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1. CHAPTER III
ELECTROCHEMISTRY

2. 3.1. Insight into Corrosion is the deterioration of metals by an
electrochemical process.
A. Forms of Corrosion
 Uniform corrosion can be defined as the attack of
the entire metal surface exposed to the corrosive
environment resulting in uniform loss of metal
from exposed surface.

3.  Galvanic corrosion occurs only when two different metals contact
each other in the presence of an appropriate electrolyte.

4.  Crevice corrosion can occur whenever small gaps exist between metal
parts

5. 13.2 Oxidation-Reduction Reactions and Galvanic Cells
A. Oxidation-Reduction and Half-Reactions
• Oxidation-reduction reaction is a type of chemical reaction that
involves a transfer of electrons between two species.
• Oxidation is the loss of electrons from some chemical species.
• Reduction is the gain of electrons.
• The most important principles of redox chemistry: the electrons lost
in oxidation must always be gained in the simultaneous reduction of
some other species.

7. 13.2 Oxidation-Reduction Reactions and Galvanic Cells
• The copper has been oxidized, and we could write an equation to
describe this change:
𝐶𝑢 𝑠 → 𝐶𝑢2+ 𝑎𝑞 + 2𝑒−
• The silver has been reduced:
𝐴𝑔+ 𝑎𝑞 + 𝑒− → 𝐴𝑔 (𝑠)
• The two equations describe are what called half-reactions for the
oxidation of copper and the reduction of silver. We can multiply the
reduction by two to make this explicit, giving us the following pair of
half-reactions:
𝐶𝑢 𝑠 → 𝐶𝑢2+ 𝑎𝑞 + 2𝑒−
2 𝐴𝑔+ 𝑎𝑞 + 2 𝑒− → 2 𝐴𝑔 (𝑠)
• Silver now gains the two electrons that the copper loses.

8. 13.2 Oxidation-Reduction Reactions and Galvanic Cells
• Adding the two half-life reactions together gives us:
𝐶𝑢 𝑠 → 𝐶𝑢2+ 𝑎𝑞 + 2𝑒−
2 𝐴𝑔+ 𝑎𝑞 + 2 𝑒− → 2 𝐴𝑔 (𝑠)
2 𝐴𝑔+ 𝑎𝑞 + 𝐶𝑢 𝑠 → 2 𝐴𝑔 𝑠 + 𝐶𝑢2+(𝑎𝑞)
• We could also write this as a molecular equation by including the
spectator ions (𝑁𝑂3
−
in this case):
2 𝐴𝑔𝑁𝑂3 𝑎𝑞 + 𝐶𝑢 𝑠 → 2 𝐴𝑔 𝑠 + 𝐶𝑢(𝑁𝑂3)2 (𝑎𝑞)
• Terminologies:
The species undergoing oxidation is referred to as a reducing agent.
The species undergoing reduction is referred to as an oxidizing agent.
: Net ionic
equation

9. Example: (Balancing Redox equations)
1. Suppose we are asked to balance the equation showing the
oxidation of Fe2+ ions to Fe3+ ions by dichromate ions (Cr2O7
2−
)
in an acidic medium. As a result, the Cr2O7
2−
ions are reduced to
Cr3+
ions.
Step 1: Write the unbalanced equation for the reaction in ionic form.
Fe2+ + Cr2O7
2−
→ Fe3+ + Cr3+
Step 2: Separate the equation into two half-reactions.
Oxidation: Fe2+
→ Fe3+
Reduction: Cr2O7
2−
→ Cr3+

10. Example: (Balancing Redox equations) Cont.:
Step 3: Balance each half-reaction for number and type of atoms and
charges. For reactions in an acidic medium, add 𝐻2𝑂 to balance the O
atoms and 𝐻+
to balance the H atoms.
Fe2+ → Fe3+ + 𝑒−
Cr2O7
2−
→ 2Cr3+ + 7H2O
• To balance the H atoms, we add 14H+ ions on the left-hand side:
14H+
+ Cr2O7
2−
→ 2Cr3+
+ 7H2O
• There are now 12 positive charges on the left-hand side and only six
positive charges on the right-hand side. Therefore, we add six
electrons on the left
14H+ + Cr2O7
2−
+ 6𝑒− → 2Cr3+ + 7H2O

11. Example: (Balancing Redox equations) Cont.:
Step 4: Add the two half-reactions together and balance the final
equation by inspection. The electrons on both sides must cancel. If the
oxidation and reduction half-reactions contain different numbers of
electrons, we need to multiply one or both half-reactions to equalize
the number of electrons.
6(Fe2+ → Fe3+ + 𝑒−)
14H+ + Cr2O7
2−
+ 6𝑒− → 2Cr3+ + 7H2O
6Fe2+ + 14H+ + Cr2O7
2−
+ 6𝑒− → 6Fe3+ + 2Cr3+ + 7H2O + 6𝑒−
• The balanced net ionic equation:
6Fe2+ + 14H+ + Cr2O7
2−
→ 6Fe3+ + 2Cr3+ + 7H2O
Step 5: Verify that the equation contains the same type and numbers of
atoms and the same charges on both sides of the equation.

12. Example: (Balancing Redox equations):
2. Write a balanced ionic equation to represent the oxidation of iodide
ion (I−) by permanganate ion (MnO4
−
) in basic solution to yield
molecular iodine (I2) and manganese (IV) oxide (MnO2).
6I− + 2MnO4
−
+ 4H2O → 3I2 + 2MnO2 + 8OH−
3. Balance the following equation for the reaction in an acidic medium
by the ion-electron method: Fe2+
+ MnO4
−
→ Fe3+
+ Mn2+

13. 13.2 Oxidation-Reduction Reactions and Galvanic Cells
A. Building a Galvanic Cell
• A galvanic cell is any electrochemical cell in which a spontaneous
chemical reaction can be used to generate an electric current.
• A salt bridge contains a strong electrolyte that allows either cations
or anions to migrate into the solution where they are needed to
maintain charge neutrality,

14. 13.2 Oxidation-Reduction Reactions and Galvanic Cells
B. Terminology for Galvanic Cells
• The electrically conducting sites at which either oxidation or
reduction take place are called electrodes.
• Oxidation occurs at the anode and reduction occurs at the cathode.
• The cell notation is a shorthand notation for representing their
specific chemistry:
Anode electrolyte of anode electrolyte of cathode cathode
Example:
Cu s Cu2+
aq 1M Ag+
aq 1M Ag(s)
• Electromotive force (EMF) or cell potential is the potential difference
between two electrodes of a galvanic or voltaic cell.
𝓌𝑚𝑎𝑥 = 𝑞𝐸

15. 13.2 Oxidation-Reduction Reactions and Galvanic Cells
B. Galvanic Corrosion and Uniform Corrosion
A tin can is usually tin-
plated steel. If the tin
coating is scratched to
expose the underlying
steel, iron in the steel
will corrode rapidly.

16. 13.3 Cell Potentials (Standard Condition)
• We can determine the standard cell potential for any pair of half-
reactions by using the equation:
𝐸 𝑐𝑒𝑙𝑙
°
= 𝐸 𝑟𝑒𝑑
°
− 𝐸 𝑜𝑥
°
• In any galvanic cell, the half-reaction with the more positive reduction
potential will be the cathode.
Standard reduction potentials for several of the half-
reactions involved in the cells.
• A large, positive value for
the standard reduction
potential implies that the
substance is reduced readily
and is therefore a good
oxidizing agent.

17. Example:
Copper and iron (generally in the form of steel) are two of the many
metals used in designing machines. (a) Using standard reduction
potentials, identify the anode and the cathode and determine the cell
potential for galvanic cell composed of copper and iron. Assume
standard conditions. (b) We can also construct a galvanic cell using
copper and silver. Confirm that the potential of the following galvanic
cell is 0.462 V:
Cu s Cu2+ 1M Ag+ 1M Ag(s)
(a)𝐸 𝑐𝑒𝑙𝑙
°
= 0.78 V
(b)𝐸 𝑐𝑒𝑙𝑙
°
= 0.462V

18. 13.3 Cell Potentials (Nonstandard Condition)
• The equation that describes cell potentials under nonstandard
conditions is called the Nernst equation:
𝐸 = 𝐸° −
𝑅𝑇
𝑛𝐹
ln 𝑄
• Where Q is the reaction quotient. Thus for general chemical reaction,
𝑎A + 𝑏B 𝑐𝐶 + 𝑑𝐷
𝑄 =
[C]𝑐
[D]𝑑
[A]𝑎[B]𝑏
• Where F is the Faraday constant, and n is the number of electrons
transferred in the redox reaction.
𝐹 = 96,485 JV−1mol−1
= 96,485 C mol−1
• We can apply the Nernst equation to estimate the potential of the
electrochemical system in the corrosion of steel at more realistic
concentrations.

19. Example:
Suppose that you work for a company that designs the drive mechanisms for large
ships. The materials in this mechanism will obviously come into contact with
environments that enhance corrosion. To estimate the difficulties that corrosion might
cause, you decide to build a model electrochemical cell using electrolyte
concentrations that might be present in your system when it is in service. Assume that
you have a cell that has an iron(II) concentration of 0.015 M and an H+ concentration
of 1 × 10−3M. The cell temperature is 38°C, and the pressure of hydrogen gas is
maintained at 0.04 atm. What would the cell potential be under these conditions?
Strategy: This problem defines nonstandard conditions that must be addressed using
the Nernst equation. Iron will be the anode, but we will need to scan the table of
standard reduction potentials to identify a possible cathode reaction. The most likely
suspect is the reduction of H1 to H2. Once we know both half-reactions, we can
calculate the standard cell potential and fill in the appropriate values in the Nernst
equation

20. Solution:
Anode reaction:Fe2+ aq + 2𝑒− → Fe(s) 𝐸° = −0.44V
Cathode reaction:2H+ aq + 2𝑒− → H2(g) 𝐸° = 0.00V
• These reactions tell us two things: first, the standard cell potential will be
𝐸°
= 0.00 V − −0.44 V = 0.44 V
• Second, there are two electrons transferred in the overall redox reaction:
Fe s + 2H+ aq → H2 g + Fe2+ aq
• These facts, plus the values given in the problem and those of the constants allow us to use the Nernst
equation to find the cell potential:
𝐸 = 𝐸°
−
𝑅𝑇
𝑛𝐹
ln 𝑄
𝐸 = 0.44 V −
8.314
J
mol ∙ K
311 K
2 × 96,485
J
V ∙ mol
ln
0.015 0.04
0.0010 2
= 0.35V (∗)
∗Use your calculator

21. 13.4 Cell Potentials and Equilibrium
• Relationship between free energy and the cell potential
∆𝐺°
= −𝑛𝐹𝐸°
Where n is the number of moles, F is the Faraday constant and 𝐸°is the
standard reduction potential.
Example:
Suppose that we wish to study the possible galvanic corrosion between zinc
and chromium, so we set up the following cell:
Cr s Cr2+
aq Zn2+
aq Zn(s)
What is the chemical reaction that takes place, and what is the standard free
energy change for that reaction?

22. Strategy: To calculate the free energy change, we must know two things: the cell
potential and the number of electrons transferred in the reaction. Then we use the
equation ∆𝐺° = −𝑛𝐹𝐸° to obtain the free energy change.
Solution: First we need the balanced chemical equation, which in this case can be written
immediately because two electrons are transferred in each half-reaction (𝑛 = 2):
Zn2+ aq + Cr s → Cr2+ aq + Zn(s)
Now if we look up the standard reduction potentials, we find
Zn2+ aq + 2𝑒− → Zn(s) 𝐸° = −0.763 V
Cr2+
aq + 2𝑒−
→ Cr s 𝐸°
= −0.910 V
According to equation 𝐸 𝑐𝑒𝑙𝑙
°
= 𝐸 𝑟𝑒𝑑
°
− 𝐸 𝑜𝑥
° ,
𝐸 𝑐𝑒𝑙𝑙
°
= −0.763 V − −0.910 V = 0.147 V
Inserting this in ∆𝐺°
= −𝑛𝐹𝐸°
,
∆𝐺°
= −𝑛𝐹𝐸°
= −2mol × 96,485
J
V ∙ mol
× 0.147 V = −2.84 × 104
J = −28.4 kJ

23. 13.4 Cell Potentials and Equilibrium
• The relationship between cell potential and the equilibrium constants:
𝐸° =
𝑅𝑇
𝑛𝐹
ln 𝐾
Where K is the equilibrium constant, R is gas constant, T temperature.
• At equilibrium, the free energy change is zero and the reaction quotient, Q,
is equal to the equilibrium constant, K.
• We can gain some important insight into electrochemical reactions in
general, and corrosion in particular, by replacing the natural logarithm with
the common (base 10) logarithm to obtain:
𝐸° =
2.303𝑅𝑇
𝑛𝐹
log 𝐾
• At 25℃ (298 K)
𝐸° =
0.0592
𝑛
log 𝐾

24. 13.5 Batteries
• A battery is a galvanic cell, or a series of combined galvanic cells, that can be
used as a source of direct electric current at a constant voltage.
A. Primary Cells – single-use batteries that cannot be recharged
Type of primary battery
a. Alkaline battery
e.g. flashlights, MP3 players
• The chemistry of an alkaline dry cell
battery:
Zn s + 2MnO2 s + H2O(𝑙) → Zn(OH)2 s + Mn2O3(s)

25. 13.5 Batteries
b. Mercury battery
e.g. heart pacemaker
• Zinc is the anode as in the
alkaline dry cell:
Zn s + 2OH−(aq) → Zn(OH)2 s + 2𝑒−
• The cathode, however, uses
mercury (II) oxide:
HgO s + H2O 𝑙 + 2𝑒−
→ Hg 𝑙 + 2OH−
(aq)

26. 13.5 Batteries
c. Zinc-air battery – these batteries are sold as single-use, long lasting
products for emergency use in cellular phones.
• The cathode reaction in this
battery is:
1
2
O2 g + H2O 𝑙 + 2𝑒− → 2OH−

27. 13.5 Batteries
B. Secondary Cells or secondary batteries – rechargeable batteries
Type of secondary battery
a. Nickel-cadmium battery – it can be expended and recharged many times,
but they are sometimes susceptible to a performance-decreasing memory
effect.
• The anode in the ni-cad battery is
cadmium, reacting according to the
following equation:
Cd s + 2OH−
(aq) → Cd(OH)2 s + 2𝑒−
• The cathode reaction is complex but is
best represented by the equation:
NiO OH s + H2O 𝑙 + 𝑒− → Ni(OH)2 s + OH−(aq)

28. 13.5 Batteries
b. Nickel-metal-hydride batteries – find use in many of the same devices as
ni-cad cells, and larger versions serve as the main batteries in hybrid cars
like the Toyota Prius.
• In this battery, the cathode reaction is
the same:
NiO OH s + H2O 𝑙 + 𝑒−
→ Ni(OH)2 s + OH−
(aq)
• But the anode reaction:
MH s + OH−(aq) → M + H2O 𝑙 + 𝑒−
Where M stands for some metal or metal
alloy.

29. 13.5 Batteries
b. Lead-acid storage batteries – most widely selling rechargeable batteries in
automobiles
• The anode reaction in this battery is:
Pb s + HSO4
−
aq → PbSO4 s + H+
aq + 2𝑒−
• And the cathode:
PbSO2 s + 3H+ aq + HSO4
−
aq + 2𝑒− → PbSO4 s + 2H2O 𝑙
Where M stands for some metal or metal alloy.

30. 13.5 Batteries
C. Fuel Cell
- it is a voltaic cell in which the reactants can be supplied continuously
and the products of the cell reaction are continuously removed.
- Unlike a battery, it can be refueled on an ongoing basis.
- The most common fuel cells are based on the reaction of hydrogen
and oxygen to produce water.
- Fuel cells are used in a variety of specialized applications, including
powering instrumentation aboard spacecraft.
• Oxygen is reduced at the cathode:
O2 + 4H+ + 4𝑒− → 2H2O
• Hydrogen is oxidized at the anode:
H2 → 2H+
+ 2𝑒−
• The overall cell reaction is simply the combination of hydrogen and oxygen to form
water:
2H2 + O2 → 2H2O

31. 13.6 Electrolysis
• Electrolysis is the process of passing an electric current through an ionic
solution or molten salt to produce a chemical reaction.
• A device used to carry out electrolysis is called an electrolytic cell.
Two categories of electrolytic cell:
1. Passive electrolysis – a process in which electrodes are chemically inert
materials that simply provide a path for electrons.
- it is used in industry to purify metals that corrode
easily.
2. Active electrolysis – It is when the electrodes are part of the electrolytic
reaction.
- it is used to plate materials to provide resistance to
corrosion.

32. 13.7 Electrolysis and Stoichiometry
I. Current and Charge
• The base unit of current, the ampere (A), is a combined unit defined as one
coulomb per second: 1A = 1 C/s
• Devices called amp-meters (or ammeters) measures current.
• Charge can be calculated using this equation:
Charge = current × time
𝑄 = 𝐼 × 𝑡
Where 𝑄 is in coulombs, 𝐼 in amperes (coulombs/second), and 𝑡 in seconds.
• If we can calculate the charge that passes through an electrolytic cell, we will
know the number of moles of electrons that pass.

33. Example:
In a process called flash electroplating, a current of 2.50 × 103
A passes
through an electrolytic cell for 5.00 minutes. How many moles of electrons are
driven through the cell?
Solution:
𝑄 = 2500 A × 5.00 min
60 s
1 min
= 7.50 × 105
C
Now use Faraday’s constant:
7.50 × 105 C ×
1mol 𝑒−
96,485 C
= 7.77mol 𝑒−
*This two-step manipulation is really a stoichiometry problem, as it allows us
to find the number of moles of something (electrons, in this case) in a
chemical reaction.

34. 13.7 Electrolysis and Stoichiometry
• Electrical power is the rate of electricity consumption, and utility charges are
based on consumption. The SI unit for power is the watt:
1 watt = 1 J/s
• To determine the amount of energy used, we multiply this rate by the time
during which it occurs.
• To obtain numbers of a convenient magnitude, electrical utilities normally
determine energy consumption in kilowatt-hours (kWh).
1kWh = 3.60 × 106
J
• Relationship between electrical potential and energy:
1 J = 1 C V

35. Example:
Suppose that a batch of parts is plated with copper in an electrolytic bath running at
0.15 V and 15.0 A for exactly 2 hours. What is the energy cost of this process if the
electric utility charges the company $0.0500 per kWh?
Strategy: We can determine the energy expended because we know the current,
time, and voltage. The current multiplied by the time gives us the charge, which
when multiplied by the voltage yields the energy. Once we know the energy
expenditure we can convert the value we calculate (in J) to kWh to obtain the cost of
the electricity.
Solution:
𝑄 = 𝐼 × 𝑡 = 15C/s × 2h𝑟
3600 s
1hr
= 1.08 × 105
C
Energy = charge × voltage = 1.08 × 105 C × 0.15 V = 1.6 × 104J
Now Convert to kWh and determine the cost.
1.6 × 104 J ×
1 kWh
3.60 × 106J
×
$0.0500
1 kWh
= $0.00023

36. Example:
An electrolysis cell that deposits gold (from Au+
(aq)) operates for 15.0
minutes at a current of 2.30 A. What mass of gold is deposited?
Solution: First write the balanced half-reaction:
Au+ aq + 𝑒− → Au(s)
Next calculate moles of electrons based on current and time:
𝑄 = 𝐼 × 𝑡 = 2.30 C/s 900 s = 2.07 × 103C
2.07 × 103C ×
1 mol 𝑒−
96,485 C
= 2.15 × 10−2mol 𝑒−
Now, we note that the mole ratio of electrons to gold is 1:1, which means that
we also have 2.15 × 10−2 mol of Au.
2.15 × 10−2 mol Au × 197g/mol = 4.23 g Au

37. 13.8 Corrosion Prevention
A. Coatings
• Most common method of corrosion protection
• Coatings are of various types:
- Metallic
- Inorganic like glass, porcelain and concrete
- Organic, paints, varnishes and lacquers
• Many methods of coating
- Electrodeposition
- Flame spraying
- Cladding
- Hot dipping
- Diffusion
- Vapour deposition
- Ion implantation
- Laser glazing

38. 13.8 Corrosion Prevention
B. Cathodic Protection
• Make the structure more cathodic by
- Use of sacrificial anodes
- Impressed currents
• Used extensively to protect marine structures, underground pipelines, water
heaters and reinforcement bars in concrete.