Electrochemistry

ELECTROCHEMISTRY
Dr.P.GOVINDARAJ
Associate Professor & Head , Department of Chemistry
SAIVA BHANU KSHATRIYA COLLEGE
ARUPPUKOTTAI - 626101
Virudhunagar District, Tamil Nadu, India

ELECTROCHEMISTRY
Definition
It is a branch of physical chemistry which deals about
 Generation of electricity by chemical reaction
 Occurring of chemical reaction by electricity
Total reaction: 2Cl- + 2 Na+ Cl2(g) + 2Na(s)
Electricity
Cell reaction : Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
At anode : Zn(s)→ Zn2+ + 2e-
At cathode : Cu2+ + 2 e- → Cu(s)
At anode : 2Cl - → Cl2(g) + 2e-
At cathode : 2Na+ + 2 e- → 2Na(s)

Definition
A state of the system in which both forward and backward processes are
balanced is called equilibrium
Types of equilibrium
There are four types of equilibrium
1. Physical equilibrium
2. Chemical equilibrium
3. Phase equilibrium
4. Ionic equilibrium (Acid – base equilibrium)
EQUILIBRIUM

EQUILIBRIUM
PCl5 ⇌ PCl3 + Cl2
ICE ⇌ WATER CH3COOH + H2O ⇌ H3O+ + CH3COO-
acid base acid base
PHASE EQUILIBRIUM IONIC EQUILIBRIUM
CHEMICAL EQUILIBRIUM
PHYSICAL EQUILIBRIUM

Definition
Substances which allow electrical current to pass through them are known
as conductors and this process is known as conduction
Types of conductors
There are two types of conductors
1.Metallic conductors
2.Electrolytic conductors
CONDUCTORS

• Metals which conduct electricity due to the mobility of outer shell electrons
of atoms present Inside them are known as metallic conductors and this process
is known as metallic conductors and this process is known as metallic conductance
(or) electronic conductance
• There is no chemical reaction in the metallic conductance
Definition
METALLIC CONDUCTORS

• Solutions of chemical compounds or in their fused state containing opposite charged
ions conduct electricity by the mobility of ions are known as Electrolytic conductors
(or) electrolytes and this process is known as electrolytic conductance (or) ionic
conductance
• The electrolytic conduction is accompanied by chemical reaction
ELECTROLYTIC CONDUCTANCE
Definition:

• There are two types of electrolytes
1. Strong electrolytes
2. Weak electrolytes
Strong electrolytes:
• Strong electrolytes are compounds which are completely dissociated (ionized) into
positive and negative charged ions when dissolved in water.
Examples:
NaCl + H2O Na+ + Cl-
Al2(SO4)3 + H2O 2Al3+ + 3 𝑆𝑂4
2−
AgNO3 + H2O Ag+ + 𝑁𝑂3
−
• The total number of positive ions and negative ions produced are equal to the formula
of the electrolytes
TYPES OF ELECTROLYTES

Weak electrolytes
• Weak electrolytes are compounds which are partially dissociated (ionized)
into positive and negative charged ions when dissolved in water
Examples:
CH3COOH + H2O ⇌ CH3COO - + H+
NH4OH + H2O ⇌ 𝑁𝐻4
+
+ OH-
• Equilibrium exists between the dissociated ions and the undissociated electrolyte
TYPES OF ELECTROLYTES

Resistance (R) and specific resistance (ρ):
• The restriction with which the electrolytic solution oppose the flow of electricity
is called as resistance of the Electrolyte
• The unit of resistance in SI system is ohm
• The resistance (R) is directly proportional to the length of the electrolytic conductor (l)
and inversely proportional to the area of cross section (A) of the electrolytic
conductor, i.e., R  l/A , R = ρ l/A and ρ = R x A / l ,
Where ρ = specific resistance
ELECTRICAL CONDUCTANCE QUANTITIES
• When l = 1m and A = 1 m2 , then the above equation
becomes ρ = R i.e., specific resistance is defined as the resistance of a
one meter cubic electrolytic Solution

Conductance (G) and Specific conductance (k):
• The ease (power) with which electricity flows through a
electrolytic solution is called as conductance of the solution
• The unit of conductance in SI system is Siemen
• Since resistance and Conductance are inverse relationship, the reciprocal of the
resistance of the solution is also called as conductance of the solution
• The unit of conductance is represented as ohm-1 or mho
• The reciprocal of specific resistance is called as specific conductance (or) specific
conductivity(k) of the electrolytic solution
i.e., G = 1 / R
i.e., k = 1 / ρ
• On substituting ρ = R x A / l in the above equation
k = 1/ R x l / A -------(1)

• The unit of specific resistance is
k =
1
𝑜ℎ𝑚
x
𝑚
𝑚2 = ohm-1 m-1 (or) Sm-1
• Since 1/ R is equal to conductance (G). The equation (1) becomes
k = conductance x l/A
When l = 1m and A = 1 m2 , then the above equation becomes
k = Conductance
• Thus, specific resistance (k) is also defined as the conductance of one meter cubic
Electrolytic solution

EQUIVALENT AND MOLAR CONDUCTANCE
Gram equivalent weight:
Equivalent weight of chemical substance expressed in grams is called as equivalent weight
1. For acid
Equivalent weight =
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
𝑁𝑜.𝑜𝑓 𝐻+
𝑟𝑒𝑙𝑒𝑎𝑠𝑒𝑑 𝑏𝑦 𝑎𝑐𝑖𝑑
2. For base
Equivalent weight =
𝑁𝑜.𝑜𝑓 𝑂𝐻−
𝑟𝑒𝑙𝑒𝑎𝑠𝑒𝑑 𝑏𝑦 𝑏𝑎𝑠𝑒
3. For reducing (or) oxidizing agent
Equivalent weight =
𝑁𝑜.𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 𝑔𝑎𝑖𝑛 𝑜𝑟 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑑𝑜𝑥 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
Formula :

Examples:
• Equivalent weight of NaOH =
1
(NaOH + H2O → Na++ OH-) = 40 / 1
Gram equivalent weight of NaOH = 40 grams
• Equivalent weight of Ca(OH)2 =
2
(Ca(OH)2 + H2O → Ca2
+ + 2OH-) = 74 / 2 = 37
Gram equivalent weight of Ca(OH)2 = 37 grams
• Equivalent weight of H2SO4 =
2
(H2SO4 + H2O → 2H+ + SO4
2-) = 98/2 = 49
Gram equivalent weight of H2SO4 = 49 grams

• Equivalent weight of KmNO4 =
𝑁𝑜.𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑔𝑎𝑖𝑛𝑒𝑑
(2KMnO4+5H2SO4→ K2SO4+2MnSO4+3H2O +5/2O2) = 158/5 = 31.6
(i.e., Mn7+ + 5e- → Mn2+)
Gram equivalent weight of KmnO4 = 31.6grams

• Molecular weight of the chemical substance expressed in grams is called Gram Molecular
Weight (or) One Mole
Example:
• Gram Molecular Weight (or) 1 Mole of NaOH = 40 grams
• Gram Molecular Weight (or) 1 Mole of Ca(OH)2 = 74 grams
• Gram Molecular Weight (or) 1 Mole of H2SO4 = 98 grams
• Gram Molecular Weight (or) 1 Mole of KMnO4 = 158 grams
Note : 1 Mole of chemical substance contain 6.023 X 1023 number of molecules.
Gram Molecular Weight (or) 1 Mole of substance:

• Equivalent conductance is defined as the conducting power of all the ions produced by
one gram equivalent weight of an electrolyte in a given solution
• Mathematically, λequ = k x V -------(1)
where k = Specific Conductance of the solution
V = Volume in cc containing 1 gram equivalent of the electrolyte
• If N is the normality then
V =
1000
𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦
and equation (1) becomes
λequ = k x
1000
• SI unit of λequ is Sieman metre square per equivalent (or) Sm2equ-1
Equivalent Conductance(λequ):

• Molar conductance is defined as the conducting power of all the ions produced by one
gram molecular weight (1 mole) of an electrolyte in a given solution
• It is represented as λm
λm = K x V -------(1)
K ----- Specific conductance of the solution
V ----- Volume in cc containing 1 mole of the electrolyte
• If M is the molarity of the solution then the equation (1) becomes
λm = K x 1000/M
• SI unit is Siemen meter square per mol (or) Sm2mol-1
Molar Conductance(λm) :

Measurement of conductance and cell constant:
Cell constant :
• Cell constant is of particular cell is determined as the ratio of the distance between
the electrodes of the cell to the area of the electrodes.
cell constant K = l/A = m-1.
k = conductance x l/A , λequ = k x
1000
& λm = K x 1000/M

EFFECT OF DILUTION ON CONDUCTANCE
Effect of dilution on conductance:
• On dilution specific conductance of the electrolytic solution decreases, but equivalent
and molar conductance increases and reaches a maximum value
Reason :
• On dilution, number of ions per unit volume decreases and hence specific conductance
decreases
• On dilution, concentration of the electrolytic solution decreases and hence λm
& λequi increases and reaches maximum value

Graphical representation:
• The variation of molar conductance (λm ) and equivalent conductance ( λequi )with square root
of concentration of electrolytic solution for weak and strong electrolyte is graphical
represented as

• For strong electrolyte, λm (or) λequi is high even at low dilution, this is due to that the strong
electrolytes are completely dissociated into ions and gives maximum number of ions at low
dilution
• Further increase in dilution, λm (or) λequi of strong electrolytes increase slowly due to
increase in the mobility of ion.
• For weak electrolytes, λm (or) λequi is low value at low dilution, this is due to the less
number of weak electrolytes dissociated at low dilution and produce less number of ions,
so that λm (or) λequi are low value at low dilution
• On increasing the dilution, dissociation of the weak electrolytes gradually increased and the
number of ions gradually increased, so that λm (or) λequi for weak electrolytes gradually
increased and steeply increased at very high dilution (or) very low concentration

• The molar conductivity of strong electrolytes at infinite dilution (λ0) can be determined
by extending this straight line to zero concentration
• However the molar conductivity of weak electrolytes increases steeply at very low
concentration and hence their molar conductivity of weak electrolytes at infinite dilution (λ0)
cannot be determined by extrapolating the to zero concentration

KOHLRAUSCH’S LAW
• At time infinite dilution, the molar conductance of an electrolyte can be expressed as
the sum of the contributions from its individual ions
i.e., λ
m = v+ λ 
+ + v- λ 
-
where, v+ and v- are the number of cations and anions per formula unit of electrolyte,
λ 
+ and λ 
- are the molar conductivities of the cation and anion at infinite dilution
For example : The molar conductivities of HCl at infinite dilution can be expressed as
λ
HCl = v H+ λ 
H+ + vCl- λ 
Cl –
For HCl, v H+ = 1 and vCl- = 1
λ
HCl = 1 x λ 
H+ + 1 x λ 
Cl –
λ
HCl = λ 
H+ + λ 
Cl –

• Kohlrausch law is used to calculate the molar conductance at infinite dilution for weak
electrolytes.
Example: The molar conductance at infinite dilution for acetic acid is obtained from the values
of 𝑚
 for HCl, CH 3COONa and NaCl by applying Kohlrauch law
𝐻𝐶𝑙
 =  + 
𝐻+ 𝐶𝑙− -------(1)
-------(2)
-------(3)
𝐶𝐻3𝐶𝑂𝑂𝑁𝑎
 =  + 
𝑁𝑎+ 𝐶𝐻3𝐶𝑂𝑂−
𝑁𝑎𝐶l
 =  + 
𝑁𝑎+ 𝐶𝑙−
Equation (1) + (2) – (3) gives

𝐻𝐶𝑙 𝐶𝐻3𝐶𝑂𝑂𝑁𝑎
+  - 𝑁𝑎𝐶𝑙
 = 𝐻+
 + 
𝐶𝐻3𝐶𝑂𝑂− 𝐶𝐻3𝐶𝑂𝑂𝐻
=  ------(4)
𝐻𝐶𝑙 𝐶𝐻3𝐶𝑂𝑂𝑁𝑎
Substituting the values of  ,  & 𝑁𝑎𝐶𝑙

𝐶𝐻3𝐶𝑂𝑂𝐻
in equation (4) gives the value of 
APPLICATION OF KOHLRAUSCH’S LAW

TRANSFERENCE NUMBER (OR) TRANSPORT NUMBER
Definition:
• The fraction of total current carried by the cation (or) anion is termed as
transport (transference) number
• Transport number (t+) for cation is given by
t+=
𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒄𝒂𝒓𝒓𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒄𝒂𝒕𝒊𝒐𝒏
= 𝒒
𝒕𝒐𝒕𝒂𝒍 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒄𝒂𝒓𝒓𝒊𝒆𝒅 𝒃𝒚 𝒃𝒐𝒕𝒉 𝒕𝒉𝒆 𝒊𝒐𝒏𝒔 𝑸
+
• Transport number (t-) for anion is given by
+
t =
𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒄𝒂𝒓𝒓𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒂𝒏𝒊𝒐𝒏
= 𝒒
𝒕𝒐𝒕𝒂𝒍 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒄𝒂𝒓𝒓𝒊𝒆𝒅 𝒃𝒚 𝒃𝒐𝒕𝒉 𝒕𝒉𝒆 𝒊𝒐𝒏𝒔 𝑸
−
Where Q = q+ + q-

• Since the quantity of current carried by the particular ion is directly proportional to the mobility
(speed) of ion (u), the equation (1) & (2) becomes
+
t =
𝑢+
𝑢 + 𝑢
+ −
• Where
u+ is the mobility (speed) of cations and it is the measure of current carried by the cations
u- is the mobility (speed) of anions and it is the measure of current carried by the anions
• Adding equation (3) & (4) we get
------(3) and -
t =
𝑢
−
𝑢++ 𝑢−
-------(4)
t+ + t- =
𝒖++ 𝒖−
𝒖++ 𝒖−
= 1
i.e., sum of the transport number of anion and cation of an electrolytic solution must be unity

Measurement of Transport number by Hittorf’s method:
• According to Hittorf, “The fall in concentration of ion
around any electrode is directly proportional to speed of
that ion moving away from it”
i.e., Fall in concentration around anode  Speed of cation
Fall in concentration around cathode  Speed of anion
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑐𝑎𝑡𝑖𝑜𝑛 (𝑢+)
= 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒
𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑎𝑛𝑖𝑜𝑛 (𝑢−) 𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛 𝑐𝑎𝑡ℎ𝑜𝑑𝑒
We know that
------(1)
+
t =
𝑢+
+
𝑢 + 𝑢−
------(2)

Substituting (2) in (1) we get
+
t =
𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒
𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒 +𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑐𝑎𝑡ℎ𝑜𝑑𝑒
+
t =
𝐹𝑎𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑎𝑛𝑜𝑑𝑒
𝑇𝑜𝑡𝑎𝑙 𝑓𝑎𝑙𝑙𝑙 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑜𝑢𝑛𝑑 𝑏𝑜𝑡ℎ 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒
---------(3)
If the concentration are measured in terms of gram equivalent, then equation (3) becomes
+
t =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑜𝑠𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑛𝑜𝑑𝑖𝑐 𝑐𝑜𝑚𝑝𝑎𝑟𝑡𝑚𝑒𝑛𝑡
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑜𝑠𝑡 𝑓𝑟𝑜𝑚 𝑏𝑜𝑡ℎ 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑎𝑟𝑡𝑚𝑒𝑛𝑡
---------(4)
• Since the total number of gram equivalent of ions lost from the anodic and cathodic
compartment is equal to the number of gram equivalents of ions discharged (or) deposited on
each electrode

So the equation (4) becomes
+
t = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑜𝑠𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑛𝑜𝑑𝑖𝑐 𝑐𝑜𝑚𝑝𝑎𝑟𝑡𝑚𝑒𝑛𝑡
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑑𝑒
---------(5)
• The number of gram equivalents deposited on each electrode must be equal to the
number of gram equivalents of copper deposited in a copper coulometer, which is
connected in series with the experimental solution, on passing same quantity of electricity
So the equation (5) becomes
+
t =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑜𝑠𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑛𝑜𝑑𝑖𝑐 𝑐𝑜𝑚𝑝𝑎𝑟𝑡𝑚𝑒𝑛𝑡
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 𝑑𝑒𝑝𝑜𝑠𝑖𝑡𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑢𝑙𝑜𝑚𝑒𝑡𝑒𝑟
---------(6)
• By determining both numerator and denominator of equation (6) experimentally
the transport number of cation (t+) can be calculated and the transport number of anion (t+)
can be calculated as
t- = 1 – t+

CONDUCTOMETRIC TITRATION
Definition:
• Conductometric titration is the titration in which the end point is determined by
measuring the electrical conductance of an electrolytic solution by conductometer
Example: Acid base titration (HCl Vs NaOH)
Principle and Experiment:
• The known volume of HCl (V2) in the conductivity vessel and the standard NaOH
(N1) is taken in the burette
• The conductance of HCl is due to the presence of H+ and Cl- ions
• As NaOH is added gradually, the H+ ions are replaced by slow moving Na+ ions as
represented below
H+
(aq) + Cl-
(aq) + Na+
(aq) + OH-
(aq) → Na+
(aq) + Cl-
(aq) + H2O (l)

• On continuous addition of NaOH, the fast moving H+ ions are gradually replaced
by the slow moving Na+ ions. So that the conductance will go on decreasing until the
acid has been completely neutralized
• Further addition of NaOH after the end point, the conductance will begin to increase due
to the introduction of fast moving OH- ions
• The end point is obtained graphically on plotting the conductance Vs the volume of NaOH
added

• Using the end point V1, the strength of HCl is
calculated using the formula
2
N =
𝑉 𝑁
1 1
𝑉2
where V1 is the volume of NaOH obtained from graph
N1 is the strength of NaOH
V2 is the volume of HCl
VOLUME OF NaOH

IONIC MOBILITY
• It is defined as the distance travelled by an ion per second under a potential gradient
of 1 volt per metre.
Mathematically,
Ionic mobility =
𝑺𝒑𝒆𝒆𝒅 𝒐𝒇 𝒂𝒏 𝒊𝒐𝒏
𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝒈𝒓𝒂𝒅𝒊𝒆𝒏𝒕
• Potential gradient is obtained by dividing the potential difference applied at the
electrode with the distance between the electrodes
Mathematically,
Potential gradient=
𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆
𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒉𝒆 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒅𝒆𝒔

• Lithium and sodium ions have comparatively low ionic mobility where as H+ ions
has high ionic mobility
IONIC MOBILITY
Explanation:
• Lithium and sodium ions have comparatively low ionic mobility. This is due to the
hydration of ions as a result of higher charge density around these ions because of
their smaller radii
• The higher charge density causes these ions to be more
highly hydrated by ion – dipole interactions than the larger
ions.
• Since a hydrated ion has to drag along a shell of water as it moves through the
solution,its mobility is less than that of an anhydrated ion

Explanation:
• The high mobility of H+ ions in hydroxylic solvents such as H2O can be
explained by Grotthus type mechanism in which a proton moves rapidly
from H3O+ to a hydrogen – bonded water molecule and is transferred further
along a series of hydrogen – bonded water molecules by a rearrangement of
hydrogen bonds, as illustrated in the following diagram
IONIC MOBILITY

APPLICATION OF CONDUCTANCE MEASUREMENTS
1. Determination of degree of dissociation () of weak electrolytes
The degree of dissociation of a weak electrolyte at any dilution can be
calculated by the following relationship
 = m / m

where
m
  is the molar conductance at infinite dilution and obtained by Kohlrausch’s law
m is the molar conductance at particular concentration and obtained by conductance
measurement using the following formula λm = K x 1000/M
where K is the specific conductance and is obtained by multiplying the conductance
measured with cell constant (K = Conductance x cell constant )

1. Determination of ionic product of water (Kw)
• Water is slightly dissociated as
H2O ⇌ H+ + OH-
• The product of the concentration of hydrogen and hydroxyl ions expressed in mol
dm-3 known as ionic product of water (Kw)
i.e., [H+][OH-] = Kw ----------(1)
• The specific conductance (K) of the purest water at 25 0C is determined
experimentally by conductance measurement
• Since the concentration of H+ ions and OH- ions are equal, the equation (1)
becomes
c x c = Kw
c2 = Kw

• We know that the relationship between specific conductance (K), molar
conductance (m) and concentration is
m
= 𝐾
𝑐
----------(3)
c = 𝐾
m
----------(4)
• The specific conductance (K) of the purest water at 25 0C is determined
experimentally by conductance measurement
• The molar conductance at infinite dilution of water is obtained by Kohlrausch’s law
2
𝑚(𝐻 𝑂)

𝐻+
𝑂𝐻−
=  +  ---------(5)

2 𝑚 𝐻2𝑂
• Assuming that 𝑚(𝐻 𝑂) differs very little from  and the concentration of H+

(or) OH- ion can be determined by substituting K and 𝑚 in equation (5)
• Then the ionic product of water is obtained by substituting the concentration of H+
and OH- ions in the equation (2)
Kw = c2

DETERMINA
TION OF SOLUBLITYAND SOLUBILITY PRODUCT OF
SPARINGLY SOLUBLE SALT
i.e., AgCl ⇌ Ag+
(aq) + Cl-
(aq)
• The solubility product of sparingly soluble salt (AgCl) is expressed as
Ksp = [Ag+][Cl-] ------------(1)
where, [Ag+] and [Cl-] are the concentration of Ag+ and Cl- ions in aqueous solution
• Let the solubility ofAgCl is x mol m-3
i.e., the concentration ofAgCl in the aqueous solution (c) = x mol m-3
• The conductance of theAgCl in the aqueous solution and water used in the preparation of
the solution are determined using conductivity meter
• When water is added to the sparingly soluble salt likeAgCl, very minute quantity
of salt will pass in solution and equilibrium with the unionizedAgCl salt
H2O

• The difference multiple with cell constant gives the specific conductance (K) of the
solution due to the dissolved salt (AgCl). Let the value be Z Sm-1
• The molar conductance of theAgCl solution is
m (AgCl) = K /C -----------(2)
and substitute ‘x’for the concentration and Z for
𝑚(𝐴𝑔𝐶𝑙)
• Assuming that m (AgCl) = 
specific conductance, equation (2) becomes
 = Z /x -----------(3)
x = Z (S m-1)/  (Sm2 mol-1)
• 𝑚(𝐴𝑔𝐶𝑙)
 is obtained from Kohlrausch’s law
𝑚(𝐴𝑔𝐶𝑙) 𝐴𝑔
+
 
= =  +  −

𝐶𝑙
DETERMINA

• Substituting the calculated values Z and in equation (3), the solubility ofAgCl ‘x’
molm-3 can be determined
• On dividing the solubility with 1000, gives the solubility (x) in terms of mol dm-3
i.e., concentration ofAgCl in aqueous solution in x mol dm-3
• On multiplying the solubility (x mol dm-3) with gram molecular weight ofAgCl
gives solubility ofAgCl in terms of g dm-3
i.e., Solubility ofAgCl = x mol dm-3 x 143.5 g mol-1
Solubility ofAgCl = y g dm-3
• The concentration ifAgCl in aqueous solution is x mol dm-3
i.e., [Ag+] = x mol dm-3
-----------(5)
[Cl-] = x mol dm-3
DETERMINA

• Solubility product (Ksp) is obtained by substituting equation (5) in equation (1)
Ksp = (x mol dm-3) (x mol dm-3)
Ksp = x2 mol2 dm-6
DETERMINA

DETERMINATION OF HYDROLYSIS CONSTANT OFA SALT
• The relationship between hydrolysis constant of the salt and degree of hydrolysis of
the salt is
Kh = cx2 ------------(1)
Where Kh is Hydrolysis constant of the salt
c is the concentration of the salt in moles per litre
x is the degree of hydrolysis
Example:
The equilibrium existed in the hydrolysis of NH4Cl is
NH4Cl + H2O ⇌ NH4OH + HCl
(1-x) moles x moles x moles
Where x is the degree of hydrolysis

• The electrical conductance of an aqueous solution of N4Cl salt is due partly to the
ions of the un hydrolysed salt and partly to the H+ and Cl- ions
i.e., The molar conductance of this solution m, as determined experimentally, will
be the sum of the conductance of (1-x) moles of N4Cl and x moles of HCl
+ x 0
m(HCl)
m = (1-x) ’m ---------(2)
where 0
m(HCl) is the molar conductance of HCl at infinite dilution
’m is the molar conductance of the unhydrolysed NH4Cl salt at the given
concentration
• The value of ’m is obtained by adding slightly excess of NH4OH to the salt solution. This
suppresses the hydrolysis of NH4Cl to such a large extent that the molar conductance of the
solution can be taken as ’m . i.e., the conductance of the unhydrolysed salt

• On rearranging the equation (2), we get
---------(3)
where
x is the degree of hydrolysis of NH4Cl
m is the molar conductance of the salt solution before adding excess of NH4OH
'm is the molar conductance of the salt solution after adding excess of NH4OH
are determined by conductance measurement
• m and 'm
• 0 is obtained by Kohlrausch’s law
m(HCl)
λ0
HCl = λ 0
H+ + λ 0
Cl –

• Substituting all the measured values in (3), we get the degree of hydrolysis (x) of NH4Cl salt
• Substituting the calculated degree of hydrolysis of NH4Cl salt (x) and the concentration of
NH4Cl salt solution in equation (1) , the hydrolysis constant (Kh) can be calculated
Kh = cx2

Electrochemistry

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