Download Flux density

KJM5120 and KJM9120 Defects and Reactions
Ch 7. Electrochemical transport
Truls Norby
Department of Chemistry
University of Oslo
Centre for Materials Science
and Nanotechnology (SMN)
FERMIO
Oslo Research Park
(Forskningsparken)
[email protected]
http://folk.uio.no/trulsn
Electrochemical transport
•
Electrochemistry is red-ox-chemistry
–
–
reduction – uptake of electrons
oxidation – loss of electrons
•
In electrochemistry the reduction and oxidation take
place at different locations.
•
Thus, the overall reaction requires transport of both
electrons and of ions (chemical species).
•
We may expose the material to a gradient in
electrical potential, and this will result in a generation
of a chemical potential gradient.
•
Correspondingly, a chemical potential gradient will
generate an electrical potential gradient.
•
We will in this chapter thus treat coexistent chemical
and electrical potential gradients and simultaneous
transport of at least two charged species.
•
We will use formalisms and relationships from the
two preceding chapters.
Electrochemical potential
•
A charged species i feels both its own chemical potential and an
electrical potential. The combined potential is its electrochemical
potential, ηi (”eta”).
•
The electrochemical potential ηi of a charged species i has a chemical
component and an electrical component:
 i= i+ zi e
•
The electrochemical potential gradient in one dimension:
d i d  i
d
=
+ zi e
dx
dx
dx
Flux equations (“Wagner theory”)
•
We shall now derive general equations for the flux and current
densities of the species i.
•
We shall then consider the result of the simultaneous transport
of all species by summing up all the current densities into a total
(net) current density.
•
Based on this we shall derive the electrical potential gradient and
the voltage over a sample exposed to a chemical potential
gradient.
•
We shall evaluate the case that the net current is zero.
•
We then derive equations for the flux density of individual
species. From this we can evaluate mass transport through the
material in several cases of applications.
•
The procedure is called Wagner theory, after Carl Wagner, who
first derived it and used it for analysis of oxidation of metals.
Flux equations (“Wagner theory”)
A couple of things to be aware of:
• Our procedure (derivation) will be a kind of recipe.
…you are kind of
expected to learn it… 
– We will do the derivation of Wagner theory first in the general case.
– Then we will repeat it for simplified example cases.
• We will use real species, not defects.
– A reason for this is that electrochemical transport involves exchange of mass
and electrons with external phases (e.g. gas phase and electrodes). It is then
easier and more consistent to deal with real species.
– We insert defects only at the end when we want to evaluate the final
expressions – e.g. their pO2 and T dependencies, or their relationship with
component or defect diffusivity.
Flux density of species i
•
The flux density of any species is ideally given by the concentration,
mobility, and the driving force, i.e. the potential gradient:
d i
d i
d Pi
d
 -ci Bi
= -ci Bi [
+ zi e ]
ji = -ci Bi
dx
dx
dx
dx
•
Via the Nernst-Einstein relations we may choose to express mobility
by diffusivity:
ji =
•
- ci Di d i - ci Di d  i
d
=
[
+ zi e ]
kT dx
kT
dx
dx
or conductivity:
-  i d i
d
[
+
e
]
ji =
zi
2
dx
dx
( z i e)
Recommendation:
Learn one or more of
these by heart!
Current density
•
The current density ii of species i is obtained by multiplying its flux
density by its charge:
d i
d

i
[
+ zi e ]
ii = z i e j i = zi e
dx
dx
Total (net) current density
•
The total (net) current density is obtained by summing over all charge
carriers k:
d k
d

k
=
e
=
[
+
e
]
j
k
itot k z k k
zk
zk e
dx
dx
•
(The reason we choose to call all charge carriers k is that we will return to one of
them, i, later on.)
•
•
Total, net current must be extracted or supplied through an external circuit.
Typical considerations of total (net) current:
–
Fuel cell:
• At open circuit, there is no net current: itot = 0
• Under load, the net current itot is non-zero.
–
Mixed ion-electron conducting gas separation membrane:
• There are no electrodes or external circuitry; no possibility of net current
–
Solid-state reaction, e.g. growth of oxide on a corroding metal:
• No electrodes or external circuitry; no possibility of net current
Solve for the potential gradient
•
We now solve the expression for total current density
d k
d
d


k
k d k
[
+ z k e ]   k
 k  k
itot = - k
zk e
dx
dx
zk e dx
dx
with respect to the electrical potential gradient. Rearrange:
d

k d k
  itot  k
k  k
dx
zk e dx
d k
d
itot

k

 k
dx
zk e k  k dx
k  k
 tot = k  k
d
itot
tk d  k
=- k
dx
 tot
z k e dx
tk =
k
k

 tot k  k
Here we use
the definitions
of total
conductivity
and transport
number
The electrical potential gradient
•
The expression we obtained
d
i
t d k
= - tot - k k
dx
 tot
z k e dx
This part represents the voltage
drop resulting from a current
through a material with a certain
conductivity (inverse of resistance).
It corresponds to Ohm’s law. It is
typically
Ohmic part
Ohmic loss
IR-loss
This part is the contribution to the potential gradient from
gradients in chemical potentials of charge carriers with
significant transport numbers tk. It is typically how a fuel cell
gets its voltage from a gradient in the chemical potential of
oxide ions or protons. Or how a battery gets its potential from
a gradient in the activity of e.g. Li+ ions.
However, we have a problem: All the species k in the summation are charge carriers, i.e.
ions or electrons. This means that their chemical potentials μk are not well-defined, and we
will therefore not directly be able to evaluate the expressions.
The next thing we need to do is thus to express chemical potentials of charged species by
chemical potentials of neutral well-defined species.
Chemical potential gradients of charged
vs neutral species
•
We need to assume and use equilibrium between charged and neutral
species. As a general example for species “S”, consider the formation
of a charged ion from the neutral atom
S = S z + ze 
(where z can be positive or negative).
•
Equilibrium in this equation along any coordinate is expressed as
d  S z + zd  e-  d  S  0
or, rearranged,
d  S z  d  S  zd  e•
In the x-direction,
d  Sz
d  ed S

z
dx
dx
dx
This expresses the chemical activity gradient for the charged
chemical species (ion) by the chemical activity gradient for the
neutral atom and for electrons.
We will now insert this for the species in the expression for the
electrical potential gradient.
Insert for chemical potential of ions
•
For all ions n (among k) we now insert for the chemical potential gradient.
d
i
t d k
= - tot - k k
dx
 tot
z k e dx
from all ions
d  Sz
dx

d  ed S
z
dx
dx
electrons
d
itot
tn d n
tn d e te d e
=- n
+ n

dx
e dx
e dx
 tot
z n e dx
d
itot
tn d  n 1 d e
=- n
+
dx
e dx
 tot
z n e dx
We have thus obtained an expression
containing chemical potentials of neutral
atoms, and their ions’ charge and transport
numbers. In addition we have a term
containing the chemical potential of electrons.
This expression contains the essential information of the electrical potential in the sample,
and will be used onwards to obtain “everything else” we want to know.
The voltage over a sample
•
In order to obtain the voltage (potential difference) over a sample, we
integrate the potential gradient
d
itot
tn d  n 1 d e
=- n
+
dx
e dx
 tot
z n e dx
over the thickness of the sample, from side I to side II:
II
II
II
II
1
tn
I d =  I  tot dx - n I z n e d  n + I e d  e
itot
II
 II   I =  
I
itot
 tot
II
dx - n 
I
II
U II  I   II   I =  
I
itot
 tot
1
tn
d  n + (  e, II   e, I )
e
zn e
II
dx - n 
I
Assume same electrode material
on both sides, so that μe,II = μe,I
tn
d n
zn e
Note ohmic part and chemical
potential part, as discussed earlier
The voltage over a sample
II
U II  I   II   I =  
I
•
 tot
II
dx - n 
I
tn
d n
zn e
In case of open-circuit, itot = 0:
II
U II  I  -  n 
I
•
itot
tn
d n
zn e
If n counts only one species, and tn can be assumed
constant, the equation becomes very simple, and is used for
e.g. open circuit voltage of fuel cells and batteries, galvanic
sensors, and for determination of transport numbers. We
shall return to specific cases later on.
If itot = constant ≠ 0 (steady state current) and σtot = constant:
U II  I  
itot X
 tot
II
- n 
I
II
II
tn
t
t
d  n   itot rtot - n  n d  n   IR - n  n d  n
e
e
zn e
I zn
I zn
Note ohmic (“IR”) part and chemical
potential part, as discussed earlier
Flux density of a particular species i
• If we wish to know the flux density of one particular species i, we
insert the obtained electrical potential gradient (from 2 slides back)
d
i
t d k
= - tot - k k
dx
 tot
z k e dx
into the initial expression of flux density for that species i
-  i d i
d
[
+ zi e
]
ji =
2
dx
dx
( z i e)
to obtain
d i
t i itot
i
tk d  k

[
 z i k
]
ji =
2
z i e ( z i e)
dx
z k dx
Flux density of a particular species
• The flux density of species i in a situation with many species (k):
The flux of species i from
total current and its own
transport number
The flux of i from its
conductivity and own
chemical driving force
The flux of i from its conductivity
and electrical potential driving
force controlled by all species
d i
t i itot
i
tk d  k

[
 z i k
]
ji =
2
z i e ( z i e)
dx
z k dx
• The flux density is given for a given position with its gradients.
Flux density of a particular species
d i
t i itot
i
tk d  k

[
 z i k
]
ji =
2
z i e ( z i e)
dx
z k dx
• We now integrate the flux density over the thickness of the
sample, from side I to side II.
II

I
ti itot
i
tk
dx  
[d

d k ]
j idx  


k
i zi
2
zi e
( z i e)
zk
I
I
II
II
Flux density - steady-state
• We further assume that the flux density of i and total (net) current
density are constant through the thickness of the sample:
II

I
i
itot
tk
t
dx

[d

d k ]
j idx = ji X 


k
i
i zi
2


zi e I
( z i e)
zk
I
II
II
• In this way, we have imposed steady state conditions.
X is total
thickness
from I to II
• If the transport number ti can be taken as constant, we have
ti itot
i
tk
ji X 
X 
[d

d k ]


k
i zi
2
zi e
( z i e)
zk
I
II
or, by dividing throughout by X,
ti itot 1
ji 

zi e X
II
i
I
i
 ( z e)
2
[d  i  z i k
tk
d k ]
zk
Note that flux
is inversely
proportional
to thickness
Allrighty!
• That concludes our first travel through “Wagner-space” for
electrochemical transport in solids, using general equations and real
species.
• The recipe was:
– Flux density for each species from diffusivity, mobility, or conductivity, and
electrochemical potential gradient.
– Current density for each species by multiplication with charge.
– Sum all current densities.
– Solve for the electrical potential gradient.
• Obtain the voltage over the sample by integration, if of interest.
– Insert electrical potential gradient back into expression for flux density of
individual species.
– Flux density obtained by integration over sample thickness assuming individual
fluxes and total current are constant (steady state conditions).
• Change from chemical potentials of ions to chemical potentials of
neutral species by assuming equilibrium in the ionisation of the
neutral species.
Specific example;
Mixed oxide ion and electron conductor
•
•
Two mobile species; oxide ions O2- (z = -2) and electrons e- (z = -1).
Flux densities:
jO2 =
•
-  O2
4e 2
d  O2
d
 2e
]
dx
dx
j e =
-  e
e2
[
d  e
d
e
]
dx
dx
Current densities:
iO 2  2e j O 2 =
•
[
O
2
[
2e
d  O 2
d
 2e
]
dx
dx
ie  e j e =
e
e

[
d  e
d
e
]
dx
dx
Total current density:
itot  iO 2  ie 
itot 
 O d O
2
2e
dx
O
2
2
2e

[
  d  e
d  O 2
d
d
 2e
] e [
e
]
dx
dx
e
dx
dx
 e d e

e
dx

 ( O 2   e )
d  O 2 d  O 2  e  d  e
d


  tot
dx
2e dx
e dx
dx
Specific example;
Mixed oxide ion and electron conductor
itot 
•
 O d O
2
2e
dx
2

 e d e

e
dx

  tot
d
dx
Solve with respect to potential gradient:
d
itot t O 2  d  O 2  t e  d  e 
=

dx
e dx
 tot 2e dx
•
•
Insert equilibrium between oxide ions, electrons, and neutral oxygen.
We choose to use oxygen molecules:
2O2 (g) + 4e  2O
•
dO2 ( g ) + 4de   2dO 2
Insertion into expression for potential gradient:
d
itot t O 2  d  O2 ( g ) t O 2  d  e t e d  e
=


dx
e dx
e dx
 tot 4e dx
Specific example;
Mixed oxide ion and electron conductor
d
itot t O 2  d  O2 ( g ) t O 2  d  e t e d  e
=


dx
e dx
e dx
 tot 4e dx
•
Sum of all transport numbers is unity:
d
itot t O 2  d  O2 ( g ) 1 d  e
=

dx
e dx
 tot 4e dx
•
Integrate to obtain the voltage:
II
U II  I
1
  IR 
t O 2  d  O2 ( g )

4e I
Specific example;
Mixed oxide ion and electron conductor
II
U II  I
•
1
  IR 
t O 2  d  O2 ( g )

4e I
Using
 O ( g )   O0 ( g )  kT ln pO
2
2
2
d  O2 ( g )  kTd ln pO2
we obtain
II
U II  I
•
kT
  IR 
t O 2  d ln pO2

4e I
If the transport number is constant, or an average value can be
assumed constant, then
U II  I   IR  t O 2 
II
kT pO2
ln I   IR  t O 2  E N
4e pO2
EN is the Nernst voltage
Specific example;
Mixed oxide ion and electron conductor
U II  I   IR  t O 2 
•
Practical use; Measurement of transport number by open circuit voltage
(OCV, I = 0) in a small gradient in oxygen partial pressure:
tO
•
2

U II  I
EN
Practical use: Galvanic oxygen sensor using the OCV over a pure oxide
ion conducting electrolyte (tO2- = 1) vs a reference pO2:
U II  Ref
•
II
kT pO2
ln I   IR  t O 2  E N
4e pO2
pOII2
kT

ln Ref
4e pO2
Practical use: Fuel cell using an oxide ion conducting electrolyte (tO2- = 1):
U II  I
II
kT pO2
  IR 
ln I   IR  E N
4e pO2
P  UI   I 2 R  IE N
Flux density of oxide ions
•
•
•
Insert potential gradient
d
i
t 2 d  O2 t e d  e
= - tot  O

dx
e dx
 tot 2e dx
into flux density of oxide ions
-  O2 d  O2
d
[

2
e
]
jO2 =
2
dx
dx
4e
and get
t O 2  itot  O 2  t e  d  O 2 
t O 2  itot  O 2  t e  d  O2 ( g )
d  e

[
2
]

jO2  
2
2e
dx
dx
2
e
dx
4e
8e 2
Integrate:
II
j
II
t O 2  itot
 O2 t e
X  
dx  
d  O2 ( g )
2
2e
8e
I
I
II
O2
dx  j O 2 
I
II
jO2
II
kT
i
  tot  t O 2  dx  2   O 2  t e  d ln pO2
2eX I
8e X I
Flux density of oxide ions
Flux due to net current
II
jO2
Flux due to chemical driving
force and mixed conductivity
II
kT
itot

t O 2  dx  2   O 2  t e  d ln pO2

2eX I
8e X I
Flux inversely proportional to thickness X
•
If the total (net, external) current is zero, we may simplify:
II
jO2
II
kT
kT

  2   O 2  t e d ln pO2   2   tot t O 2  t e  d ln pO2
8e X I
8e X I
Alternative representation of
mixed conduction
Assume a simple defect model
II
j O 2
•
II
kT
kT
  2   O 2 t e  d ln pO2   2   tottO 2 t e d ln pO2
8e X I
8e X I
Assume oxygen deficiency, like in MaOb-y:
n  2[vO ]  (2 K v/ )1/ 3 pO12 / 6
O
•
Assume electronic conductivity dominates because of higher mobility
of electrons than of oxygen vacancies:
t e  1
•
Oxygen ion conductivity:
 O   v  2e[v ]uv   0 p
2
jO2

O

O
kT 0
 2
8e X
II

O
1 / 6
p
 O2 d ln pO2
I
1 / 6
O2
σ0 is the oxide
ion conductivity
at pO2 = 1 bar
Assume a simple defect model
jO2
•
II
1 / 6
p
 O2 d ln pO2
I
1
In order to do the integration, we substitute d ln pO2 
dpO2
pO2
to obtain
jO2
jO2
•
kT
 2 0
8e X
kT 0
 2
8e X
II
7 / 6
p
 O2 d pO2
I
6kT 0
II 1 / 6
I
1 / 6

(
p
)

(
p
)
O2
O2
8e 2 X


The smallest pO2 (at the reducing side) will dominate the parenthesis
(because of the negative power) and the highest pO2 (at the oxidising
side) may be neglected to a first approximation. For pO2I>>pO2II:
j O 2
6kT 0 II 1/ 6
 2 ( pO2 )
8e X
Ambipolar conductivity and diffusivity
•
The term for mixed conduction that we used in the example case is
called ambipolar conductivity, and can be expressed in several ways:
 amb =  tot t O t e =  O
2-
-
O e

te
O e
2-
2-
-
2-
•
-
-
We can correspondingly express the mixed transport as ambipolar
diffusivity
D amb = DO 2- t e-
Chemical diffusion coefficient in the case of mixed
oxygen vacancy and electron transport
~ dci
ji = - D
i
dx
•
We have seen that Fick’s 1st law can be useful, but what is the
diffusion coefficient – generally referred to as the chemical diffusion
coefficient - that enters?
•
We will analyse the chemical diffusion coefficient of oxygen in our
example material.
•
In order to approach that, we need to get hold of the concentration
gradient of oxygen in the material. We (may) know the concentration
gradient of oxygen vacancies, so we write and manipulate:
0
O2-
cO 2- =c - cvO..
dcvO..
dcO 2 =dx
dx
dcvO.. d ln pO 2
dcvO.. d ln pO 2
dcO 2 ==dx
dx d ln pO 2
d ln pO 2 dx
Chemical diffusion coefficient in the case of
mixed oxygen vacancy and electron transport
dcvO..
dcO 2 =dx
dx
0
O2-
cO 2- =c - cvO..
•
dcvO.. d ln pO 2
dcvO.. d ln pO 2
dcO 2 ==dx
dx d ln pO 2
d ln pO 2 dx
Insert into flux equation:
jO2  
 O te d  O (g)
2
8e 2

2
dx

 O t e kT d ln pO
2

8e 2
2
dx

 O t e kT d ln pO dcO
2

8e 2
2
dcv
O
•
Compare with Fick’s 1st law:
~ dci
ji = - D
i
dx
and obtain:
~ 2 -    O 2  t e  kT d ln pO2
DO
dcv
8e 2
O
2
dx
Chemical diffusion coefficient in the case of
mixed oxygen vacancy and electron transport
~ 2 -    O 2  t e  kT d ln pO2
DO
dcv
8e 2
O
•
Use Nernst-Einstein and
DO2 cO2- = DV O.. cV O..
to obtain


D c
D
Dv t e
O
~ 2 -   DO 2  cO 2  t e d ln pO2   vO vO t e d ln pO2   vO t e d ln pO2  
DO
d ln cv
2
dcv
2
dcv
2 d ln cv
O
O
O
O
2
d ln pO2
•
The chemical diffusion coefficient is:
– proportional to the self diffusivity of the oxygen defect (here the vacancy),
– proportional to the electronic transport number (but which is often unity),
– enhanced by the term d ln pO 2 /2d ln cV O which for various limiting cases of
simplified defect situations takes on values of 1, 3 or 4, and
– fully forwards and backwards transformable into defect diffusivities (and in
turn self diffusivities) if we know the transport numbers and how defect
concentrations vary with pO2.
Surface and electrode kinetics limitations
•
For electrochemical cells with electrodes and external circuitry, we
often express the voltage over the external circuit (load) as consiting of
the Nernst voltage minus the IR drop and the electrode overpotentials:
E = E N - IRi - (  c + a )
or
EN = E  IRi   c + a
ηc
EN
EOCV = EN
EN
E < EN
ηa
•
Membranes without electrodes may have surface kinetics limitations;
the effect is the same, but the electrode overpotentials must then be
replaced by chemical potential drops
Typical redox reactions
Example of individual steps (cathode):
O2 (g)  O2,ads
O 2,ads + e  O 2,ads
Cathode
O 2,ads + e  2 O ads
2O2 (g) + 4 e  2 O
2+

O
e Oads
ads
Oads  OO
2-
2-
EN
Anode
2H 2 (g) + O  H 2 O(g) + 2 e
EOCV = EN
Interface kinetics – exchange flux and
current densities
•
Equilibrium exchange flux density back and forth across the interface:
j0 = k i c = r0 si c
•
where the exchange rate coefficient ki is the equivalent of D in bulk
diffusion:
k i = r 0 si
•
Dr = 16 r 0 s 2
Multiply with charge to get exchange current density:
i0 = j0 ne = k i nec = r0 si nec
Interface kinetics: Apply a force
•
A force is applied from a potential step dP, and a net flux over the
interface is obtained:
ki
j = -c
dP
kT
•
Current density:
i = jne = cne
•
ki
i
dP =  0 dP
kT
kT
If dP = -ηne we obtain the (linear, ohmic, small) overpotential as
kT i
=
 iRe
ne i0
•
Re is called the charge transfer resistance:
Re =

i
=
Note that the
interface
thickness has
dropped out of
the expressions
and is generally
not an essential
parameter
kT
kT
= 2 2
ne i0 n e k i c
Intermezzo
•
We have made a short analysis of interface processes
•
r0, ki, j0, i0, and Re were all measures of the random exchange going on
at the interface. i0, and Re are the most common to specify.
•
j, i, and η are the result of force and net flux. They are thus not intrinsic
properties of the interface, but proportional to the force.
•
Next we will couple bulk and interface kinetics
Bulk and surface limitations
•
Steady state conditions:
•
The flux through bulk and through surfaces must be the same
•
Continuity demands that the chemical and electrical potentials
match at the interfaces between bulk and surfaces, and that
gradients sum up.
•
But this is difficult to do mathematically and analytically
•
We resort to chemical diffusion and assume that species flow in
concentration gradients both at the surface and in bulk:
- D(
dc
)bulk = -k  cinterface
dx
Example: Oxygen permeation through a mixed
oxide ion and electron conductor; Bulk first
•
Flux density in bulk:
1  O2 t e
 
d  O2 ( g )
2
L I 8e
II
jO2
•
Introduce ambipolar transport and assume them constant. Integrate:
Damb cO2-   O2  amb   O2
=
jO2 =
2kT
L
8 e2 L
•
Rearrange to obtain the chemical potential drop over the bulk:
 O2 =
- 2 jO 2 kTL
Damb cO 2-
=
- j O 2 8 e2 L
 amb
…then the interface (surface):
•
Flux given from earlier equatons for interfaces:
jO 2 - =
•
- cO 2- k i   Ointerface
2
2kT
=
-   Ointerface
2
8 e2 R e
and by rearrangement:

interface
O2
=
- 2 jO 2 kT
k i cO 2-
=- jO 2 8 e2 Re
The sum of bulk and interface (surface):
•
The total chemical potential drop is distributed over bulk
and interfaces (surfaces):
interface
  Otot2 =   Obulk
+
2


O2
2
•
Insert expressions for the individual potential drops and
solve w.r.t. flux:
These expressions are
for the case of 2
surfaces
-   Otot2 cO 2- /2kT -   Otot2 / 8 e2
=
jO 2 - =
L
2
L
+
+ 2 Re
 amb
Damb k i
•
Critical thickness when surface(s) and bulk contribute
equally much:
Lcrit = D/k (for one surface) or Lcrit = 2D/k for two surfaces.
Typical critical
thickesses: 100 μm
Can be affected by
surface roughening
and catalysts
Back to bulk transport;
Anions and cations in a binary oxide MaOb;
First some considerations of chemical potential gradients
•
The formation of the binary oxide
aM zcat  bO zan  M a Ob
•
has equilibrium condition:
ad M zcat  bd O z an  d M aOb (s )
•
By inserting
az cat  bz an
and
d M aOb ( s )  0
Since MaOb is
present as a pure,
condensed phase
we obtain
d cat
b d an z cat d an


dx
a dx
z an dx
Important and useful relationship
between gradients in the chemical
potential of anions and cations.
Note the opposite signs!
Anion and cation chemical and
electrochemical gradients
• From previous slide: d cat   b d an  z cat d an
dx
a dx
z an dx
•
•
We add
z cat e
d
dx
on both sides and rearrange to get:
d cat z cat d an

dx
z an dx
This may be inserted into the total ionic current density
iion  ian  icat  z an ej an  z cat ejcat
  an d an   cat d cat


z an e dx
z cat e dx
to obtain
iion 
 ion d cat
z cat e dx

 ion d an
 z an e dx
The total ionic current can thus be obtained from a
single chemical potential gradient.
Note that anions and cations go in opposite
directions, but contribute to the same sign of current.
Minority cation diffusion of oxide ion
conducting membranes
•
Cations diffuse towards low chemical
potential of metal, i.e. towards high
chemical potential of oxygen
•
This leads to chemical creep
(membrane walkout)
•
If there are more than one cation,
different diffusivities may additionally
lead to
–
Demixing
–
Decomposition (even if the compound is stable
per se under both sets of conditions)
Examples of determination of
minor cation diffusion in La2NiO4
• Solid state reaction
80
70
60
Thickness / m
– Self-diffusion coefficient by diffusion couples
– Wagner’s parabolic rate law
– In our studies: La2O3+NiO = La2NiO4 yielded DNi
50
40
30
20
10
0
-50
0
50
100 150 200 250 300 350 400
Time / hours
La
• Inter-diffusion
80
c / weight%
– Inter-diffusion coefficient by diffusion couples
– Fick’s second law
– In our studies: Nd2NiO4 or La2CuO4 vs La2NiO4
60
40
20
0
-60
-40
-20
0
20
40
x / m
7
• Tracer diffusion
Sr
6
Tracer-diffusion coefficient by tracer isotope
Radio tracer (radiation count) or chemical tracer (SIMS)
Fick’s second law
In our studies: Chemical tracers Pr and Co
-1
log (I / c s )
–
–
–
–
88
Region 1
5
58
138
3
2
59
Region 2
1
0
2
4
6
8
10
Depth x (m)
From Thesis work of Nebojsa Cebasek, UiO
Ni
4
12
14
La
Co
16
60
La
High temperature oxidation of metals:
Wagner oxidation theory
•
Metal oxidises and forms a dense oxide layer
a M(s) + b/2 O2(g) = MaOb(s)
•
If the oxide forms a dense scale, then the rate of
growth dx/dt should be inversely proportional to
the scale thickness d:
dx
1
 k *p
dt
x
•
In integrated form:
x 2  2k *p t  C0  k p t  C0
•
kp = 2kp* and kp* are parabolic rate constants.
Currents in the oxide scale on a metal
•
Ionic current has contributions from anions and cations:
iion  
•

z an e dx
 ion d O
2e
2
dx
Electronic current:
iel  
•
 ion d an
 el d el
 1e dx

 el d el
e
dx
Sum itot = iion + iel = 0 inserted and solve with respect to one current:
iion 
 iont el d O ( g )
2
4e
dx

 iont el kT d ln pO
2
4e
dx
From ionic current to scale growth rate
•
Total ionic current:
iion 
•
 iont el d O ( g )
2
4e
dx

 iont el kT d ln pO
2
4e
dx
Growth rate of MaOb:
iion
 iont el d O2 ( g )
 iont el kT d ln pO2
dn 1 dnO




2
dt b dt
 2eb
dx
dx
8e b
8e 2 b
•
Integrate over the scale at steady state:
pO 2


dn  kT
 1
  2  σ iont el d ln pO2 
dt  8e b pi
Δx

O
2


o
Parabolic rate constant
pO 2


dn
kT

 1
• From the previous slide:
  2  σ iont el d ln pO2 
dt  8e b pi
Δx

O2


o
 kT p O 2

σ iont el d ln pO2 
The term 
2

 8e b piO 2

o
•
is a form of the parabolic rate constant in the expression
•
•
•
dn
1
 kt
dt
Δx
The term σiontel is the ambipolar conductivity. It can be rewritten
in many forms, e.g. σeltion. Often the material is either mainly an
electronic or an ionic conductor such that a transport number
can be set equal to unity.
Often the rate limiting conductivity (ionic or electronic) is
dominated by one species (cations or anions). Moreover, often
one defect mechanism prevails (vacancies or interstitials,
electrons or holes).
In these cases we can integrate the expression analytically if the
prevailing defect structure is known or can be anticipated.
3 important
paragraphs!
More detail
in the text.
Example: Growth of Ma-yOb
How did we get here?:
pO

dx b dn 
 1 2 z cat
 1
* 1


(
D

D
)
d
ln
p

k

M
O
O2
p
dt c0 dt  2 pi
2
Δx
Δx

O2


How do we get here?:
dx 
z
  cat
dt  4

o
•
•

 1
* 1
D
d
ln
p

k

M
O
p
i
2
Δx
Δx

pO 2

p oO 2
pO

dx 
 zcat DvM/ 2 /
 1
* 1

[v
]
d
ln
p

k

M
O
p
2
dt  4[M Mx ] pi
Δx
Δx

O2


o
•
Introduce cation vacancies:
•
Defect chemistry:
•
Equilibrium constant + electroneutrality gives:
/
[v M ]  
1
 1
1
 1
/
b
b
O2 ( g )  vM /  h   OOx
2a
a
b
2 a (  1)
KV pO2
M
This part
done
differently
than in
the text
ln[ vM / ]  11 ln(  1 KV  / )  2 a (b 1) ln pO2
and then we are in the position to integrate!
M