CHAPTER 3: ELECTROCHEMISTRY

Module 1 : Conductance of Electrolytic Solutions

Learning Outcomes
• Understand conductors and its types
• Know about the factors affecting conductance in solution
• Define conductivity and molar conductivity and solve numericals on this
• Understand the experimental setup for the measurement of conductivity
• Understand the variation of conductivity and molar conductivity with
concentration
• State Kohlrausch’s Law and its applications and practice numericals.

CONDUCTORS
There are substances which allow the passage of electricity through
them. They are broadly classified as metallic conductors &
electrolytes.

ELECTROLYTIC CONDUCTORS
METALLIC CONDUCTORS (ELECTROLYTES)
These are metallic substances which These are substances which allow
allow the passage of electricity the passage of electricity, through
through them without undergoing their molten state or aqueous
any chemical change. solution and also undergo chemical
The conduction is due to movement decomposition at the same time.
of electrons. They are also called Here the conduction is due to the
electronic conductors. movement of ions produced by the
Conductivity decrease with increase electrolyte is aqueous solution,
in temperature. Conductivity increase with increase
in temperature.

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 nature and structure of the metal.

number of valence electrons per atoms.

The electronic conductance(metallic
conduction) depends on

temperature -It decrease with increase in

temp due to vibration of positively charged
metal atoms which hinders flow of electrons.

(a) Nature of the electrolyte added - Greater the

interaction between the ions furnished by the
electrolyte in solution (solute – solute interaction),
smaller will be the conductivity.

(b) Solvation of ions - Greater the interaction between

ions of the electrolyte and molecules of solvent
(solute solvent interaction). Greater is the extent of
solvation and lesser will be its conductivity.
The conductivity of electrolytic
solution depends on:
(c) Viscosity of solvent - Greater the interaction
between the solvent molecules (solvent – solvent
interaction), more will be the viscosity, and lesser
will be the conductivity.

(d)Temperature- As temp of electrolytic solution

increases dissociation increases and K.E. of ions
increases hence the conductivity increases.

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 ELECTRICAL RESISTANCE & CONDUCTANCE

CONDUCTANCE (G):It is a measure of

RESISTANCE (R): It is obstruction to the ease with which current flows
flow of current.R =
𝑉 through a conductor.
𝐼 1
Unit for R is  or ohms. G=
𝑅
S.I. unit is ohm-1 or mho or
Siemen’s(S)

RESISTIVITY (ρ ): Resistivity is
the
resistance offered by a conductor of CONDUCTIVITY (K)-It is inverse of
resistivity,also called specific
length 1 cm & area of cross section
conductance
equal to 1cm2 or it is the resistance
1 1 𝑙
offered by 1cm3 of the conductor. K= =𝑅*𝐴
𝜌
𝜌𝑙
R= It is defined as the conductance of a
𝐴
𝑅𝐴 solution of 1 cm length and having 1
= 𝑙
sq. cm as the area of cross section or
it is the conductance of 1 cm3 of a
 = resistivity (specific resistance) solution of an electrolytic.
Unit = ohm – cm or  - m Unit is -1 cm-1. or Sm-1

Molar Conductivity (m) of a solution is the conductance of all the ions produced by
dissolving 1 mole of the electrolyte in V cm3 of solution when the electrodes are 1cm
apart and the area of electrodes are so large the whole of the solution is contained
between them.
Relation between conductivity & Molar conductivity
m = k x v
Where v is the volume of solution in cm3 containing one gram mole of electrolyte.
If M is the concentration of solution in mol/litre

m = k 1000
M
Unit is S cm2 mol-1

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Measurement of conductivity of ionic solutions
https://youtu.be/z6C2_mIKm0A

• We know that conductivity is the product of the cell constant and the conductance.
1 𝑙
• The conductivity of the solution can be calculated as k = . .
𝑅 𝐴
• l/a is called the cell constant which is defined as the ratio of distance between the
two electrodes in a conductivity cell and its area of cross section.
• We will measure conductance(1/R) and cell constant separately and then multiply
them.

• There are two problem in measuring the resistance of an ionic solution.

(i) passing direct current (DC) changes the composition of the solution. So Ac
should be used.
(ii) A galvanometer cannot be used to detect the flow of current as it does
not work with ac.
(iii) Solution cannot be connected to the bridge like a metallic wire or other
solid conductor.
• To overcome these problems, we use an alternating current (AC) source of
power and a specially designed vessel called conductivity cell as shown in fig

(a) Measurement of conductance

We know G = 1/R
The experimental determination of conductance of a solution involves the
measurement of its resistance.
The solution whose resistance has to be measured is taken in a conductivity cell and it
is made one arm of Wheatstone bridge.

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 The bridge is connected to the source of a.c. when current flows through the circuit,
the resistance R1, R2, & R3 are adjusted to get a null point(no flow of current) as
indicated by indicator.
R1 R3
At the null point =
R2 R4
R1 R4
R2 =
R3

The reciprocal of the resistance gives the value of conductance.

(b) To determine cell constant
A solution whose conductivity is known (std KCl) is placed in the conductivity cell
and the conductance is measured as described earlier
conductivity
cell constant = .
conduc tan ce
(c) By knowing the cell constant and the conductance, the conductivity of the solution
1 𝑙
can be calculated as k = . .
𝑅 𝐴

Variation of conductivity (k) & molar conducitivty (m) with concentration.

Conductivity (K )is the conductance of 1 cm3 of solution. On dilution, the no. of ions in
1cm3 of solution decreases; consequently the conductivity of electrolytic solution also
decreases with dilution.
Variation of m with c.
The molar conductivity of a solution decreases with
concentration or increases with dilution.
m = k V

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The variation of m with concentration for KCl (strong electrolyte) and CH3COOH
(weak electrolyte) are shown below.
Strong electrolyte
Strong electrolytes are completely ionized at all concentrations. The increase in the
molar conductivity with dilution is not due to the increase in the number of ions
rather due to increase in the ionic mobility.
When the conc. approaches zero (at infinite dilution) m reaches a certain limiting
value m0 which is called limiting molar conductivity.
The relation between m & m0 for strong electrolyte is given by Debye Huckel
Onsager equation as
m = °m - AC. (This eqn hold good at low conc.)
A plot of m against c gives a straight line with slope –A & y intercept 0m.
A → Constant depending on the nature of the electrolyte.
As C approaches zero, m = °m
Weak electrolyte
The degree of ionization of weak electrolyte is very less & hence they have less m .As
dilution increases, the degree of ionization of weak electrolyte also increases,
increasing the number of ions. There is a large increase in m especially near infinite
dilution. ( α=√K/c)
Note: From the graph it is seen that, for strong electrolyte, the value of 0m is
determined by extrapolating it to zero concentration.
But for weak electrolyte, the when the concentration approaches 0, the graph
becomes almost parallel to y axis.  m0 cannot be obtained by extrapolating the
graph to zero concentration. It is obtained indirectly by using Kohlrausch’s law.
Kohlrausch’s law https://youtu.be/heDl2qI-S9U
At infinite dilution, when the dissociation of electrolyte is complete each ion makes a
definite contribution towards molar conductivity of electrolyte, irrespective of the
nature of the other ion with which it is associated.
OR
The limiting molar conductivity of an electrolyte is the sum of the limiting molar
conductivities of the cation and anion each multiplied by the number of ions present in
one formula unit of the electrolyte.
eg. m° NaCl =  oNa+ +  Clo −
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m0 BaCl2 =  oBa2+ + 2 Cl0 −
Application of Kohlrausch’s law-
(a) It helps in calculation of m0 of weak electrolytes.
m° of CH3COOH can be calculated by knowing m° of NaCl & m° of HCl
m° CH3COOH = m° CH3COO- + m°H+.
 m CH3COOH = °CH3COO- + ° Na+ + ° H+ + ° Cl- - ° Na+ - ° Cl-.
Or °mCH3COOH = °m CH3COONa + °m (HCl) - °m NaCl.
(b) Calculation of degree of dissociation of weak electrolyte 
m c
=
m 0
where mc & m° are the molar conductance at any concentration c & at infinite
dilution respectively.
(c) Calculation of dissociation constant of weak electrolyte
AB → A+ + B-
C 0 o
At. eqn. C (1-) c c
c 2
k= ,  = mc/m0
1− 
C  (m c / m) C (m c ) 2
K = =
1 − (m c / m) m 0 (m 0 − m c )

Solved examples
1. What is meant by limiting molar conductivity?
Ans Molar conductivity of an electrolyte at infinite dilution.

2. Give the relationship between molar conductivity and conductivity ?

Ans Λm= k (1000\M)
3.Why does conductivity decrease with dilution?
Ans: On dilution the number of ions per cm3 of the solution decreases ,resulting in
decrease in conductivity.
4. Why does molar conductivity increase with dilution?

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Ans: m = k V,
On dilution , Volume increases and increase in V is more than decrease in
conductivity K , so m increase with dilution.

5..If HCl is used instead of KCl, then would you expect the value of m values to be
more or less than those per KCl for a given concentration. Justify.
Ans: m values will be more for HCl, as H+ has greater than mobility than K+.

6. Give one point of difference and one point of similarity between the variation of
m with dilution for a strong and weak electrolyte.

Ans: Similarity: For both strong and weak electrolyte m increase with dilution or
decrease with concentration.
Difference: For KCl m increases gradually with dilution ,whereas for weak
electrolyte m values increases steeply with dilution. Also mo (limiting molar
conductivity) for strong electrolytes can be obtained graphically but not for weak
electrolytes.
7. Why is not possible to find the limiting molar conductivity of weak electrolyte
graphically.?
Ans; For weak electrolyte, the when the concentration approaches 0, the graph
becomes almost parallel to y axis and does not intercept the y axis.

8.Solutions of two electrolyte A and B are diluted. Molar conductivity (m ) of B

increases 1.5 times while that of A increases 25 times. Which one is the a strong
electrolyte and why?

Ans: B is a strong electrolyte because in case of strong electrolyte, dissociation is

complete and on dilution there is no increase in the number of ions. So molar
conductivity increases slowly.
Solved numericals.
1. Conductivity of 0.3 M KCl solution is 3.72 X 10-2 S cm-1 . Calculate its molar
conductivity.
Ans Λm= k (1000\M) = 3.72 X 10-2 X 1000 / 0.3 = 124 S cm2/mol

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2. The molar conductivity of 0.025 M solution of methanoic acid is 46.15 S cm2/mol.
Calculate degree of dissociation and dissociation constant. Given λ°H+ = 349.6 and
λ°HCOO- = 54.6
Ans Λ°HCOOH = 349.6 + 54.6 = 404.2
Degree of dissociation = α = Λcm/ Λ°m = 46.15 / 404.2 = 0.1140
Dissociation Constant = K = Cα2/ 1-α = 0.025 X (0.1140)2/ 1-0.1140
= 3.667 X 10-4
5. Given molar conductivities at infinite dilution for Ba(OH)2 , BaCl2 and NH4Cl are
517.6 S cm2 mol-, 240.6 S cm2 mol- and 129.8 S cm2 mol- respectively. Calculate m°
for NH4OH
Ans °Ba(OH)2 = °Ba2+ + 2x °OH-
°BaCl2 = °Ba2+ + 2x °Cl-
°NH4Cl = °NH4+ + °Cl
° NH4OH = °NH4Cl + ½ ° Ba(OH)2 – ½ ° BaCl2
= 129.8 + ½ X 517.6 – ½ X 240.6
= 268.3 S cm2 mol-
Assignment
1. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its
molar conductivity.

2. Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar
conductivity if Λ° = 390.6 also calculate dissociation constant.

3. Conductance of 0.001 M acetic acid is 4.95 X 10-5 S /cm. Calculate it molar

conductivity and degree of dissociation if Λ° = 390.5

4. Calculate Λ° for calcium chloride and magnesium sulphate if

λ°Ca2+ = 119 λ°Mg2+ = 106 λ°Cl- = 76.3 λ°sulphate = 160

5. How does molar conductivity varies with dilution for strong and weak electrolyte.
Explain
6.
7. State Kohlrausch’s Law

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 8. What modification are made in the Wheatstone’s bridge to find the conductivity
of electrolytic solution.

9. Solve N.C.E.R.T Q no. 3.7, 3.8, 3.9, 3.10, 3.11

MCQ
1. An increase in molar conductivity of strong electrolyte on dilution is due to
a. Increase in number of ions
b. Increase in ionic mobility of ions
c. 100% ionization of electrolyte on dilution
d. both a and b
2. The molar conductivity of 0.5M AgNO3 solution having conductivity 5.76 × 10-3
is
a. 2.88 S cm2 mol- b. 11.52 S cm2 mol- c. 0.086 S cm2 mol-
d. 28.8 S cm2 mol-
3.. Resistance of 0.2 M solution of an electrolyte is 50Ω. Specific conductance of the
solution is 1.4 S/cm. The resistance of 0.5 M solution of same electrolyte is 280Ω.
Molar conductivity of 0.5M solution in S cm2 mol- is
a. 500 b. 0.0005 c. 0.005 d. 5000
Ans 1. B 2.b 3. a

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