Chapter 19: Electrochemistry
Renee Y. Becker
Valencia Community College
1
Oxidation–Reduction Reactions
• Redox reactions are those involving the oxidation
and reduction of species.
• Oxidation and reduction must occur together.
They cannot exist alone.
Fe2+ + Cu0  Fe0 + Cu2+
Reduced: Iron gained 2 electrons
Oxidized: Copper lost 2 electrons
Fe2+ + 2 e  Fe0
Cu0  Cu2+ + 2 e
• Remember that electrons are negative so if you gain
electrons your oxidation # decreases and if you lose
electrons your oxidation # increases
2
Oxidation–Reduction Reactions
Fe2+ + Cu0  Fe0 + Cu2+
• Fe2+ gains electrons, is reduced, and we call it an
oxidizing agent
– Oxidizing agent is a species that can gain
electrons and this facilitates in the oxidation of
another species. (electron deficient)
• Cu0 loses electrons, is oxidized, and we call it a
reducing agent
– Reducing agent is a species that can lose
electrons and this facilitates in the reduction of
another species. (electron rich)
3
Example 1: Oxidation–Reduction Reactions
Which is a reduction half reaction?
1. Fe  Fe2+ + 2e
2. Fe2+  Fe3+ + 1e
3. Fe  Fe3+ + 3e
4. Fe3+ + 1e  Fe2+
4
Example 2: Oxidation–Reduction Reactions
For each of the following, identify which species is
the reducing agent and which is the oxidizing
agent.
A) Ca(s) + 2 H+(aq)  Ca2+(aq) + H2(g)
B) 2 Fe2+(aq) + Cl2(aq)  2 Fe3+(aq) + 2 Cl–(aq)
C) SnO2(s) + 2 C(s)  Sn(s) + 2 CO(g)
5
Oxidation–Reduction Reactions
• Assigning Oxidation Numbers: All atoms have
an “oxidation number” regardless of whether it
carries an ionic charge.
1. An atom in its elemental state has an oxidation
number of zero.
Elemental state as indicated by single elements with no
charge. Exception: diatomics H2 N2 O2 F2 Cl2 Br2 and I62
Oxidation–Reduction Reactions
2. An atom in a monatomic ion has an oxidation
number identical to its charge.
7
Oxidation–Reduction Reactions
3. An atom in a polyatomic ion or in a molecular
compound usually has the same oxidation number it
would have if it were a monatomic ion.
A. Hydrogen can be either +1 or –1.
B. Oxygen usually has an oxidation number of –2.
In peroxides, oxygen is –1.
C. Halogens usually have an oxidation number of –1.
• When bonded to oxygen, chlorine, bromine, and
iodine have positive oxidation numbers.
8
Oxidation–Reduction Reactions
4. The sum of the oxidation numbers must be zero
for a neutral compound and must be equal to the
net charge for a polyatomic ion.
A. H2SO4 neutral atom, no net charge
SO42- sulfate polyatomic ion
[SO4]2- [Sx O42-] = -2
X + -8 = -2
X = 6 so sulfur has an oxidation # of +6
9
Oxidation–Reduction Reactions
B. ClO4– , net charge of -1
[ClO4]-1 [Clx O42-] = -1
X + -8 = -1
X = 7 so the oxidation number of chloride is +7
10
Example 3: Oxidation–Reduction Reactions
Assign oxidation numbers to each atom in the
following substances:
A. CdS
G. V2O3
B. AlH3
H. HNO3
C. Na2Cr2O7
I. FeSO4
D. SnCl4
J. Fe2O3
E. MnO4–
K. H2PO4F. VOCl3
11
Example 4:
What is the oxidation number of arsenic in AsO43- ?
1. 8
2. 1
3. 5
4. -1
12
Activity Series of Elements
Lithium, Li can
reduce anything
under it. Lithium is
a very strong
reducing agent and
this is in part why
lithium batteries
work so well. Light
weight and produce a
high voltage
Anode: Li  Li+ + e
Cathode: MnO2 + Li+
+ e  LiMnO2 13
Activity Series of Elements
Activity series looks at the
relative reactivity of a free
metal with an aqueous cation.
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Cu(s) + 2 Ag+(aq)  2 Ag(s) + Cu2+(aq)
Mg(s) + 2 H+(aq)  Mg2+(aq) + H2(g)
14
Example 5: Activity Series of Elements
• Given the following three reactions, determine the
activity series for Cu, Zn, & Fe.
1. Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
2. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
3. Fe(s) + Zn2+ (aq)  NR
15
Balancing Redox Reactions
• Half-Reaction Method: Allows you to focus on
the transfer of electrons. This is important when
considering batteries and other aspects of
electrochemistry.
• The key to this method is to realize that the overall
reaction can be broken into two parts, or halfreactions. (oxidation half and reduction half)
16
Balancing Redox Reactions
Balance for an acidic solution:
MnO4–(aq) + Br–(aq)  Mn2+(aq) + Br2(aq)
1. Determine oxidation and reduction half-reactions:
Oxidation half-reaction: Br–(aq)  Br20(aq)
Reduction half-reaction: MnO4–(aq)  Mn2+(aq)
2. Balance for atoms other than H and O:
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq)  Mn2+(aq)
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Balancing Redox Reactions
3. Balance for oxygen by adding H2O to the side
with less oxygen
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq)  Mn2+(aq) + 4 H2O(l)
4. Balance for hydrogen by adding H+ to the side
with less hydrogens
Oxidation: 2 Br–(aq)  Br2(aq)
Reduction: MnO4–(aq) + 8 H+(aq)  Mn2+(aq) + 4 H2O(l)
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Balancing Redox Reactions
5. Balance for charge by adding electrons (e–):
Oxidation: 2 Br–(aq)  Br2(aq) + 2 e–
Reduction: MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4 H2O(l)
6. Balance for numbers of electrons by multiplying:
Oxidation: 5[2 Br–(aq)  Br2(aq) + 2 e–]
Reduction: 2[MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4 H2O(l)]
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Balancing Redox Reactions
7. Combine and cancel to form one equation:
Oxidation: 10 Br–(aq)  5 Br2(aq) + 10 e–
Reduction: 2 MnO4–(aq) + 16 H+(aq) + 10 e– 2 Mn2+(aq) + 8 H2O(l)
2 MnO4–(aq) + 10 Br–(aq) + 16 H+(aq) 2 Mn2+(aq) + 5 Br2(aq) + 8 H2O(l)
20
Example 6: Balancing Redox Reactions
Balance the following in an acidic sol’n
NO3–(aq) + Cu(s)  NO(g) + Cu2+ (aq)
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Balancing Redox Reactions
Balance for an basic solution:
MnO4-(aq) + SO32-(aq)  MnO2(s) + SO42-(aq)
1. Determine oxidation and reduction half-reactions:
Oxidation half-reaction: SO32–(aq)  SO42-(aq)
Reduction half-reaction: MnO4–(aq)  MnO2(s)
2. Balance for atoms other than H and O:
Oxidation: SO32– (aq)  SO42- (aq)
Reduction: MnO4– (aq)  MnO2(aq)
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Balancing Redox Reactions: Basic Sol’n
3. Balance for oxygen by adding H2O to the side with
less oxygen
Oxidation: H2O(l) + SO32–  SO42Reduction: MnO4–  MnO2 + 2 H2O(l)
4. Balance for hydrogen by adding H+ to the side with
less hydrogens
Oxidation: H2O(l) + SO32–  SO42- + 2H+(aq)
Reduction: 4H+(aq) + MnO4–  MnO2 + 2 H2O(l)
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Balancing Redox Reactions: Basic Sol’n
5. Balance for charge by adding electrons (e–):
Oxidation: H2O + SO32–  SO42- + 2H+ + 2 eReduction: 3 e- + 4H+ + MnO4–  MnO2 + 2 H2O
6. Balance for numbers of electrons by multiplying:
Oxidation: 3[H2O + SO32–  SO42- + 2H++ 2 e-]
Reduction: 2[3 e- + 4H+ + MnO4–  MnO2 + 2 H2O]
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Balancing Redox Reactions: Basic Sol’n
7. Combine and cancel to form one equation:
Ox: 3H2O(l) + 3SO32–(aq)  3SO42-(aq) + 6H+(aq) + 6 eRed: 6 e- + 8H+(aq) + 2MnO4–(aq)  2MnO2(s) + 4 H2O(l)
2
1
2H+(aq) + 3SO32-(aq) + 2MnO4-(aq)  3SO42-(aq) + 2MnO2(s) + H2O(l)
So far same as acidic sol’n, 2 more steps for basic sol’n
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Balancing Redox Reactions: Basic Sol’n
8. Note # of H+ ions in the equation. Add this # of OHto both sides of equation
2OH- + 2H+ + 3SO32- + 2MnO4-  3SO42- + 2MnO2 + H2O + 2OH-
9. Simplify equation by combining H+ and OH- to make
H2O, reduce to lowest terms
2H2O + 3SO32- + 2MnO4-  3SO42- + 2MnO2 + H2O + 2OH1
Write balanced equation
26
Balancing Redox Reactions: Basic Sol’n
H2O(l) + 3SO32-(aq) + 2MnO4-(aq) 
3SO42-(aq) + 2MnO2(s) + 2OH-(aq)
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Electrochemistry
2 types of electrochemical cells
1. Galvanic cells (voltaic cells)
Spontaneous chemical rxn generates an
electric current
2. Electrolytic cells
An electric current drives a non-spontaneous
rxn
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Overall rxn
Zn + Cu2+  Zn2+ + Cu
½ ox. rxn
Zn  Zn2+ + 2e½ red. rxn
Cu2+ + 2e-  Cu
1.
Cu2+ is reduced and it is the
oxidizing agent
2.
Zn is oxidized and it is the
reducing agent
3.
Electrons are transferred
directly from Zn to Cu2+
Review of Redox
4. Enthalpy of rxn is lost to the
surroundings as heat
29
Same rxn with electrochemical cell (galvanic)
Some of the chemical energy released by the rxn is converted
to electrical energy which can be used to light a light bulb
Apparatus
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Apparatus
Electrodes- strips of zinc and copper connected by an
electrically conductive wire (electrons transferred through
wire make current)
Anode electrode is negative, oxidation takes place Zn strip
Cathode electrode reduction takes place Cu strip
Salt bridge- U-shaped tube that contains a gel permeated with a
sol’n of inert electrolyte (will not react)
Anode: Ox ½
Cathode: Red ½
Overall Rxn
Zn  Zn2+ + 2eCu2+ + 2e-  Cu
Zn + Cu2+  Zn2+ + Cu
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Apparatus
Why is the salt bridge necessary?
1. Completes the electrical circuit
Without it the anode cell would become positively charged
as Zn2+ ions appear the sol’n in the cathode beaker would
become negatively charged as Cu2+ ions are removed
2. Because of the charge imbalance the elctrode rxns would
quickly stop and electron flow through the wire stops
3. With the salt bridge the electrical neutrality is maintained in
both beakers by a flow of ions
Anions- SO42- flow through salt bridge from cathode to anode
Cations- from anode to cathode
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Overview
Anode: Oxidation occurs, electrons are
produced, anions move toward, neg. sign
Cathode: Reduction occurs, Electrons are
consumed, cations move towards, positive
sign
Remember: An ox ate a red cat
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Shorthand Notation for Galvanic Cells
Example: Zn(s) + Cu2+  Zn2+ + Cu(s)
Shorthand notation:
Zn(s)  Zn2+(aq)  Cu2+(aq)  Cu(s)
1. , single vertical line represents a phase boundary (between a solid electrode
and an aqueous solution
2. , double vertical line denotes a salt bridge
3. The anode half-cell is always on the left of the salt bridge, with the solid
electrode to the far left
4. The cathode half-cell is always on the right of the salt bridge, with the solid
electrode on the far right
5. The reactants in each half cell are always written first, followed by the products
6. Electrons move through the external circuit from left to right (from anode to
cathode)
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Example 7:
Write the shorthand notation for a galvanic cell that
uses the reaction
Fe(s) + Sn2+(aq)  Fe2+(aq) + Sns)
35
Cell Potentials and Free-Energy Changes for Cell Reactions
Electrons move through the external circuit from the zinc anode to
the copper cathode because they have lower energy when on
copper than on zinc. The driving force that pushes the negatively
charges electrons away from the anode (- electrode) and pulls
them toward the cathode (+ electrode) is an electrical potential
called the electromotive force (emf) also known as cell
potential (E) or the cell voltage. SI units in volts (V)
36
Cell Potentials and Free-Energy Changes for Cell Reactions
1 J = 1C x 1V
G = -nFE
n = number of moles of electrons transferred in the reaction
F = faraday constant 9.648534 x 104 C/mol
G = gibbs free energy
G = -nFE
Standard free-energy change and standard cell potential
Because both are directly proportional a voltmeter can be
regarded as a “free-energy meter”
When a voltmeter measures E, it is also indirectly measuring
G
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Example 8:
The standard cell potential at 25C is 0.92 V for the
reaction
Al(s) + Cr3+(aq)  Al3+(aq) + Cr(s)
What is the standard free-energy change for this
reaction at 25C in kJ?
38
Standard Reduction Potentials
The standard potential of any galvanic cell is the sum
of the standard half-cell potentials for oxidation at
the anode and reduction at the cathode:
Ecell = Eox + Ered
H2 gas is oxidized to
H+ ions at the anode
and Cu2+ ions are
reduced to copper
metal at the cathode
39
Standard Reduction Potentials
Anode (oxidation):
Cathode (reduction):
Overall:
H2(g)  2 H+(aq) + 2 e
Cu2+(aq) + 2 e  Cu(s)
H2(g) + Cu2+(aq)  2 H+(aq) + Cu(s)
Ecell = Eox + Ered = EH2H+ + ECu2+  Cu
= .34 V
To come up with a table of standard potentials we must first
chose a reference half-cell to measure all others with.
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Standard Reduction Potentials
Standard Hydrogen Electrode (SHE) (last cell we
considered!!) E = 0 V
.34 V = 0 V + .34 V
Because the Cu2+/Cu half-reaction is a reduction the
half-cell potential is called a standard reduction
potential
Cu2+ + 2 e  Cu
E = .34 V
Cu  Cu2+ + 2 e
E = -.34 V
41
42
Standard Reduction Potential Table
1.
The half-reactions are all written as reductions.
Oxidizing agents and electrons are on the left side of
each half-reaction and reducing agents are on the right
side
2.
The listed half-cell potentials are standard reduction
potentials, also known as standard electrode potentials
3.
The half-reactions are listed in order of decreasing standard
reduction potential(decreasing tendency to occur in the
forward direction; increasing tendency to occur in the reverse
direction) The strongest oxidizing agents are located in
the upper left of the table (F2, H2O2, MNO4-) the strongest
reducing agents are found in the lower right of the table (Li,
Na, Mg
43
Using Standard Reduction Potentials
Table arranges oxidizing or reducing agents in order
of increasing strength, this allows us to predict the
spontaneity or nonspontaneity of thousands of
redox rxns.
Let’s calculate E for the oxidation of Zn(s) by
Ag+(aq):
2 Ag+(aq) + Zn(s)  2 Ag(s) + Zn2+(aq)
44
Using Standard Reduction Potentials
Step 1: Find half-rxns in Table and write them in the
appropriate direction
Step 2: Multiply the Ag+/Ag half-reaction by a factor of 2 so
that the electrons cancel. Do not multiply the E by 2
because electric potential is an intensive property,
which does not depend on the amount of substance.
E = -G / nF
G will double and n will double so that E
will remain constant
Step 3: Change E for oxidations since the table is based on
reductions
Step 4: Add the half-reactions to get the overall reaction.
45
Using Standard Reduction Potentials
Reduction:
Oxidation:
2 x [Ag+ + e-  Ag]
Zn  Zn2+ + 2e-
Overall reaction:
2 Ag+(aq) + Zn(s)  2 Ag(s) + Zn2+(aq)
E = 0.80 V
E = -(-0.76 V)
E= 1.56 V
E = -G / nF
G = -E(nF) = -3.01 x 105
Because E is positive and G is negative, oxidation of zinc by
Ag+ is a spontaneous reaction under SS conditions. Note: Ag+
can oxidize any reducing agent that lies below it in the table.
Can’t oxidize a reducing agent that appears above it on the table.
Because E will have a negative value.
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Example 9:
Arrange the following oxidizing agents in
order of increasing strength under SS
conditions:
Br2(l) , Fe3+(aq) , Cr2O72-(aq)
47
Example 10:
Arrange the following reducing agents in
order of increasing strength under SS
conditions:
Al(s) , Na(s) , Zn(s)
48
Example 11:
Predict from the Reduction table whether each of the
following reactions can occur spontaneously under
SS conditions:
a)
2 Fe3+(aq) + 2 I-(aq)  2 Fe2+(aq) + I2(s)
b)
3 Ni(s) + 2 Al3+(aq)  3 ni2+(aq) + 2 Al(s)
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Cell potentials and composition of the rxn mixture: The Nernst Equation
Cell pot. and gibbs free energy depend on temp. and
composition of rxn mixture
-Concentration of solutes
-partial pressures of gases
G = G + RT ln Q
nFE = -nFE + RT ln Q
Q = Rxn quotient
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The Nernst Equation
Nernst Equation: E = E - RT ln Q
nF
or E = E - 2.303 RT ln Q
nF
or E = E - 0.0592 V log Q
in Volts @ 25C
n
This allows us to calculate cell potentials under nonstandard state conditions
51
Example 12:
Consider a galvanic cell that uses the reaction
Zn(s) + 2 H+(aq)  Zn2+(aq) + H2(g)
Calculate the cell potential @ 25C when [H+] = 1.0 M,
[Zn2+] = 0.0010 M, and PH2 = 0.10 atm
Going to use: E = Eox + Ered
Going to use: E = E - 0.0592 V
n
log Q
52
Example 13:
Consider a galvanic cell that uses the reaction
Cu(s) + 2 Fe3+(aq)  Cu2+(aq) + 2 Fe2+(aq)
What is the potential of a cell at 25C that has the following ion
concentrations
[Fe3+] = 1.0 x10-4 M
[Cu2+] = 0.25 M
[Fe2+] = 0.20 M
Going to use: E = Eox + Ered
Going to use: E = E - 0.0592 V
n
log Q
53
Standard Cell potentials and equilibrium constants
E = - 0.0592 V log K in Volts @ 25C
n
+ E = + log K therefore K>1
- E = - log K
therefore K<1
Remember: when K is large rxn essentially to 100%
completion
When K is very small rxn doesn’t
proceed at all
54
Standard Cell potentials and equilibrium constants
Now we have 3 ways to find an equilibrium constant
1. aA + bB  cC + dD from concentration or partial pressures
K = [C]c [D]d
[A]a [B]b
2.
G = -RT ln K,
3.
E = RT ln K
nF
ln K = -G
RT
From thermochemical data
or ln K = nFE
RT
or
log K =
nE
.0592 V
From electrochemical data
55
Example 14:
Use the standard reduction potentials in Table
to calculate the equilibrium constant at
25C for the reaction
6 Br-(aq) + Cr2O72-(aq) + 14 H+(aq)  3 Br2(l) + 2 Cr3+(aq) + 7 H2O(l)
56
Example 15:
Use the data in the standard reduction table to
calculate the equilibrium constant @ 25C
for the reaction
4 Fe2+(aq) + O2(g) + 4 H+(aq)  4 Fe3+(aq) + 2 H2O(l)
57
Batteries
• Batteries are the most important practical
application of galvanic cells.
• Single-cell batteries consist of one galvanic cell.
• Multicell batteries consist of several galvanic cells
linked in series to obtain the desired voltage. (car
batteries)
58
Batteries
• Lead Storage Battery: A typical 12 volt battery consists of six
individual cells connected in series.
– Anode: Lead grid packed with spongy lead.
– Cathode: Lead grid packed with lead oxide.
– Electrolyte: 38% by mass sulfuric acid.
– Cell Potential: 1.924 V
59
Batteries
• Zinc Dry-Cell: Also called a Leclanché cell, uses a viscous
paste rather than a liquid solution.
• Anode: Zinc metal can on outside of cell.
• Cathode: MnO2 and carbon black paste on graphite.
• Electrolyte: NH4Cl and ZnCl2 paste.
• Cell Potential: 1.5 V but deteriorates to 0.8 V with use.
60
Batteries
• Alkaline Dry-Cell: Modified Leclanché cell which
replaces NH4Cl with NaOH or KOH.
– Anode: Zinc metal can on outside of cell.
– Cathode: MnO2 and carbon black paste on
graphite.
• Electrolyte: NaOH or KOH, and Zn(OH)2
paste.
– Cell Potential: 1.5 V but longer lasting, higher
power, and more stable current and voltage.
61
• Mercury Dry-Cell: Modified
Leclanché cell which replaces
MnO2 with HgO and uses a steel
cathode.
Batteries
– Anode: Zinc metal can on
outside of cell.
• Cathode: HgO in contact
with steel.
• Electrolyte: KOH, and
Zn(OH)2 paste.
– Cell Potential: 1.3 V with small
size, longer lasting, and stable
current and voltage.
62
Batteries
• Nickel–Cadmium Battery: Modified Leclanché cell
which is rechargeable.
– Anode: Cadmium metal.
• Cd(s) + 2 OH–(aq)  Cd(OH)2(s) + 2 e–
– Cathode: Nickel(III) compound on nickel metal.
• NiO(OH) (s) + H2O(l) + e–  Ni(OH)2(s) + OH–
(aq)
– Electrolyte: Nickel oxyhydroxide, NiO(OH).
– Cell Potential: 1.30 V
63
Batteries
• Nickel–Metal–Hydride (NiMH):
• Replaces toxic Cd anode with
a hydrogen atom impregnated
ZrNi2 metal alloy.
• During oxidation at the anode,
hydrogen atoms are released
as H2O.
• Recharging reverses this reaction.
64
Batteries
• Lithium Ion (Li–ion): The newest rechargeable battery is
based on the migration of Li+ ions.
– Anode: Li metal, or Li atom impregnated graphite.
• Li(s)  Li+ + e–
– Cathode: Metal oxide or sulfide that can accept Li+.
• MnO2(s) + Li+(aq) + e–  LiMnO2(s)
– Electrolyte: Lithium-containing salt such as LiClO4, in
organic solvent. Solid state polymers can also be used.
– Cell Potential: 3.0 V
65
Fuel Cell
• Fuel Cell: Uses externally fed CH4 or H2, which react to form
water. Most common is H2.
– Anode: Porous carbon containing metallic catalysts.
• 2 H2(s) + 4 OH–(aq)  4 H2O(l) + 4 e–
– Cathode: Porous carbon containing metallic catalysts.
• O2(s) + 2 H2O(l) + 4 e–  4 OH–(aq)
– Electrolyte: Hot aqueous KOH solution.
– Cell Potential: 1.23 V, but is only 40% of cell capacity.
66
Fuel Cell
• Fuel cells are not
batteries because they
are not self-contained.
• Fuel cells typically have
about 40% conversion to
electricity; the remainder
is lost as heat.
• Excess heat can be used
to drive turbine
generators.
67
Corrosion
1. Corrosion is the oxidative deterioration of a
metal, such as iron to rust.
2. To prevent corrosion first we have to understand
how it occurs
a) Rusting requires both oxygen and water
b) Rusting results from tiny galvanic cells
formed by water droplets
68
Oxidation: Fe(s)  Fe2+(aq) + 2 e–
Corrosion
Reduction: O2(g) + 4 H+(aq) + 4 e–  2 H2O(l)
Overall:
2 Fe(s) + O2(g) + 4 H+(aq)  2 Fe2+(aq) + 2 H2O(l)
• This electrochemical mechanism for corrosion also explains why
automobiles rust more rapidly in parts of the country where road
sat is used to melt snow and ice. Dissolved salts in the water
droplet greatly increase the conductivity of the electrolyte, thus
69
accelerating the pace of corrosion.
Prevention of corrosion
• Galvanizing: is the coating of iron with zinc. Zinc
is more easily oxidized than iron, which protects
and reverses oxidation of the iron.
• Cathodic Protection: is the protection of a metal
from corrosion by connecting it to a metal (a
sacrificial anode) that is more easily oxidized.
• All that is required is an electrical connection to
the sacrificial anode (usually magnesium or zinc).
70
Prevention of corrosion
• Cathodic Protection with Zinc Layer
71
Prevention of corrosion
• Cathodic Protection with Magnesium Anode
72
Electrolysis
• Electrolysis: is the process in which electrical
energy is used to drive a nonspontaneous chemical
reaction.
• An electrolytic cell is an apparatus for carrying
out electrolysis.
• Processes in an electrolytic cell are the reverse of
those in a galvanic cell.
73
Applications of Electrolysis
• Manufacture of Sodium (Downs Cell):
74
Applications of Electrolysis
• Manufacture of Cl2 and NaOH (Chlor–Alkali):
75
Applications of Electrolysis
• Manufacture of Aluminum (Hall–Heroult):
76
Applications of Electrolysis
• Electrorefining and Electroplating:
77