Thermodynamics and Electrode Potential
ME 472-062
Copyright ©
Dr. Zuhair M. Gasem

Dr. Z. Gasem
ME 472 KFUPM
Corrosion Science and Engineering 2
Corrosion
Science
Engineering: corrosion forms, and controlling methods
Chpater2
Thermodynamics
•Predict corrosion tendency
•Answer whether corrosion is possible or not
Chapter 3
Kinetics
•Predict speed of corrosion
•If corrosion is possible, how fast?

Dr. Z. Gasem
ME 472 KFUPM
Objective 3
• Given a metal in an electrolyte:
– Anodic reaction is
• M → M +n + ne -
– Cathodic reaction is
• N n+ + n e → N
– Will the metal corrode?
– Corrosion is possible if the spontaneous direction of the reaction is
• M + N n+ → M +n + N
• We need to determine the spontaneous direction of an electrochemical reaction.

Dr. Z. Gasem
ME 472 KFUPM
Electrode Potential
• An electrode is a metal surface allowing charge transfer in or out of an ionic solution.
• Metal conductors consist of free electrons and cations
(+ve ions).
• Water molecules are polar.
• Water molecules polarize the metal surface and attract free electrons of the metal.
• Excess electrons exist at the surface of a metal in water.
• This creates the electrical double layer which prevents easy charge transfer.
Electrode potential
4

Dr. Z. Gasem
ME 472 KFUPM
Electrode Potential 5
• The electrical double layer results in the formation of an electrical potential difference at the metal-solution interface.
• This potential is called the electrode potential or electrochemical potential.
• The electrode potential indicates the tendency of a metal towards electrochemical reactions:
– anodic tendency (M → M +n + ne )
– cathodic tendency (M +n + ne → M)
• The electrode potential depends on the electrolyte, ions concentration, and temperature.

Dr. Z. Gasem
ME 472 KFUPM
Reversible Electrode Potential 6
• The presence of the electrical double layer drives the electrode to reach equilibrium condition where the net current crosses the double layer is zero.
• At equilibrium, there is still charge transfer between the metal and the electrolyte but the anodic current (M →
M +n + ne ) is equal to the cathodic current (M +n + ne →
M) and the net current is zero (i =0).
reversible electrode potential.
net
= i anodic
– i cathodic
• The potential of a reaction at equilibrium is called the
• The reaction at equilibrium is written as M ↔ M +n + ne or M = M +n + ne -

Dr. Z. Gasem
ME 472 KFUPM
Standard Hydrogen Electrode
(S.H.E.).
• The absolute electrode potential can not be measured experimentally.
• The electrode potential is measured relative to a reference electrode such as the standard hydrogen electrode (S.H.E.).
• Assume that the reaction 2H +
+ 2e ↔ H
2 has a reversible potential of 0 when [H+] = 1
M and P
H2
= 1 atm, and the temperature is 25 C.
• This is called the standard condition
2H+ + 2e↔ H2
7

Dr. Z. Gasem
ME 472 KFUPM
Emf Series 8
• We will measure the reversible electrode potentials of all important reactions relative to the S.H.E.
• This produces the Electromotive Force Series
(emf Series) which gives the reversible electrode potential of a reaction at standard conditions
(ion concentration is 1 M, pressure of any gas at
1 atm, and a temperature of 25 C)
– [M +n ] = 1 M, P = 1 atm, T= 25 C.

Dr. Z. Gasem
ME 472 KFUPM
Standard emf
(Reduction
Potentials)
9

Dr. Z. Gasem
ME 472 KFUPM
Standard emf (Reduction Potentials) 10
• Note the followings:
– The potential of a reaction is called the halfcell electrode potential
– The reactions in the emf series are all written as reduction reactions
– Noble metals such as Au and Pt have positive reduction potentials
– Active metals such as Mg and Na have negative reduction potentials

Dr. Z. Gasem
ME 472 KFUPM
Example: half-cell potential for Zn
• Measure the standard reversible half-cell reaction for Zn electrode relative to S.H.E.
• e o
Zn/Zn+2
= -0.76 V vs.
S.H.E.
• Note that the +ve terminal of the voltmeter is connected to the SHE.
• Hence, SHE electrochemical potential is more +ve than that for
Zn/Zn 2+
+
11

Dr. Z. Gasem
ME 472 KFUPM
Example: half-cell potential for Zn
• Note that when the
-ve terminal of the voltmeter is connected to the SHE, the voltmeter reads
-0.76 V vs. SHE.
• This indicates that
Zn/Zn 2+ is –ve towards SHE
+
-0.76
-
12

Dr. Z. Gasem
ME 472 KFUPM
Galvanic Cell Potentials
• A galvanic cell consists of two electrodes: an anode and a cathode.
• Corrosion always involves two half-cell reactions.
• We will use the emf series to predict which half-cell reaction will act as the cathode and which will act as the anode when the two are connected.
13

Dr. Z. Gasem
ME 472 KFUPM
Cell Potentials 14
• Thermodynamics states that a spontaneous reaction is associated with a decrease in free energy.
• Free energy of an electrochemical reaction is given by:
– ∆ G cell
= -n*F*E cell
• n = number of electrons exchanged
• F = Faraday’s constant = 96500 Coulomb per mole of electrons
• E cell
= (e c
- e a
): the potential difference b/w the cathode and the anode in the cell.

Dr. Z. Gasem
ME 472 KFUPM
Faraday’s Constant 15
• One mole of electrons, contains
Avogadro’s number (6.023x10
23 ) electrons
• Charge on one electron is 1.6
10 -19
Coulomb
• Hence the charge on one mole of electrons=6.023x10
23 *1.6x10
-19 =96500 C
• This is known as
– F=96500 C/(mole of e)

Dr. Z. Gasem
ME 472 KFUPM
Example: Cu-Zn Galvanic Cell
• A galvanic cell is at equilibrium (no current flow), consists of Zn and Cu electrodes at standard conditions.
• Which one will be the anode (the corroding metal) and which one is the cathode if we allow current to flow?
• Before current flows:
• the two reactions are at equilibrium and at standard conditions (we can use the emf series directly):
‰ Zn +2 + 2e ↔ Zn e o = -0.763 V
‰ Cu +2 + 2e ↔ Cu e o = 0.337 V
16

Dr. Z. Gasem
ME 472 KFUPM
Cu-Zn Galvanic Cell 17
• After current flows:
• Assume initially Zn will react in the cathodic direction
( Zn +2 + 2e → Zn) , and Cu will react in the anodic direction ( Cu → Cu +2 + 2e ) (in essence we are assuming that Cu will corrode)
• Net reaction is Cu + Zn +2 → Cu +2 + Zn
– e a
– e c
= 0.337 V
= -0.763 V
– E cell
= e c
- e a
= -0.763-0.337= -1.1 V
– ∆ G = -n*F*E cell
( ∆ G is +ve)
• Hence Cu + Zn +2 → Cu +2 + Zn is not the spontaneous direction of the reaction)

Dr. Z. Gasem
ME 472 KFUPM
Cu-Zn Galvanic Cell 18
• Assume Cu will react in the cathodic direction ( Cu +2 +
2e → Cu) , and Zn will react in the anodic direction ( Zn
→ Zn +2 + 2e ) (in essence we are assuming that Zn will corrode)
• Net reaction is Cu +2 + Zn → Cu + Zn +2
– e
– e
– E a c cell
= -0.763 V
= 0.337 V
= e c
- e a
– ∆ G = -n*F*E
= 0.337 – (-0.763) = 1.1 V cell
( ∆ G is –ve)
– Hence Cu +2 + Zn → Cu + Zn +2 is the spontaneous direction of the reaction)
• Conclusion: Zn will act as the anode (Zn will corrode) when coupled with Cu to form a galvanic cell (Daniel cell) at the standard condition.

Dr. Z. Gasem
ME 472 KFUPM
Cu-Zn Galvanic Cell
• C: ( Cu +2 + 2e → Cu)
• A: ( Zn → Zn +2 + 2e )
• Overall reaction is the addition of A and C:
Cu +2 + Zn → Cu + Zn +2
• The equilibrium cell potential is
1.1 V
• This can act as a battery supplying current.
• The maximum work available from
• the battery is (
∆ G cell
∆ G cell
= -n*F*E
= -2*96500*1.1
cell
)
= -212.300 KJ
19

Dr. Z. Gasem
ME 472 KFUPM
Example: Cu-Ag 20
• A silver electrode is immersed in silver nitrate
(AgNO
3
) and a copper electrode is immersed in a copper nitrate Cu(NO
3
)
2 to form a galvanic cell at the standard condition . Which electrode will act as the cathode and which is the anode?
– Cu +2 + 2e → Cu (e o = 0.337 V vs. SHE)
– 2Ag + + 2e → 2Ag (e o = 0.799 V vs. SHE)
– E cell
– E cell
= e c
– e a
– ∆ G = -n*F*E cell
= 0.799-0.337=0.462 V

Dr. Z. Gasem
ME 472 KFUPM
Example: Galvanic cell Cu-Ag
• In galvanic cell such as a battery:
• The anode is (-ve) where anodic reaction takes place
• The cathode is
(+ve) where cathodic reaction takes place
Cu → Cu 2+ +2e -
21
2Ag + + 2e → 2Ag

Dr. Z. Gasem
ME 472 KFUPM
Current circuit in galvanic cell 22
• Salt bridge replaces a porous barrier
• Salt bridge allows ions flow
•Note the flow of negative current:
•Electrons in the wire
•Anions in solution

Dr. Z. Gasem
ME 472 KFUPM
Determination of the Spontaneous
Direction of a reaction
23
• Example: Indicate the spontaneous direction of the reaction (assume standard conditions):
– 3Pb + 2Al 3+ ↔ 3Pb 2+ + 2Al
• Assume the reaction goes from left to right
– A: 3Pb → 3Pb 2+ + 6e e o a
– C: 2Al 3+ +6e → 2Al e o c
= -0.126 V
= -1.662 V
– E cell
= e c
- e a
= -1.536 V (note that e is used for halfcell potential while E for the overall cell)
– ∆ G cell
= -n*F*E cell
= +ve (hence the spontaneous direction of the reaction is from right to left or
• 3Pb 2+ + 2Al → 3Pb + 2Al 3+ and E cell
= 1.536 V

Dr. Z. Gasem
ME 472 KFUPM
Concentration Effects on Reversible
Standard Potential
24
• The emf series lists the electrode potential at standard condition ([M +n ] = 1 M, P = 1 atm, T=
25 C).
• What happen if we change the ions concentration or the temperature or pressure?
• It would be impossible to measure the electrode potential for all conditions.
• Nernst ’ s equation calculates the electrode potential at conditions different than the standard conditions.

Dr. Z. Gasem
ME 472 KFUPM
Nernst’s Equation 25
• For a reaction written in the reduction direction :
• Nernst’s Equation:
• e = e o + 2.3 RT/(nF)* Log ([Reactant]/[Product])
– R : gas constant = 8.314 J/(mol*K)
– T : temperature in K
– n: number of electron mole
– F: Farady’s constant
– Log [solid] = 1
• at room temperature 2.3RT/F=2.3*8.314*298/96500=
0.059 V
• Nernst’s equation at room temp is written as:
– e = e o + 0.059/n* Log ([Reactant]/[Product])

Dr. Z. Gasem
ME 472 KFUPM
Nernst’s Equation 26
• Example: what would be the half-cell electrode potential of Fe when [Fe 2+ ] = 1x10 -6 M, at RT?
¾ Fe +2 + 2 e ↔ Fe e o = -0.447 V vs. SHE
¾ e = -0.447 + 0.059/2* log(1x10 -6 /1) = -0.617 V
• Example: what would be the half-cell electrode potential for hydrogen reduction when [H + ] = 1x10 -3 , P
H2
=0.1 atm; at 100 C?
¾ 2H + + 2e ↔ H
2 e o = 0 V
¾ e = e o + 2.3*8.314*373/nF* Log ([H + ] 2 /P
¾ e = 0 + 0.074/2*(2*log 1x10 -3 -log 0.1)
H2
)
¾ e = 0.074/2*(-6- (-1)) = -0.185 V

Dr. Z. Gasem
ME 472 KFUPM
Nernst’s Equation 27
• For a general reaction described by aA + mH + + ne -
bB + dH
2
O
The reversible potential for a given concentration different than the standard conditions are given by Nernst
s equation as: e = e o + 2.3RT/nF log {[A] a [H + ] m /[B] b [H
2
O] d }

Dr. Z. Gasem
ME 472 KFUPM
Derivation of Nernst’s Equation 28
• The free energy of a reaction in equilibrium as a function of composition is given by:
¾ ∆ G = - RT ln (K/Q)
► K is the equilibrium constant of the reaction
► Q is the reaction quotient (Q=[Product]/[Reactant])
¾ ∆ G = -RT ln K + RT ln Q
► but ∆ G o = -n F E o , ∆ G = -n F E
¾ -n F E = -n F E o + RT ln Q
¾ E = E o – RT/nF ln Q and (ln x = 2.3 log x)
¾ E = E o – 2.3 RT/nF log Q
Nernst’s equation for a reduction reaction is written as:
E = E o + 2.3 RT/nF log {[Reactant]/[Product]}

Dr. Z. Gasem
ME 472 KFUPM
Example
• Write the reaction
(Zn +2 + Fe ↔ Zn + Fe +2 ) in the spontaneous direction and
Calculate the cell potential :
[Fe +2 ] = 0.1 M and [Zn +2 ]=1.5 M.
►
Calculate the reduction potential for each half-cell reaction.
►
►
►
►
Fe +2 + 2e ↔ Fe ; e o = -0.447V
e = -0.447+
0.059/2*log(0.1/1)= -0.477 V
Zn +2 + 2e ↔ Zn ; e o = -0.762V
e= -0.762+0.059/2*log(1.5/1)=
-0.757 V
Fe/Fe +2
(FeSO
4
)
29
Zn/Zn +2
(ZnSO
4
)

Dr. Z. Gasem
ME 472 KFUPM
Example
►
►
►
►
►
Assume that the reaction has a tendency to proceed from L to R
Cathode: Zn +2 + 2e → Zn e = -0.757V
Anode : Fe → Fe +2 + 2e e = -0.477 V
E cell
= -0.757 +0.477 = -0.28 V
Since ∆ G is +ve, the spontaneous direction is reversed (from R to L)
►
The spontaneous direction
Zn + Fe +2 → Zn +2 + Fe (Zn is anode and Fe is cathode)
Thus, Zn has the tendency to corrode in this cell.
E cell
= -0.477+0.757 = 0.28 V
30

Dr. Z. Gasem
ME 472 KFUPM
Predicting The Spontaneous
Direction by Inspection
31
M +m + N ↔ M + N +n
• Calculate the potential for each half cell reaction as written in the reduction direction.
• Use Nernst equation if not in standard conditions.
►
►
M +m + me → M e = e
M
N +n + ne → N e = e
N
• Note that the electrode potential in the cathodic direction becomes:
– More –ve than e o if the concentration is < 1 M
– More +ve than e o if the concentration is > 1 M
• Then
– The spontaneous cathode in the cell is the more +ve
– The spontaneous anode in the cell is the more –ve
– The E cell
= e c
- e a

Dr. Z. Gasem
ME 472 KFUPM
Example 32
– Cu +2 + 2e → Cu; e o = 0.342 v, when [Cu +2 ]
=1M
– If [Cu +2 ] >1M, the electrode potential becomes more +ve. For [Cu +2 ] = 2M
• e = e o + 0.059/n* Log ([Reactant]/[Product])
• e = 0.342+0.059/2*log(2/1)= 0.351 V
– If [Cu +2 ] <1M, the electrode potential becomes more – ve. For [Cu +2 ] = 0.01M
• e = 0.342+0.059/2*log(0.01/1)= 0.283 V

Dr. Z. Gasem
ME 472 KFUPM
Concentration Cell
• Use the inspection method to predict which electrode will corrode?
• Left electrode:
Fe +2 + 2e ↔ Fe e = -0.447+0.059/2*log(0.01/1)
= -0.506 V
• Right electrode e = -0.447+0.059/2*log(0.1/1) =
-0.477 V
Since the more negative electrode acts as anode; then the left electrode is the anode and the right one is the cathode.
33

Dr. Z. Gasem
ME 472 KFUPM
Concentration Cell
• Use the inspection method to predict which electrode will corrode?
34

Dr. Z. Gasem
ME 472 KFUPM
Example: 35
• Use the inspection method to write the following reaction in the spontaneous direction.
Cd + Fe +2 ↔ Cd +2 + Fe
[Cd +2 ] = 1x10 -6 M and [Fe +2 ] = 1.5 M
– Which electrode is the anode?
– Which electrode is the cathode?
– Draw a diagram for the galvanic cell and show the flow of electrons, anions, and cations?
– Which metal is corroding?
– Calculate E cell
?
– Calculate the maximum energy available?
Fe/Fe +2 Cd/Cd +2
Fe in a solution of FeSO
4 and Cd in a solution of
CdSO
4

Dr. Z. Gasem
ME 472 KFUPM
Example Problem 36
• Determine the thermodynamic tendency for tin (Sn) to corrode in de-aerated sulfiric acid with [H + ]=10 -2 M. Assume [Sn +2 ]=10 -6 M and P
H2
= 1 atm.
1.
Identify the anodic and cathodic reactions.
– Sn +2 +2e ↔ Sn e o = -0.138 V
– 2H +2 + 2e ↔ H
2
2.
Calculate the half-cell potential for each reaction
•
• e
Sn e e o = 0 V
= -0.138+0.059/2*log(10
= - 0.118 V
-6 /1)= - 0.315 V
= 0.0+0.059/2*log([10 -2 ] 2 /1)=0.059*log10 -2 = - 0/059*2
3.
By inspection, Sn is the anode and H
2
4.
E cell
= e c
-e a
= -0.118 – is the cathode
(-0.315) = 0.197 V ( ∆ G is
5.
Spontaneous reaction is Sn + 2H +2 → Sn +2 + H
2
– ve)
6.
Conclusion: Sn has a tendency to corrode (will corrode) in the given acidic solution.

Dr. Z. Gasem
ME 472 KFUPM
Galvanic cell application1: Batteries
• A battery is a galvanic cell which produces direct current from the electrochemical reactions of two electrodes
(similar to corrosion or useful corrosion)
• Lead storage battery
– Anode reaction (-ve)
• Pb (s) +HSO
(e o = -0.35 V)
4
→ PbSO
4
+ H + + 2e -
– Cathode reaction (+ve)
• PbO
2
+ HSO
(e o =1.69 V )
4
+ 3H + + 2e → PbSO
4
+ 2H
2
O
– Pb+ PbO
(E cell
2
+ 2HSO
4
+2H +
=1.69 +0.35 = 2.04 V)
→ 2PbSO
4
(s) + 2H
2
O
• Note that sulfuric acid (H
2
SO
4
) is consumed during discharging and the battery must recharged (reactions are reversed by external current) to reproduce the acid.
37

Dr. Z. Gasem
ME 472 KFUPM
Galvanic cell application2:Fuel Cells
• A fuel cell produces electricity from a continuous external supply of fuel (on the anode) and oxidant (on the cathode) in the presence of an electrolyte.
• A hydrogen fuel cell uses H oxidant:
2 as fuel and O
2 as
– Anode:
• 2H
2
+ 4OH → 4H
2
O + 4e -
– Cathode:
• O
2
+ 2H
2
O + 4e → 4OH -
– Net reaction:
• 2H
2
• Application
(g) + O
2
(g) → 2H
2
O (l)
– Electrical power source (a plant in Tokyo)
– Electric battery to power electric motor in electric and hybrid vehicles.
38

Dr. Z. Gasem
ME 472 KFUPM
Water Chemistry 39
• Many metals corrode in water medium.
• Any water solution at 25 C, dissociate into [H + ] and [OH ] ions such that:
[H + ][OH ] = 1x10 -14
– The solution is called neutral if [H + ] = [OH ]
– The solution is called acidic if [H + ] > [OH ]
– The solution is called basic if [H + ] < [OH ]
– Examples of
• acidic solutions: HCl, HNO
3
, H
2
SO
4
• Basic Solutions: NaOH, KOH

Dr. Z. Gasem
ME 472 KFUPM
Water Chemistry
• The pH scale provides a convenient way to represent solution acidity: pH = - log [H + ]
• Thus, for a neutral solution where [H + ] =
1x10 -7 pH = -log (1x10 -7 ) = 7
• Note that the pH decreases as the [H + ] increases.
• A solution of pH 3 has a [H + ] concentration 10 times higher than pH of
4 and 100 times that of pH 5.
40

Dr. Z. Gasem
ME 472 KFUPM
Water Chemistry 41
• The water pH and the level of dissolved
O
2 determine the reduction reaction that takes place during corrosion of metals in water solution:
I.
Acid solutions with no dissolved O
2
‰ 2H + + 2e ↔ H
2
‰ For [H + ] = 1M (pH=0) @ RT ; e o = 0
(1)
‰ Apply Nernst ’ s equation for reversible potential at different [H + ] (assume @ RT and P
H2
=1)
‰ e = 0 + 0.059/2 log {[H + ] 2 /P
H2
} = 0.059/2 *2 log [H + ]
‰ The potential of the reaction at RT varies according to e = -0.059 pH (a)

Dr. Z. Gasem
ME 472 KFUPM
Water Chemistry 42
II. Neutral or alkaline solutions
‰ Add [OH ] to both sides of reaction (1) to neutralize it yields
‰
‰ 2H + + 2OH + 2e ↔ H
2H
2
O + 2e ↔ H
2
2
+ 2OH -
+ 2OH (2)
‰ For neutral solution ; [H + ] = 1x10 -7 M @ RT
‰ e = -0.059*7 = -0.413 V
‰ For alkaline solution ; [H + ] = 1x10 -14 M @ RT
‰ e = -0.059*14 = -0.826 V
‰ Plotting Equation (a) as a function of pH yields the e-pH (Pourbaix) diagram

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water 43

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water 44
III. Acid solutions with dissolved O
2
‰ Add [O
2
] to both sides of reaction (1):
‰ 2H + + 2e + O
2
↔ H
2
+ O
2
‰ 4H + + O
2
+ 4e ↔ 2H
2
O (3)
‰ For [H + ] = 1 M (pH=0) , P
O2
‰ e o = 1.229 V
= 1 atm @ RT
‰ Apply Nernst ’ s equation for reversible potential at different [H + ] (assume P
O2
= 1 atm ; @RT):
‰ e = 1.229 + 0.059/4 log {[H + ] 4 P
O2
} = 1.229-0.059 pH
‰ The potential of the reaction at RT varies according to e = 1.229-0.059 pH (b)

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water 45
IV. Alkaline solutions with dissolved O
2
‰ Add [OH ] to both sides of reaction (3):
‰ 2OH + 2H + + 2e + O
2
↔ H
2
+ O
2
+ 2OH
‰ 2H
2
O + O
2
+ 4e ↔ 4OH (4)
‰ Apply Nernst ’ s equation for reversible potential at different [H + ] (assume P
O2
= 1 atm ; @RT):
‰ e = 1.229 + 0.059/4 log {P
O2
/[OH ]} = 1.229-0.059 pH
‰ The potential of the reaction at RT varies according to
e = 1.229-0.059 pH (b)
‰ Plotting Equation (b) as a function of pH yields the e-pH (Pourbaix) diagram

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water 46

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water 47
• Note the followings:
– Pourbaix diagram is a plot of reversible potential vs. pH for reactions in pure water.
– If we apply a more -ve potential than the eq value:
• 2H + + 2e ↔ H
2
• 2H
2
O + 2e ↔ H
2
+ 2OH -
• Hydrogen is generated or evolved
– If we apply a more +ve potential than the eq value:
• 4H + + O
2
• 2H
2
O + O
2
+ 4e ↔ 2H
2
O
+ 4e ↔ 4OH -
• Oxygen is evolved
– The region b/w lines a and b is the region where water is stable.

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water 48
O
2
H
2

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Metals 49
• If we consider all possible reactions of a metal in water gives three types of reactions:
– Reactions dependent only on potential such as
Fe +2 + 2e ↔ Fe. These reactions are represented as horizontal line in Pourbaix diagram (independent of pH)
– Reactions dependent only on pH such as
Fe +2 + 2OH ↔ Fe(OH)
2
. These reactions are represented as vertical line in Pourbaix diagram (independent of potential)
– Reactions dependent on both pH and potential, i.e.
Fe(OH)
2
+ 2H + + 2e ↔ Fe + 2H
2
O

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water
For Al in water, the possible reactions:
1. Al 3+ + 3e ↔ Al e 0 = -1.662 V
– Nernst ’ s equation is written as
– e = -1.662+0.059/3*log([Al 3+ ]/1)
• For [Al 3+ ]=1x10 -6 M, e = -1.76 V
Al +3
50
Al

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water
2. Al
2
O
3
+ 6H + + 6e ↔ Al + 3H
– e = -1.55- 0.059 pH
2
O; e 0 =-1.55V
51
Al
2
O
3

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water
3. Al
2
O
3
+ 3H
2
+ 6H + ↔ 2Al +3
• At equilibrium; K =
(H + ) 6 /(Al +3 ) 2 =10 -11.4
• For [Al 3+ ]=1 M, pH=1.9
• For [Al 3+ ]=1x10 -6 M, pH=3.9
52

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water
4. AlO -
2
+ 2H
2
+ 4H + + 3e ↔ Al
O
– e = -1.262 + 0.02 log (AlO -
2
) - 0.079 pH
53

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Water
5. Al
•
2
O
3
+ H
2
O ↔ AlO
2
+ 2H
At equilibrium; K = (AlO
2
) (H + ) 2 =10 -14.6
+
•
• For [AlO -
2
] = 1 M, pH = 14.6
For [AlO -
2
] = 1 M, pH = 8.6
54

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Al
Ions con =1x10 -6 M
• The Pourbaix diagram shows the ranges of stability for all major phases of a metal in water
• The diagram shows graphical representations of electrochemical equilibra .
• Note that Al can corrode in low pH and high pH but not in neutral solutions.
• Two possible anodic reactions of
Al (reactions 1 and 4) are below the water lines.
• Pourbaix diagram shows the stable phases:
– Metal
– Soluble ions
– Oxide films
AL
55

Dr. Z. Gasem
ME 472 KFUPM
Effect of Concentration
Ions con =1x10 -6 M
Different ions concentrations
56
[AlO
2
]=10 -2 M

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix diagram and Corrosion
• Pourbaix diagram shows the stable phases:
– Metal
– Soluble ions
– Oxide films
• We call these regions as:
– Immune region (where the metal is stable and corrosion is not possible )
– Corrosion region (where soluble ions are stable and corrosion is possible )
– Passive region (where oxide films are stable and the metal is protected by a passive film )
57

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Iron
• The shaded region shows the stable region for soluble ions where Fe can corrode.
• Two possible anodic reactions of Fe are below the water lines.
58

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Gold
• Note that the reversible anodic potential for pure gold is above the oxygen line for all pH.
• Water can not oxidize gold.
• Thus, gold is immune from corrosion in water at all pH.
59

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Copper
• Note that corrosion reactions of copper are above the deaerated water.
• Hence, copper does not corrode in deaerated water.
• However, copper does corrode in aerated water.
60

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Titanium 61
• Ti is highly corrosion resistant due to its wide range of passivity.

Dr. Z. Gasem
ME 472 KFUPM
Limitation of Pourbaix diagrams 62
• A Pourbaix diagram is potential-pH diagram which shows all the thermodynamic equilibria for a given metal in pure water at 25 C.
• Engineers can control the potential or the pH of the solution to control corrosion.
– The vertical axis is varied by:
• Varying the oxidizer concentration in the solution
• Applying an external potential to the metal
– The pH can be varied by changing the acidity of the solution.
• Limitations:
– The diagrams are drawn for equilibrium condition, and practical corrosion is far from equilibrium
– The diagrams are for pure metals in dilute solutions but in practice there are other ions which may affect equilibria
– The diagram can tell us whether corrosion is possible or not (no indication for corrosion rates)

Dr. Z. Gasem
ME 472 KFUPM
Cathodic Protection of Iron
• Iron in neutral water (pH 7) develops a potential of -0.5 V, which is inside the corrosion region (Fe +2 ).
• To protect a piece of iron, we need to force the metal to be in the immune region (reduce the potential).
• The potential should be reduced to less than -0.6 V
(vs. SHE) using an external current supply
• This is the basic principle of cathodic protection.
63

Dr. Z. Gasem
ME 472 KFUPM
Pourbaix Diagram for Chromium
• Cr is added to steel in large amounts
(>11%) to make stainless steels.
• Cr has wide passive region which will make stainless steel more corrosion resistance than steel alloys.
64

Dr. Z. Gasem
ME 472 KFUPM
Reference electrodes
• The SHE is complicated setup.
• There are several other practical reference electrodes:
• The saturated calomel electrode (SCE)
• Hg
2
Cl -
Cl
2
+ 2e ↔ 2 Hg + 2
• e = 0.268 – 0.059 log (Cl )
• e = 0.241 V vs. SHE
• Mainly used in chloride containing solutions.
65

Dr. Z. Gasem
ME 472 KFUPM
Reference electrodes
• The Silver-silver chloride electrode
• AgCl + e ↔ Ag + + Cl -
• e = 0.222 + 0.059 pH
• The copper-saturated copper sulfate electrode
– Cu +2 + 2e↔ Cu
– e = 0.340 + 0.0295 log (Cu +2 )
Lead wire to voltmeter
Brass washer
Brass nut
Threaded plastic cap
Plastic tube
Electrode
(copper rod)
Electrolyte
(saturated copper sulfate solution)
Undissolved copper sulfate crystals
Porous plug
Mud or moist soil
66

Dr. Z. Gasem
ME 472 KFUPM
Conversion of potentials b/w different reference electrodes
67
• Example: If the corrosion potential of steel in seawater is -0.5 V (SCE), what is the potential
– in the SHE: -0.5+0.241= -0.259 V (SHE)
– in the Ag/AgCl: -0.5-(0.241-0.197)= -0.456 V (Ag/AgCl)
• Example: If the corrosion potential of Zn in seawater is -0.8 V (SHE), what is the potential
– in the SCE: -0.8-0.241= -1.041 V (SCE)
– in the Ag/AgCl: -0.8-0.197= -0.997 V (Ag/AgCl)

Dr. Z. Gasem
ME 472 KFUPM
Electrolysis 68
• We can use external electric current to drive an electrochemical reaction in the reverse of its spontaneous direction.
• The applied voltage must be greater than the spontaneous E cell
.
• In electrolytic cell, the reduction reaction occurs at the –ve terminal of the power supply and the anodic reaction at the +ve terminal.
Galvanic Cell
Electrolytic Cell
+
-
+