{i(t)}{nFA} = k_fO(x=0,t) + k_bR(x=0,t)\]We can translate these physical varaibles into simulation variables:
\[\frac{i(k)}{nFA} = k_f(k)O(k,j=1) + k_b(k)R(k,j=1)\]Here, $ O(k,j=1) $ and $ R(k,j=1) $ are the concentrations of $ O $ and $ R $ at timestep $ k $ in box $ 1 $, which we take to be the surface. To calculate the current, then, we need to know $ O(k,j=1) $ and $ R(k,j=1) $. We can obtain these using four equations for flux, $ J $. The first two come from Fick’s first law:
\[J_O(x,t) = D_O\frac{\partial C_O(x,t)}{\partial x}\] \[J_R(x,t) = D_R\frac{\partial C_R(x,t)}{\partial x}\]Again, we assume $ D_O = D_R = D $. For box 1 ($ j = 1 $), we can translate these equations using simulation terms:
\[J_O(k,1) = D\frac{O(k,2)  O(k,1)}{\Delta x} \Rightarrow O(k,1) = O(k,2)  J_O\frac{\Delta x}{D}\] \[J_R(k,1) = D\frac{R(k,2)  R(k,1)}{\Delta x} \Rightarrow R(k,1) = R(k,2)  J_R\frac{\Delta x}{D}\]At the surface, we can define the current by the flux of $ O $ (pg 792 & B.4.19). This relationship gives us another equation for the electrodeelectrolyte interface:
\[\frac{i(k)}{nFA} = J_O(k) = k_f(k)O(k,j=1) + k_b(k)R(k,j=1)\]Lastly, at the interface, the flux of $ O $ is equal and opposite to the flux of $ R $:
\[J_O(k) = J_R(k)\]We can now substitute variables into the third equation:
\[J_O(k) = k_f(k)\left( O(k,2)  J_O\frac{\Delta x}{D} \right) + k_b(k)\left( R(k,2)  J_R\frac{\Delta x}{D} \right)\] \[J_O(k) = k_f(k)O(k,2)  k_f(k)J_O\frac{\Delta x}{D}  k_b(k)R(k,2)  k_b(k)J_O\frac{\Delta x}{D}\] \[J_O(k)\left(1 + k_f(k)\frac{\Delta x}{D} + k_b(k)\frac{\Delta x}{D}\right) = k_f(k)O(k,2)  k_b(k)R(k,2)\]With one more rearrangement, we obtain:
\[J_O(k) = \frac{k_f(k)O(k,2)  k_b(k)R(k,2)} {1 + k_f(k)\frac{\Delta x}{D} + k_b(k)\frac{\Delta x}{D}}\]That’s a lot of $ k $’s! We use this equation to calculate the surface flux of $ O $.
At the end, we calculate current density, $ Z $, from flux:
\[\frac{i(k)}{A} = nFJ_O\]Surface concentrations
We derived equations for the surface (box 1) concentrations of $ O $ and $ R $ in the previous section from a discrete approximation of Fick’s first law. We account for changes in concentration due to diffusion, electrochemical reaction, and, in the case of $ R $, the subsequent chemical reaction.
\[J_O(k,1) = D\frac{O(k,2)  O(k,1)}{\Delta x} \Rightarrow O(k,1) = O(k,2)  J_O\frac{\Delta x}{D}\] \[J_R(k,1) = D\frac{R(k,2)  R(k,1)}{\Delta x} \Rightarrow R(k,1) = R(k,2) + J_O\frac{\Delta x}{D}  \left( k_c \Delta t \right) R(k1,1)\]Plot results
%% PLOT RESULTS %%
% Sometimes length(eta) = length(Z) + 1. If this is the case, truncate last value
if length(eta) > length(Z)
eta = eta(1:end1);
end
plot(eta,Z)
xlabel('Overpotential (V)'), ylabel('Current density (mA/cm^2)')
Under some conditions of etai
and etaj
, eta
is one element larger than
Z
. If so, we chop off the last value of eta
.
For $ L = 500 $, this operation will have a negligible effect on the output.
With that, we’re done! Plot eta
vs Z
to see the final result.
The Bard and Faulkner current convention is counterintuitive to me, so
I plot Z
instead of Z
.
The results of a simulation using the default values is below: