9. Electron Transfer Kinetics
Through space hope
Through space hope
Collisional frequency, work
Collisional frequency, work
precursor
Nuclear frequency factor
Solution self exchange
electrode
11


 Aox  Aox*  Ared
k
*
red
A
Self Exchange constant
Homogeneous electron
transfer
k  K A, A*n el exp
Precursor equilibrium
Constant

*
 A*    A
Ared
 Aox 
ox
 red


  G f 


 RT 


Activation energy
Electron tunneling (related to distance
The electron can “hop”

Nuclear frequency factor = frequency of attempts on
The energy barrier associated with bond vibrations and
Solvent motion
Can relate the homogeneous self exchange rate constant to the heterogeneous rate
constant
11


 Aox  Aox*  Ared
k
Self Exchange constant
Homogeneous electron
transfer
*
red
A
ko
Aox  e  Ared

Heterogeneous
Electron transfer
 k11 

k o  Z el 
Z
 so ln 
Z so ln 
kT
2m
1/ 2
Where Zel is a collision frequency factor
Relating to the number of electrode surface
Collisions required to produce a successful
Orientation, generally taken to be 103 to 104 cm/s
Zsoln is a solution collisional factor estimated
From the thermal velocity of the reaction molecules
And their effective or reduced mass. Typical values are 1011 to
1012 1/Ms
Two formulations for ket
1. Classical
Butler-Volmer Eq.
the alpha parameter- a measure of the shape of the energy
wells for the electron transfer event
The Tafel plot to get the alpha parameter
2. Marcus Theory
inner sphere (bond length) changes
outer sphere (solvation) changes
k
f



 Re d
Ox  e 
kg
rate  k f  Ox  surface  k f CO , x  0
To relate this rate to current we first note:
q, f  nFN ox
Where n is the number of electrons per mole of compound
F is the coulombs per mole of electrons
N is the number of moles oxidized or reduced

  coulombs 
moles e

 mole compound 
C 
mole
compound
mole
e



if 
dq , f
Equate the two expressions for rate:
dt
if
The rate of the reaction is
 qf 

d
if
 nF 
dN ox
1 dq f
rate 



dt
dt
nF dt
nF
The rate of the reaction per unit area is
rate forward 
if
nFA
rate  k f CO, x  0,t 
nFA
Depends upon the potential
i f  nFAk f CO, x  0,t
When the free energy for the reaction is zero
There is an activation energy of  G 
0
1    nF E
 nF E
nF E
The free energy for the reaction  Grx
Can be changed by changing the
Potential of the reactant or product
A change in the reaction free energy changes
The activation energy
 G f   G0  nF E  1   nF E
Grx   nF E  E o 
 G f   G0  nF E   nF E  nF E 
 GF   G0  nF E  nF E  nF E
 G f   G0  nF  E  E 0 
 G f   G0   Grx
Classical Theory (Butler-Volmer)
G /Jmol-1
a

x
b
FdE
c

y
c
q /pm

y
x y
x
y
 c
tan 
tan 
x  tan 
G /Jmol-1

y
tan 
y
tan 

tan 
tan   tan 
y
 y
tan 

x
FdE
q /pm
c

y
In a bond breaking mechanism the nuclear coordinate vs energy map
shows that the bond is broken (the atom is not contained in a well)
b
d
a
c
Adding the reactant energy profile we get
Alpha is a fraction between 0 and 1 which scales the amount of the change in the
Reaction energy to the portion which contributes to the change in the activation barrier
In this image it can be seen that the line ab is less than ½ of the line cd, where cd represents
The change in the reaction energy, therefore for bond breaking alpha is less than 1/2
V
-0.60
-0.40
-0.40
-0.20
-0.20
-0.5
-0.4
-0.3
-0.2
-0.1
0.20
0
0.40
to
0.60
  0.5
0.80
0
-0.6
0.00
Current
-0.6
0.00
Current
V
-0.60
-0.5
-0.4
-0.3
-0.2
0.20
  0.5
0.40
to
0.60
1
0.80
Ox  e  Re d
Ox  e  Re d
Conclusion: Alpha =1 facilitates forward (reduction) reaction
Alpha = 0 facilitates reverse (oxidation) reaction
Alpha = 0
Reverse easier than
forward, as would be
expected in a bond
breaking mechanism
Alpha = 1
Forward easier
than reverse
-0.1
0
The general relationship between activation energy and a rate constant is:
k f  Ze

  G f
RT
Where Z is a collision factor
We found that the activation energy is
Alanah write
Out explicitly
Ko here
 G f   G0  F  E  E 0 
Plugging this into the general rate constant expression we get:
k f  Ze
k f  Ze


G0  F E  E o
G0

RT

RT
e

F  E  E o 
RT
similarly
e
a b
e e
a b
k f  k oe

F  E  E o 
kb  k e
o
RT

1   F  E  E o 
RT
Notice ko contains
The collisional
Terms and hopping
terms
How can we measure either ko or alpha?
Need to relate these parameter current and separate them from
The potential effects
This leads to
The exchange current
the Butler-Volmer formulation
And
The Tafel plot
The exchange current is independent of potential because it is
The current that occurs when the system is at equilibrium
(net zero current)
Similar equation can be written for the reverse reaction
o 
i f  nFAk e
nF  E  E o 
RT
CO, x  0,t
ib  nFAk o e
1 nF  E  E o 
RT
CR , x  0,t
inet  i0  e nf  e 1 nf 
Simplify and
Create Butler Volmer
As per p. 96 bard
Ex[ress in terms of k0
See equation 3.3.11
The Exchange Current
when E  E o
Equilibrium prevails, the current is zero, and the
Nernst eq. applies
Eeq  E o 




 nF  E   Eeq 
i f  nFAk e
o
i f  nFAk o e

RT  CR 

ln
nF  CO 
or E o  Eeq 
RT  CR   
ln 
 
nF  CO   
RT
RT  CR  
  nF  

ln 
  E  Eeq 

 RT  
nF  CO  
i f  nFAk e
Simplify and
Create Butler Volmer
As per p. 96 bard
CO , x  0,t
CO , x  0,t
   nF 
  nF 
  nF  RT  CR  
 
E

E

ln 


 eq 








RT
RT
RT
nF
C
o 
 O 
RT  CR 

ln
nF  CO 
CO , x  0,t
   nF 
  nF 
  nF  RT  CR  
 
ln 
 E 
 E eq  








RT
RT
RT
nF
C
o 
 O 
i f  nFAk e
   nF 
 CR  
 
 E  E eq   ln 



RT
C

o
 O 

i f  nFAk e

CO , x  0,t
eab  e a eb
let
  E  E eq
CO , x  0,t
and

   nF 
 CR  
 

  RT  E  Eeq  ln  C  
 O 
o 

i f  nFAk e
i f  nFAk o e nf e



 CR  
  ln 
 

 CO  

F
f 
RT
CO, x  0,t
log100  2
10 2  100
so 10 log100  100
or
CO, x  0,t

i f  nFAk e
o
  nf
o  nf
i f  nFAk e
 CR 
  CO , x  0,t
 CO 


CR CO, x  0,t

1 
e ln x  x
For an exercise for yourself
Derive the corresponding eq:
o 1  nf
ib  nFAk e


CR CO, x  0,t

1 
o  nf
i f  nFAk e


CR CO, x  0,t

1 
o 1  nf
ib  nFAk e


CR CO, x  0,t
  E  E eq
when
E  E eq
The “exchange current”
 0


i f  nFAk CR CO, x  0,t
o

1 


 i0  ib  nFAk CR CO, x  0,t
o

1 
inet  i f  ib


inet  nFAk CR CO, x  0,t
 e
o nf
inet  nFAk e
o



CR CO, x  0,t
1 
1 
nf
inet  i0  e nf  e 1 nf 
o 1  nf
 nFAk e


CR CO, x  0,t

1 
 e 1 nf 
Current Overpotential equation

1 
The Butler-Volmer Equation is used
1. When you don’t have modeling packages
2. When you need to know how the reaction occurs at the
microscopic level so that you can change the activation complex
3. By people in the more technical fields
a)
batteries
b)
corrosion
c)
plating
inet  i0  e nf  e 1 nf 
E-Eo, V
-10
-8
net normalized current
-6
-4
-2
-0.6
0
2
4
6
8
10
-0.4
-0.2
0
0.2
0.4
0.6
This is the current that is flowing
Even with no NET current!!
inet  i0  e
For
 nf
e
1  nf

Butler Volmer Equation
  0.008 to  0.008
exp x  1  x x  0




inet  i0 A 1   nf  1  1    nf
inet  i0 A 1  nf  1  nf  nf
inet  i0 A1  nf  1  nf  nf
inet  i0 Anf
Max Volmer
1885-1965
  120mV / n
E-Eo, V
-10
-8
net normalized current
-6
-4
-2
-0.6
0
-0.4
-0.2
2
4
6
8
0
0.2
0.4
0.6
inet  i0 Anf
inet  i0 Anf 
 
inet  R p 
E  IR
E
I
R
10
Corrosion literature refers to this as
polarization resistance, Rp
A “charge transfer resistance”

inet  i0 A e  nf  e 1 nf

Butler Volmer Equation
  120mV / n
o


nFAk CR CO, x  0,t
inet  i0 Ae  nf
inet   i0 Ae 1 nf
loginet   logi0 A  n f
Tafel Plot
Must get a system
Where only electron exchange
Is limiting (diffusion must not be limiting)

1 
 i0
R p  Anf io 
Julius Tafel
1862-1982
How can we get currents that are independent of diffusion and are only
Kinetically limited for the Tafel plot?
Levich equation for a RDE
 0.62nFAD 2 / 3  *
ic  
 COx 
1/ 6



 0.62nFAD 2 / 3 
m 

1/ 6



iC  mC ;
*
Ox

i  mC
*
Ox
iC
m *
COx
 Cox , x  0


ic
*
i  * COx
 Cox , x  0
COx
*
iCOx
*
 COx
 Cox , x  0
ic

*
Cox , x  0  COx
*
iCOx

ic
Must conform to
i k  FAk f Cox , x  0
*
 * i COx


i k  nFk f  COx 
ic 

i 

*
i k  nFk f COx  1  
ic 


 1  1
iD   1   1 
*
i k      nFk f COx  1      
ic   i D   i k 
 i D   ik 

 1 1  1
 1
*
   nFk f COx     
 ik 
 i k ic   i k 
Not quite right!!!!!!!
1

i
ka
1


 Do   nFAk c
 Do  


nFA
nFA
4
4 kc
. x10 
. x10 
 161
 161
1
1

i
Veniamin G. Levich
1917-1987
1
1
From a plot of
vs
i

kc  ka
1

 Do   nFAk c

nFA
4
. x10 
 11
Jaroslav Koutecky
1922-2005
the magnitude of the reduction rate constant can be obtained as predicted from
the KL equation.
Koutecky-Levich equation
1 1
1


i i K .62nFADO2 / 3 1/ 2 1/ 6 CO*
i,c  0.62nFADO2/ 3 1/ 2 1/ 6 CO*
Diffusion related current depends on rotation rate
  F  E  E o '  
RT
 CO*
i K  FAk f CO*  FA e



Electron transfer related current independent of rotation
Jaroslav Koutecky
1922-2005
1 1
1


i i K i,c
1 1
1


i i K .62nFADO2 / 3 1/ 2 1/ 6 CO*
Koutecky-Levich equation
Veniamin G. Levich
1917-1987
Change to the stuff you made an excel problem of
Two different systems (ket 104 and ket 0.01) are plotted for each rotation rate
Tablulate the current at a selected potential and plot
versus rotation rate
-3.00E-02
-2.00E-02
-1.00E-02
0.00E+00
Current
1.00E-02
2.00E-02
3.00E-02
4.00E-02
5.00E-02
6.00E-02
7.00E-02
-0.5
-0.45
-0.4
-0.35
-0.3
V
-0.25
-0.2
-0.15
-0.1
-0.05
0
1 1
1


i i K .62nFADO2 / 3 1/ 2 1/ 6 CO*
40
y = 485.91x + 2.509
R2 = 0.9981
35
y = 377.44x + 10.239
R2 = 0.9911
30
25
  F  E  E o '  
RT
 CO*
i K  nFAk f CO*  nFAk o  e



1/I
20
For an exercise use this data
To calculate kf and ket:
KL plot at -0.35 V, Eo at -0.25 V
A=1cm2, Co* = 1 M/L
15
10
5
0
-0.02
-0.01
0
0.01
0.02
0.03
-5
1/sqrt(w)
0.04
0.05
0.06
0.07
0.08
  F  E  E o '  
RT
 CO*
i K  nFAk f CO*  nFAk o  e



KL plot at -0.35 V, Eo at -0.25 V
A=1, Co* = 1, alpha=0.5, 1/ik = 10.239
Calculate kf and ko
i K  FAk f CO*
1 C
 1mole   1L 
 9.648 x10 4 Ccm2 

k
 L   cm3  f
10.239 s
1 C
cm
10.239 s
 k f  101
. x10  3
1mole
1L
s
9.648x104 Ccm2  L   1000cm3 
For an exercise finish the calculation for ko
What parameters control ko and alpha?
ko is related to the reorganization energy defined in the
Following graph
The model was developed by Hush and Marcus, for
Which Marcus got the Nobel prize. It applies to
Outer sphere electron transfer reactions in which
No bond breaking is allowed.
(Gray and Ellis review chapter)
H AB   
k et 
   RT 
2
  G 


o 2
1/ 2
e
4  RT
The Marcus model
HAB describes electronic coupling between reactants & products at the transition state.

o

d

d
o
2
H AB  H AB e


Rate of decay with distance
Related to the protein medium
Close contact position
HAB magnitude depends
1.
upon Donor (reductant)- A (oxidant) separation
2.
orientation
3.
nature of intervening medium
This energy model assumes that no work is required to bring the reactants together or to move
the products apart
R
O
  k q O  q R 
Free Energy
1
2
G
2
λ
G#
*
O
G R*
qO
q#
Nuclear Coordinate
qR
G Aeq  21  2  X C  X A   GDeq  21  2  X C  X D 
2
2
algebra
XC 
XC 
1
2
1
2
X
X
A
A
 XD 
 XD 
GDeq  G Aeq
 2X A  XD
 G0
 2X A  XD

Gactivation  GDeq  21  2  X C  X D 
Gactivation 
Gactivation
1
2
 2 XC  X D 
0



G
2 1
1
 2   2 XA  XD  2
 XD
 XA  XD


Gactivation

2 1
1
 2   2 XA  XD 


G

XD  2
  X A  X D  
0
2
2
2
2
2
2
 G
eq
D
Gactivation
Gactivation


 G0
2 1
1

 2   2 XA  XD  2
  X A  X D  

2
 
 21  2  X  X  2

 G0
A
D
2
1

 2
 2
2
  XA  XD
  X A  X D  

Gactivation
 21  2  X  X  2   G 0 
A
D

 21  2 
2


 XA  XD


2
2
b1/ 2 c 2  b c 2
Gactivation 
Gactivation
1
2

4
X
2
A
 XD


2
1
 21 
2
  1
 2   2  X  X 2
A
D
1
2

2
X
A
  X
1
2
 XD  G
2
A

 XD  G
2
2
0
2
0

2
1
2

2
X
D
 X A
2
Gactivation
1
 21 
2
  1
 2   2  X  X 2
A
D
Gactivation 
 
1
2

2
1
2
X
Gactivation 

D
2
X
1
4
A
 X A
 XD
  X


2
1
2
1
2

2
X
A
 XD  G
2
A
0
0


2
2
2
1
4
   G 

0 2
  G 


0 2
Gactivation
 XD  G
2
2
4
Here lambda is the total reorganization
energy
At the exchange current position, deltaG0=0
So
 Intrinsic Barrier
#
G 
4
See also the derivation at: http://www.life.uiuc.edu/croftsbioph354/lect19.html
  i  o
Compare Butler-Volmer to Marcus
 G f   G0  nF  E  E 0 
F E  E o 
  

 G     1 
 4 




f
2
   G 

0 2
 G#
4
G 
#

4
Change to delta with standard
Outer sphere reorganization
relates to polarizability of the “solvent”
  e 2  1
1
1  1
1


 
o 



4  o  2r1 2r2 r12   Dop Ds 
e  1602
.
x 1019 C
F
C2
or
m
N m2
 o  8.85x10
e is the charge transferred from donor to acceptor
r1 and r2 are the radii of the two reactants when in contact 1
1

r12=r1+r2
 D  D  ~ 0.5 for most solvents
 op
s
Dop is the square of the refractive index of the local medium
Ds is the static dielectric constant
ε is the permittivity of space
12
At an electrode surface the outer sphere reorganization energy is
  e 2  1
1 1
1

   
o 

8 o  r1 R   Dop Ds 
Where R is twice the distance from the
center of the molecule
To the electrode
For membrane or protein associated electron transfer a major issue is
Determining the appropriate estimate for the dielectric constant which can
Vary through space due to the membrane or protein structure
  e 2  1
1 1
1

   
o 

8 o  r1 R   Dop Ds 
Try a Calculation
e  1602
.
x 1019 C
C2
12 F
 o  8.85x10
or
m
N m2
 1
1

 ~ 0.5 for most solvents

 Dop Ds 
1cal  4.184 J
Calculate the outer sphere
Reorganization energy for the
Oxidation of hexa-aquo cobalt when
It is 7 A from the electrode
When it is 14 A from the electrode
#$%@ units
kg m
N 2
s
kg m 2
J
s2
2
2
kg
m
kg
m
1
m




Nm2     2 

 J
 m  s  m
s2
  e 2  1
1
1  1
1


 
o 



4  o  2r1 2r2 r12   Dop Ds 
As r goes up expect reorganization
Energy to go down
Work can be reduced by spreading the charge over a wide radius
-Hence evolution appears to have selected cytochromes and chlorophylls (with a basic porphyrin
ring structure) for electron transfer
-Charge is spread through the entire pi conjugated systems
-This means that the distance for an electron hop is from ring edge to ring edge, as opposed to
some particularly localized spot (such as a metal center)
When there is proton coupled to the electron transfer the energetics change and may
Become dependent on the energy of proton transfer as opposed to the energetics of the
Electron transfer process.
They relate the proton driving force to
 G proton transfer  2.303RT  pK D  pK A 
Cukier, R. I. and D. G. Nocera, 1998, Proton-coupled electron transfer, Ann.Rev. Physical Chemistry, 49,
337-369
i 
1
2
k
H
Q
2
j
Inner sphere reorganization is related to the bond length changes
j
Related to the vibrational mode energy
In the absence of work terms
What is the exchange current density for the oxidation of hexaaquo cobalt (present in
1 mM concentrations in both the reduced and oxidized forms)
At a 1 cm2 electrode, assuming D of cobalt is on the order of 1x10_5 cm2/s, and the
Temperature is 25C

#
G

What would the
1 
4

o
Current be if the potential
nFAk CR CO, x  0,t
 i0
Is 200 mV positive of the formal potential?
1 
200 mV negative of the formal potential?

o  nf

i f  nFAk e
CR

C
O , x  0 ,t
Assume, initially,
That alpha is 1/2
Correct the notation

What happens if the reactants and products
Experience electrostatic repulsion?


f
F  E  E o   wo  wr 

  1 

i  o

 i  o  
G  

4
2
a A 
 za z pe2   ea D
e
rDA

( w p  wr )  U r  

e

 40   1  a D 1  a A 




The larger kappa the smaller the activation energy, the closer
Ions can approach each other without work
What is the meaning of alpha in the Marcus formulation?
Free Energy
Free Energy
Nuclear Coordinate
Free Energy
Free Energy
Nuclear Coordinate
Nuclear Coordinate
Nuclear Coordinate
Raising the Energy of the Reactant increases the speed up to……the inverted region


f
 G

 nFE


f
o
 G
1 F  E  E    wo  wR 

 
 nFE 2
2i  o 
This eq. predicts alpha is dependent on reaction energy.
Rudolph Marcus
1923Nobel Prize:
If the activation barrier at equilibrium is large (dissociative reactions) then one needs
to go to very large over potentials to make the reaction work
The equation predicts that under these conditions alpha will be small
“When electron transfer is rate limiting the electron transfer coefficient is a sensitive diagnostic
probe of mechanism. The electron-transfer coefficient is directly related, in the context of
Marcus theory, to the intrinsic barrier for electron transfer. If bond breaking is occurring in the
transition state , as would be the case for a concerted pathway, considerable structural
reorganization is occurring (i.e. high intrinsic barrier) and alpha is characteristically low.”
Tanko, JACS, 2007 129, 4181
Consider an electron transfer reaction that involves bond breaking as the
Electron is transferred
A   B  e A  B



1
  
2
F  E  E o    wo  wR 
2 i   o 
For this reaction we expect
the internal reorganization energy  i should be very large
the potential required to rise over the activation barrier will be big
E  E o
such that
 E  E o   wO  wR    E  E o 
under these conditions
o
F
E

E

 1
1
  

2
2 i
2
1  E ( wo  wr )
  

2 2
2
If the term
Then
And at
( wo  wr )     E
1 E
  
2 2
E  0
1
 
2
And alpha is linear with delta E
1  E ( wo  wr )
  

2 2
2
If the term
( wo  wr )     E
For alpha not be ½ when delta E = 0
( wo  wr )
 1
  0.05   ~ 0.025
 2
2
( wo  wr )  0.05
Assignment:
Compare the shape of the activation energy barriers obtained in
Classical Butler-Volmer formulation (alpha is related to geometry)
And with Marcus theory for outer sphere electron transfer (alpha is
Related to potential)
Excel sheets and instructions can be found on line.
Example of using alpha to understand a reaction
Tanko, JACS, 2007 129, 4181
Their Question: What is the mechanism?
et
Ox  e 
Re d

k
kf
Re d 
 products
kb
If an et controlled mechanism then rate
proportional to applied potential, Ep
invariant with scan rate
If an EC mechanism (Module 2) CV peak
potential varies with scan rate (-29.6 mV/log
scan rate)
In the linear region
 RT 
 RT   k  RT  
n E p  E1/ 2   
 0.780  
 ln 

 nF 
 2nF     nF  

Module 2

Tanko, JACS, 2007 129, 4181
et
Ox  e 
Re d

k
kf
Re d 
 products
kb
If an EC mechanism (Module 2) CV peak potential
varies with scan rate (-29.6 mV/log scan rate), with
a peak width (Ep-Ep/2) ~ 50 mV and invariant with
scan rate
If ET is rate limiting then it should be -29.6 mV/alph
And peak width should be 1.85RT/alphaF
Conclude that compounds 5&6 are et controlled,
Notice the “smaller” values of alpha
Donkers et al JACS 1999 121 7239
di-tert-butyl-peroxide DTBP
Dicumyl peroxide DCP
Di-n-butyl peroxide DNBP

A  B  e
A

B


A  B   A*  B 
*

A  B  e
A

B

Step wise
Concerted (more reorganization in the excited state)
Notice the small values for alpha for these reactions, even
When measured in two entirely different experiments
• Calculating the Rates of Outer Sphere Cross Exchange
Rates
We have shown that the rate of a reaction is proportional the potential difference
Cross Exchange Reactions
Fitch (Chapter 6)
k
f
A  e 
 A
b
B  
B  e
E1o
Red cat
 E 2o
k
An ox
12 
A  B  
A B

k
Erx   Erx  E1o  E2o
o
Erx  Ered
 Eoxo
A  e  A
o
Ered
,A
D  D   e
 E oxo , D
 A   D 
A  D 

k12
Red acceptor
Donor ox
o
o
or  Ecat
 Ean
You may see this written as:
Organic chemists
o
o
Erx  Ered

E
,A
ox , D
This is “1 and 2” not “12” to indicate the reaction is between 1 and a 2nd
compound
Electron Transfer Rate Constants Modulated by Potential
LUMO, donor
HOMO, donor~Eox,D
LUMO, acceptor~Ered,A
HOMO, donor
12 
A  D 
A  D 

k
o
o
 Erx  Ered
, A  E ox , D  1   G  0, spon tan eous
F  E  E o   wo  wr 
 i  o

G  
 1

4 
i  o



f
2
   i   i2
 Grxo 
 

G    1
 4 
 


f
For the cross reaction
Marcus assumes
2
A  B  12 A  B
k


 G12o
1

*
* 

*
*
 G11   G22   4 G11   G22   o
*
 G12 

 G12
2
2
From this assumption math works (eventually) to:
For outer sphere e.t. only!
log k12   (log k11  log k 22  16.9 E  log f )
o
12
1
2
Zsoln is a solution collisional factor
estimated
From the thermal velocity of the
reaction molecules
And their effective mass. Typical
values are 1011 to 1012 1/Ms
log f 
log K12 
2
 k11 k 22 
4 log

 Z 
2
Where f = 1 there is a simple prediction that the cross exchange
Reaction increases with a larger difference between in standard
Potential for the two reactants:
 A   D 
A  D 

k12
o
o
 Erx  Ered

E
,A
ox , D
If E,red,A>Eox,D large rate constant
Another form of this equation often seen is:
k12 
k
k K12 f 
11 22
log k12   (log k11  log k 22  16.9 E  log f )
o
12
1
2
Try a calculation
log f 
log K 
12
k k 
4 log 11 22 
 Z 
3 

2
 RT ln K   G   nF E o
From the Fitch Appendix we find that cobalt en has a self exchange constant of
~10-3 1/Ms; with a formal potential (NHE) of -0.24 V. Ru(bpy)32+/3+ has a formal potential
Of 1.26 V vs NHE, and a self exchange rate constant of 3x109 1/Ms. What is the rate
Constant for the assumed outer sphere cross reaction?
Ru(bpy) 23  Co en 3
2
Ru(bpy) 33  Co en 3
2
log k12   21 (log k11  log k 22  16.9 E12o  log f )
Early applications of the cross reaction system to metal complexes
Michael J. Weaver and Edmund L. Yee, Activation Parameters for Homogeneous Outer-Sphere ElectronTransfer Reactions. Comparisons between Self-Exchange and Cross Reactions Using Marcus’ Theory,
Inorg. Chem., 1980, 19, 1936-1945.
Mei Chou, Carol Creutz, and Norman Sutin, Rate Constants and Activation Parameters for Out-sphere
electron-transfer reactions and comparisons with the predictions of Marcus Theory, J. Phys. Chem. 1977
Conclude the “fit” between calc.
And measured is “reasonably” good
To a factor of 25. But if used
To estimate unknown k11 then
The error increases to 252 or 103!
A few slides ago we said that we had to (maybe) worry about work
red1  ox2  ox1  red 2
Started with both inorganic and organic and with uncharged and charged
reactants to see if Marcus theory with work terms to account for charge
works
Z=0
Z=+1
H3C
CH3
Z=-3,-2,+3
H3C
N
+
CH3
N .
Fe CN  6
Fe CN  6
3
N
H3C
CH3
TMPPD 
4
RnOH2  6
Co NH3  6
3
RnOH2  6
N
2
H3C

3
CH3
TMPPD

MnO4


 Co NH3  6
H3C
+
2
CH3
N
TMPDD 2 
Z=+2
+
N
Gramp, J. Chem. Soc., Perkin Trans, 2, 2002, 178-180
H3C
CH3
Fe CN  6
3
I
II
RnOH 2  6
2
III Co NH 3  6
3
To do this calculation
He has to include k12calc  W12 k11 k 22 K12 f 12
A work term
W12  exp
IV MnO4
ln f 12
k12  Z12 exp

Z12  d 2 N L
*
 G12
RT
8k B T

  reducedmass
d  rox  rred
Gramp, J. Chem. Soc., Perkin Trans, 2, 2002, 178-180

w12  w21  w11  w22
2 RT
w12  w21 

 ln K12 


1
RT 

4  k11 k 22  w11 w22
ln

 Z2 
RT
zi z j eo2 N L 
1 


wij 
4 o  s d  1   D d 
Work related
To each encounter
D 
2 N L eo2 I
4 o  s d
I  ionic strength
2
Using the cross exchange reaction
To get at a self exchange constant
Electron transfer reactions between copper(II) porphyrin complexes
and various oxidizing reagents in acetonitrile† by Masahiko Inamo,
Hideto Kumagai, Ushio Harada, Sumitaka Itoh, Satoshi Iwatsuki, Koji
Ishiharac and Hideo D. Takagi, Dalton Transactions 2004, 1703-1707
Inamo, Masahiko et al: Electron transfer reactions between copper(II) porphyrin complexes
and various oxidizing reagents in acetonitrile, Dalton Trans. 2004, 1703-1707,
Proposition: e.t. of Fe2+/3+ and Co2+/3+ depend on the kinetics of the metal center site
and have been shown to be 107 to 108 1/Ms for Fe and 1x10-3 to 1x104 1/Ms for Co.
These are smaller than porphyrin driven e.t. (for example copper porphyrin) due to the
N-M bond length changes experienced by the metal that increase the activation barrier.




11
Cu porphyrin  Cu porphyrin

k

Porphyrin driven e.t. expected to be much faster (harder to study) so have to get at them
indirectly using
log k12   21 (log k11  log k 22  16.9 E12o  log f )
log k12   21 (log k11  log k 22  16.9 E12o  log f )
H
N
Bis(1,4,7-triazacyclononane)nicle(III) = Ni(tacn)2
NH
K22 (Ni) known
NH
F
F
Tris(hexafluoroacetylacetonato)ruthenate(II)=Ru(hfac)3
F
F
F
K22 (Ru) known
F
O
O
 Ru
 Ni
II
III
hfac
3
  Cu
 tacn 2 
2
aq


12 Ru III hfac  Cu 
aq
3

k


12


 Ni II  tacn  Cu 2 
 Cu
2
aq

k

aq
0.78 vs Fc
0.57 V Fc
Use to get an average value for
Cu
2
aq
11


 Cu 2   Cu 
 Cu
aq
aq


aq
Cu porphyrin  Cu
2
aq
k


 Cu porphyrin

0.66 vs Fc

 Cuaq
 Ni
III
 tacn 2 




12


 Ni II  tacn  Cu 2 
 Cu
2
aq

k

aq
3.5 1/MS
 Ru
II
hfac
3
  Cu
2
aq
12 Ru III hfac  Cu 
aq
3

k
1.6x104 1/Ms
log k12   21 (log k11  log k 22  16.9 E12o )
Cu
2
aq
11


 Cu 2   Cu 
 Cu
aq
aq


aq
k
Gives: 1.5x10-5 and 1.2 x10-4 1/Ms
CH3
X
CH3
H3C
N
HN
Cu
NH
H3C
CH3
N
X
N
HN
Cu
NH
X
N
CH3
H3C
H3C
λi= 39 kJ/mol
k11 =1011.5 (1/Ms)
For Fe
and Co
λi =74 kJ/mol
k11= 109.5 (1/Ms)
X
107 to 108
10-3 and 104
Conclude small λ values indicate
that the e.t. is ligand centered not
metal centered, consistent
with Cu-N bond change
of only 1.988 vs 1.9817 Å
Cu porphyrin  Cu
2
aq


 Cu porphyrin

 Final argument – if Cu is
 Cuaq
Use average of 1.5x10-5 and 1.2 x10-4 1/Ms
log k12   21 (log k11  log k 22  16.9 E12o )
Involved requires d8 to d9
electron conversion which
would involve very large Cu-N
bond distance change
Marcus theory can also be applied to proton transfer reactions
And, in such cases, work terms are important
Standard free energy of reaction
o



G
R
r
*

 G*  w   Go  1 
4  Go* 

2
Intrinsic barrier
Work term for
Forward reaction


f
F  E  E o   wo  wr 
 Grxo 
  
     1 

  1 



i  o
4




 i  o  
G  

4
2
2