Lecture 6

Chemistry For Engineers

Electrochemistry
ELECTROCHEMISTRY

All chemical reactions are

fundamentally electrical in nature since
electrons are involved ( in various ways) in
all types of chemical bonding.
Electrochemistry is a branch of chemistry
that deals with the relationship between
electricity and chemical reactions.
ELECTROCHEMISTRY

The relations between chemical

change and electrical energy
have theoretical as well as
practical importance.
ELECTROCHEMISTRY

Chemical reactions can be

used to produce electrical
energy ( in cells that are called
voltaic or galvanic cells).
ELECTROCHEMISTRY

Electrical energy can be used to

bring about chemical
transformations ( in electrolytic
cells).
ELECTROCHEMISTRY

The study of electrochemical

processes leads to an
understanding as well as to
systematization of oxidation-
reduction phenomena that take
place outside cells.
ELECTROCHEMISTRY

ELECTRICITY CHEMISTRY

The study of electricity and how it relates to chemical

reactions.

Certain chemical reactions can produce electricity.

Electricity can be used to make certain chemical

reactions.
 ELECTROCHEMISTRY

Electricity
The flow or movement of electrons in a conductor.
 e-  e-  e-  e-  e-  e-  e-  e-
Chemical Reactions: Oxidation and Reduction Reactions
Electrons move between atoms.

e-

A B
ELECTROCHEMISTRY

Certain chemical reactions can produce electricity.

e-

A B
SPONTANEOUS REACTION

A e-  B
ELECTROCHEMISTRY

Electricity can be used to make certain chemical reactions.

e-

A REACTION DO NOT
NORMALLY HAPPEN
B

A e-  BATTERY
e-  B
PULL PUSH
 ELECTROCHEMISTRY

O
OXIDATION R
REDUCTION
IIs IIs
LOSING of e-
L GAINING of e-
G
REDUCING
R OXIDIZING
O
AGENT
A AGENT
A
REMEMBER: electron have a NEGATIVE CHARGE
Need to know the oxidation numbers of each atom in a chemical
reaction.
 ELECTROCHEMISTRY

The oxidation state of the uncombined element is zero.

Example: Be0, N20, Cs0
The sum of the oxidation state of a neutral compound is always equal to zero
Example: CaO Ca+2 O-2
1(+2) + 1(-2) = 0
MgBr2 Mg+2 Br-1
1(+2) + 2(-1) = 0
Li2CO3 Li+1 (CO3)-2
2(+1) + 1(-2) = 0
 The Group A elements number indicates its oxidation state
 ELECTROCHEMISTRY

 The oxidation number of a polyatomic ion is equal to its

charge. Oxygen oxidation is -2 except in peroxides
Examples: C in CO3-2
oxidation of C + oxidation of O = -2
C + 3(-2) = -2
C + (-6) = -2
C = -2 + 6
C = +4
Cr in KCrO4-2
oxidation of K + oxidation of Cr + oxidation of O = -2
1(+1) + Cr + 4(-2) = -2
1 + Cr + (-8) =-2
Cr =-2 + 7
Cr = +5
 ELECTROCHEMISTRY

 Hydrogen + non-metal, hydrogen oxidation

state +1
Example: HI H+1 I-1
H2SO4 H+1 (SO4)-2
 Metal + Hydrogen, hydrogen oxidation state -1
Example: LiH Li+1 H-1
ELECTROCHEMISTRY

a. Be
b. Cr in KCrO4-2
c. C in Li2CO3
d. P in Zn3PO4
e. oxidation number of group VIA elements
 ELECTROCHEMISTRY

2 Mg(s) + O2(g)  2 MgO(s)

0 0 +2 -2
Reducing Agent
Mg 0  +2 Loss 2 e-
Oxidized
Oxidizing Agent
O2 0  -2 Gains 2 e-
Reduced
 ELECTROCHEMISTRY

Zn + 2H+  Zn+2 + H2
0 +1 +2 0
Zn 0  +2 Loss 2 e-
Oxidized
H2 +1  0 Gains 1 e-
Reduced
ELECTROCHEMISTRY
 ELECTROCHEMISTRY

When balancing redox reactions, make

sure that the number of electrons lost by
the reducing agent equals the number of
electrons gained by the oxidizing agent
•Two methods can be used:
1.Oxidation number method
2.Half-reaction method
 ELECTROCHEMISTRY

Method 1: Oxidation number method

1.Assign oxidation numbers to all elements in the reaction
2.From the changes in O.N., identify the oxidized and reduced
species
3.Compute the number of electrons lost in the oxidation and
gained in the reduction from the O.N. changes
4.Multiply one or both of these numbers by appropriate factors
to make the electrons lost equal the electrons gained, and use
the factors as balancing coefficients
5.Complete the balancing by inspection, adding states of matter
ELECTROCHEMISTRY
 ELECTROCHEMISTRY
METHOD A
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METHOD A
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METHOD B
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METHOD B
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ELECTROCHEMISTRY
 ELECTROCHEMISTRY
METHOD A
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METHOD A
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METHOD A
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METHOD B
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METHOD B
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METHOD B
 ELECTROCHEMISTRY
METHOD B
ELECTROCHEMISTRY

METALLIC CONDUCTION
- conduction of electricity through
metal by electron displacement
- An electric current is the flow of
electric charge. In metals this charge is
carried by electrons
ELECTROCHEMISTRY

ELECTRICAL UNITS
I - electric current
- rate of flow of an electric charge
- Unit: Ampere (A)
ELECTROCHEMISTRY

ELECTRICAL UNITS
C - quantity of an electric charge measured in
coulombs ( C )
- quantity of electricity carried past a point in
one second by a current of 1 A
1 A = 1 C/ s
- from a point of view of fundamental particles:
1 proton = 1 electron = 1.602 x 10 -19 C
ELECTROCHEMISTRY

ELECTRICAL UNITS
V - potential difference - force through a circuit
measured in volts
1 V = 1 J/ C
 OHM’s LAW - expressed the quantitative relation between
voltage and current
V = IxR , R = resistance in ohms (Ω )
P = I x V
P in J/ sec = watt
ELECTROCHEMISTRY

ELECTROLYTIC CONDUCTION
- Conduction of electricity by the movement of
ions through a solution or a molten salt. A
sustained current requires that chemical
changes at the electrodes also occur.
- cations move towards the cathode and anions
move towards the anode
ELECTROCHEMISTRY

FARADAY’S LAWS OF ELECTROLYSIS

The mass of any substance liberated or deposited at an
electrode is proportional to the electrical charge.
mα Q or m α It
where :
m= mass
I= current
t= time
Michael Faraday
ELECTROCHEMISTRY

The mass of different substances liberated or

deposited by the same quantity of electricity are
proportional to the equivalent weights of various
substance.
1𝐹
1.602 𝑥10−19 𝐶 6.023 𝑥10−23 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
= 𝑥
𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑚𝑜𝑙
= 9.65 𝑥 104 𝐶/𝑚𝑜𝑙
ELECTROCHEMISTRY

Electrolysis
Electrolysis is a process by which electric current is passed
through a substance to effect a chemical change
The chemical change is one in which the substance loses
or gains an electron (oxidation or reduction).
Electrolysis is used extensively in metallurgical processes,
such as in extraction (electrowinning) or purification
(electrorefining) of metals from ores or compounds and
in deposition of metals from solution (electroplating).
ELECTROCHEMISTRY

Electrolysis
- use of electric
current to bring about
oxidation-reduction (
REDOX) reaction
- electric charge in
electrolytic
conduction is carried
by CATIONS (+)
moving toward the
cathode and anion
ELECTROCHEMISTRY

For a complete circuit , electrode reactions must accompany the movement

of ions. At the cathode, come chemical species ( not necessarily the charge
carrier) must accept electrons and can be reduced. At the anode , electrons
must be removed from some chemical species which as a consequence is
oxidized.
In the given diagram, Na ions are reduced at the cathode:
Na+ + e- → Na0
And Cl ions are oxidized at the anode:
2Cl- → Cl2 + 2e-
Proper addition of these partial equations gives the reaction for the entire
cell:
2NaCl (l) → 2Na + Cl2 (g )
ELECTROCHEMISTRY

VOLTAIC CELL

- A cell that is used as a source of electrical energy. It is also called a

GALVANIC CELL after ALESSANDRO VOLTA or LUIGI GALVANI,
who first experimented with the conversion of chemical energy or
electrical energy.

- In voltaic cells, half reactions are made to occur at different electrodes

so that the transfer of electrons takes place through the external circuit
rather than directly between the metal electrodes and the ions.
ELECTROCHEMISTRY

The cell diagram in the

figure is designed to make
use of this reaction to
produce an electric current.
The cell on the LEFT
contains a zinc metal
electrode and ZnSO4
solution. The cell on the
RIGHT consists of a copper
metal electrode in a
solution of CuSO4. The
ZnSO4 is connected ZnSO4
ELECTROCHEMISTRY

VOLTAIC CELL or Galvanic Cell

A device in which spontaneous oxidation-reduction occurs in the

passage of electrons through an external circuit.

Oxidation – a process in which a substance loses electrons

- occurs at the anode
Reduction – a process in which a substance gains electrons
- occurs at the cathode
ELECTROCHEMISTRY

Certain chemical reactions can produce electricity.

Reduction- Oxidation Reaction

Example:
ELECTROCHEMISTRY

e- WIRE
ZnSO4 Solution CuSO4 Solution

Salt bridge

Copper metal
Zinc Metal

Moving e-

ELECTRICITY
 WIRE
ELECTROCHEMISTRY
STANDARD
Zn Cu
Zn REDUCTION Cu
Zn POTENTIAL
Zn Cu
Cu
Zn
Zn Cu
Zn has weaker Cu
Zn Cu+2 has stronger
Zn pull of e- pull of e-
Cu
Cu
Zn
Zn Cu
Cu
Zn Neutral atoms
Zn Zn+2 Cu+2 Cu
normally make up Cu
Zn Zn a solid metal. Cu+2 Cu
Zn+2 Cu
Zn Zn+2 Cu
Cu
Zn Metal ions can Cu+2 Cu
Zn+2 Cu+2 Cu
Zn usually dissolve in Cu
Zn Cu
Zn+2 water
Zn+2
ELECTROCHEMISTRY
WIRE

Zn Cu

Copper metal
Zinc Metal

ZnSO4 Solution CuSO4 Solution

 WIRE
ELECTROCHEMISTRY Zn
Zn Cu
Cu
Zn
Zn Cu
OXIDATION - loss e- Cu
Zn
Zn Cu
Zn
OXIDATION REDUCTION – gains e- REDUCTION Cu
Zn Cu
Zn - ANODE Cu- CATHODE Cu
Zn
Zn ANODE – site of Cu
Oxidation half cell Cu
Zn oxidation Reduction half cell
Zn Cu
Cu
Zn CATHODE – site of Cu
Cu+2
Zn+2 reduction Cu
Zn Cu
Zn Cu
Zn Cu Cu
Zn Cu

Half Reactions Cu+2 (aq) + 2e-  Cu(s)

Zn(s)  Zn+2 (aq) + 2e-

NET IONIC EQUATION: Zn(s) + Cu+2  Zn+2 (aq) + Cu(s)

ELECTROCHEMISTRY

Oxidation at Anode (ANOX)

Zn(s)  Zn+2 (aq) + 2e-

Reduction at Cathode (REDCAT)

Cu+2 (aq) + 2e-  Cu(s)
ELECTROCHEMISTRY
WIRE
Salt bridge – filled with NaCl solution

Na+ Cl- Na+ Na+ Cl- Na+ Na+Cl- Na+ Na+

Cl-
Zn Cl- Cu
Zn Na+ Cu
Cl- Na+
Zn
Zn Cl- Cu
Cl- Cu
Zn
Zn Na+ Cu
Cl- Cu
Zn Cl-
Zn Na+ Cu
Cu
Zn Build up of
Zn Cu
positive charge Cu
Zn Zn+2 Zn+2
SO4-2 Cu
SO4-2 Cu
Zn SO4-2 Cu
Cu
Zn+2 Build up of SO4 -2
Cu
Cu
Zn Zn+2 Cu
Zn SO4 -2
negative charges SO4 -2
Cu
Zn Cu Cu
Zn Zn+2 SO4 -2 Cu
Zn+2 Cu
ELECTROCHEMISTRY

Notation of Voltaic Cell

A useful notation that describes the components of a cell
𝐴𝑛𝑜𝑑𝑒 //𝐶𝑎𝑡ℎ𝑜𝑑𝑒
Example: 𝑍𝑛 𝑠 + 𝐶𝑢+2 𝑎𝑞 → 𝑍𝑛+2 𝑎𝑞 + 𝐶𝑢(𝑠)
Half Reactions:
𝐴𝑛𝑜𝑑𝑒 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 : 𝑍𝑛 𝑠 + → 𝑍𝑛+2 𝑎𝑞 + 2𝑒ҧ
Cath𝑜𝑑𝑒 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 : 𝐶𝑢+2 𝑎𝑞 + 2𝑒ҧ → 𝐶𝑢(𝑠)
Voltaic Cell: 𝑍𝑛/𝑍𝑛+2 // 𝐶𝑢+2 /𝐶𝑢
ELECTROCHEMISTRY

Standard half cell Potentials

Cell Potential - difference between two electrode potentials, one associated with
the cathode and the other associated with the anode.
𝐸 0 𝑐𝑒𝑙𝑙 = 𝐸 0 𝑟𝑒𝑑 𝑎𝑛𝑜𝑑𝑒 + 𝐸 0 𝑟𝑒𝑑 𝑐𝑎𝑡ℎ𝑜𝑑𝑒
𝐸 0 𝑟𝑒𝑑 - Standard reduction potential (obtained from the table)

For the cell 𝑁𝑖/𝑁𝑖 +2 // 𝐶𝑙2 /𝐶𝑙 −

𝑁𝑖 → 𝑁𝑖 +2 + 2𝑒ҧ 𝐸 0 = +0.23 𝑉 from the table
𝐶𝑙2 + 2𝑒ҧ → 2𝐶𝑙 − 𝐸 0 = +1.36 𝑉 from the table
𝑁𝑖 + 𝐶𝑙2 → 𝑁𝑖 +2 + 2𝐶𝑙 − 𝐸 0 𝑐𝑒𝑙𝑙 = +1.59 𝑉
ELECTROCHEMISTRY

Standard Electrode Potentials

in Aqueous Solution at 25°C
ELECTROCHEMISTRY

Standard Electrode Potentials

in Aqueous Solution at 25°C
ELECTROCHEMISTRY

Standard Electrode Potentials

in Aqueous Solution at 25°C
ELECTROCHEMISTRY
CELL NOTATION is shorthand that expresses a certain
reaction in an electrochemical cell.
 solidAqueousAqueoussolid

 Anode on the left (oxidation)Cathode on the right (reduction)

 Single line different phases.

 Double line porous disk or salt bridge.

 If all the substances on one side are aqueous, a platinum electrode is

indicated.

 If non standard concentration of the aqueous solution was used,

indicate the concentration enclosed with an open and close
parenthesis
ELECTROCHEMISTRY
 A galvanic cell have a reduction reaction (in
cathode half-cell) and an oxidation reaction (in
anode half-cell).

 Electrons flows through a wire from the anode

half-cell to the cathode half-cell.

 The driving force that allows electrons to flow is

called the electromotive force (emf) or the cell
potential (Ecell).
 The unit of electrical potential is volt (V).
1 V = 1 J/C of charge transferred.
ELECTROCHEMISTRY
Determining Standard State Cell Potentials

A cell's standard state potential is the potential of the cell under standard state
conditions, which is approximated with concentrations of 1 mole per liter (1 M)
and pressures of 1 atmosphere at 25oC.

To calculate the standard cell potential for a reaction

•Write the oxidation and reduction half-reactions for the cell.

•Look up the reduction potential, Eoreduction, for the reduction half-reaction in a

table of reduction potentials

•Look up the reduction potential for the reverse of the oxidation half-reaction
and reverse the sign to obtain the oxidation potential. For the oxidation half-
reaction, Eooxidation = - Eoreduction.

•Add the potentials of the half-cells to get the overall standard cell potential.
Eocell = Eoreduction + Eooxidation
ELECTROCHEMISTRY

E0cell > 0 Spontaneous

E0cell < 0 Not Spontaneous

E0cell = 0 Equilibrium
ELECTROCHEMISTRY
Find the standard cell potential for an electrochemical cell and write its cell
notation:
Zn(s) + Cu+2  Zn+2 (aq) + Cu(s)

Reduction: Cu+2 (aq) + 2e-  Cu(s) E°= +0.34V

Oxidation: Zn(s)  Zn+2 (aq) + 2e- E°= +0.76V

Overall: Zn(s) + Cu+2  Zn+2 (aq) + Cu(s) E°= +1.10V

Spontaneous reaction
A galvanic cell runs spontaneously in the direction that gives a
positive value for E°cell
Cell Notation: Oxidation Anode || Reduction Cathode
Zn(s) | Zn+2 (aq) || Cu+2 (aq) | Cu (s)
Find the standard cell potential for an electrochemical cell with the following
cell reaction and write its cell notation.

Given the following reduction potentials:

Fe3+ (aq) + e–  Fe2+ (aq) E° = 0.77 V
Cu2+ (aq)+ 2e–  Cu (s) E° = 0.34 V
Solution:
Reduction: 2Fe3+ (aq) + 2e–  2Fe2+ (aq) E° = 0.77 V
Oxidation: Cu(s)  Cu2+(aq)+ 2e– E° = – 0.34 V
Overall Reaction: Cu + 2Fe3+  Cu2+ + 2Fe2+ E° = 0.43 V
Cell Notation: Oxidation Anode || Reduction Cathode
Cu(s) | Cu+2 (aq) || Fe+3 (aq), Fe+2(aq) | Pt (s)
Find the standard cell potential for an electrochemical cell with the following
cell reaction and write its cell notation.

Given the following reduction potentials:

Fe3+ (aq) + e–  Fe2+ (aq) E° = 0.77 V
Cu2+ (aq)+ 2e–  Cu (s) E° = 0.34 V
Solution:
Reduction: 2Fe3+ (aq) + 2e–  2Fe2+ (aq) E° = 0.77 V
Oxidation: Cu(s)  Cu2(aq)+ 2e– E° = – 0.34 V
Overall Reaction: Cu + 2Fe3+  Cu2+ + 2Fe2+ E° = 0.43 V
Cell Notation: Oxidation Anode || Reduction Cathode
Cu(s) | Cu+2 (aq) || Fe+3 (aq), Fe+2(aq) | Pt (s)
Given the following reduction potentials:
Ag+(aq) + e-  Ag(s); E° = 0.80 V
Zn2+(aq) + 2e-  Zn(s); E° = -0.76 V

Calculate the cell potential for the following reaction and

predict whether the reaction will take place. If it will take
place write the cell notation.

Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)

Given the following reduction potentials:
Cu2+(aq) + 2e-  Cu(s); E° = 0.34 V
Ni2+(aq) + 2e-  Ni(s); E° = -0.23 V

Predict whether the following reaction will take place. If it

will take place write the cell notation.
Cu(s) + Ni2+(aq)  Cu2+(aq) + Ni(s)
 To determine the cell potential when the conditions are other than
standard state (concentrations not 1 molar and/or pressures not 1 atmosphere).
Use the NERNST EQUATION.
Ecell = Eocell - (RT/nF) ln Q
Ecell = cell potential at non-standard state conditions
Eocell = standard state cell potential
R = constant (8.31 J/mole K)
T = absolute temperature (Kelvin scale)
F = Faraday's constant (96,485 C/mole e-)
n = number of moles of electrons transferred in the balanced equation for the
reaction occurring in the cell
Q = reaction quotient for the reaction.

If the temperature of the cell remains at 25oC, the equation simplifies to:
Ecell = Eocell - (0.0257/n) ln Q
or in terms of log10

Ecell = Eocell - (0.0592/n) log Q

Calculate the cell potential of the following system:
Zn(s) | Zn2+ (0.01M) || Cu2+ (1.0M) | Cu(s)
Solution: Ecell = Eocell - (RT/nF) ln Q
Assume the T=25°C: Ecell = Eocell - (0.0257/n) ln Q
Eocell :
Reduction: Cu+2 (aq) + 2e-  Cu(s) E°= +0.34V

Oxidation: Zn(s)  Zn+2 (aq) + 2e- E°= +0.76V

Overall: Zn(s) + Cu+2  Zn+2 (aq) + Cu(s) E°= +1.10V

n = 2 mole e-
Q = [Zn+2] = [0.01M]
[Cu+2] [1.0M]

Ecell = 1.10V - (0.0257/2) ln [0.01M]

[1.0M]
Ecell = 1.16V
Calculate the cell potential of the following system:
Fe(s) | Fe2+ (0.01M) || Cu2+ (3.0M) | Cu(s)
Solution: Ecell = Eocell - (RT/nF) ln Q
Assume the T=25°C: Ecell = Eocell - (0.0257/n) ln Q
Eocell :
Reduction: Cu+2 (aq) + 2e-  Cu(s) E°= +0.34V

Oxidation: Fe(s)  Fe+2 (aq)+ 2e- E°= +0.44V

Overall: Fe(s) + Cu+2  Fe+2 (aq) + Cu(s) E°= +0.78V

n = 2 mole e-
Q = [Fe+2] = [0.01M]
[Cu+2] [3.0M]

Ecell = 0.78V - (0.0257/2) ln [0.01M]

[3.0M]
Ecell = 0.85V
The process that returns metals
to their oxidized state.

Involves oxidation of the metal.

 Apply coating (such as paint or metal
plating)
 Galvanizing
 Alloying
 Cathodic Protection
 Protects steel in buried fuel tanks and
pipelines.
Electricity can be used to make certain chemical
reactions. This is a process where electricity is used to
drive an otherwise non spontaneous reaction.

NaCl  Na + Cl2 (unbalanced reaction)

H2O  H2 + O2 (unbalanced reaction)

Electrolysis of Sodium Chloride
Cl is oxidized, loss e-
Does not happen on
its own. This is not a
NaCl  Na + Cl2 spontaneous reaction.

+1 -1 0 0
Na is reduced, gains e-

An + -
electrical
energy
will force BATTERY
the
reaction
PULL e- PUSH e-
to
happen.
 + -
BATTERY

Electrode

Molten NaCl

Electrolytic cell
 + -
e- MOVES e- MOVES
BATTERY

ANODE Cl- Cl- Na+ Cl- NaNa +

+
Cl- ClNa- + CATHODE
NaNa Na+
+
Na+ Cl -
+
PULLS IN Na+ Na+ Na+ Na+ PUSHES OUT
Cl -
Na+
ELECTRONS Na+ Na +
Na+ + ELECTRONS
Na- +Cl Na
-
Cl- Cl NaNa + Na + + Cl
- Cl-

Na + Na +
Na + - Na
Cl- Cl Na ++ -
Cl Na +
Cl Na-- +
- Cl
Cl Na + NaCl
- + + - Cl
- Na Na+ NaNa+
+ +
Na Cl
Na
Cl
-
+ Cl
- Cl -
OXIDATION TAKES ClNa
- +
Na+ NaNa
+
NaCl + + - Na Na
+ + REDUCTION TAKES
PLACE IN ANODE Na+ Na + Na +
PLACE IN CATHODE
Cl - Na+
Cl- Cl Na- + Na
NaNa
+
+ + Cl- Cl-
Na+ - Na Na
Na
+
++ -
Cl Cl Cl- Cl- Cl -
Cl - Cl
Na+
-

Na++ - Na
+
Cl- Na+ Na
Cl -
Na+Cl
NaNa Na
+ +
+
Na+ Na+ Cl- Na+
 + -
e- MOVES e- MOVES
BATTERY

ANODE Cl CATHODE
Cl
PULLS IN PUSHES OUT
ELECTRONS Cl- ELECTRONS
Na+ Na+
Cl-
OXIDATION TAKES Cl-
Na+ REDUCTION TAKES
PLACE IN ANODE PLACE IN CATHODE
Na+ Na+
Cl-

Na+
 + -
e- MOVES
e- MOVES BATTERY

ANODE Cl CATHODE
Cl
PULLS IN PUSHES OUT
ELECTRONS ELECTRONS

OXIDATION TAKES Cl- Na+

REDUCTION TAKES
PLACE IN ANODE PLACE IN CATHODE
Na+
2Cl -  Cl2+ 2e- Cl- Na+  Na + e-

HAlF REACTION
 COMBINE THE HAlF REACTION

REDUCTION: 2Na+ + 2e-  2Na

OXIDATION: 2Cl -  Cl2+ 2e-

OVERALL: 2Na+ + 2Cl-  2Na + Cl2

2NaCl  2Na + Cl2

 Cl is oxidized, loss e-

2H2O  2H2 + O2
+1 -2 0 0
H is reduced, gains e-

REDUCTION: 4H2O + 4e-  2H2 + 4OH-

OXIDATION: 2H2O  O2 + 4H+ + 4e-

OVERALL: 4H2O + 2H2O  2H2 + O2 + 4OH- + 4H+

OVERALL: 6H2O  2H2 + O2 + 4H2O
OVERALL: 2H2O  2H2 + O2
Production of aluminum
Purification of metals
Metal plating
Electrolysis of sodium chloride
Production of chlorine and sodium hydroxide
In all aspects of our lives we are surrounded by
products with electroplated surfaces.
Whether we are looking at a silver-plated watch
through gold-plated glasses, watching television,
using the washing machine, getting into a car or
boarding a plane: electroplating plays an
important part in all of these situations.
The objective is to prevent corrosion and wear
,produce hardness and conductivity, and give
products an attractive appearance.
The principle: thin
metallic layers with
specific properties
are deposited on base
materials including
steel, brass,
aluminium, plastic
and die-cast parts.
Silver electroplating
was the first large
scale use of
electrolysis for
coating base metal
objects with a higher
value decorative
finish.
ELECTROCHEMISTRY

PROBLEMS:
1. A lamp draws a current of 2.0 A. Find the charge in coulombs
used by the lamp in seconds
2. Compute the time required to pass 36,000 C through an
electroplating bath using a current of 5A.
3 A dynamo delivers 15 A at 120 V.
a. Compute the power in KW supplied by the dynamo
b. How much electric energy in KW- HR , is supplied by the
dynamo in 2 hours?
c. What is the cost of this energy at
4. How many electrons per second pass through a cross section of
copper wire carrying 10-16 A?
ELECTROCHEMISTRY

5. Exactly 0.2000 mol of electrons is passed through 3

electrolytic cells in series . One contains silver ion,
one zinc ion, and one ferric ion. Assume that the only
cathode reaction in each cell is the reduction of the
ion to the metal. How many grams each metal will be
deposited?
Atomic weights( g/mol):
Ag = 107.9 Zn = 65.39 Fe = 55.85
ELECTROCHEMISTRY

Solution: 1 mole of electron liberates 1 g-eq wt of an element.

Equivalent weights: 65.39 𝑔
𝑍𝑛+2 is = 32.69 .
2 𝑚𝑜𝑙 𝑒ҧ
107.9 𝑔 55.85 𝑔
𝐴𝑔+1 is = 107.9 . 𝐹𝑒 +3 is = 18.62 𝑚𝑜𝑙 𝑒ҧ .
1 𝑚𝑜𝑙 𝑒ҧ 3

𝑔
𝐴𝑔𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 = 0.2000 𝑚𝑜𝑙 𝑒ҧ 107.9 = 21.58 𝑔
𝑚𝑜𝑙 𝑒ҧ
𝑔
𝑍𝑛𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 = 0.2000 𝑚𝑜𝑙 𝑒ҧ 32.69 = 6.538 𝑔
𝑚𝑜𝑙 𝑒ҧ
𝑔
𝐹𝑒𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 = 0.2000 𝑚𝑜𝑙 𝑒ҧ 18.62 = 3.724 𝑔
𝑚𝑜𝑙 𝑒ҧ
ELECTROCHEMISTRY

6. A current of 5 A flowing for exactly 30.0 minutes deposits 3.048 g

of zinc at the cathode. Calculate the equivlent weight of zinc from
this information.
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶 = 5.00 𝐶/𝑠 30 × 60 𝑠 = 9.00 × 103 𝐶
9.00 × 103 𝐶
𝑛 𝑒ҧ 𝑢𝑠𝑒𝑑 = = 0.0933 𝑚𝑜𝑙 𝑒ҧ
4 𝐶
9.65 × 10
𝑚𝑜𝑙 𝑒ҧ
3.048 𝑔
𝐸𝑞𝑢𝑖𝑣 𝑤𝑡 = 𝑚𝑎𝑠𝑠 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 𝑏𝑦 1 𝑚𝑜𝑙 𝑒ҧ =
0.0933 𝑚𝑜𝑙 𝑒ҧ
𝑔
= 32.7 ൗ𝑔 − 𝑒𝑞𝑢𝑖𝑣
ELECTROCHEMISTRY

7. Find the charge in coulomb on 1 g-ion of N3-.

Solution:
Charge on one ion of N3-
= 3 × 1.6 × 10-19 coulomb

Thus, charge on one g-ion of N3-

= 3 × 1.6 10-19 × 6.02 × 1023
= 2.89 × 105 coulomb
ELECTROCHEMISTRY

8. How much charge is required to reduce (a) 1 mole of

Al3+ to Al and (b)1 mole of Mn4- to Mn2+ ?
Solution:
(a) The reduction reaction is
Al3+ + 3e- → Al
Thus, 3 mole of electrons are needed to reduce 1 mole of Al3+
Q = 3 × F = 3 × 96500 = 289500 coulomb
ELECTROCHEMISTRY

(b) The reduction is

Mn4-+ 8H+ 5e- → Mn2+ + 4H2O
1 mole 5 mole
Q = 5 × F = 5 × 96500 = 48500 coulomb
9. How much electric charge is required to
oxidize (a) 1 mole of H2O to O2 and (b)1 mole
of FeO to Fe2O3?
ELECTROCHEMISTRY

Solution:
(a) The oxidation reaction is
H2O → 1/2 O2 + 2H+ + 2e-
Q = 2 × F = 2 × 96500 =193000 coulomb

(b) The oxidation reaction is

FeO + 1/2 H2O → 1/2 Fe2O3 + H+ + e-
Q = F = 96500 coulomb
ELECTROCHEMISTRY

9. Exactly 0.4 faraday electric charge is passed through

three electrolytic cells in series, first containing AgNO3,
second CuSO4 and third FeCl3 solution. How many
grams of each metal will be deposited assuming only
cathodic reaction in each cell?
10. An electric current of 100 ampere is passed
through a molten liquid of sodium chloride for 5 hours.
Calculate the volume of chlorine gas liberated at the
electrode at normal temperature and pressure (NTP).
ELECTROCHEMISTRY

11. A 100 watt, 100 volt incandescent lamp is connected in series

with an electrolytic cell containing cadmium sulphate solution.
What mass of cadmium will be deposited by the current flowing
for 10 hours?
We know that
Watt = ampere × volt
Solution:
100 = ampere × 110
Ampere = 100/110
ELECTROCHEMISTRY

Quantity of charge = ampere × second

= 100/110×10×60×60 coulomb
The cathodic reaction is
Cd2+ + 2e- → Cd
Mass of cadmium deposited by passing
100/110×10×60×60
Coulomb charge =
112.4/(2×96500)×100/110×10×60×60=19.0598 g
ELECTROCHEMISTRY

12. In an electrolysis experiment, a current was

passed for 5 hours through two cells connected in
series. The first cell contains a solution gold salt and
the second cell contains copper sulphate solution.
9.85 g of gold was deposited in the first cell. If the
oxidation number of gold is +3, find the amount of
copper deposited on the cathode in the second cell.
Also calculate
the magnitude of the current in ampere.
ELECTROCHEMISTRY

13. How long has a current of 3 ampere to be

applied through a solution of silver nitrate to
coat a metal surface of 80 cm2 with 0.005 cm
thick layer? Density of silver is 10.5 g/cm3.
14. What current strength in ampere will be
required to liberate 10 g of chlorine from
sodium chloride solution in one hour?
ELECTROCHEMISTRY

15. 0.2964 g of copper was deposited on passage of a

current of 0.5 ampere for 30 minutes through a solution
of copper sulphate. Calculate the atomic mass of
copper. (1 faraday = 96500 coulomb)
16. 19 g of molten SnCI2 is electrolysed for some time
using inert electrodes until 0.119 g of Sn is deposited at
the cathode. No substance is lost during electrolysis.
Find the ratio of the masses of SnCI2 : SnCI4 after
electrolysis.
ELECTROCHEMISTRY

17. A current of 2.68 ampere is passed for one hour through an

aqueous solution of copper sulphate using copper electrodes.
Calculate the change in mass of cathode and that of the anode.
(At. mass of copper = 63.5).
18. An ammeter and a copper voltameter are connected
in series through which a constant current flows. The
ammeter shows 0.52 ampere. If 0.635 g of copper is
deposited in one hour, what is the percentage error of
the ammeter? (At. mass of copper = 63.5)
ELECTROCHEMISTRY

19. A current of 3.7 ampere is passed for 6

hours between platinum electrodes in 0.5 litre
of a 2 M solution of Ni(NO3)2. What will be the
molarity of the solution at the end
of electrolysis?
What will be the molarity of solution if nickel
electrodes are used? (1 F = 96500 coulomb;
Ni = 58.7)
ELECTROCHEMISTRY

20. An acidic solution of Cu2+ salt containing

0.4 g of Cu2+ is electrolysed until all the
copper is deposited. The electrolysis is
continued for seven more minutes with
volume of solution kept at 100 mL and the
current at 1.2 amp. Calculate the gases
evolved at NTP during the entire electrolysis.
ELECTROCHEMISTRY

21. Consider the reaction,

2Ag+ + Cd → 2Ag + Cd2+
The standard electrode potentials for Ag+ --> Ag and
Cd2+ --> Cd couples are 0.80 volt and -0.40 volt,
respectively.
(i) What is the standard potential Eo for this reaction?
(ii) For the electrochemical cell in which this reaction
takes place which electrode is negative electrode?
ELECTROCHEMISTRY

22. Calculate the electricity that would be required to reduce

12.3 g of nitrobenzene to aniline, if the current efficiency for the
process is 50 per cent. If the potential drop across the cell is
3.0 volt, how much energy will be consumed?
23. After electrolysis of a sodium chloride solution with inert
electrodes for a certain period of time, 600 mL of the solution
was left which was found to be 1 N in NaOH. During the same
period 31.75 g of copper was deposited in the copper
voltameter in series with the electrolytic cell. Calculate the
percentage theoretical yield of NaOH obtained.
ELECTROCHEMISTRY

24. To find the standard potential of M3+/M electrode, the

following cell is constituted:
Pt|M|M3+(0.0018 mol-1L)||Ag+(0.01 mol-1L)|Ag
The emf of this cell is found to be 0.42 volt. Calculate the
standard potential of the half reaction M3+ + 3e- M3+. = 0.80 volt.
25. Cadmium amalgam is prepared by electrolysis of a solution
of CdCl2 using a mercury cathode. Find how long a current of 5
ampere should be passed in order to prepare 12% Cd-Hg
amalgam on a cathode of 2 g mercury. At mass of Cd =
112.40.
ELECTROCHEMISTRY

26. What volume of O2 (g) at 0oC and 760 mmHg is produced when
5 A of current is passed through a dilute aqueous solution of KCl for
5 minutes?
27. Molecular masses can be determined through electroplating.
Determine the molecular mass and identity of a +2 metal, X, that
plates 46.3g of X in 6.75 hours at a current of 2 A.
28. How long will it take for 1.25 L of a 1.0 M CuSO4 solution
being electrolyzed with a current of 3.40 A to reach a
concentration of 0.25 M?
29. What is the Difference Between a Battery and a Fuel Cell?
https://www.youtube.com/watch?v=7uIIq_Ofzgw
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https://www.youtube.com/watch?v=OdpvTr-7bYI
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ap04_chem_standard_9793.pdf Standard Reduction
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w.portnet.org/cms/lib6/.../Oxidation%20and%20Reduction%20Reactions.ppt;
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chemconnections.org/.../Redox/Balancing%20Redox%20Eqs.../Balancing%20Redox...
https://web.iit.edu/sites/web/files/departments/academic.../pdfs/RedOx_Rxns.pdf
End of Lecture 6