The document covers various aspects of electrochemistry, including types of electrodes (metal-metal ion, metal-amalgam, metal-insoluble salt, gas, and red-ox electrodes), their applications, and methods for determining pH and EMF. It explains Faraday's laws of electrolysis and details the electrolysis of sodium chloride and copper sulfate solutions. Additionally, the document addresses ionic strength and its calculation in different solutions.

1. Electrochemistry
(Types of Electrodes, Applications of EMF, Determination of pH)
(Faradays law of Electrolysis, Electrolysis of NaCl, Ionic Strength)
Dr.S.SURESH
Assistant Professor
Email:avitsureshindia@gmail.com

2. Types of Electrodes

3. Different Types of Electrodes
• An electrochemical cell consists of two electrodes, positive
and negative. Each electrode along with the electrolyte
constitutes a half cell. The commonly used electrodes in
different electrochemical cells are
• Metal-Metal ion electrodes
• Metal-Amalgam electrodes
• Metal insoluble metal salt electrodes
• Gas electrodes
• Oxidation-reduction electrodes

4. Metal-Metal ion electrodes
An electrode of this type consists of a metal rod (M)
dipping into a solution of its metal ions(Mn+
). This is
represented as
M/Mn+
Example:
Cu rod dipping in Copper sulphate solution (Cu/Cu2+
)

5. Metal-Metal ion electrodes
The electrode reaction may be represented as
Mn+
+ ne‒
⇌ M
If the metal rod behaves as negative electrode (i.e
the reaction involves oxidation) the equilibrium
will shift to the left and hence the concentration of
the Metal ions in the solution will increase.
On the other hand, if the metal rod behaves as
positive electrode (i.e the reaction involves
reduction), and the equilibrium will shift to the
right and hence the concentration of the metal
ions will decrease.

6. Metal-Amalgam Electrodes
The activities and the electrode potentials of highly
reactive metals such as sodium, potassium, etc. are
difficult to measure in aqueous solution. Hence, the
activity of the metal is lowered by dilution with
mercury.
M(Hg)/Mn+
Example
Na, Hg (C1)/Na+
(C2)

7. Metal-Insoluble metal salt electrodes
• This type of electrode consists of a metal (M)
covered by a layer of sparingly soluble salt (MX)
dipping into a solution containing a common anion
(X‒
). These electrodes are represented as
M/MX // X‒
(a)
Example:
• (a)Calomel Electode: It consists of Mercury-
Mercurous chloride in contact with a solution of
potassium chloride (Hg/Hg2Cl2, KCl)

8. Gas Electrodes
• In a gas electrode hydrogen gas is continuously
bubbled through a 1M solution of the acid. A inert
metal like platinum is used, since it is not attacked
by acid. The inert metal in electrode does not
participate in the electrode reaction but helps in
making electrical contact. Let the gas bubbled be X2,
then the electrode is denoted as
Pt, X2(1atm) /X+
(a)
Example:
Standard hydrogen electrode: Pt, H2 (1atm)/H+
(1M)

9. Oxidation-Reduction electrodes
(or) Red-Ox electrodes
• In this type of electrode the potential is developed
due to the presence of ions of the same substance
in two different valence(i.e. oxidation) states.
• This electrode is set up by inserting an inert metal
like platinum in an appropriate solution containing
a mixture of ferrous (Fe2+
) and Ferric (Fe3+
) ions.
• In this electrode the potential is due to the
tendency of the ions to change from one oxidation
state to the other more stable oxidation state.
These electrodes are represented as
Pt/Mn
1
+
(a1), Mn
2
+
(a2)

10. Applications of EMF measurements

11. Applications of EMF
measurements
(i) Determination of valency of ion in doubtful cases
• The valency of mercurous ion can be determined by
determining the EMF of a concentration cell of the type given
below.
Hg/ Hg2(NO3)2(C1) // Hg2(NO3)2(C2) / Mercury
• The EMF of the cell, E, is given by the expression
E =
• It was found that when C2/C1 was 10, (i.e. C2 = 1M and C1
= 0.1M) the EMF was 0.0295V. Therefore, the valency of
mercurous ion is 2, and it should be represented as .
1
2
C
C
log
n
0.059
+2
2Hg

12. Applications of EMF
measurements
(ii) Determination of solubility product of sparingly soluble
salt:
The solubility product constant of a sparingly soluble salt is a
kind of equilibrium constant. Consider the salt MX in
equilibrium with its ions in a saturated solution.
MX(s) +⇌
The solubility product of the salt is given by
Ksp = [M+
] [X‒
]
+
)(aqM −
)(aqX

13. The cell is represented as M, M+
X‒
(sat.sol.) // MX(s), M
R.H.E MX(s) + e‒
M + X⇌ ‒
L.H.E M + e⇌ ‒
Overall reaction MX(s) +⇌
E° =
We know that ∆G° = nFE°‒
And ∆G° = 2.303 RT log K‒ sp
log Ksp =
E° =
=
+
)(aqM
+
)(aqM −
)(aqX
ο
L
ο
R E-E
RT2.303
nFE
spKlog
nF
RT2.303
spKlog
n
0.059

14. (iii)Free energy, enthalpy and entropy changes in
electrochemical reactions:
The standard free energy can be calculated as follows
∆G° = nFE°‒
n= number of electrons, F = 96500 C, E° = EMF of the cell.
• By knowing the standard EMF of a cell we can calculate
the entropy change using the equation
∆S° = n F
• Then the enthalpy change can be calculated using the
equation
∆G° = ∆H° T∆S°‒
pT
E






∂
∂

15. Determination of pH

16. Determination of pH
By using a glass electrode:
It consists of thin walled glass bulb (made of
special type of glass having low melting point
and high electrical conductivity) containing a
Pt wire in 0.1M HCl. The thin walled glass bulb
functions as an ion-exchange resin, and an
equilibrium is set up between the sodium ions
of glass and hydrogen ions in solution. The
potential difference varies with the hydrogen
ion concentration, and it given by
pH]0592.0[E0
G +

17. Determination of pH by using a glass electrode
The cell may be represented as
Pt, 0.1M HCl/ Glass// KCl(Satd.Soln.)/ Hg2Cl2(s). Hg
The EMF of the cell is measured by
E°Cell =
E°Cell =
= 0.2422 V ‒
= 0.2422 V ‒
pH =
ο
L
ο
R E-E
GlassCalomel E-E
pH]0592.0[E0
G +
pH0592.0E0
G −
V0.0592
E-E-V0.2422 Cell
0
G

18. Faraday’s Law of Electrolysis

19. Faraday’s Law of Electrolysis
First Law: According to it “during electrolysis, the
amount of any substance deposited or evolved at
any electrode is proportional to the quantity of
electricity passed”
The quantity of electricity (Q) is equal to the
product of the current strength and the time for
which it is passed.
Q = current strength × time

20. Faraday’s Law of Electrolysis
If W is the weight of a substance
liberated/deposited at an electrode during
electrolysis, then from first law, we get:
W α Q
But Q = it
W α it
W = Zit

21. Faraday’s Law of Electrolysis
Second Law: According to it “the weights of
different substance evolved/deposited by the
passage of same quantity of electricity are
proportional to their chemical equivalent weights
W α E
Where
W = Weight of the substance liberated or
deposited.
E = Chemical equivalent weight of the substance
liberated/deposited.

22. Electrolysis of aqueous sodium
chloride
When sodium chloride is dissolved in water, it
ionises as
NaCl Na⇌ +
+ Cl‒
Water also dissociates as
H2O H⇌ +
+ OH‒
When electric current is passed through aqueous
sodium chloride solution using platinum electrodes

23. Electrolysis of aqueous sodium
chloride
H+
ions move towards the cathode. The H+
ions gain
electrons and change into neutral atoms. Hydrogen
atom is unstable and combines with another atom
to form stable hydrogen molecule.
Hydrogen atom is unstable and combines with
another atom to form stable hydrogen molecule
H+
+ e‒
H (atom unstable)
H + H H2 (stable)
→
→

24. Electrolysis of aqueous sodium
chloride
Cl‒
ions move towards anode. These Cl‒
ions lose electrons
and change into neutral atoms, chlorine atom is unstable
and combines with another atom to form stable chlorine
molecule.
At Anode:
Cl‒
e‒ ‒
Cl atom (unstable)
Cl + Cl Cl2 (stable)
Hence in the electrolysis of aqueous solution of sodium
chloride, hydrogen is liberated at cathode while chlorine is
liberated at anode.
→
→

25. Electrolysis of aqueous copper
sulphate solution (using pt
electrode)
When copper sulphate is dissolved in water, it ionizes as
CuSO4 Cu⇌ 2+
+ SO4
2‒
H2O H⇌ +
+ OH‒
(slightly ionized)
When electric current is passed through copper sulphate
solution using platinum (Pt) electrodes:
(a) Cu2+
ions move towards cathode. These Cu2+
ions gain
electrons and change into neutral atoms and get deposited
at cathode.
Cu2+
+ 2e‒
Cu (deposited)→

26. At Cathode
OH‒
ions move towards anode. These OH‒
ions lose
electrons and change into neutral hydroxyl groups.
At Anode:
2OH‒
e 2OH (neutral)‒
The neutral hydroxyl groups being unstable react with other
neutral OH‒
groups to form water and oxygen.
2OH H2O + O
O + O O2
Hence during electrolysis of copper sulphate solution using
platinum electrodes, copper and oxygen are liberated.
→
→
→

27. Ionic Strength

28. Ionic strength of solutions
Ionic strength is a measure of the
concentration of ions in that solution. Ionic
strength may be expressed as
2
iiZC
2
1
μ ∑=

29. Ionic strength
Calculate the ionic strength of 0.1M solution of NaCl
For Na+
C = 0.1 and Z = 1
For Cl‒
C = 0.1 and Z = 1
2
iiZC
2
1
μ ∑=
0.1)11.011.0(
2
1
μ 22
=×+×=

30. Ionic strength
Calculate the ionic strength of 0.1M solution of
Na2SO4, MgCl2 and MgSO4 respectively.
Na2SO4
MgCl2
MgSO4
0.3)21.012.0(
2
1
μ 22
=×+×=
0.3)12.021.0(
2
1
μ 22
=×+×=
0.4)21.021.0(
2
1
μ 22
=×+×=

31. Ionic strength
• The ionic strength of a solution containing
more than one electrolyte is the sum of the
ionic strength of the individual salts
comprising the solution.
• Hence, the ionic strength of a solution
containing Na2SO4, MgCl2 and MgSO4 each at
a concentration of 0.1M is 1.0.