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Published by g-18018924, 2020-03-30 09:50:34

Application of Electrochemistry

Electrochemistry

Keywords: electrochemistry

20/3/2020

2.6 Applications of electrochemistry
Principle of electrochemistry is widely applied in
industries and in our daily life. A few occurrences that will
be discussed in this section includes rusting of iron and
ways of prevention ; extraction of aluminium in industries
and ways to thickening the protective oxide layer of
aluminium ; extraction of chlorine in industries ; treatment
of industrial effluent by electrolysis to remove Ni2+, Cr3+
and Cd2+ ; and electroplating in plastic industries.

2.6.1 Corrosion of iron
1. Rusting of iron is a deterioration of iron metal to iron (III) oxide, that

occur naturally under continuous exposure to air and water. In such
case, iron (III) oxide caused the iron to become brittle hence
deteriorate the iron metal.
a) The air around atmosphere is mostly acidic (especially in industrial
area) which hasten the rusting process as the following chemical
reactions take place

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 In acidic condition

At anode : At cathode :

Fe (s)  Fe2+ (aq) + 2e- Eo = +0.44 V O2(aq) + 4 H+(aq) + 4e− → 2H2O(l)

Eo = + 1.23 V

2 Fe (s) + O2 (aq) + 4 H+  2 Fe2+ + 2 H2O Ecell = + 1.67 V

At anode : At cathode :

Fe2+ (aq)  Fe3+ (aq) + e- O2(g) + 4 H+(aq) + 4e− → 2H2O(l)

Eo = -0.77 V Eo = + 1.23 V

4 Fe2+ (aq) + O2 (g) + 4 H+(aq) → 4 Fe3+ (aq) + 2H2O(l) Ecell = + 0.46 V

Forming rust : 2 Fe3+ (aq) + 4 H2O (l)  Fe2O3.H2O (s) + 6 H+ (aq)

In alkaline / neutral condition

At anode : At cathode :

Fe (s)  Fe2+ (aq) + 2e- Eo = +0.44 V O2(aq) + 2H2O(l) + 4e− → 4OH−(aq)

Eo = + 0.40 V

2 Fe (s) + O2 (aq) + 2 H2O  2 Fe2+ + 4 OH– [or 2 Fe(OH)2] Ecell = + 0.84 V

Fe(OH)2 (aq) + OH–  Fe(OH)3 + e – O2(aq) + 2H2O(l) + 4e− → 4OH−(aq)

Eo = +0.56 V Eo = + 0.40 V

4 Fe(OH)2 (aq) + O2(aq) + 2H2O(l)  4 Fe(OH)3 (aq) Ecell = + 0.96 V

Forming rust : 2 Fe(OH)3 (s)  Fe2O3.x H2O + (3 – x) H2O

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2. Corrosion of iron can be prevented by using the following solutions
a) Coating the metal surface with paint ~ This is to prevent water to directly in

contact with metal. However, if the paint is scratched, pitted, or dented to
expose even the smallest area of bare metal, rust will form under the paint
layer.
b) The surface of iron metal can be made inactive by a process called
passivation. A thin oxide layer is formed when the metal is treated with a
strong oxidizing agent such as concentrated nitric acid. A solution of sodium
chromate is often added to cooling systems and radiators to prevent rust
formation.
c) Galvanizing iron metal ~ by applying zinc as a coating of the surface of
iron, it is protected by zinc plate hence prevent corrosion
d) Cathodic protection ~ a process in which the metal that is to be protected
from corrosion is made the cathode in what amounts to a galvanic cell.
Rusting of underground iron pipes and iron storage tanks can be prevented
or greatly reduced by connecting them to metals such as zinc and
magnesium, which oxidize more readily than iron. Therefore, metal with
greater reducing strength will be reacted first

2.13 Industrial Electrolysis :

 In this Chapter, we shall discussed the manufacturing of aluminium and
chlorine gas using the principle of electrolysis

Part 1 : Getting pure aluminium oxide (alumina) from bauxite.

 1st step: Removal of impurities from the ore by dissolving powdered
bauxite in hot concentrated sodium hydroxide solution.
Al2O3 + 2 NaOH + 3 H2O  2 NaAl(OH)4
SiO2 + 2 NaOH  Na2SiO3 + H2O

 2nd step: Insoluble impurities are filtered off. Filtrate contain aluminium
and silicon ions. Aluminium ion is precipitated as aluminium hydroxide
which is filtered out later as white gelatinous precipitate.

Use acid : 2 [Al(OH)4]- + 2 H+  2 Al(OH)3 + 2 H2O
Or use CO2 : 2 [Al(OH)4]- + CO2  2 Al(OH)3 + H2O + CO32-

 3rd step: Aluminium hydroxide is filtered, washed, dried and finally
heated out to 12000C to produce pure aluminium oxide (alumina), Al2O3
2 Al(OH)3  Al2O3 + 3 H2O

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Part 2 : Extracting aluminium out from aluminium oxide
 Hall-Heroult process:- A process of electrolysing aluminium oxide (alumina)

to extract out aluminium.
 Aluminium metal is extracted by the cell electrolytic reduction of alumina.

Melting point of alumina is 20300C. To lower the temperature of the
electrolyte, alumina is dissolved in molten cryolite (Na3AlF6), to maintain
a temperature at about 9600C.
 When alumina dissolve in molten cryolite :
 Al2O3 (s)  2 Al3+ (l) + 3 O2– (l)

 Electrolyte mixture is then placed in carbon-lined iron vat (cathode).
The heating effect of the electric current melts the electrolyte
mixture, producing Na+, Al3+, O2- and F- ions.

Na+ + e-  Na E0 = – 2.71 V F2 + 2 e-  2 F- E0 = + 2.87 V
Al3+ + 3e-  Al E0 = – 1.66 V O2 + 4e-  2O2- E0 = + 1.++V

Half equation occur at cathode Half equation occur at anode

Al3+ + 3e-  Al E0 = – 1.66 V 2O2-  O2 + 4e- E0 = + 1.++V

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2. Anodising is an electrolytic passivation process used to
increase the thickness of the natural oxide layer on the
surface of metal parts, especially for aluminium. The process
is called "anodising" because the part to be treated forms
the anode electrode of an electrical circuit. Anodising
increases corrosion resistance and wear resistance, and
provides better adhesion for paint primers and glues than
does bare metal. Anodic films can also be used for a number
of cosmetic effects, either with thick porous coatings that can
absorb dyes or with thin transparent coatings that
add interference effects to reflected light.

a) Aluminium alloys are anodized to increase corrosion
resistance, to increase surface hardness and to
allow dyeing (coloring), improved lubrication, or
improved adhesion. The anodic layer is non-conductive.

b) When exposed to air at room temperature, or any other gas
containing oxygen, pure aluminium self-passivates by forming
a surface layer of amorphous aluminium oxide which
provides very effective protection against corrosion.

c) Aluminium alloy parts are anodized to greatly increase the thickness of this
layer for corrosion resistance. The corrosion resistance of aluminium alloys is
significantly decreased by certain alloying elements or impurities
copper, iron, and silicon, tend to be most susceptible.

d) By making an aluminium the anode of cell in which dilute sulphuric acid is
the electrolytes, it is possible to produce a thicker and harder film of
aluminium oxide on the surface of metal.

AC Half equation occur at anode (aluminium plate) :
l
2 H2O (l) → 4 H+ (aq) + 4e- + O2 (g)
The oxygen formed at anode will then react with
aluminium plate to form a thick layer of Al2O3.

2 Al (s) + 3 O2 (g) → Al2O3 (s)

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3. Environmental pollution arises as cans are littered everywhere. The best
solution of environmental pollution is recycling. The benefits of recycling can be
seen by comparing the energy consumed in the extraction of aluminium from the
bauxite or using Hall process with that consumed when aluminium is recycle.

Process Heat Energy(kJ/mol) Electrical energy(kJ/mol) Total energy(kJ/mol)
Purification 30 5 25
Electrolysis 70 272
Total energy 100 192 297
197

Pure aluminium has a rather low melting print of 6600C, thus requiring only 26.1
kJ mol-1 of energy. On comparison between the Hall process and recycling,

Energy used in recycling = 26.1  297  100%
= 8.8%

 This means that about 91% of the energy is saved for every 1 mole of the
aluminium produced through recycling

2.6.3 Chlor-alkali cell

1. Chlorine gas, Cl2, is prepared industrially by the electrolysis of
molten NaCl or by the chlor-alkali process, the electrolysis of a
saturated aqueous NaCl solution (called brine). (Chlor denotes
chlorine and alkali denotes an alkali metal, such as sodium.)

a) The set-up of apparatus for the chlor-alkali cell is described in
diagram below :

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i. Since saturated NaCl (brine) is used as electrolyte in the anode
compartment, therefore, chloride ion is discharged according to the half
equation where

Anode : 2 Cl- (aq) → Cl2 (g) + 2e-

ii. In the other hand, sodium ion crossed to the cathode compartment which
contain mainly water. A selectivity occur between Na+ (aq) and H2O (l). By
referring to their standard electrode potential :

Na+ (aq) + e- → Na (s) E0 = - 2.71 V

2 H2O (l) + 2 e- → 2 OH- + H2 (g) E0 = - 0.41 V

The substance discharged at cathode is water. The half equation is

Cathode : 2 H2O (l) + 2 e- → 2 OH- + H2 (g)
[hydrogen gas is released]

iii. Note that OH- is also a side product, which react with Na+ in the
cathode compartment to form sodium hydroxide, NaOH, another
useful substance especially in the production of soap and detergent.
Equation : Na+ (aq) + OH- (aq) → NaOH (aq)

b) The asbestos diaphragm is permeable to the ions but not to the
hydrogen and chlorine gases and so prevents the gases from mixing.
During electrolysis a positive pressure is applied on the anode side
of the compartment to prevent the migration of the OH- ions from the
cathode compartment

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Electrorefining and electroplating
1. Electrolysis is another important purification technique. The copper

metal obtained by roasting copper sulphide (CuS) usually contains
impurities such as zinc, iron, silver, and gold.

The more electropositive metals are
removed by an electrolysis process in
which the impure copper acts as the
anode and pure copper acts as the
cathode in a sulphuric acid solution
containing Cu2+ ions. The reactions are
Anode (oxidation) :

Cu (s) → Cu2+ (aq) + 2 e
Cathode (reduction) :

Cu2+ (aq) + 2 e → Cu (s)

Reactive metals in the copper anode, such as iron and
zinc, are also oxidized at the anode and enter the
solution as Fe2+ and Zn2+ ions. They are not reduced at
the cathode, however. The less electropositive metals,
such as gold and silver, are not oxidized at the anode.
Eventually, as the copper anode dissolves, these metals
fall to the bottom of the cell. Thus, the net result of this
electrolysis process is the transfer of copper from the
anode to the cathode. Similar technique can also be
used to purify silver and gold

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Through this method, we can also use to determine percentage purity
of copper, Faradays' constant or Avogadro's constant

Example : Calculate the Faradays' constant, hence Avogadro's

constant, when 0.600 g of pure copper is formed at cathode by

electrolysing impure copper using 1.50 A of current for 20.0 minutes.

[Given charge per electron = 1.60 x 10-19 C]

Solution : Equation for the reaction occur at cathode :

Cu2+ (aq) + 2 e- → Cu (s)

Calculate the mol of Cu where mol Cu = mass / AR ;
mol of Cu = 0.600 / 63.5 ; mol Cu = 9.45 x 10-3 mol

Since 1 mol Cu = 2 mol e-, therefore mol of e- = mol of Cu x 2 ;

mol of e- = 9.45 x 10-3 mol x 2 ; mol of e- = 0.0189 mol

Quantity of charge used in the cell Q=It ;

Q = (1.50A) (20.0 x 60) ; Q = 1800 C

Therefore Faradays' constant F = Q / mol e- ;

F = 1800 / 0.0189 Faradays' constant based on experiment = 95250 C

Using Avogadro's constant, NA NA = Q / e

NA = 95250 / 1.60 x 10-19 NA = 5.95 x 1023

2.6.4 Treatment of industrial effluent by electrolysis to
remove Ni2+, Cr3+ and Cd2+

1. Heavy metals released into the environment have posed a
significant threat to the environment and public health
because of their toxicity and persistence in environment. The
contamination of wastewaters and surface waters by toxic
heavy metals is a worldwide environmental problem. These
toxic metal ions commonly exist in process waste streams from
mining operations, metal plating facilities, power generation
facilities, electronic device manufacturing units, and tanneries.

a) Heavy metal such as Ni2+, Cr3+ and Cd2+ are especially
harmful to human as they accumulate in human body to form
carcinogen, hence triggered cancerous cell in human body.

b) One of the most common method to remove these heavy
metal is by applying electrolytic membrane, discharging the
metal ions to form metal (Mn+ + ne-  M), and absorbed by
membrane layer

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