Notes by MIT Student (and MZB)
10.626
05/09/2014
Linear Sweep Voltammetry
Linear Sweep Voltammetry → V ( t ) =
± St where S is the scan rate (change in voltage per time).
Definition of Current: I =
dQ
where Q is charge
dt
Definition of an ideal Capacitor: Q = CV where C = C (V ) is the capacitance which can be a function
of voltage but not time
Substitution: I =
d ( CV )
dV
= C
= ±CS
dt
dt
The following plot is obtained when performing cyclic voltammetry if C is a constant with respect to
voltage, there is no resistance, and the scan has voltage bounds of Vlow and Vhigh :
If linear sweep voltammetry is performed at multiple scan rates on a system of constant capacitance,
the magnitude of the current will increase with scan rate.
Ideal Solution/Lattice Gas Model for Lithium Ion Intercalation
Li + + e − → Li( sol )
O + e− → R
1
Notes by MIT Student (and MZB)
Let x =
10.626
05/09/2014
CLi+
Cmax, Li+
It may help to think of the lithium ion intercalation as a binary mixture of empty spaces that could
contain lithium ions and spaces that contain lithium ions.
CLi+
Definition of activity: aR ≈
CLi+
Cmax, Li+ − CLi+
C
+
x
= max, Li =
CLi+
1− x
1−
Cmax, Li+
a=
1.
Assuming a=
O
e
From the Nernst Equation the following relationship can be obtained:
∆φ θ −
Interfacial Voltage at Cathode = ∆φeq =
kT  aR 
kT  x 
∆φ θ −
ln 
ln 
=

e  aO ae 
e  1− x 
Solving for x,
∆φeq =
∆φ θ −
2kT
tanh −1 ( 2 x − 1)
e
Note: The definition of tanh
−1
( y) =
→
 ( ∆φ θ − ∆φeq ) e  1
1
+
x = tanh 

 2
2
2kT


1  1+ y 
y +1
ln 
it
 . If we define a new variable y = 2 x − 1 → x =
2  1− y 
2
 x 
 1− x 
−1
becomes easier to see how=
ln 
 2 tanh ( 2 x − 1) .
 y +1 

 1  1 + y 
 1+ y 
x


2 =
=
ln 
ln 
2*  ln 
2 tanh −1 ( y ) =
2 tanh −1 ( 2 x − 1)


 =
 → ln 
+
y
1
 1− x 
 1− y 
 2  1 − y 
 1−


2 
Charge Equation:
=
Q
=
( charge of Li ) * ( amount of Li
)
+
+
=
eC
eCmax, Li+ x
Li +
2
 ∆φ θ − ∆φeq ) e 
dQ
dx e Cmax, Li+
2 (

From Pseudocapacitance: CF =
sech 
eCmax, Li+
−
=
=


2kT
2kT
dV
dV


sech
=
( z)
1
2
= z
cosh ( z ) e + e − z
2
Notes by MIT Student (and MZB)
10.626
05/09/2014
The figure above is the result of only faradaic pseudocapacitance assuming there is a finite amount of
reactant and therefore a finite reaction capacity. Peak occurs at equilibrium potential, V0. Also, size
increases linearly with increasing scan rate.
For an infinite reservoir, it would have an infinite capacity and would go to infinity as voltage goes to
infinity.
3
Notes by MIT Student (and MZB)
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05/09/2014
Generally, when doing cyclic voltammetry every peak corresponds to a reaction taking place in the
system. The system in Figure 4 is indicative of three reactions in a given system. The reactions take place
at approximately -1, 1, and 4 volts.
The slower the scan rate, the more equilibrated the system becomes at each point in time. Therefore,
the peaks will more closely line up at each V0 the slower the scan is performed. Increasing the scan rate
leads to more transient behavior observed in the reaction and in the transport.
4
Notes by MIT Student (and MZB)
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05/09/2014
To reiterate, if the points in the CV are generated near equilibrium ( S → 0 ) it will create a symmetric
peak shift that linearly decreases in size with scan rate. At low scan rates this behavior is indicative of a
capacitor because minimal resistance is present while the system is at equilibrium. If the scan rate is
large, resistance comes into play and asymmetries begin to occur.
 ( ∆φ θ − ∆φeq ) e 

sech 2 


2kT
2kT


 ( ∆φ θ − ∆φeq ) e 
e 2Cmax, Li+

lim CF ≈
exp 


2kT
2kT
( ∆φθ −∆φeq )→0


CF =
e 2Cmax, Li+
If CV was performed at 0 Kelvin, there would not be any width to the peak. The width is related to
prefactor in the exponential,
kT
e
≈ 26mV .
. At room temperature
e
2kT
5
Notes by MIT Student (and MZB)
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05/09/2014
Regular Solution Model
∆=
=
Ω
S M k *ln(Ω) where
number of configurations =
Ns !
N
∏n !
i =0
i
∆H M ≠ 0
G= H − TS
s=
G
S
H
and h =
and g =
Ns
Ns
Ns
Using Stirling’s Approximation, the intensive entropy can be calculated.
(
s
)
 N + + N Hole ! 

k ln  Li
 N Li + !* N Hole ! 
N Li+

 → −k x ln( x) + (1 − x=
) ln(1 − x) ) where x
(
N Li+ + N Hole
N Li+ + N Hole
The assumed form of the enthalpy of mixing is shown below.
h= h0 ( x )(1 − x )= h0 ( x − x 2 ) where h 0 is the pair interaction energy
h0 > 0 : Attractive forces between particles → each pair lowers the energy → −h0 x 2
Can also be thought of as a repulsive force between a hole/vacancy and a particle on a lattice.
Combining these two expressions yields the intensive Gibbs energy of mixing.
g= h0 ( x − x 2 ) + Tk ( x ln( x) + (1 − x) ln(1 − x) ) + µ0 x
A plot of g is shown below.
6
Notes by MIT Student (and MZB)
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At high temperatures, the enthalpy term becomes negligible compared to the entropy term. Physically
speaking, interactions between particles become meaningless compared to their arrangement.
This figure below was derived at a different temperature. It shows that the Gibbs energy can be
minimized by separating into two stable phases, one that is lithium ion rich, and another that is lithium
ion deficient.
The derivative of g with respect to x yields the diffusional chemical potential.
µ=
dg
 x 
= kT ln 
 + h0 (1 − 2 x ) + µ0
dx
 1− x 
7
Notes by MIT Student (and MZB)
10.626
05/09/2014
The second term on the right hand side in the above equation is the interaction energy.
 x 
kT ln 
 = kT ln ( x ) − kT ln (1 − x )
 1− x 
The first term on the right hand side of the equation above is the particle entropy. The second term is
the hole/vacancy entropy.
Finally, I can write down an expression for the open circuit voltage as a function of mean filling fraction
using this final relationship.
V=
V0 −
O
VO = V 0 +
µ ( x)
e
h0
Tk
( 2 x −1) + ( ln(1− x) − ln(x) )
e
e
8
MIT OpenCourseWare
http://ocw.mit.edu
10.626 Electrochemical Energy Systems
Spring 2014
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