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Electrochemical Thermodynamics and Concepts
Sensitivity of electrochemical measurements
Measurements of electrochemical processes are made by measuring electrical currents and voltages.
The currents results from the flow of charged ions (positive or negative) and or electrons. Current
measurement can be extremely sensitive as small as 10-15 amps (coulombs per second). Let’s get
“feeling” for the sensitivity of such a measurement.
monolayer: ≈1.5 x1015 atoms/cm2 . Suppose each atom on the
surface dissolves as a single charged positive ion (i.e., a cation).
M Þ M + + e-
A
+1
1 cm
1 cm
corresponding to 1.6 x 10-19 C/atom of charge. Today we can
easily measure at the nanoamp scale. Suppose we measure a
current density of 10-6 A cm-2 for 1 s or 10-6 C. A full monolayer of
metal dissolving as a +1 cation results in a charge of 1.6 x 10-19
C/atom x 1.5 x 1015 atoms/cm2 or ≈ 2.4 x 10-4 C cm-2. Then 10-6 C
corresponds to only of order 0.01 ML.
An electrochemical reaction is a chemical reaction involving electron
transfer.
An ordinary chemical reaction does not involve electron transfer
Chemical reactions: Note the mass balance
N2 +3H2 = 2NH3
synthesis of ammonia
Electrochemical Reactions: Note the mass and charge balance
2H+ +2e- = H2
Hydrogen ion reduction
Zn = Zn2+ + 2e-
Zn oxidation
Since electrochemical reactions involve electron transfer we can measure
the rate of these processes by designing an appropriate electrical circuit and
measuring the flow of an electric current.
Faraday’s Law relates the flow of electric current to the mass of metal
reacting.
Ita
m
nF
m=
t =
I =
a =
n =
F =
mass of metal reacting (gm)
time (s, min, hours, years)
current (Amps)
atomic weight of metal
number of electrons transferred in the reaction
96,500 C; Faraday’s constant which is equal to the
charge associated with 1 mole of electrons.
If , for example, we divide Faraday’s law by the time and area of
A dissolving surface, we obtain a relation describing the rate of this process.
m ia
r 
tA nF
A =
i =
r=
surface area of corroding metal (cm2)
I/A: the current density (A/cm2)
Rate measured as a weight per unit area per
unit time (gm/cm2-s).
Examples
Consider the case for steel or iron corrosion. Suppose we make a measurement
of the current associated with the corrosion of 100 cm2 of a steel surface and
find a current of 0.1A. Then dividing through by the area of the sample,
0.1 A/100 cm2 = 0.001 A/cm2 = 1 mA/cm2 current density.
m ia
r 
tA nF
a =
1 mA/cm2
55.85 g
F =
96,500 C
n =
2
i =
Fe  Fe2  2e
0.001A / cm 2  55.85 g
 2.89 107 gm / cm 2 s
2  96,500C
We can convert this to a penetration rate (cm/s) by dividing by the density of Fe.
 Fe  7.88 g / cm3
0.001A / cm 2  55.85 g
r
 2.89 107 gm / cm 2 s
2  96,500C
p
r
 Fe
2.89 107 gm / cm 2 s
8


3.67

10
cm / s
3
7.88 gm / cm
Next consider:
Electrochemical Cells
V
{Zn2+} =1
{H+} =1
Zn
Pt
We measure a potential
difference of - 0.762 V
H2 at 1Atm
porous barrier
Pt is inert and acts as a catalyst:
Zn is oxidized:
Total reaction:
2H+ + 2e- = H2
Zn = Zn2+ +2eZn + 2H+ = Zn2+ + H2
2H+ + 2e- = H2
E0 =
Zn2+ +2e- = Zn
E0 = - 0.762 V (NHE) sign taken to be negative
since Zn is oxidized.
0.000 V
(defined as zero by convention)
Electrochemical Cells
V
{Cu2+} =1
{H+} =1
Cu
Pt
We measure a potential
difference of 0.342 V
H2 at 1Atm
porous barrier
Pt is inert and acts as a catalyst:
Cu plates;
Total reaction
H2 = 2H+ + 2e- : hydrogen is oxidized
Cu2+ +2e- = Cu
Cu2+ + H2 = Cu + 2H+
2H+ + 2e- = H2
E0 =
Cu2+ +2e- = Cu
E0 = 0.342 V (NHE) This sign is positive since
Cu is reduced.
0.000 V
(defined as zero by convention)
Electrochemical Cells
+1.104 V
1 M CuSO4
1 M ZnSO4
If we construct this “cell” we observe the following:
Zn dissolves;
Zn = Zn2+ +2e-; oxidation occurs at the anode
Cu plates;
Cu2+ +2e- =Cu
reduction occurs at the cathode
If we connect a voltmeter as shown we measure a voltage of 1.104 V.
Daniell Cell
+1.104 V
Zn dissolves;
Cu plates;
Zn = Zn2+ +2e-;
Cu2+ +2e- =Cu
oxidation occurs at the anode
reduction occurs at the cathode
For the Cu-Zn cell we measured a voltage difference of 1.104 V.
Zn2+ +2e- = Zn
E0 = - 0.762 V (NHE)
Cu2+ +2e- = Cu
E0 = +0.342 V (NHE)
This difference is 1.104 V
Standard Reduction Potentials: EMF Series
Volts (NHE): E0
Au 3+ +3e- =Au
+1.498
O 2 +4H + +4e- =2H 2 0 (pH = 0)
+1.229
Ag + +1e- =Ag
+0.798
Fe3+ +1e- =Fe 2+
+0.771
Cu 2+ +2e- =Cu
+0.342
2H + +2e- =H 2
+0.000
Pb 2+ +2e- =Pb
-0.126
Sn 2+ +2e- =Sn
-0.138
Ni 2+ +2e- =Ni
-0.250
Co 2+ +2e- =Co
-0.277
Cd 2+ +2e- =Cd
-0.403
Fe 2+ +2e- =Fe
-0.447
Zn 2+ +2e- =Zn
-0.762
Al3+ +3e- =Al
-1.662
Noble
Cathode
Anode
Active
Standard conditions
Standard states:
•
For a solid; a =1: pure metal, metal oxide, etc.
•
For a gas, 1 Atm pressure is taken as unit activity.
•
For dilute solutes typically found in most instances of corrosion,
activity is reasonably approximated by the concentration in M. The
standard state is 1 M.
•
Temperature is taken as 25ºC = 298 K
definition: pH = -log [H+]
Thermodynamics of Chemical Equations
aA + bB
xX + yY
A, B … reactants
X, Y … products
a, b, …., x, y,…. Stoichiometric Coefficients defining how
many moles of A and B produce moles of X and Y
The chemical reaction can proceed as indicated if the energy change is
negative. That is if the energy of the products is less then that of the
reactants. Mathematically we say
Gproducts  Greact  0
Gi  Gi0  RT ln ai
OR
G  0
Activity or concentration “correction”
-non standard state
Energy of the standard state
Chemical Equilibrium
aA + bB + … ® xX + yY + …
A, B … reactants
X, Y … products
a, b, …., x, y,…. Stoichiometric Coefficients
Equilibrium is defined by the condition
(
DG = xGx + yGy +
Since
Gi  G  RT ln ai
0
i
(
DG = xGx0 + yG y0 +
) - ( aG
A
+ bGB +
)=0
G ≣ standard free energy change /mole
) - ( aG
0
A
+ bG B0 +
)
+RT (x ln ax + y ln a y + ) - RT (aln aA + bln aB + ) = 0
Solving for
G
Chemical Equilibrium
DG = -RT ln
0
x y
X Y
a b
A B
a a
aa
= -RT ln K
K is the equilibrium constant for the reaction.
The equilibrium constant is defined by
G 
K  exp(
)
RT
When components are not in standard state:
DG = DG 0 + RT ln
a Xx aYy
aAaaBb
Electrochemical Equilibrium
DG = DG 0 + RT ln
a Xx aYy
aAaaBb
Free energy is measured in units of kcal or kJ.
1 calorie
1Joule
=
=
4.186 Joules
6.24 x 1018 eV
Since 1 Volt = Joule/Coulomb
G  QE
DG = -nFE
where Q is the total charge. The total charge transferred
in an electrochemical reaction will be equal to the number
of moles of electrons participating in the reaction times the
charge/mole, Q = nF
By convention a minus sign relates ΔG and E.
Electrochemical Equilibrium
If we flip the numerator and denominator in the argument of the ln function,
Electrochemical Equilibrium
Nernst Equation
Formal Potential
Electrochemical Equilibrium: pH effects
We can write a general metal dissolution reaction in the following way:
aA  mH   ne  bB  dH 2 0
Here A could correspond to some oxidized species such as Fe2+ and B
would be its reduced form, metallic Fe. Alternatively A could be an oxide such as
FeO and B its reduced form, also metallic Fe.
The Nernst equation for this general reaction is
a
m
éH ù
é
ù
A
RT
ë
û
ë û
E = E0 +
ln
nF é B ùb é H O ù d
ë û ë 2 û
+
Electrochemical Equilibrium and Corrosion
a
m
éH ù
é
ù
A
RT
ë
û
ë û
E = E0 +
ln
nF é B ùb é H O ù d
ë û ë 2 û
+
It’s convenient at this stage to convert the natural log (Ln) to base 10.
In general:
Ln (x) = 2.303 Log (x)
Also
RT 2.303  8.31J / moleK  298K

 0.059 V
F
96,500C / mole
Electrochemical Equilibrium and Corrosion
The Nernst Equation above can now be rewritten as:
a
éë Aùû m
0.059
+
é
ù
E=E +
log
+
0.059log
H
b
ë
û
n
n
éë B ùû
0
a
éë Aùû m
0.059
E=E +
log
- 0.059 pH
b
n
n
éë B ùû
0
Electrochemical Equilibrium and Corrosion
a
éë Aùû m
0.059
0
E=E +
log
- 0.059 pH
b
n
n
éë B ùû
Examples:
(1) Consider the reaction Fe  Fe
2
 2e
This type of reaction is favored at low values of pH.
First note that for this and all simple metal dissolution reactions there is no
H+ in the reaction and no dependence of the reaction on pH, m = 0
The Fe2+ is equivalent to A. Also a =1. The quantity [Fe2+ ] is the
concentration of Fe2+ in the electrolyte.
Fe is equivalent to B. Since Fe is a pure metal [Fe ] = 1
Then:
E = E 0 + 0.0295log[Fe 2+ ] = -0.447 + 0.0295log[Fe 2+ ]; (V)
Electrochemical Equilibrium and Corrosion
ΔE
(V)
[Fe2+]
M
-0.477
1
-0.506
0.1
-0.534
0.01
-0.565
0.001
-0.595
10-4
-0.624
10-5
-0.654
10-6
E = -0.447 + 0.0295log[Fe 2+ ]; (V)
For every decade change in ferrous cations
there is ~ 30 mV decrease in the equilibrium
potential.
What is the difference between an open
circuit potential (corrosion potential) and an
Equilibrium Potential????? – Lab.
Electrochemical Equilibrium and Corrosion
a
éë Aùû m
0.059
E=E +
log
- 0.059 pH
b
n
n
éë B ùû
0
Examples:


(2) Consider Al2 O3  6 H  6e  2 Al  3H 2 O
This type of reaction occurs at some intermediate value of pH.
Both Al and Al2O3 are solids with unit activity, A=1, B=1.
o
On a complete EMF series one could find that E Al
/ Al2 O3  1.55 V.
Also note that m = 6, n = 6 so,
E Al/ Al O = -1.55 - 0.059 pH
2 3
The equilibrium potential decreases by ~ 60 mV per pH unit.
Electrochemical Equilibrium and Corrosion
a
éë Aùû m
0.059
E=E +
log
- 0.059 pH
b
n
n
éë B ùû
Examples:
0
(3) One other type of corrosion reaction that can occur involves the formation
of a soluble metal oxide anion at some high pH,
AlO2  4 H   3e   Al  2 H 2O
o
E Al/ AlO - = E Al/
+
AlO 2
2
0.059
4
log éë AlO2- ùû - 0.059 pH
3
3
E Al/ AlO- = -1.262 + 0.020log éë AlO2- ùû - 0.079 pH
2
The ΔE for this reaction is a function of both the dissolved ion content
and the pH of the electrolyte.
Cathode reactions Supporting Corrosion
Hydrogen reduction
+
-
2H + 2e = H2
EH

/ H2
 EH0

/ H2
 0.059 pH
The above reaction occurs in an acid solution.
An equivalent reaction in neutral or alkaline solutions is:
2H2O + 2e- = H2 + 2OH -
EH

/ H2
 EH0

/ H2
EH  / H  0  0.059 pH
2
 0.059 pH
Cathode reactions Supporting Corrosion
Oxygen reduction
O2 + 4H + + 4e- ® 2H2O EO
2
/ H 2O
 EO02 / H 2O  0.059 pH
The above reaction occurs in an acid solution.
An equivalent reaction in neutral or alkaline solutions is:
O2  2 H 2 O  4e   4OH 
EO2 / H 2O  EO02 / H 2O  0.059 pH
EO2 / H 2O  1.229  0.059 pH
Potential/pH (Pourbaix) Diagram
It turns out to be very convenient to represent the
results of all these anodic metal oxidation
processes and cathodic reduction process on a
Map-like “phase diagram”.
The map considers the parameters of voltage and
solution pH and shows the possible
electrochemical/corrosion reactions that can
occur for a particular metal such as Fe, Al, Cd,
Zn, … and water.
Potential/pH (Pourbaix) Diagram
Water
oxygen evolution
and acidification
EO2 / H2O  1.229  0.059 pH
EH  / H  0  0.059 pH
2
hydrogen evolution
and alkalization
Potential/pH (Pourbaix) Diagram
Generic diagram for a metal
Corrosion-soluble ions of the
metal are stable
1.2
POTENTIAL, E(V)
0.8
0.4
Passivation- oxides are stable
0.0
-0.4
Corrosion
Immunity-reduced form of the
metal is stable
-0.8
Passivation
-1.2
Corrosion
-1.6
-2.0
Immunity
-2.4
-2
0
2
4
6
pH
8
10
12
14
Potential/pH (Pourbaix) Diagram
Pourbaix diagram for Al
Reaction 1: Metal oxidizes to
aqueous cations
M = M n + + neAl = Al 3+ + 3eeAl / Al 3  e0 
 A
a
0.059
m
log

0.059 pH
b
n
 B n
eAl / Al 3  1.662 
0.059
log  Al 3 
3
Independent of pH since
no H+ is involved.
Only depends on Al3+
activity
Potential/pH (Pourbaix) Diagram
Pourbaix diagram for Al
Reaction 2: Metal reacts to metal
hydroxide or oxide
M + nH 2O = M (OH ) n + nH + + ne2Al + 3H 2O = Al2O3 + 6H + + 6e-
EAl / Al2O3  1.55  0.059 pH
At higher pH Al2O3 is formed. At lower pH Al2O3 chemically
dissolves to Al3+
Intersection depends depends on Al3+ activity (dashed lines are
portions of the reactions with no significance)
Potential/pH (Pourbaix) Diagram
Pourbaix diagram for Al
The reaction rate constant for
2Al 3+ + 3H2O = Al2O3 + 6H +
is known
H )
(
K=
( Al )
+ 6
3+ 2
= 10-11.4
For (Al3+)=10-6,
log K  6log  H    2log  Al 3   6 pH  2log  Al 3 
log  Al 3  106   0.5log(10 11.4 )  3 pH
6  5.7  3 pH
pH  3.9
Independent of potential
Potential/pH (Pourbaix) Diagram
Pourbaix diagram for Al
Reaction 3: Metal reacts to form
soluble aqueous anions
Al  2H 2 O  AlO2  4H   3e 
EAl / AlO  1.262  0.020log  AlO2   0.079 pH

2
At higher pH, Al2O3 dissolves to AlO2
2
Al2O3  H 2O  2 AlO  2H
pH  14.6  log[ AlO2 ]

K   H    AlO2   1029.2
2
2