This document provides an overview of electrochemistry concepts including resistance, conductance, conductivity, cell constant, molar conductivity, and their relationships. It discusses how these properties are affected by factors like electrolyte type, concentration, and temperature. Numerical problems demonstrate calculations of conductance, conductivity, and molar conductivity. The variation of molar conductivity with concentration is explained by the Debye-Huckel-Onsager equation and Kohlrausch's law of independent migration of ions. Limiting molar conductivity and its applications are also summarized.

1. ELECTROCHEMISTRY (PART 2)
CLASS XII
BY: ARUNESH GUPTA
PGT (CHEMISTRY)
KENDRIYA VIDYALAYA, BARRACKPORE (AFS)
Sub-topics: Resistance (R) & conductance (G) of a solution of an electrolyte, Conductivity (k) of solution,
factors affecting conductivity, Cell constant (G*) & its unit, Molar conductivity (Λm) & its variation with
concentration & temperature Numerical problems based on G, k, & Λm , Debye Huckel Onsager equation.
Limiting molar conductivity, Kohlrausch’s law of independent migration of ions, Its application & numerical
problems.

2. ELECTROCHEMISTRY
• Contents:
• Resistance (R) & conductance (G) of a solution of an electrolyte
• Conductivity (k) of solution,
• factors affecting conductivity,
• Cell constant (G*) & its unit,
• Molar conductivity (Λm) & its variation with concentration & temperature,
• Numerical problems based on G, k, & Λm ,
• Debye Huckel Onsager equation.
• Limiting molar conductivity,
• Kohlrausch’s law of independent migration of ions,
• Its application & numerical problems.

3. Conductance (G) (or electrolytic conductance)
It is the ease of flow of electric current through the conductor.
It is reciprocal of resistance (R).
G =
1
𝑅
Unit of G is ohm-1, mho, S, 𝛺−1
. (S denotes seimens)
(i) Ohm’s law:” the strength of current(I) is directly proportional to the potential
difference applied across the conductor & inversely proportional to the
resistance of the conductor. We can directly write: V = I R
(ii) Resistivity ( 𝝆) of a conductor is its resistance of 1 cm length & having area of
cross section to 1 cm2.
R = 𝝆
𝒍
𝒂
or. 𝝆 = 𝑹
𝒂
𝒍
Unit of resistivity = ohm cm or 𝜴 𝒄𝒎
SI unit of resistivity = 𝜴 𝒎

4. Conductivity (k) is defined as the conductance of a Solution if 1 cm length and having
1 sq. cm as the area of cross section. (k  kappa)
Conductivity is the conductance of 1cm3 of a solution of an electrolyte.
Conductivity of a solution is the reciprocal of resistivity of a solution of an electrolyte.
K =
𝟏
𝝆
or k =
𝟏
𝑹
𝟏
𝒂
𝒍
k = G.G* where cell constant G* =
𝒍
𝒂
.
Unit of cell constant is cm-1 or m-1 (SI unit)

5. Conductivity
The unit of conductivity (k) of a solution in S cm-1 or S m-1
Here 1 S m-1 = 10-2 S cm-1.
Different materials have different conductivity (k) values.
Conductance (G) of a solution increases on dilution (ie. by adding water) but its
conductivity (k) decreases, as number of ions per unit volume decreases on
dilution.
Molar Conductivity (Λm)
The conductivity of all the ions produced when 1 mole of an electrolyte is
dissolved in V mL of solution is known as molar conductivity.
It is the conductance of 1 mole of an electrolyte placed between two electrodes 1
cm apart having V cm2 area of cross section.
It is related to conductance as Λm =
𝟏𝟎𝟎𝟎 𝒌
𝑴
where M is the molarity of solution
of electrolyte in mol L-1. It units are Ω-1 cm2 mol-1 or S m2 mol-1 (in SI unit).
Also, 1 S m2 mol-1 = 102 S cm2 mol-1

6. # We know that conductance of a solution of 1 cm3 is G
& conductance of 1 mole of V cm3 solution is Vk which is the definition of molar
conductivity.
Hence 𝛬 𝑚 = Vk.
Let the concentration of solution is M molar,
so, M mole of electrolyte is dissolved in 1 L = 1000 cm3 solution.
Hence, 1 mole of electrolyte is dissolved in
1000
𝑀
cm3 of solution = V cm3 of solution.
So, 𝛬 𝑚 = V.k or, 𝜦 𝒎 =
𝟏𝟎𝟎𝟎 𝒌
𝑴
# Unit of 𝜦 𝒎 =
𝑺𝒎−𝟏
𝒎𝒐𝒍 𝑳−𝟏 =
𝑺𝒎−𝟏
𝒎𝒐𝒍 𝒎−𝟑 𝟏𝟎 𝟑 So, S.I. unit of 𝜦 𝒎 is S m2 mol-1.
# With the increase of temperature, G, k & 𝛬 𝑚 increases.
# With the decrease in concentration on dilution, G increases, k decreases but
𝛬 𝑚 increases.
Molar conductivity

7. Numerical Problems:
Examples: (1) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is
1500 Ω. What is the cell constant f conductivity of 0.001 M KCl solution at 298 K is 0.146 x 10-3 S
cm-1?
Solution: Given k = 0.146 x 10-3 S cm-1 & R = 1500 Ω We know, G = 1/R & k = GG* =
𝐺∗
𝑅
or,
G* = k.R = 0.146 x 10-3 x 1500 Hence G* = 0.219 cm-1 (Ans)
(2) A 0.05 M NaOH solution : a resistance of 31.6 ohm in a conductivity cell at 298 K. If area of
plates of conductivity cell is 3.8 cm2 & distance between them is 1.4 cm, calculate the molar
conductivity of NaOH solution.
Solution: Cell constant G* = l/a =
1.4 𝑐𝑚
3.8 𝑐𝑚2 = 0.368 cm-1. Given concentration = 0.05M, R = 31.6
ohm,
So, conductivity k = GG* =
𝐺∗
𝑅
=
0.368
31.6
– 0.0116 S cm-1. Hence, 𝛬 𝑚 =
1000 𝑘
𝑀
=
1000 𝑥 0.0116
0.05
= 232 S cm2
mol-1 (Ans)
(3) A conductivity cell when filled with 0.01 M KCl has a resistance of 747.5 ohm at 298K. When
the same cell was filled with an aqueous solution of 0.005 M CaCl2 solution, the resistance was
876 ohm. Calculate (i) conductivity and (ii) molar conductivity of CaCl2 solution. ( conductivity of
0.01 M KCl solution is 0.14114 S/m.
Solution: (a) For KCl solution: R = 747.5 ohm k = 0.14114 S/m
Hence Cell constant G* = R. k = 747.5 x 0.14114 = 105.5 m-1.
(b) For CaCl2 solution, conductivity cell is same. So cell constant (G*) is same.
Hence, conductivity k =
𝑐𝑒𝑙𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
=
105.5
= 0.1204 S/m

8. Numerical problems for practice:
(1) The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2mol-
1. Calculate the conductivity of this solution. [Ans: 0.2083
S/cm]
(2) The conductivity of a solution containing 1.0 g of anhydrous BaCl2 in 200 mL of the
has found to 0.0058 Scm-1. Calculate the molar conductivity of the solution. (At. No. Ba = 137,
Cl = 35.5)
[ Ans241.67 Scm2mol-1]
(3) The resistance of a conductivity cell with 0.1 M KCl solution is found to be 200 ohm at 298
K. When the same cell was filled with 0.02 M NaCl solution, the resistance at the at the same
temperature is found to be 1100 ohm. Calculate (i) cell constant of the cell in m-1.
. (ii) the molar
conductivity of 0.02 M NaCl solution in S m2 mol-1. (k for 0.1 M KCl solution at 298 K = 1.29
S/m)
[Ans: 258 m-1, 1.175 x 10-2 S m2 mol-1]
(4) The molar conductance of 0.05 M solution of MgCl2 is 194.5 S cm2 mol-1 at 298 K. A cell
electrodes having 1.50 cm2 surface area and 0.50 cm apart is filled with 0.05 M solution of
MgCl2. How much current will flow when the potential difference between the electrodes is 5
V? [Ans. 0.146 A]

9. Variation of molar conductivity with concentration for strong electrolytes.
In case of a strong electrolyte, molar conductivity increases slowly on dilution. On further
dilution till infinite dilution when concentration tends to zero , molar conductivity value
becomes a constant value for a particular electrolyte.
The molar conductivity when the concentration approaches zero is called molar conductivity
infinite dilution (𝛬 𝑚
𝑜
) or limiting molar conductivity
We can say, 𝛬 𝑚 = 𝛬 𝑚
𝑜
when molar concentration C tends to zero.
Debye-Huckel Onsager equation :
It gives a relation between molar conductivity, Λm at a particular concentration
and molar conductivity at infinite dilution 𝛬 𝑚
𝑜 . Mathematically, Λm = Λ0
m –
A√C
where, A is a constant. It depends upon the nature of solvent and
temperature.

10. 𝛬 𝑚
𝑜
𝛬 𝑚
𝑜
Debye-Huckel Onsager equation
Here, 𝛬 𝑚 = -A√C - 𝛬 𝑚
𝑜 (compare with
straight line eqn. y = mx + c
where Slope = -A and
y-intercept = 𝛬 𝑚
𝑜
,
The limiting value, Λ0
m or Λ∞
m.
(the molar conductivity at zero
concentration (or at infinite dilution)
can be obtained extrapolating the
graph. (Λ0
m is called limiting molar
conductivity).

11. Factors Affecting Conductivity :
(i) Nature of electrolyte
The strong electrolytes like KNO3 KCl. NaOH. etc. are completely ionised
in aqueous solution and have high values of molar conductivity.
The weak electrolytes are ionised to a lesser extent in aqueous solution
and have lower values of molar conductivity.
ii) Concentration of the solution
The concentrated solutions of strong electrolytes have significant
interionic attractions. which reduce the speed of ions and lower the
value of Λm.
The dilution decreases such attractions and increase the value of Λm.

12. Variation of molar conductivity (Λ0
m) with concentration of solution of weak electrolytes:
In case of weak electrolytes, the degree of ionisation
increases dilution which increases the value of Λ m. The
limiting value Λ0
m (limiting molar conductivity) cannot be
obtained by extrapolating the graph. The limiting value,
Λ0
m for weak electrolytes is obtained by Kohlrausch law
of independent migration of ions:
“At infinite dilution, the limiting molar conductivity of an
electrolyte is the sum of the limiting ionic conductivities of
all the cations and anions.”
e.g., for AxBy  x Ay+ + y Bx-
Here 𝚲 𝐦
𝐨
𝐀 𝐱 𝐁 𝐲 = 𝐱𝛌 𝐀 𝐲+
𝐨
+ 𝐲𝛌 𝐁 𝐲−
𝐨
For weak electrolyte CH3COOH → CH3COO- + H+
𝚲 𝐦
𝐨 (𝐂𝐇 𝟑 𝐂𝐎𝐎𝐇) = 𝛌 𝐂𝐇 𝟑 𝐂𝐎𝐎−
𝐨
+ 𝛌 𝐇+
𝐨

13. Variation of molar conductivity with Temperature:
The increase of temperature decreases inter-
attractions of ions in the solution of an electrolyte and
increases kinetic energy of ions and their speed. Thus,
molar conductivity (Λm ) increase with temperature.

14. Applications of Kohlrausch law of independent migration of ions:
(i) We can determine the molar conductivities of weak electrolytes at infinite dilution, e.g.,
𝛬 𝑚
𝑜 𝐶𝐻3 𝐶𝑂𝑂𝐻 = 𝛬 𝑚
𝑜 𝐶𝐻3 𝐶𝑂𝑁𝑎 + 𝛬 𝐻𝐶𝑙
𝑜
− 𝛬 𝑁𝑎𝐶𝑙
𝑜
𝛬 𝑚
𝑜
(𝑁𝐻4 𝑂𝐻) = 𝛬 𝑚
𝑜
𝑁𝐻4 𝐶𝑙 + 𝛬 𝑁𝑎𝑂𝐻
𝑜
− 𝛬 𝑁𝑎𝐶𝑙
𝑜
(ii) Determination of degree of dissociation (α) of an electrolyte at a given dilution.
𝝰 =
𝒎𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒊𝒕𝒚 𝒂𝒕 𝒂 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝑪
𝒎𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒊𝒕𝒚 𝒂𝒕 𝒊𝒏𝒇𝒊𝒏𝒊𝒕𝒆 𝒅𝒊𝒍𝒖𝒕𝒊𝒐𝒏 (𝒍𝒊𝒎𝒊𝒕𝒊𝒏𝒈 𝒎𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒊𝒗𝒊𝒕𝒚)
or 𝝰 =
𝜦 𝒎
𝜦 𝒎
𝒐
The dissociation constant (Kc) of the weak electrolyte (of type AB) at concentration C of
the solution can be calculated by using the formula Kc =
𝑪𝜶 𝟐
𝟏− 𝜶
where, α is the degree of dissociation of the electrolyte.

15. Applications of Kohlrausch law of independent migration of ions:
(iii) Salts like BaSO4, PbSO4, AgCl, AgBr, AgI etc which do not dissolve
to a large extent in water are called sparingly soluble salts.
The solubility of a sparingly soluble salt can be calculated
as 𝜦 𝒎
𝒐 =
𝟏𝟎𝟎𝟎 𝒌
𝑺
where S is the solubility of a salt in mol/L.
Example: (1) The limiting molar conductivities of NaCl, NaAc & HCl are 126.4, 425.9 and 91.0 S
cm2 mol-1 respectively. Calculate the limiting molar conductivity of AcH.
Solution: ΛAcH
o
= λAc−
o
+ λH+
o
= λAc−
o
+ λNa+
o
+ λH+
o
+ λCl−
o
− λNa+
o
− λCl−
o
= ΛNaAc
0
+
ΛHCl
0
− ΛNaCl
0
= 425.9 + 91.0 – 126.4 = 226.0 S cm2 mol-1 (Ans)
Example (2) The conductivity (k) of 0.001028 M acetic acid is 4.95 x 10-5 S cm-1. Calculate its
dissociation constant if limiting molar conductivity is acetic acid is 390.5 S cm2 mol-1.
Solution: We know, Λm =
1000 𝑘
𝑀
=
1000 𝑥 4.95 𝑥 10−5
0.001028
= 48.15 S cm2mol-1.
Degree of dissociation 𝝰 =
𝛬 𝑚
𝛬 𝑚
𝑜 =
48.15
390.5
= 0.1233
And dissociation constant Kc =
𝐶𝛼2
1− 𝛼
0.001028 𝑥 (0.1233)2
1−0.1233
= 1.78 x 10-5 mol L-1 (Ans)

16. Numerical problems for practice: (from Kohlrausch’s law)
(1) Suggest a way to determine limiting molar conductivity of water.
[Ans: 𝛬 𝐻2 𝑂
0
= 𝛬 𝑁𝑎𝑂𝐻
0
+ 𝛬 𝐻𝐶𝑙
0
− 𝛬 𝑁𝑎𝐶𝑙
0
]
(2) The molar conductivity of of 0.025 M HCOOH is 46.1 S cm2 mol-1.
Calculate its degree of dissociation & dissociation constant.
(Given: 𝜆 𝐻+
𝑜
= 349.6 S cm2 mol-1 & 𝜆 𝐻𝐶𝑂𝑂−
𝑜
= 54.6 S cm2 mol-1.
[Ans: 0.114, 3.67 x 10-4]
(3) The molar conductivity at infinity dilution of aluminium sulphate is 858
S cm2 mol-1. Calculate the limiting molar ionic conductivity of Al3+ ion.
(Given 𝜆 𝑆𝑂4
2−
0
= 160 S cm2 mol-1. [Ans. 189 Scm2
(4) The limiting molar conductivity of NaOH, NaCl and BaCl2 at 298K are
2.481 x 10-2, 1.265 x 10-2 and 2.80 x 10-2 S cm2 mol-1 respectively. Calculate
𝛬 𝑚
𝑜 for Ba(OH)2. [Ans 5.23 x 10-2 S
mol-1]

17. THANK YOU
Arunesh Gupta
PGT (Chemistry) KV BKP (AFS)