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CMT552
ELECTROCHEMISTRY AND
CORROSION SCIENCE
ELECTROLYTE
CONDUCTANCE
What is electrolyte?
 Any substance that produce ions when dissolved
in a solvent (usually water) is an electrolyte.
 It is the electrically conductive solution that
must be present for corrosion to occur.
Types of electrolytes
Strong electrolyte
Weak electrolyte
Non-electrolyte
Strong Electrolytes
 Strong electrolytes are substances that only
exist as ions in solution.
 They completely dissociate to their ions when
dissolved in solution.
 Ionic compounds are typically strong
electrolytes.
 Strong acids, strong bases and salts are strong
electrolytes.
 They conduct electricity when molten or in
aqueous solution.
 Example: Hydrochloric acid, Sodium chloride
HCl  H 2O  H 3O   Cl 

NaCl  H 2O  Na  Cl

Weak Electrolytes
 A weak electrolyte only partially dissociates in
solution and produces relatively few ions (exist
in water as a mixture of individual ions and
incontact molecules).
 Polar covalent compounds are typically weak
electrolytes.
 Weak acids and weak bases are weak
electrolytes.
 They conduct electricity weakly.
 Example: Acetic acid, ammonia

CH 3COOH  H 2O  CH 3COO  H

NH3  H 2O  NH4  OH


Non-electrolytes
 A non-electrolyte does not dissociate at all
(present entirely as intact molecules) in
solution and therefore does not produce any
ions.
 Non-electrolytes are typically polar covalent
substances that do dissolve in water as
molecules instead of ions.
 They do not conduct electricity at all.
 Example: Sugar
C12 H 22O11  H 2O  C12 H 22O11
Acids
 Are molecular compounds which ionize or turn
into ions in water.
 The properties of acids were due to the
presence of hydrogen ions, H+.
 All acids are soluble in water
 Some acids are strong electrolytes and some
are weak electrolytes.
 No acids are non-electrolytes.
Bases
 Can be molecular compounds or ionic
compounds.
 Some bases are soluble and some are not.
 The soluble bases ionize or dissociate into ions
in water.
 The properties of bases were due to the
presence of hydroxide ions, OH-.
 All of the ionic bases which are soluble are also
strong electrolytes.
Salts
 Are ionic compounds which are not acids or
bases.
 In other words, the cation is not hydrogen and
the anion is not hydroxide.
 Some salts are soluble in water and some are
not.
 All of the salts which are soluble are also
strong electrolytes.
Electricals Terms
SI Term
SI Symbol
SI Unit
Electrical Current
I
Ampere (A)
Quantity of Electricity
Q
Coulomb (C)
Electric Potential
V
Volt(V)
Electric Resistance
R
Resistivity

Ohm()
m
Conductance
G
Conductivity

Siemens (S);ohm-1
Sm-1; -1m-1; -1cm-1
Molar conductivity

Sm2mol-1
Molar Conductivity of Ion

Sm2mol-1
Electric Mobility of Ion
u
m2V-1s-1
Transport Number of Ion
t
 Other Symbols and Terms
Symbol
Term
C
Molar Concentration, mol dm-3 (with :mol m-3)

Degree of dissociation
l
Length
A
Area
Kcell
Cell Constant
°
Molar conductivity at infinite dilution or Limiting
Molar Conductivity
Electrolytic conductance
 Electrolytic conductance occurs when a voltage is
applied to the electrode dipped into an
electrolyte solution, ions of the electrolyte move
and electric current flows through the electrolytic
solution.
 This power of the electrolyte to conduct electricity
is known as conductance or conductivity.
 Electrolytic solution also obey Ohm’s Law just like
metallic conductor.
 Ohm’s Law: It states that the current flowing
through a conductor is directly proportional to
the potential difference across it:
V=IR
where,
V = applied potential (V)
I = current measured (A)
R = solution resistance () between the two
electrodes
Solution Resistance (R)
 The increase the [ions] presence in the
solution, the lower the solution resistance, R,
will be.
 A strong electrolyte like KCl is dissolve in
water, the no. of ions per unit volume increase
and the solution resistance, R, is lowered,
thus increasing the current measured for a
particular applied potential.
 Thus, current can be related to the [ions] in a
particular solution.
 However,
 the distance between the electrodes,
 the surface area of the electrodes and
 the identity of the ions
also affect the solution resistance, R.
Solution Conductance (G)
 The reciprocal of solution resistance (1/R) is
called Conductance, G.
 Conductance is expressed as Siemens (S) or
ohm-1 (-1) or mho.
1
A
G
R

l
 Where, A = surface area of each electrode
l = distance btwn electrode
 = conductivity
Values of conductivity,, increased with T and
concentration.
The conductivity of a solution of water is highly
dependent on its concentration of dissolved
salts and sometimes other chemical species
which tend to ionize in the solution.
 Electrical conductivity of water samples is
used as an indicator of how salt free or
impurity free the sample is; the purer the
water, the lower the conductivity.
Solution
Electric Conductivity
(Sm-1)
Seawater
5
Drinking water
0.0005 to 0.05
Deionized water
5.5 x 10-6
Molar Conductivity ()
 Defined by:
 

C
Units: Sm-1
mol m-3
Example 1:
Molar conductivity of 0.005 M KCl is 144 Scm2
mol-1. Calculate its electrolytic conductivity in SI
units (Sm-1).
*(Hint: 1m2 = 104 cm2; mol/L or mol/dm3 convert to
mol m-3).
2
1
m
2
1
144 Scm 2 mol 1  4

0.144
Sm
mol
10 cm 2
0.005 mol
1 dm 3
3


5
mol
m
dm 3
(110 1 m) 3
  c
 0.0144  5
 0.072 Sm -1
Measurement of Conductivity
 The conductivity of a solution is measured in a
cell called conductance cell or conductivity cell.
1 l 
   
R  A
 Since l and A are difficult to measure, the usual
procedure is to treat l as a cell constant, K cell
A
 Therefore,
1
  K cell   GKcell
R
Example 2
In a certain conductivity cell, the resistance of
a 0.01 M KCl solution is 150 . The known
molar conductivity of the solution is 141.27 -1
cm2 mol-1. Calculate the cell constant (Kcell).
*(Kcell unit is cm-1)
kcell  R
 cR
 141.27  0.01150 10
 0.2119cm
1
3
Exercise 1
Using the same conductance cell as in
example 2, a student measured the resistance
of a 0.10 M NaCl solution to be 19.9 .
Calculate the experimental value of the molar
conductivity of this solution.
 Use the same value of kcell
kcell  R
 cR
0.2119    0.119.9
  0.106485 10
3
 106.48 Scm mol
2
1
Ex
 In order to determine the molar conductivity of a
0.05 M solution of AgNO3, you need to measure
the solution resistance in a conductivity cell and
found that R = 75.8 . Then, in the same cell, a
0.02 M KCl solution had a resistance of 157.9 .
Given that the accepted molar conductivity of the
KCl solution is 0.013834 -1 m2 mol-1, calculate
the molar conductivity of the AgNO3 solution.
KCl :
k cell  cR
 0.013834  0.02  157.9  10-3
 43.7 m -1
AgNO 3 : k cell  ΛcR
43.7 m -1  Λ  0.05  75.8
Λ  11.53  10-3 Sm 2 mol 1
 0.01153 Sm 2 mol 1
Variation of Molar Conductivity with
Concentration
 Molar conductivity () of electrolytes increases
with dilution.
 The variation is different for strong and weak
electrolytes.
Strong electrolytes
Fully ionized in solution
  increases slowly with dilution and there is a
tendency for  to approach a certain limiting value
when the concentration approaches zero(i.e. When
dilution is infinite).

 The  when the concentration approaches zero
(infinite dilution) is called molar conductivity at
finite dilution or limiting molar conductivity
(°).
 = °
when C → 0 (at infinite dilution)
 For strong electrolytes molar conductivity
increase slowly with dilution and can be
represented by:
   C

DEBYE HUCKEL ONSAGER
equation
   C

 = Molar conductivity at a given concentration
° = Molar conductivity at infinite dilution
 = constant
 From the graph, it has been noted that the
variation of molar conductivity () with
concentration (C) is small so that the plot can
be extrapolated to zero concentration.
 The intercept is equal to (°) and the slope is .
b) Weak electrolytes
 Not fully ionized in solution
 In weak electrolyte like acetic acid they have
low degree of dissociation as compared to
strong electrolyte.
 However, the variation of molar conductivity ()
with concentration (C) is very large and we
can’t obtain molar conductivity at infinite
dilution (°) by extrapolation of  versus C
plots.
Explanation for the variation of Molar
Conductivity with concentration
1. Conductance behaviour of strong electrolyte:
 No increase in the no. of the ions with the
dilution ( completely ionized in the solution at
all concentration).
 In concentrated solution:
 strong inter-ionic forces
 Molar conductivity is low
 In dilute solution:
Inter-ionic forces low
Molar conductivity increases with dilution
When concentration very low, inter-ionic
interaction becomes almost negligible and
molar conductance approaches the limiting
value, °.
2. Conductance behaviour of weak electrolyte:
 The no. of ions produced in solution depends
upon the degree of dissociation with dilution.
 Higher the degree of dissociation, larger is the
molar conductance.
 With increase in dilution
 Degree of dissociation increases as a result
molar conductivity increases.
 At infinite dilution, the electrolyte is
completely dissociated so that the degree
of dissociation become one.
 = °
(at C → 0)
Thus, if
 = Molar conductivity at a given concentration
° = Limiting molar conductivity or molar
conductivity at infinite dilution
Then, degree of dissociation

 

Ostwald Dilution Law & Dissociation
Constant of Weak Electrolyte
 Consider an aqueous solution of a weak binary
electrolyte, AB, of concentration C mol dm-3 and
degree of dissociation of .
Initial/mol dm-3
Equilibrium/mol dm-3
 At equilibrium:
AB (aq) ↔ A+ (aq) + B- (aq)
C
0
0
C(1-)
C
C
[ A ][ B  ]
K
[ AB]
(C )(C )
Kc 
C (1   )
Therefore, the dissociation constant can be
expressed as:
2
C
Kc 
1
Ostwald Dilution Law
 However, for weak electrolyte;  is very small.
 Hence, (1- )  1
 Therefore,
K a  C

2
Ka
C
 Since H+ = C

[H+
Ka
] = C = C
C
[ H  ]  K a C
KOHLRAUSCH’S LAW
 At infinite dilution the ions act completely
independently, and the ° obeys a rule of
additivity:
  AX     AY    BX    BY 




where AX, AY, BX and BY are strong electrolytes.
 ° for a weak electrolyte can be deduced from
° values obtained from strong electrolytes.
 For example, consider CH3COOH denoted as
HAc’:
 HAc    HX    MAc    MX 




where HX, Mac and MX are strong electrolytes.
Table 1: Limiting Molar Conductivity,
°, of some strong electrolytes
Electrolyte
HCl
° (S cm2 mol-1)
426.16
HBr
NaCl
KBr
428.10
126.45
151.80
KCl
NaNO3
KNO3
149.86
121.55
144.96
NH4Cl
KHCO3
149.70
118.00
Exercise 2
Calculate ° for a weak electrolyte NH4OH
from the ° values for these strong
electrolytes: NH4Cl: 149.7; NaCl: 126.5 and
NaOH: 248.10
( NH 4OH )  ( NH 4Cl )  ( NaOH )  ( NaCl)
 149.7  248.10  126.5
 271.3
 Kohlrausch also stated at infinite dilution when
the dissociation complete,
 each ion makes a definite contribution towards
molar conductance of the electrolyte irrespective
of the nature of the other ion with which it is
associated.
 It means that the molar conductivity at infinite
dilution for a given salt can be expressed as the sum of
the individual contributions from the ions of
the electrolyte.
  v    v




where
v+ and v-: stoichiometric coefficients for the
cation and anion in the electrolyte.
°+ and °-: ionic conductance of individual
ions (cation and anion)
Example 3
For NH4OH electrolyte: v+ = 1 and v- = 1
Since 1NH4+ ion present for each OH- ion present
in solution.
Example 4
For K4Fe(CN)6 electrolyte: v+ = 4 and v- = 1
Since there are 4K+ ions present for each
Fe(CN)4-6 ion present in solution.
 Thus, the limiting ionic conductivities represent
the contributions to the total solution
conductivity made per mole of each ion present
in a dilute solution.
Exercise 3
Calculate the ° of the following electrolytes:
1) Acetic acid
2) Hydrochloric acid
3) Potassium Chloride
Ionic Conductivities at Infinite
Dilution at 25°C
Cation
°+ / Scm2mol-1
Anion
°- / Scm2mol-1
H+
349.6
OH-
197.8
Li+
38.7
Cl-
76.4
Na+
50.1
Br-
78.2
K+
73.5
I-
76.8
Fe2+
108.0
CH3COO-
40.9
Fe3+
204
CO2-3
138.6
NH4+
73.4
NO-3
71.5
Ba2+
127.3
SO2-4
160.0
 (CH 3COOH )  v    v 


 1(349.6)  1(40.9)
 390.5 Scm mol
2
1


Example 5
Molar conductivity for 0.10 M NaCl is 107 Scm2
mol-1. Calculate the degree of dissociation for the
Solution.
1) Calculate the limiting molar conductivity for
NaCl

2) Use formula
 

( NaCl)       
 1(50.1)  1(76.4)
 126.5



107

126.5
 0.846
Exercise 4
At 25° C,  = 3.40  10-3 Sm-1 for 0.001 M
NH4OH. Values of ° are NH4Cl = 0.01497,
NaOH = 0.02481, NaCl = 0.01265 Sm2mol-1.
Calculate the dissociation constant, K, of
ammonium hydroxide.
( NH 4OH )  ( NH 4Cl )  ( NaOH )  ( NaCl )
 0.01497  0.02481  0.01265
 0.02713


c
3.40 10 3

10 3  3.40 10 3
0.001
 3.40 10 3


 0.1253

0.02713