B.Sc II sem III Chemistry

1. .
SHRI R.R.LAHOTI SCIENCE COLLEGE,MORSHI
E- learning material
CLASS – BSc II sem III
CHEMISTRY
ELECTROCHEMISTRY
Mr G.D.Rawate
Department of chemistry

2. Electrochemistry is branch of chemistry which deals with production of electricity from energy
released during spontaneous chemical reaction and use of electric energy to bring about
nonspontaneous transformation.
Conductors
Substances that allow electric current to pass through them are known as conductors, Conductor
are two types
1) Metallic Conductors or Electronic Conductors
Substances which allow the electric current to pass through them by the movement of electrons are
called metallic conductors, e.g.. metals.
2)Electrolytic Conductors or Electrolytes
Substances which allow the passage of electricity through their fused state or aqueous solution and
undergo chemical decomposition are called electrolytic conductors, e.g., aqueous solution of acids.
bases and salt

3. Electrolytes are of two types:
Strong electrolytes:-
The electrolytes that completely dissociate or ionise into
ions are called strong electrolytes. e.g., HCl, NaOH,
K2SO4
Weak electrolytes :-
The electrolytes that dissociate partially (ex < 1) are
called weak electrolytes, e.g., CH3COOH, H2CO3,
NH4OH,H2S, etc

4. Conductance of electrolyte solution
The conductance of electrolyte (C) is reciprocal of its resistance
(R)
C = 1/R
It is measured in Ohm-1 OR mho OR Siemens
Resistance of any uniform conductor varies directly as its length
and inversely to area of its cross section.
R α l/a = ρ l/a
where ρ = specific resistace or resistivity
l = length of conductor
a = Cross sectional area of conductor

5. Specific conductance (K)
The conductance of one centimeter cube or one cubic meter solution of an electrolyte is
known as specific conductance, It is denoted by K
specific conductance is reciprocal of specific resistance
K = 1/ ρ
we know R = ρ x l/a
substituting R = 1/C , ρ = 1/K
1/C = 1/Kxl/a
K = C. l/a
When l = 1m & a = m3
K = C
Thus for electrolyte specific conductance is the
conductance of 1 m3 or 1cm3 of solution.
l = length of conductor , a = cross sectional area of conductor
In electrolytes conductance of one meter cube of solution
In CGS unit is ohm-1cm-1 or Scm-1
SI unit is ohm-1 m-1 or Sm-1

6. Molar conductance
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 related to specific conductance as Λm = (k x 1000/M)
where. M = molarity.
It units are Ω-1 cm2 mol-1 or S cm2 mol-1.
Equivalent conductivity (Λm)
The conducting power of all the ions produced when 1 g-equivalent of an
electrolyte is dissolved in V mL of solution, is called equivalent conductivity. It is
related to specific conductance as
Λm = (k x 1000/N) where. N = normality.
Its units are ohm-1 cm2 (equiv-1) or mho cm2 (equiv-1) or S cm2 (g-equiv-1).

7. Variation of specific and equivalent
conductance with dilution
The solution of electrolyte conducts electricity due to the
presence of ions.
According to ionic theory the number of ions increases as
the solution of electrolyte progressively diluted. Therefore ,
equivalent conductivity increases on dilution.
Specific conductivity decreases on dilution because on
dilution the number of ions per dm3 of solution decreases in
spite of increase in dissociation.
 The equivalent and molar conductance increases because
these are the product of conductivity and volume containing

9. Determination of Conductance of electrolyte solution
Conductance of solution is reciprocal of resistance of solution.
Measurement of conductance done indirectly by determing the
resistance of the solution.
 Resistance is measured by Wheatstone AC bridge method.
DC current is not used in this process ,this will give wrong result.
1)Change in concentration due to electrolysis.
2) Change in resistance due to polarization at electrode.
 Difficulties is removed by using AC current within audio frequency
range
and galvanometer is replaced by headphone.

10. Determination of
Conductance of
electrolyte solution
 The solution whose conductance is determined is
taken suitable conductivity cell C .
 When current is flowing known resistance R is
introduced through resistance box.
 The sliding contact is moved along the wire AB of
uniform thickness.
 Untill a point X of minimum sound is detected in
headphone T.
Resistance of solution Length BX
Resistance from resistance box Length AX
Thus resistance of solution is determined , from
resistance of solution, conductance can be determined.

11. Determination of cell constant
Defination of cell constant
Defined as the ratio of length between the electrode l and area of cross
section a of electrode.
Cell constant = l/a
In order to determine cell constant, it is necessary to determine l and a of
electrode, but actually it is not possible.
Indirect method based on measurement of conductance of standard
KCL solution is employed
Specific conductance = Observed conductance x Cell constant

12. Conductometric titration
The principle of conductometric titration is based
on the fact that during the titration, one of the ions
is replaced by the other and invariably these two
ions differ in the ionic conductivity with the result
that conductivity of the solution varies during the
course of titration. The equivalence point may be
located graphically by plotting the change in
conductance as a function of the volume of titrant
added.

13. Some example of conductometric
titration
Acid with a Strong Base, e.g. HCl
with NaOH:
Before NaOH is added, the
conductance is high due to the
presence of highly mobile hydrogen
ions. When the base is added, the
conductance falls due to the
replacement of
hydrogen ions by the added cation as
H+ ions react with OH- ions to form
undissociated water. This decrease in the
conductance continues till the equivalence
point. At the equivalence point, the solution
contains only NaCl. After the equivalence
point, the conductance increases due to the
-
Titration of Strong acid Vs Strong
base

14. Weak acid against Strong base
Acid with a Strong Base, e.g.
acetic acid with NaOH:
Initially the conductance is low due to the feeble
ionization of acetic acid. On the addition of base,
there is decrease in conductance not only due to
the replacement of H+ by Na+ but also suppresses
the dissociation of acetic acid due to common ion
acetate. But very soon, the conductance increases
on adding NaOH as NaOH neutralizes the un-
dissociated CH3COOH to CH3COONa which is
the strong electrolyte. This increase in
conductance continues raise up to the equivalence
point. The graph near the equivalence point is
curved due the hydrolysis of salt CH3COONa.
Beyond the equivalence point, conductance
increases more rapidly with the addition of NaOH

15. Strong Acid against weak Base
Acid with a Weak Base, e.g.
sulphuric acid with dilute ammonia:
Initially the conductance is high and then
it decreases due to the replacement of
H+. But after the endpoint has been
reached the graph becomes almost
horizontal, since the excess aqueous
ammonia is not appreciably ionised in
the presence of ammonium sulphate
(Fig.).

16. Weak Acid against Weak Base
Acid with a Weak Base:
CH3COOH vs NH4OH
The nature of curve before the
equivalence point is similar to the
curve obtained by titrating weak
acid against strong base. After
the equivalence point,
conductance virtually remains
same as the weak base which is

17. Precipitation Titration
Precipitation Titration
Ex-AgNO3 Vs KCl
The titration of AgNO3 against
KCl involve precipitate formation.
Since mobility of Ag+ and K+ ions
are nearly same, the conductance
remain almost constant till the
equivalence point. After
equivalence point, the added KCl,
increases the conductance rapidly.

18. Advantages of Conductometric titration
 Small quantity of solution is required for titration.
As end point determined graphically , no special precaution is
necessary.
 Indicator is not required so used in colored and turbid solution.
 Conductometric titration are used for analysis of dilute solution
as well for weak acids.
Conductometric titration can be applied to mixture of
acids,precipitation and other type of titration.
 Conductometric titration gives more accurate result.

19. Migration of ions under the influence of
electric field
On passing electric current through electrolyte solution,
ions migrate and discharged oppositely charged electrodes
the migration of ions can be demonstrated by simple
experiment.

20. .The lower portion of U tube is filled with 5% agar agar solution in
water with small quantity of CuCr2O7( Obtained by mixing
equimolar quantities of K2Cr2O7 +CuSO4).It is allowed to set by
cooling as dark green jelly
 Some charcoal powder is sprinkled in both limbs.
Solution of KNO3 and agar agar is placed in each limb and allowed
set as jelly.
Solution of KNO3 in water is filled in each limb and platinum
electrode are placed as shown in figure.
When electric current is passed Cu 2+ ions migrate towards
cathode (-ve electrode) due to this blue colour appear in cathode
side.
Yellow colour in anode side by Cr2O72- ions

21. Hittorfs theoretical device
Hittorfs method

22. Hittorfs theoretical device
Although most of ions differ in their mobilities, the total number of ions
discharged at electrodes on electrolysis is same. This can be explained by
Hittorfs theoretical device as shown in figure. It consists of an electrolytic cell
containing same number of positive and negative ions with same valency. The
electrolytic cell is divided into three compartments by porous participant B& C.
Metal electrodes A and D represent cathode and anode respectively
I. Represents initial state of electrolyte solution before electrolysis in which
equal number of positive and negative ions are present. on passing electric
current , on passing electric current, three cases may arise
II. Suppose, only anions are migrating and cations remains stationary.
If speed of anions ( v=2 )and speed of cation (u=0).Then two anions migrate
from cathode to anode compartment and are discharged at anode. The unpaired
cations in cathode compartments is two.
III. Suppose, both ions are moving with same speed.
Let u=v =2,then the number of ions discharged at respective electrode is four

23. Suppose ,both ions are moving with same
speeds(u=1,v=2)In this case, total number of ions are
migrating with different speeds(u=1,v=2).In this
case, total number of ions discharged at respective
electrode is three
Following Conclusions can be drawn about the process of electrolysis.
1) During electrolysis, ions are discharged in equivalent amounts,
irrespective of their speeds of migration.
2) Concentration of electrolyte around electrode changes as a result of
migration of ion which moves away from that electrode.
Fall in concentration around anode speed of Cation
Fall in concentration around cathode speed of anion

24. u/r =r (Speed ratio)
Transference number or transport number or Hittorffs number of ions
During electrolysis current is carried by cations and anions The fraction of total
current carried by ionic species is called its transference number or transport
number The transport number of ions ions is proportional to their absolute
velocities
The transport number of ions is proportional to their absolute velocities.
Transport no.of cation (t+) and anion (t-) is given as
t+ = u/u+v t- = v/u+v
u= speed of cation v= speed of anion
Sum transport no.of two ions
t+ + t- =1
If the speed ratio is r =u/v =t+/t-

25. Hittorfs method

26. Hittorfs method
This method is based on the fact that change in concentration
around the electrodes is due to migration of ions.
In the apparatus it consists of two vertical glass tubes connected
through U tube ,all three tubes are provided withstop cock at
bottom.
 Cathode is a piece of freshly silvered silver foil.A silver anode is
connected to copper or silver voltmeter in series.
The apparatus is connected to copper or silver voltmeter in
series.The app is filled with standard solution of silver nitrate.
 A steady current of 0.01 ampere is passed for nearly 2 to 3 hrs at
the end of this period,it is then titrated potassium thiocynate
solution to determine amount of silver present in it.

27. Precaution
1) Steady current should passed
2) There should be no change in concentration
Observation and Calculation
Since Change in concentration accompanied by change
in volume, loss of material should be determined with
reference to definite weight of solvent present after
current has passed. Two different case may arise
Case I –When electrode are non attackable

28. .
After passing electric current
Let the weight of anodic solution taken out = a gm
Weight of AgNO3 present in it by titration = b gm
Weight of water = (a-b) gm
BEFORE PASSING ELECTRIC CURRENT:
Let the weight of AgNO3 in (a-b) gm of water before passing electric
current = c gm
Fall in concentration = (c-b) gm of Ag NO3
= {c-b/170} gm equi of AgNO3
= d gm
Let the weight of silver deposited in silver coulometer = w1 gm
= w1/108 gm equi ofAg
= W gm equi of Ag

29. .
Fall in concentration around anode
Transport number of Ag+, t+ =
Amount of Ag deposited in gm
equivalent
= d/ W
Transport no. of NO3- , t- = 1- d/W
= W-d/W

30. When electrode are attackable(Ag Electrode)
In this case b › c because NO3- ion react with Ag anode to produce
AgNO3.Thus the concentration of AgNO3 or Ag is increased in anode
compartment.
Increase in concentration of anodic solution = (b-c) gm of AgNO3
={b-c/170} gm equivalent of
AgNO3
= e
If Ag+ ion migrated from the anode, the increase in concentration
ofAg+ ions would have been equal to W
Fall in conc due toi migration of Ag+ = (W-e)
Hence transport no. of Ag+, t+ = W-e/W
And transport no.of NO3-,t- = 1-{W-e/W}

31. Moving Boundary method
 Hittorfs method does not give accurate result for dilute
solution.
 Small changes in concentration due to passage of electric
current may lead to experimental error.
Direct observation method is called Moving boundary method for
determination of transport number.
Three condition of determination of transport number
1. Cation of indicator electrolyte shoud not move faster than
cation whose transport no.is to be determined.
2. Both have same anion
3. Indicator electrolyte should have more density.

32. Moving Boundary method

33. .
 It consist of electrolytic cell with vertical tube of uniform cross section with two
electrode at two ends, at lower side anode is made of cadmium rod and cathode
is platinum foil.
 If we have to determine transport no.of cation H+ in HCl ,we have choose
indicator electrolyte
1 .Having common anion Cl-
2. The cation of indicator must be slow moving as compared to cation whose
transport no.is to be determined( Speed of Cd 2+ is less than H+)
3. The indicator electrolyte placed in lower half overthis solution of HCl,
produces sharp boundary B1 between two solution.
4. Constant current is passed to this apparatus for 5-6 Hrs ,the H+ ion move
towards cathode followed by Cd2+ ions, the boundary gradually move upword
upto B2 ; through a distance l meter.
Hence quantity of current carried by H+ ions = t+ Q
Amount of H+ ios migrated from B1 to B2 = t+.Q/F kg equivalent ------1
If X square meter is cross sectional area of tube,then volume between B1 to B2
= X.l dm3

34. .
If c is the concentration of H+ ions in Kg equivalent per liter,The
amount of H+ in given volume.
= x.l.c/1000 kg equivalent ----------2
From eq (1) and (2) ,
t+.Q/F = x.l.c/1000
OR t+ = x.l.c.F/1000
n = Q/F
t+ = x.l.c/1000.n
and t- = 1- t+
Where
Q = Quantity of electric current passed = ampere x
second
n = Number of faradays of electric current passed
t+ = transport number of H+.

35. Application of Conductivity Measurement
1.Determination of λꝏ for weak electrolyte
Weak electrolyte do not ionize to a sufficient extent in solution
and not being completely ionized even at very great dilution.The
practical determination of equivalent conductance at infinite
dilution λꝏ in such cases is therefore, not possible. However, It
can be calculated with the help of Kohlrauschs law.
Example ,The equivalent conductance at infinite dilution of
CH3COOH (weak electrolyte) can be obtained from the
equivalent conductance at infinite dilution of HCl,CH3COONa
and NaCl (all of which are strong electrolytes)

36. .

37. .

38. 2.Determination of Degree of Dissociation(α) of weak
electrolyte
Degree of dissociation (α ) of weak electrolyte at any
dilution can be calculated by the relationship
α = λc/ λꝏ
Where λc = equivalent conductance at given
concentration
λꝏ = equivalent conductance at infinite
dilution

39. 3. Determination of Dissociation Constant (Ka)
weak electrolyte
The dissociation constant (Ka) is defined as equilibrium constant
for dissociation of an electrolyte obtained by applying law of
mass action at given temperature.
HA = H+ + A-
Applying law of mass action
Ka = [H+] [A-]/[HA]
For strong electrolyte Ka has higher value i.e. higher value of Ka
indicates greater degree of dissociation (α) can be given as

40. .
HA = H+ + A-
c α α
(1- α )C αC α C
Ka = [H+] [A-]/[HA]
Ka = α2C/(1- α) ------(1)
But in case of weak electrolyte ka has lower value i.e the degree of dissociation(α)
is negligible hence dissociation constant weak electrolyte can be given as
Ka = α2C -------------(2)
From the value of equivalent conductance at given concentration(λc) and
equivalent conductance at infinite dilution (λꝏ) degree of dissociation can be
determined.
α = λc/λꝏ
Hence dissociation constant of weak electrolyte can be determined.