EC Why Electron Transfer Lecture - 4213

Electroanalytical
Chemistry
Lecture #4
Why Electrons Transfer?
The Metal Electrode
Ef = Fermi level;

highest occupied
electronic energy
EF level in a metal
E
Why Electrons Transfer
Reduction Oxidation
EF Eredox
E E
Eredox E
F
•Net flow of electrons from M •Net flow of electrons from

to solute solute to M
•Ef more negative than Eredox •Ef more positive than Eredox
•more cathodic •more anodic
•more reducing •more oxidizing
The Kinetics of Electron
Transfer
Consider:
kR
O+ ne- =R
ko
Assume:
O and R are stable, soluble
Electrode of 3rd kind (i.e., inert)
no competing chemical reactions occur
Equilibrium for this Reaction
is Characterised by...
The Nernst equation:

Ecell = E0 - (RT/nF) ln (cR*/co*)
where:
cR* = [R] in bulk solution
co* = [O] in bulk solution
So, Ecell is related directly to [O] and [R]

Equilibrium (cont’d)
At equilibrium,
no net current flows, i.e.,
E = 0  i = 0
However, there will be a dynamic
equilibrium at electrode surface:
O + ne- = R
R - ne- = O
both processes will occur at equal rates
so no net change in solution composition
Current Density, I
Since i is dependent on area of electrode,
we “normalize currents and examine
I = i/A
we call this current density
So at equilibrium, I = 0 = iA + iC
 ia/A = -ic/A = IA = -Ic = Io
which we call the exchange current
density
Note: by convention iA produces positive
current
Exchange Current Density
Significance?
Quantitative measure of amount of
electron transfer activity at equilibrium
Io large  much simultaneous ox/red
electron transfer (ET)
 inherently fast ET (kinetics)
Io small  little simultaneous ox/red
electron transfer (ET)
 sluggish ET reaction (kinetics)
Summary: Equilibrium
Position of equilibrium characterized

electrochemically by 2 parameters:
Eeqbm - equilibrium potential, Eo
Io - exchange current density
How Does I vary with E?
Let’s consider:
case 1: at equilibrium
case 2: at E more negative than Eeqbm
case 3: at E more positive than Eeqbm
Case 1: At Equilibrium
E = Eo - (RT/nF)ln(CR*/CO*)
E - E0 = - (RT/nF)ln(CR*/CO*)
E = Eo so, CR* = Co*
I = IA + IC = 0 no net current flows
IA
G O R
IC
Reaction Coordinate
Case 2: At E < Eeqbm
E - Eeqbm = negative number
= - (RT/nF)ln(CR*/CO*)
 ln(CR*/CO*) is positive
 CR* > CO*  some O converted to R
 net reduction
 passage of net reduction current
IA O
G R
IC
I = IA + I C < 0 Reaction Coordinate

Case 2: At E > Eeqbm
E - Eeqbm = positive number
= - (RT/nF)ln(CR*/CO*)
 ln(CR*/CO*) is negative
 CR* < CO*  some R converted to O
 net oxidation
 passage of net oxidation current
IA
R
G
IC O
I = IA + I C > 0 Reaction Coordinate

Overpotential, 
fast slow
Current, A
Cathodic

Eeqbm Edecomp Cathodic Potential, V
Fast ET = current rises almost vertically

Slow ET = need to go to very positive/negative
potentials to produce significant current
Cost is measured in overpotential,  = E - Eeqbm
Can We Eliminate ?
What are the Sources of 
 = A + R + C
A, activation
an inherently slow ET = rate determining step
R, resistance
due to finite conductivity in electrolyte
solution or formation of insulating layer on
electrode surface; use Luggin capillary
C, concentration
polarization of electrode (short times, stirring)
Luggin Capillary
Reference electrode
placed in glass
Reference
capillary containing
test solution
Working
Narrow end placed Electrode
close to working
electrode Luggin
Exact position Capillary
determined
experimentally
The Kinetics of ET
Let’s make 2 assumptions:

both ox/red reactions are first order
well-stirred solution (mass transport plays no
role)
Then rate of reduction of O is:

- kR co*
where kR is electron transfer rate constant
The Kinetics of ET (cont’d)
Then the cathodic current density is:

IC = -nF (kRCO*)
Experimentally, kR is found to have an
exponential (Arrhenius) potential
dependence:
kR = kOC exp (- CnF E/RT)
where C = cathodic transfer coefficient
(symmetry)
kOC = rate constant for ET at E=0 (eqbm)
, Transfer Coefficient
 - measure of symmetry of O
activation energy barrier G R
 = 0.5  activated complex
halfway between reagents/
products on reaction coordinate; Reaction Coordinate
typical case for ET at type III
M electrode
The Kinetics of ET (cont/d)
Substituting:
IC = - nF (kR co*) =
= - nF c0* kOC exp(- CnF E/RT)
Since oxidation also occurring

simultaneously:
rate of oxidation = kA cR*
IA = (nF)kACR*
kA = kOA exp(+ AnF E/RT)

So, substituting
IA= nF CR* kOA exp(+AnF E/RT)
And, since I = IC + IA then:
I = -nF cO*kOC exp(- CnF E/RT) +
nF cR*kOA exp(+ AnF E/RT)
I = nF (-cO*kOC exp(- CnF E/RT) +
cR*kOA exp(+ AnF E/RT))
At equilibrium (E=Eeqbm), recall

I o = IA = - I C
So, the exchange current density is given
by:
nF cO*kOC exp(- CnF Eeqbm/RT) =
nF cR*kOA exp(+ AnF Eeqbm/RT) = I0
We can further simplify this expression by

introducing  (= E + Eeqbm):
I = nF [-cO*kOC exp(- CnF ( +
Eeqbm)/RT) + cR*kOA exp(+ AnF ( +
Eeqbm)/ RT)]
Recall that ea+b = eaeb
So,
I = nF [-cO*kOC exp(- CnF /RT) exp(-
CnF Eeqbm/RT) + cR*kOA exp(+ AnF /
So,
I = nF [-cO*kOC exp(- CnF /RT) exp(-
CnF Eeqbm/RT) + cR*kOA exp(+ AnF /
RT) exp(+ AnF Eeqbm/ RT)]
And recall that IA = -IC = I0
So,
I = Io [-exp(- CnF /RT) +
exp(+ AnF / RT)]
This is the Butler-Volmer equation
The Butler-Volmer
Equation
I = Io [- exp(- CnF /RT) +

exp(+ AnF / RT)]
This equation says that I is a function of:


I0
C and A
The Butler-Volmer
Equation (cont’d)
For simple ET,

C + A = 1 ie., C =1 - A
Substituting:
I = Io [exp((A - 1)nF /RT) + exp(AnF
/ RT)]
Let’s Consider 2 Limiting
Cases of B-V Equation
1. low overpotentials, < 10 mV
2. high overpotentials,  > 52 mV

Case 1: Low Overpotential
Here we can use a Taylor expansion to
represent ex:
ex = 1 + x + ...
Ignoring higher order terms:
I = Io [1+ (A nF /RT) - 1 - (A- 1)nF /
RT)] = Io nF/RT
I = Io nF/RT
so total current density varies linearly with
 near Eeqbm
Case 1: Low Overpotential
(cont’d)
I = (Io nF/RT) 
intercept = 0
slope = Io nF/RT
Note: F/RT = 38.92 V-1 at 25oC

Case 2: High
Overpotential
Let’s look at what happens as  becomes

more negative then if IC >> IA
We can neglect IA term as rate of
oxidation becomes negligible then
I = -IC = Io exp (-CnF /RT)
So, current density varies exponentially
with 
Case 2: High
Overpotential (cont’d)
I = Io exp (-CnF /RT)

Taking ln of both sides:
ln I = ln (-IC) = lnIo + (-CnF/RT) 
which has the form of equation of a line
We call this the cathodic Tafel equation
Note: same if  more positive then

ln I = ln Io + A nF/RT 
we call this the anodic Tafel equation
Tafel Equations
Taken together the equations form the

basis for experimental determination of
Io
c
A
We call plots of ln i vs.  are called Tafel
plots
can calculate  from slope and Io from y-
intercept
Tafel Equations (cont’d)
Cathodic: ln I = lnIo + (-CnF/RT) 

y = b + m x
If C = A = 0.5 (normal),

for n= 1 at RT
slope = (120 mV)-1
Tafel Plots
ln |i|
High overpotential:
ln I = lnIo + (AnF/RT) 
Cathodic Anodic
Mass transport ln Io Low overpotential:

limited current I = (Io nF/RT) 
_
Eeqbm + , V
In real systems often see large negative deviations

from linearity at high  due to mass transfer
limitations
EXAMPLE:
Can distinguish simultaneous vs.
sequential ET using Tafel Plots
EX: Cu(II)/Cu in Na2SO4
If Cu2+ + 2e- = Cu0 then slope = 1/60 mV
If Cu2+ + e- = Cu+ slow ?
Cu+ + e- = Cu0 then slope = 1/120 mV
Reality: slope = 1/40 mV
viewed as n = 1 + 0.5 = 1.5
Interpreted as pre-equilibrium for 1st ET
followed by 2nd ET
Effect of  on Current
Density
A = 0.75 oxidation is favored
C = 0.75 reduction is favored

Homework:
Consider what how a Tafel plot changes

as the value of the transfer coefficient
changes.

EC Why Electron Transfer Lecture - 4213

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Electroanalytical

Ef = Fermi level;

•Net flow of electrons from M •Net flow of electrons from

The Nernst equation:

So, Ecell is related directly to [O] and [R]

Position of equilibrium characterized

I = IA + I C < 0 Reaction Coordinate

I = IA + I C > 0 Reaction Coordinate

Eeqbm Edecomp Cathodic Potential, V

Fast ET = current rises almost vertically

Let’s make 2 assumptions:

Then rate of reduction of O is:

Then the cathodic current density is:

Since oxidation also occurring

kA = kOA exp(+ AnF E/RT)

At equilibrium (E=Eeqbm), recall

We can further simplify this expression by

I = Io [- exp(- CnF /RT) +

This equation says that I is a function of:

For simple ET,

1. low overpotentials, < 10 mV

2. high overpotentials,  > 52 mV

Note: F/RT = 38.92 V-1 at 25oC

Let’s look at what happens as  becomes

I = Io exp (-CnF /RT)

Note: same if  more positive then

Taken together the equations form the

Cathodic: ln I = lnIo + (-CnF/RT) 

If C = A = 0.5 (normal),

Mass transport ln Io Low overpotential:

In real systems often see large negative deviations

A = 0.75 oxidation is favored

C = 0.75 reduction is favored

Consider what how a Tafel plot changes

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