Electrochemistry is the study of production of
electricity from energy released during spontaneous
chemical reactions and the use of electrical energy
to bring about non-spontaneous chemical
transformations. The subject is of importance both
for theoretical and practical considerations. A large
number of metals, sodium hydroxide, chlorine,
fluorine and many other chemicals are produced by
electrochemical methods. Batteries and fuel cells
convert chemical energy into electrical energy and are
used on a large scale in various instruments and
devices. The reactions carried out electrochemically
can be energy efficient and less polluting. Therefore,
study of electrochemistry is important for creating new
technologies that are ecofriendly. The transmission of
sensory signals through cells to brain and vice versa
and communication between the cells are known to
have electrochemical origin. Electrochemistry, is
therefore, a very vast and interdisciplinary subject. In
this Unit, we will cover only some of its important
elementary aspects.
After studying this Unit, you will be
able to
• describe an electrochemical cell
and differentiate between galvanic
and electrolytic cells;
• apply Nernst equation for
calculating the emf of galvanic cell
and define standard potential of
the cell;
• derive relation between standard
potential of the cell, Gibbs energy
of cell reaction and its equilibrium
constant;
• define resistivity (ρ), conductivity
(κ) and molar conductivity (
L
m
) of
ionic solutions;
• differentiate between ionic
(electrolytic) and electronic
conductivity;
• describe the method for
measurement of conductivity of
electrolytic solutions and
calculation of their molar
conductivity;
• justify the variation of
conductivity and molar
conductivity of solutions with
change in their concentration and
define
°
m
Λ
(molar conductivity at
zero concentration or infinite
dilution);
• enunciate Kohlrausch law and
learn its applications;
• understand quantitative aspects
of electrolysis;
• describe the construction of some
primary and secondary batteries
and fuel cells;
• explain corrosion as an
electrochemical process.
Objectives
Chemical reactions can be used to produce electrical energy,
conversely, electrical energy can be used to carry out chemical
reactions that do not proceed spontaneously.
3
Electrochemistry
Unit
Unit
Unit
Unit
Unit
3
Electrochemistry
2020-21
66Chemistry
Cu
E
ext
>1.1
e
–
Current
Cathode
+ve
Anode
–ve
Zn
Fig. 3.2
Functioning of Daniell
cell when external
voltage E
ext
opposing the
cell potential is applied.
In Class XI, Unit 8, we had studied the construction and functioning
of Daniell cell (Fig. 3.1). This cell converts the chemical energy liberated
during the redox reaction
Zn(s) + Cu
2+
(aq) → Zn
2+
(aq) + Cu(s) (3.1)
to electrical energy and has an electrical
potential equal to 1.1 V when concentration
of Zn
2+
and Cu
2+
ions is unity (1 mol dm
–3
)
*
.
Such a device is called a galvanic or a
voltaic cell.
If an external opposite potential is applied
in the galvanic cell [Fig. 3.2(a)] and increased
slowly, we find that the reaction continues to
take place till the opposing voltage reaches
the value 1.1 V [Fig. 3.2(b)] when, the reaction
stops altogether and no current flows through
the cell. Any further increase in the external
potential again starts the reaction but in the
opposite direction [Fig. 3.2(c)]. It now functions
as an electrolytic cell, a device for using
electrical energy to carry non-spontaneous
chemical reactions. Both types of cells are
quite important and we shall study some of
their salient features in the following pages.
*Strictly speaking activity should be used instead of concentration. It is directly proportional to concentration. In dilute
solutions, it is equal to concentration. You will study more about it in higher classes.
3.13.1
3.13.1
3.1
ElectrochemicalElectrochemical
ElectrochemicalElectrochemical
Electrochemical
CellsCells
CellsCells
Cells
Fig. 3.1: Daniell cell having electrodes of zinc and
copper dipping in the solutions of their
respective salts.
salt
bridge
Zn
Cu
anode
cathode
current
ZnSO
4
CuSO
4
E
<
ext
1.1
V
e
-ve +ve
I=0
Zn Cu
ZnSO
4
CuSO
4
E =
ext
1.1
V
When E
ext
< 1.1 V
(i) Electrons flow from Zn rod to
Cu rod hence current flows
from Cu to Zn.
(ii) Zn dissolves at anode and
copper deposits at cathode.
When E
ext
= 1.1 V
(i) No flow of
electrons or
current.
(ii) No chemical
reaction.
When E
ext
> 1.1 V
(i) Electrons flow
from Cu to Zn
and current flows
from Zn to Cu.
(ii) Zinc is deposited
at the zinc
electrode and
copper dissolves at
copper electrode.
(a)
(b)
(c)
2020-21
67
Electrochemistry
As mentioned earlier (Class XI, Unit 8) a galvanic cell is an
electrochemical cell that converts the chemical energy of a spontaneous
redox reaction into electrical energy. In this device the Gibbs energy of
the spontaneous redox reaction is converted into electrical work which
may be used for running a motor or other electrical gadgets like heater,
fan, geyser, etc.
Daniell cell discussed earlier is one such cell in which the following
redox reaction occurs.
Zn(s) + Cu
2+
(aq) → Zn
2+
(aq) + Cu(s)
This reaction is a combination of two half reactions whose addition
gives the overall cell reaction:
(i) Cu
2+
+ 2e
–
→ Cu(s) (reduction half reaction) (3.2)
(ii) Zn(s) → Zn
2+
+ 2e
–
(oxidation half reaction) (3.3)
These reactions occur in two different portions of the Daniell cell.
The reduction half reaction occurs on the copper electrode while the
oxidation half reaction occurs on the zinc electrode. These two portions
of the cell are also called half-cells or redox couples. The copper
electrode may be called the reduction half cell and the zinc electrode,
the oxidation half-cell.
We can construct innumerable number of galvanic cells on the pattern
of Daniell cell by taking combinations of different half-cells. Each half-
cell consists of a metallic electrode dipped into an electrolyte. The two
half-cells are connected by a metallic wire through a voltmeter and a
switch externally. The electrolytes of the two half-cells are connected
internally through a salt bridge as shown in Fig. 3.1. Sometimes, both
the electrodes dip in the same electrolyte solution and in such cases we
do not require a salt bridge.
At each electrode-electrolyte interface there is a tendency of metal
ions from the solution to deposit on the metal electrode trying to make
it positively charged. At the same time, metal atoms of the electrode
have a tendency to go into the solution as ions and leave behind the
electrons at the electrode trying to make it negatively charged. At
equilibrium, there is a separation of charges and depending on the
tendencies of the two opposing reactions, the electrode may be positively
or negatively charged with respect to the solution. A potential difference
develops between the electrode and the electrolyte which is called
electrode potential. When the concentrations of all the species involved
in a half-cell is unity then the electrode potential is known as standard
electrode potential. According to IUPAC convention, standard
reduction potentials are now called standard electrode potentials. In a
galvanic cell, the half-cell in which oxidation takes place is called anode
and it has a negative potential with respect to the solution. The other
half-cell in which reduction takes place is called cathode and it has a
positive potential with respect to the solution. Thus, there exists a
potential difference between the two electrodes and as soon as the
switch is in the on position the electrons flow from negative electrode
to positive electrode. The direction of current flow is opposite to that of
electron flow.
3.2 Galvanic Cells3.2 Galvanic Cells
3.2 Galvanic Cells3.2 Galvanic Cells
3.2 Galvanic Cells
2020-21
68Chemistry
The potential difference between the two electrodes of a galvanic
cell is called the cell potential and is measured in volts. The cell
potential is the difference between the electrode potentials (reduction
potentials) of the cathode and anode. It is called the cell electromotive
force (emf) of the cell when no current is drawn through the cell. It
is now an accepted convention that we keep the anode on the left and
the cathode on the right while representing the galvanic cell. A galvanic
cell is generally represented by putting a vertical line between metal
and electrolyte solution and putting a double vertical line between
the two electrolytes connected by a salt bridge. Under this convention
the emf of the cell is positive and is given by the potential of the half-
cell on the right hand side minus the potential of the half-cell on the
left hand side i.e.,
E
cell
= E
right
– E
left
This is illustrated by the following example:
Cell reaction:
Cu(s) + 2Ag
+
(aq) → Cu
2+
(aq) + 2 Ag(s) (3.4)
Half-cell reactions:
Cathode (reduction): 2Ag
+
(aq)
+ 2e
–
→ 2Ag(s) (3.5)
Anode (oxidation): Cu(s) → Cu
2+
(aq) + 2e
–
(3.6)
It can be seen that the sum of (3.5) and (3.6) leads to overall reaction
(3.4) in the cell and that silver electrode acts as a cathode and copper
electrode acts as an anode. The cell can be represented as:
Cu(s)|Cu
2+
(aq)||Ag
+
(aq)|Ag(s)
and we have E
cell
= E
right
– E
left
= E
Ag
+
Ag
– E
Cu
2+
Cu
(3.7)
The potential of individual half-cell cannot be measured. We can
measure only the difference between the two half-cell potentials that
gives the emf of the cell. If we arbitrarily choose the potential of one
electrode (half-cell) then that of the other can be determined with respect
to this. According to convention, a half-cell
called standard hydrogen electrode (Fig.3.3)
represented by Pt(s) H
2
(g) H
+
(aq), is assigned
a zero potential at all temperatures
corresponding to the reaction
H
+
(aq) + e
–
→
1
2
H
2
(g)
The standard hydrogen electrode consists
of a platinum electrode coated with platinum
black. The electrode is dipped in an acidic
solution and pure hydrogen gas is bubbled
through it. The concentration of both the
reduced and oxidised forms of hydrogen is
maintained at unity (Fig. 3.3). This implies
that the pressure of hydrogen gas is one bar
and the concentration of hydrogen ion in the
solution is one molar.
3.2.1
Measurement
of Electrode
Potential
Fig. 3.3: Standard Hydrogen Electrode (SHE).
2020-21
69
Electrochemistry
At 298 K the emf of the cell, standard hydrogen electrode second
half-cell constructed by taking standard hydrogen electrode as anode
(reference half-cell) and the other half-cell as cathode, gives the reduction
potential of the other half-cell. If the concentrations of the oxidised
and the reduced forms of the species in the right hand half-cell are
unity, then the cell potential is equal to standard electrode potential,
E
J
R
of the given half-cell.
E
J
= E
J
R
– E
J
L
As E
J
L
for standard hydrogen electrode is zero.
E
J
= E
J
R
– 0 = E
J
R
The measured emf of the cell:
Pt(s) H
2
(g, 1 bar) H
+
(aq, 1 M) Cu
2+
(aq, 1 M) Cu
is 0.34 V and it is also the value for the standard electrode potential
of the half-cell corresponding to the reaction:
Cu
2+
(aq, 1M) + 2 e
–
→ Cu(s)
Similarly, the measured emf of the cell:
Pt(s) H
2
(g, 1 bar)
H
+
(aq, 1 M) Zn
2+
(aq, 1M) Zn
is -0.76 V corresponding to the standard electrode potential of the
half-cell reaction:
Zn
2+
(aq, 1 M) + 2e
–
→ Zn(s)
The positive value of the standard electrode potential in the first
case indicates that Cu
2+
ions get reduced more easily than H
+
ions. The
reverse process cannot occur, that is, hydrogen ions cannot oxidise
Cu (or alternatively we can say that hydrogen gas can reduce copper
ion) under the standard conditions described above. Thus, Cu does
not dissolve in HCl. In nitric acid it is oxidised by nitrate ion and not
by hydrogen ion. The negative value of the standard electrode potential
in the second case indicates that hydrogen ions can oxidise zinc (or
zinc can reduce hydrogen ions).
In view of this convention, the half reaction for the Daniell cell in
Fig. 3.1 can be written as:
Left electrode: Zn(s) → Zn
2+
(aq, 1 M) + 2 e
–
Right electrode: Cu
2+
(aq, 1 M) + 2 e
–
→ Cu(s)
The overall reaction of the cell is the sum of above two reactions
and we obtain the equation:
Zn(s) + Cu
2+
(aq) → Zn
2+
(aq) + Cu(s)
emf of the cell = E
J
cell
= E
J
R
– E
J
L
= 0.34V – (– 0.76)V = 1.10 V
Sometimes metals like platinum or gold are used as inert electrodes.
They do not participate in the reaction but provide their surface for
oxidation or reduction reactions and for the conduction of electrons.
For example, Pt is used in the following half-cells:
Hydrogen electrode: Pt(s)|H
2
(g)| H
+
(aq)
With half-cell reaction: H
+
(aq)+ e
–
→ ½ H
2
(g)
Bromine electrode: Pt(s)|Br
2
(aq)| Br
–
(aq)
2020-21
70Chemistry
With half-cell reaction: ½ Br
2
(aq)
+ e
–
→ Br
–
(aq)
The standard electrode potentials are very important and we can
extract a lot of useful information from them. The values of standard
electrode potentials for some selected half-cell reduction reactions are
given in Table 3.1. If the standard electrode potential of an electrode
is greater than zero then its reduced form is more stable compared to
hydrogen gas. Similarly, if the standard electrode potential is negative
then hydrogen gas is more stable than the reduced form of the species.
It can be seen that the standard electrode potential for fluorine is the
highest in the Table indicating that fluorine gas (F
2
) has the maximum
tendency to get reduced to fluoride ions (F
–
) and therefore fluorine
gas is the strongest oxidising agent and fluoride ion is the weakest
reducing agent. Lithium has the lowest electrode potential indicating
that lithium ion is the weakest oxidising agent while lithium metal is
the most powerful reducing agent in an aqueous solution. It may be
seen that as we go from top to bottom in Table 3.1 the standard
electrode potential decreases and with this, decreases the oxidising
power of the species on the left and increases the reducing power of
the species on the right hand side of the reaction. Electrochemical
cells are extensively used for determining the pH of solutions, solubility
product, equilibrium constant and other thermodynamic properties
and for potentiometric titrations.
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
3.1 How would you determine the standard electrode potential of the system
Mg
2+
|Mg?
3.2 Can you store copper sulphate solutions in a zinc pot?
3.3 Consult the table of standard electrode potentials and suggest three
substances that can oxidise ferrous ions under suitable conditions.
3.33.3
3.33.3
3.3
NernstNernst
NernstNernst
Nernst
EquationEquation
EquationEquation
Equation
We have assumed in the previous section that the concentration of all
the species involved in the electrode reaction is unity. This need not be
always true. Nernst showed that for the electrode reaction:
M
n+
(aq) + ne
–
→ M(s)
the electrode potential at any concentration measured with respect to
standard hydrogen electrode can be represented by:
( ) ( )
n n
M / M M / M
E E
+ +
=
V
–
RT
nF
ln
[M]
[M ]
n+
but concentration of solid M is taken as unity and we have
( ) ( )
n n
M / M M / M
E E
+ +
=
V
–
RT
nF
ln
n+
1
[M ]
(3.8)
( )
n
M / M
E
+
V
has already been defined, R is gas constant (8.314 JK
–1
mol
–1
),
F is Faraday constant (96487 C mol
–1
), T is temperature in kelvin and
[M
n+
] is the concentration of the species, M
n+
.
2020-21
71
Electrochemistry
F
2
(g) + 2e
–
→ 2F
–
2.87
Co
3+
+ e
–
→ Co
2+
1.81
H
2
O
2
+ 2H
+
+ 2e
–
→ 2H
2
O 1.78
MnO
4
–
+ 8H
+
+ 5e
–
→ Mn
2+
+ 4H
2
O 1.51
Au
3+
+ 3e
–
→ Au(s) 1.40
Cl
2
(g) + 2e
–
→ 2Cl
–
1.36
Cr
2
O
7
2–
+ 14H
+
+ 6e
–
→ 2Cr
3+
+ 7H
2
O 1.33
O
2
(g) + 4H
+
+ 4e
–
→ 2H
2
O 1.23
MnO
2
(s) + 4H
+
+ 2e
–
→ Mn
2+
+ 2H
2
O 1.23
Br
2
+ 2e
–
→ 2Br
–
1.09
NO
3
–
+ 4H
+
+ 3e
–
→ NO(g) + 2H
2
O 0.97
2Hg
2+
+ 2e
–
→ Hg
2
2+
0.92
Ag
+
+ e
–
→ Ag(s) 0.80
Fe
3+
+ e
–
→ Fe
2+
0.77
O
2
(g) + 2H
+
+ 2e
–
→ H
2
O
2
0.68
I
2
+ 2e
–
→ 2I
–
0.54
Cu
+
+ e
–
→ Cu(s) 0.52
Cu
2+
+ 2e
–
→ Cu(s) 0.34
AgCl(s) + e
–
→ Ag(s) + Cl
–
0.22
AgBr(s) + e
–
→ Ag(s) + Br
–
0.10
2H
+
+ 2e
–
→ H
2
(g) 0.00
Pb
2+
+ 2e
–
→ Pb(s) –0.13
Sn
2+
+ 2e
–
→ Sn(s) –0.14
Ni
2+
+ 2e
–
→ Ni(s) –0.25
Fe
2+
+ 2e
–
→ Fe(s) –0.44
Cr
3+
+ 3e
–
→ Cr(s) –0.74
Zn
2+
+ 2e
–
→ Zn(s) –0.76
2H
2
O + 2e
–
→ H
2
(g) + 2OH
–
(aq) –0.83
Al
3+
+ 3e
–
→ Al(s) –1.66
Mg
2+
+ 2e
–
→ Mg(s) –2.36
Na
+
+ e
–
→ Na(s) –2.71
Ca
2+
+ 2e
–
→ Ca(s) –2.87
K
+
+ e
–
→ K(s) –2.93
Li
+
+ e
–
→ Li(s) –3.05
Table 3.1: Standard Electrode Potentials at 298 K
Ions are present as aqueous species and H
2
O as liquid; gases and solids are shown by g and s.
Reaction (Oxidised form + ne
–
→→
→→
→ Reduced form) E
JJ
JJ
J
/V
Increasing strength of oxidising agent
Increasing strength of reducing agent
1. A negative E
J
means that the redox couple is a stronger reducing agent than the H
+
/H
2
couple.
2. A positive E
J
means that the redox couple is a weaker reducing agent than the H
+
/H
2
couple.
2020-21
72Chemistry
In Daniell cell, the electrode potential for any given concentration of
Cu
2+
and Zn
2+
ions, we write
For Cathode:
( )
2
Cu / Cu
E
+
=
( )
2
Cu / Cu
E
+
V
–
RT
F2
ln
( )
2
1
Cu aq
+
(3.9)
For Anode:
( )
2
Zn / Zn
E
+
=
( )
2
Zn / Zn
E
+
V
–
RT
F2
ln
( )
2
1
Zn aq
+
(3.10)
The cell potential,
E
(cell)
=
( )
2
Cu / Cu
E
+
–
( )
2
Zn / Zn
E
+
=
( )
2
Cu / Cu
E
+
V
–
RT
F2
ln
2+
1
Cu (aq)
–
( )
2
Zn / Zn
E
+
V
+
RT
F2
ln
2+
1
Zn (aq)
=
( )
2
Cu / Cu
E
+
V
–
( )
2
Zn /Zn
E
+
V
–
RT
F2
( ) ( )
2+ 2+
1 1
ln – ln
Cu aq Zn aq
E
(cell)
=
( )
cell
E
V
–
RT
F2
ln
[ ]
+
[ ]
2
Zn
2
Cu
+
(3.11)
It can be seen that E
(cell)
depends on the concentration of both Cu
2+
and Zn
2+
ions. It increases with increase in the concentration of Cu
2+
ions and decrease in the concentration of Zn
2+
ions.
By converting the natural logarithm in Eq. (3.11) to the base 10 and
substituting the values of R, F and T = 298 K, it reduces to
E
(cell)
=
( )
cell
E
V
–
0 059
2
2
2
. [ ]
[ ]
log
Zn
Cu
+
+
(3.12)
We should use the same number of electrons (n) for both the
electrodes and thus for the following cell
Ni(s) Ni
2+
(aq) Ag
+
(aq) Ag
The cell reaction is Ni(s) + 2Ag
+
(aq)
→ Ni
2+
(aq)
+ 2Ag(s)
The Nernst equation can be written as
E
(cell)
=
( )
cell
E
V
–
RT
F2
ln
[Ni ]
[Ag ]
2+
2+
and for a general electrochemical reaction of the type:
a A + bB
ne
–
cC + dD
Nernst equation can be written as:
E
(cell)
=
( )
cell
E
V
–
RT
nF
1nQ
=
( )
cell
E
V
–
RT
nF
ln
[C] [D]
[A] [B]
c d
a b
(3.13)
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73
Electrochemistry
If the circuit in Daniell cell (Fig. 3.1) is closed then we note that the reaction
Zn(s) + Cu
2+
(aq) → Zn
2+
(aq) + Cu(s) (3.1)
takes place and as time passes, the concentration of Zn
2+
keeps
on increasing while the concentration of Cu
2+
keeps on decreasing.
At the same time voltage of the cell as read on the voltmeter keeps
on decreasing. After some time, we shall note that there is no change
in the concentration of Cu
2+
and Zn
2+
ions and at the same time,
voltmeter gives zero reading. This indicates that equilibrium has been
attained. In this situation the Nernst equation may be written as:
E
(cell)
= 0 =
( )
cell
E
V
–
2.303
2
log
[Zn ]
[Cu ]
2
2
RT
F
+
+
or
( )
cell
E
V
=
2
2
2.303 [Zn ]
log
2
[Cu ]
RT
F
+
+
But at equilibrium,
[ ]
[ ]
Zn
Cu
2
2
+
+
= K
c
for the reaction 3.1
and at T = 298K the above equation can be written as
( )
cell
E
V
=
0 059
2
. V
log K
C
= 1.1 V (
( )
cell
E
V
= 1.1V)
log K
C
=
(1.1V × 2)
37.288
0.059 V
=
K
C
= 2 × 10
37
at 298K.
In general,
( )
cell
E
V
=
2.303RT
nF
log K
C
(3.14)
Thus, Eq. (3.14) gives a relationship between equilibrium constant
of the reaction and standard potential of the cell in which that reaction
takes place. Thus, equilibrium constants of the reaction, difficult to
measure otherwise, can be calculated from the corresponding E
J
value
of the cell.
3.3.1 Equilibrium
Constant
from Nernst
Equation
Example 3.1Example 3.1
Example 3.1Example 3.1
Example 3.1
Represent the cell in which the following reaction takes place
Mg(s) + 2Ag
+
(0.0001M) → Mg
2+
(0.130M) + 2Ag(s)
Calculate its E
(cell)
if
( )
cell
E
V
= 3.17 V.
The cell can be written as Mg
Mg
2+
(0.130M) Ag
+
(0.0001M) Ag
( )
cell
E
=
( )
2
cell
2
Mg
RT
– ln
2F
Ag
E
+
+
V
= 3.17 V –
0 059
2 0 0001
2
.
log
( . )
V
0.130
= 3.17 V – 0.21V = 2.96 V.
SolutionSolution
Solution
Solution
Solution
2020-21
74Chemistry
The standard electrode potential for Daniell cell is 1.1V. Calculate
the standard Gibbs energy for the reaction:
Zn(s) + Cu
2+
(aq) → Zn
2+
(aq) + Cu(s)
∆
r
G
J
= – nF
(cell)
E
V
n in the above equation is 2, F = 96487 C mol
–1
and
( )
cell
E
V
= 1.1 V
Therefore, ∆
r
G
J
= – 2 × 1.1V × 96487 C mol
–1
= – 21227 J mol
–1
= – 212.27 kJ mol
–1
Example 3.3Example 3.3
Example 3.3Example 3.3
Example 3.3
SolutionSolution
SolutionSolution
Solution
Electrical work done in one second is equal to electrical potential
multiplied by total charge passed. If we want to obtain maximum work
from a galvanic cell then charge has to be passed reversibly. The
reversible work done by a galvanic cell is equal to decrease in its Gibbs
energy and therefore, if the emf of the cell is E and nF is the amount
of charge passed and ∆
r
G is the Gibbs energy of the reaction, then
∆
r
G = – nFE
(cell)
(3.15)
It may be remembered that E
(cell)
is an intensive parameter but ∆
r
G
is an extensive thermodynamic property and the value depends on n.
Thus, if we write the reaction
Zn(s) + Cu
2+
(aq) → Zn
2+
(aq) + Cu(s) (3.1)
∆
r
G = – 2FE
(cell)
but when we write the reaction
2 Zn (s) + 2 Cu
2+
(aq) → 2 Zn
2+
(aq) + 2Cu(s)
∆
r
G = – 4FE
(cell)
If the concentration of all the reacting species is unity, then
E
(cell)
=
( )
cell
E
V
and we have
∆
r
G
J
= – nF
(cell)
V
E
(3.16)
Thus, from the measurement of
( )
cell
E
V
we can obtain an important
thermodynamic quantity, ∆
r
G
J
, standard Gibbs energy of the reaction.
From the latter we can calculate equilibrium constant by the equation:
∆
r
G
J
= –RT ln K.
3.3.2 Electro-
chemical
Cell and
Gibbs
Energy of
the Reaction
Calculate the equilibrium constant of the reaction:
Cu(s) + 2Ag
+
(aq) → Cu
2+
(aq) + 2Ag(s)
( )
cell
E
V
= 0.46 V
( )
cell
E
V
=
0 059
2
. V
log K
C
= 0.46 V or
log K
C
=
0 46 2
0 059
.
.
V
V
×
= 15.6
K
C
= 3.92 × 10
15
Example 3.2Example 3.2
Example 3.2Example 3.2
Example 3.2
SolutionSolution
Solution
Solution
Solution
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75
Electrochemistry
It is necessary to define a few terms before we consider the subject of
conductance of electricity through electrolytic solutions. The electrical
resistance is represented by the symbol ‘R’ and it is measured in ohm (Ω)
which in terms of SI base units is equal to (kg m
2
)/(S
3
A
2
). It can be
measured with the help of a Wheatstone bridge with which you are
familiar from your study of physics. The electrical resistance of any object
is directly proportional to its length, l, and inversely proportional to its
area of cross section, A. That is,
R ∝
l
A
or R = ρ
l
A
(3.17)
The constant of proportionality, ρ (Greek, rho), is called resistivity
(specific resistance). Its SI units are ohm metre (Ω m) and quite often
its submultiple, ohm centimetre (Ω cm) is also used. IUPAC recommends
the use of the term resistivity over specific resistance and hence in the
rest of the book we shall use the term resistivity. Physically, the resistivity
for a substance is its resistance when it is one metre long and its area
of cross section is one m
2
. It can be seen that:
1 Ω m = 100 Ω cm or 1 Ω cm = 0.01 Ω m
The inverse of resistance, R, is called conductance, G, and we have
the relation:
G =
1
R
=
ρ
κ
=
A A
l l
(3.18)
The SI unit of conductance is siemens, represented by the symbol
‘S’ and is equal to ohm
–1
(also known as mho) or Ω
–1
. The inverse of
resistivity, called conductivity (specific conductance) is represented by
the symbol,
κ
(Greek, kappa). IUPAC has recommended the use of term
conductivity over specific conductance and hence we shall use the term
conductivity in the rest of the book. The SI units of conductivity are
S m
–1
but quite often,
κ
is expressed in S cm
–1
. Conductivity of a
material in S m
–1
is its conductance when it is 1 m long and its area
of cross section is 1 m
2
. It may be noted that 1 S cm
–1
= 100 S m
–1
.
3.43.4
3.43.4
3.4
ConductanceConductance
ConductanceConductance
Conductance
of Electrolyticof Electrolytic
of Electrolyticof Electrolytic
of Electrolytic
SolutionsSolutions
SolutionsSolutions
Solutions
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
3.4 Calculate the potential of hydrogen electrode in contact with a solution
whose pH is 10.
3.5 Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag
+
(0.002 M) → Ni
2+
(0.160 M) + 2Ag(s)
Given that
(cell)
E
V
= 1.05 V
3.6 The cell in which the following reaction occurs:
(
)
(
)
(
)
( )
3 2
2
2Fe 2I 2Fe Iaq aq aq
s
+ − +
+ → +
has
0
cell
E
= 0.236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the
cell reaction.
2020-21
76Chemistry
* Electronically conducting polymers – In 1977 MacDiarmid, Heeger and Shirakawa discovered that acetylene gas can be
polymerised to produce a polymer, polyacetylene when exposed to vapours of iodine acquires metallic lustre and
conductivity. Since then several organic conducting polymers have been made such as polyaniline, polypyrrole and
polythiophene. These organic polymers which have properties like metals, being composed wholly of elements like
carbon, hydrogen and occasionally nitrogen, oxygen or sulphur, are much lighter than normal metals and can be used
for making light-weight batteries. Besides, they have the mechanical properties of polymers such as flexibility so that
one can make electronic devices such as transistors that can bend like a sheet of plastic. For the discovery of conducting
polymers, MacDiarmid, Heeger and Shirakawa were awarded the Nobel Prize in Chemistry for the year 2000.
It can be seen from Table 3.2 that the magnitude of conductivity
varies a great deal and depends on the nature of the material. It also
depends on the temperature and pressure at which the measurements
are made. Materials are classified into conductors, insulators and
semiconductors depending on the magnitude of their conductivity. Metals
and their alloys have very large conductivity and are known as conductors.
Certain non-metals like carbon-black, graphite and some organic
polymers
*
are also electronically conducting. Substances like glass,
ceramics, etc., having very low conductivity are known as insulators.
Substances like silicon, doped silicon and gallium arsenide having
conductivity between conductors and insulators are called
semiconductors and are important electronic materials. Certain materials
called superconductors by definition have zero resistivity or infinite
conductivity. Earlier, only metals and their alloys at very low temperatures
(0 to 15 K) were known to behave as superconductors, but nowadays a
number of ceramic materials and mixed oxides are also known to show
superconductivity at temperatures as high as 150 K.
Electrical conductance through metals is called metallic or electronic
conductance and is due to the movement of electrons. The electronic
conductance depends on
(i) the nature and structure of the metal
(ii) the number of valence electrons per atom
(iii) temperature (it decreases with increase of temperature).
Table 3.2: The values of Conductivity of some Selected
Materials at 298.15 K
Material Conductivity/ Material Conductivity/
S m
–1
S m
–1
Conductors Aqueous Solutions
Sodium 2.1×10
3
Pure water 3.5×10
–5
Copper 5.9×10
3
0.1 M HCl 3.91
Silver 6.2×10
3
0.01M KCl 0.14
Gold 4.5×10
3
0.01M NaCl 0.12
Iron 1.0×10
3
0.1 M HAc 0.047
Graphite 1.2×10 0.01M HAc 0.016
Insulators Semiconductors
Glass 1.0×10
–16
CuO 1×10
–7
Teflon 1.0×10
–18
Si 1.5×10
–2
Ge 2.0
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Electrochemistry
As the electrons enter at one end and go out through the other end,
the composition of the metallic conductor remains unchanged. The
mechanism of conductance through semiconductors is more complex.
We already know (Class XI, Unit 7) that even very pure water has
small amounts of hydrogen and hydroxyl ions (~10
–7
M) which lend it
very low conductivity (3.5 × 10
–5
S m
–1
). When electrolytes are dissolved
in water, they furnish their own ions in the solution hence its conductivity
also increases. The conductance of electricity by ions present in the
solutions is called electrolytic or ionic conductance. The conductivity
of electrolytic (ionic) solutions depends on:
(i) the nature of the electrolyte added
(ii) size of the ions produced and their solvation
(iii) the nature of the solvent and its viscosity
(iv) concentration of the electrolyte
(v) temperature (it increases with the increase of temperature).
Passage of direct current through ionic solution over a prolonged
period can lead to change in its composition due to electrochemical
reactions (Section 3.4.1).
We know that accurate measurement of an unknown resistance can be
performed on a Wheatstone bridge. However, for measuring the resistance
of an ionic solution we face two problems. Firstly, passing direct current
(DC) changes the composition of the solution. Secondly, a solution cannot
be connected to the bridge like a metallic wire or other solid conductor.
The first difficulty is resolved by using an alternating current (AC) source
of power. The second problem is solved by using a specially designed
vessel called conductivity cell. It is available in several designs and two
simple ones are shown in Fig. 3.4.
3.4.1 Measurement
of the
Conductivity
of Ionic
Solutions
Connecting
wires
Platinized Pt
electrodes
Platinized Pt electrode
Platinized Pt electrode
Connecting
wires
Fig. 3.4
Two different types of
conductivity cells.
Basically it consists of two platinum electrodes coated with platinum
black (finely divided metallic Pt is deposited on the electrodes
electrochemically). These have area of cross section equal to ‘A’ and are
separated by distance ‘l’. Therefore, solution confined between these
electrodes is a column of length l and area of cross section A. The
resistance of such a column of solution is then given by the equation:
R = ρ
l
A
=
κ
l
A
(3.17)
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78Chemistry
Table 3.3: Conductivity and Molar conductivity of KCl solutions
at 298.15K
mol L
–1
mol m
–3
S cm
–1
S m
–1
S cm
2
mol
–1
S m
2
mol
–1
1.000 1000 0.1113 11.13 111.3 111.3×10
–4
0.100 100.0 0.0129 1.29 129.0 129.0×10
–4
0.010 10.00 0.00141 0.141 141.0 141.0×10
–4
Concentration/Molarity Conductivity Molar Conductivity
The quantity l/A is called cell constant denoted by the symbol, G*.
It depends on the distance between the electrodes and their area of
cross-section and has the dimension of length
–1
and can be calculated
if we know l and A. Measurement of l and A is not only inconvenient
but also unreliable. The cell constant is usually determined by measuring
the resistance of the cell containing a solution whose conductivity is
already known. For this purpose, we generally use KCl solutions whose
conductivity is known accurately at various concentrations (Table 3.3)
and at different temperatures. The cell constant, G*, is then given by
the equation:
G* =
l
A
= R
κ
(3.18)
Once the cell constant is determined, we can use
it for measuring the resistance or conductivity of
any solution. The set up for the measurement of the
resistance is shown in Fig. 3.5.
It consists of two resistances R
3
and R
4
, a variable
resistance R
1
and the conductivity cell having the
unknown resistance R
2
. The Wheatstone bridge is
fed by an oscillator O (a source of a.c. power in the
audio frequency range 550 to 5000 cycles per
second). P is a suitable detector (a headphone or
other electronic device) and the bridge is balanced
when no current passes through the detector. Under
these conditions:
Unknown resistance R
2
=
1 4
3
R R
R
(3.19)
These days, inexpensive conductivity meters are
available which can directly read the conductance or resistance of the
solution in the conductivity cell. Once the cell constant and the resistance
of the solution in the cell is determined, the conductivity of the solution
is given by the equation:
cell constant G*
R R
κ
= =
(3.20)
The conductivity of solutions of different electrolytes in the same
solvent and at a given temperature differs due to charge and size of the
Fig. 3.5: Arrangement for measurement
of resistance of a solution of an
electrolyte.
2020-21
79
Electrochemistry
ions in which they dissociate, the concentration of ions or ease with
which the ions move under a potential gradient. It, therefore, becomes
necessary to define a physically more meaningful quantity called molar
conductivity denoted by the symbol
Λ
m
(Greek, lambda). It is related
to the conductivity of the solution by the equation:
Molar conductivity =
Λ
m
=
c
κ
(3.21)
In the above equation, if
κ
is expressed in S m
–1
and the concentration,
c in mol m
–3
then the units of
Λ
m
are in S m
2
mol
–1
. It may be noted that:
1 mol m
–3
= 1000(L/m
3
) × molarity (mol/L), and hence
Λ
m
(S cm
2
mol
–1
) =
κ
−
− −
1
3 1
(S cm )
1000 L m × molarity (mol L )
If we use S cm
–1
as the units for
κ
and mol cm
–3
, the units of
concentration, then the units for
Λ
m
are S cm
2
mol
–1
. It can be calculated
by using the equation:
Λ
m
(S cm
2
mol
–1
) =
1 3
(S cm ) × 1000 (cm / L )
molarity (mol / L)
−
κ
Both type of units are used in literature and are related to each
other by the equations:
1 S m
2
mol
–1
= 10
4
S cm
2
mol
–1
or
1 S cm
2
mol
–1
= 10
–4
S m
2
mol
–1
.
Resistance of a conductivity cell filled with 0.1 mol L
–1
KCl solution is
100 Ω . If the resistance of the same cell when filled with 0.02 mol L
–1
KCl solution is 520 Ω , calculate the conductivity and molar conductivity
of 0.02 mol L
–1
KCl solution. The conductivity of 0.1 mol L
–1
KCl
solution is 1.29 S/m.
The cell constant is given by the equation:
Cell constant = G* = conductivity × resistance
= 1.29 S/m × 100 Ω = 129 m
–1
= 1.29 cm
–1
Conductivity of 0.02 mol L
–1
KCl solution = cell constant / resistance
=
*
G
R
=
–1
129 m
520
Ω
= 0.248 S m
–1
Concentration = 0.02 mol L
–1
= 1000 × 0.02 mol m
–3
= 20 mol m
–3
Molar conductivity =
=
m
c
κ
Λ
=
–3 –1
–3
248 × 10 S m
20 mol m
= 124 × 10
–4
S m
2
mol
–1
Alternatively,
κ
=
–1
1.29 cm
520
Ω
= 0.248 × 10
–2
S cm
–1
Example 3.4Example 3.4
Example 3.4Example 3.4
Example 3.4
SolutionSolution
SolutionSolution
Solution
2020-21
80Chemistry
and
Λ
m
=
κ
× 1000 cm
3
L
–1
molarity
–1
–2 –1 3 –1
–1
0.248×10 S cm ×1000 cm L
=
0.02 mol L
= 124 S cm
2
mol
–1
The electrical resistance of a column of 0.05 mol L
–1
NaOH solution of
diameter 1 cm and length 50 cm is 5.55 × 10
3
ohm. Calculate its
resistivity, conductivity and molar conductivity.
A = π r
2
= 3.14 × 0.5
2
cm
2
= 0.785 cm
2
= 0.785 × 10
–4
m
2
l = 50 cm = 0.5 m
=
l
R
A
ρ
or
ρ
× Ω ×
= =
3 2
5.55 10 0.785 cm
50 cm
RA
l
= 87.135 Ω cm
Conductivity =
κ
ρ
1
=
=
1
87.135
S cm
–1
= 0.01148 S cm
–1
Molar conductivity,
m
Λ
=
× 1000
c
κ
cm
3
L
–1
=
–1 3 –1
–1
0.01148 S cm ×1000 cm L
0.05 mol L
= 229.6 S cm
2
mol
–1
If we want to calculate the values of different quantities in terms of ‘m’
instead of ‘cm’,
ρ
=
RA
l
=
3 –4 2
5.55 × 10 × 0.785×10 m
0.5 m
Ω
= 87.135 ×10
–2
Ω m
1
=
κ
ρ
=
100
m
87.135
Ω
= 1.148 S m
–1
and
=
m
c
κ
Λ
=
–1
–3
1.148 S m
50 mol m
= 229.6 × 10
–4
S m
2
mol
–1
.
Example 3.5Example 3.5
Example 3.5
Example 3.5
Example 3.5
SolutionSolution
Solution
Solution
Solution
Both conductivity and molar conductivity change with the
concentration of the electrolyte. Conductivity always decreases with
decrease in concentration both,
for weak and strong electrolytes.
This can be explained by the fact that the number of ions per unit
volume that carry the current in a solution decreases on dilution.
The conductivity of a solution at any given concentration is the
conductance of one unit volume of solution kept between two
3.4.2 Variation of
Conductivity
and Molar
Conductivity
with
Concentration
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81
Electrochemistry
platinum electrodes with unit area of cross section and at a distance
of unit length. This is clear from the equation:
= =
A
G
κ
κ
l
(both A and l are unity in their appropriate units in
m or cm)
Molar conductivity of a solution at a given concentration is the
conductance of the volume V of solution containing one mole of
electrolyte kept between two electrodes with area of cross section A and
distance of unit length. Therefore,
κ
Λ κ
= =
m
A
l
Since l = 1 and A = V ( volume containing 1 mole of electrolyte)
Λ
m
=
κ
V (3.22)
Molar conductivity increases with
decrease in concentration. This is
because the total volume, V, of solution
containing one mole of electrolyte also
increases. It has been found that decrease
in
κ
on dilution of a solution is more
than compensated by increase in its
volume. Physically, it means that at a
given concentration,
Λ
m
can be defined
as the conductance of the electrolytic
solution kept between the electrodes of a
conductivity cell at unit distance but
having area of cross section large enough
to accommodate sufficient volume of
solution that contains one mole of the
electrolyte. When concentration
approaches zero, the molar conductivity
is known as limiting molar
conductivity and is represented by the
symbol
Λ
°
m
. The variation in
Λ
m
with
concentration is different (Fig. 3.6) for
strong and weak electrolytes.
Strong Electrolytes
For strong electrolytes,
Λ
m
increases slowly with dilution and can be
represented by the equation:
Λ
m
=
Λ
°
m
– A c
½
(3.23)
It can be seen that if we plot (Fig. 3.6)
Λ
m
against
c
1/2
, we obtain a straight line with intercept equal to
Λ
°
m
and slope
equal to ‘–A’. The value of the constant ‘A’ for a given solvent and
temperature depends on the type of electrolyte i.e., the charges on the
cation and anion produced on the dissociation of the electrolyte in the
solution. Thus, NaCl, CaCl
2
, MgSO
4
are known as 1-1, 2-1 and 2-2
electrolytes respectively. All electrolytes of a particular type have the
same value for ‘A’.
Fig. 3.6: Molar conductivity versus c½ for acetic
acid (weak electrolyte) and potassium
chloride (strong electrolyte) in aqueous
solutions.
2020-21
82Chemistry
The molar conductivity of KCl solutions at different concentrations at
298 K are given below:
c/mol L
–1
ΛΛ
ΛΛ
Λ
m
/S cm
2
mol
–1
0.000198 148.61
0.000309 148.29
0.000521 147.81
0.000989 147.09
Show that a plot between
Λ
m
and c
1/2
is a straight line. Determine the
values of
Λ
°
m
and A for KCl.
Taking the square root of concentration we obtain:
c
1/2
/(mol L
–1
)
1/2
ΛΛ
Λ
Λ
Λ
m
/S cm
2
mol
–1
0.01407 148.61
0.01758 148.29
0.02283 147.81
0.03145 147.09
A plot of
Λ
m
( y-axis) and c
1/2
(x-axis) is shown in (Fig. 3.7).
It can be seen that it is nearly a straight line. From the intercept
(c
1/2
= 0), we find that
Λ
°
m
= 150.0 S cm
2
mol
–1
and
A = – slope = 87.46 S cm
2
mol
–1
/(mol/L
–1
)
1/2
.
Example 3.6Example 3.6
Example 3.6Example 3.6
Example 3.6
SolutionSolution
SolutionSolution
Solution
Fig. 3.7: Variation of
Λ
m
against c½.
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83
Electrochemistry
Kohlrausch examined
Λ
°
m
values for a number of strong electrolytes
and observed certain regularities. He noted that the difference in
Λ
°
m
of
the electrolytes NaX and KX for any X is nearly constant. For example
at 298 K:
Λ
°
m
(KCl)
–
Λ
°
m
(NaCl)
=
Λ
°
m
(KBr)
–
Λ
°
m
(NaBr)
=
Λ
°
m
(KI)
–
Λ
°
m
(NaI)
Y 23.4 S cm
2
mol
–1
and similarly it was found that
Λ
°
m
(NaBr)
–
Λ
°
m
(NaCl)
=
Λ
°
m
(KBr)
–
Λ
°
m
(KCl)
Y 1.8 S cm
2
mol
–1
On the basis of the above observations he enunciated Kohlrausch
law of independent migration of ions. The law states that limiting
molar conductivity of an electrolyte can be represented as the sum of
the individual contributions of the anion and cation of the electrolyte.
Thus, if
λ
°
Na
+
and
λ
°
Cl
–
are limiting molar conductivity of the sodium and
chloride ions respectively, then the limiting molar conductivity for sodium
chloride is given by the equation:
Λ
°
m
(NaCl)
=
λ
°
Na
+
+
λ
°
Cl
–
(3.24)
In general, if an electrolyte on dissociation gives ν
+
cations and ν
–
anions then its limiting molar conductivity is given by:
Λ
°
m
= ν
+
λ
°
+
+ ν
–
λ
°
–
(3.25)
Here,
λ
°
+
and
λ
°
–
are the limiting molar conductivities of the cation
and anion respectively. The values of
λ
°
for some cations and anions at
298 K are given in Table 3.4.
Table 3.4: Limiting Molar Conductivity for some
Ions in Water at 298 K
Weak Electrolytes
Weak electrolytes like acetic acid have lower degree of dissociation at
higher concentrations and hence for such electrolytes, the change in
Λ
m
with dilution is due to increase in the degree of dissociation and
consequently the number of ions in total volume of solution that contains
1 mol of electrolyte. In such cases
Λ
m
increases steeply (Fig. 3.6) on
dilution, especially near lower concentrations. Therefore,
Λ
°
m
cannot be
obtained by extrapolation of
Λ
m
to zero concentration. At infinite dilution
(i.e., concentration c → zero) electrolyte dissociates completely (α =1),
but at such low concentration the conductivity of the solution is so low
that it cannot be measured accurately. Therefore,
Λ
°
m
for weak electrolytes
is obtained by using Kohlrausch law of independent migration of ions
(Example 3.8). At any concentration c, if α is the degree of dissociation
Ion
λλ
λλ
λ
0
/(S cm
2
mol
–1
) Ion
λλ
λλ
λ
0
/(S cm
2
mol
–1
)
H
+
349.6 OH
–
199.1
Na
+
50.1 Cl
–
76.3
K
+
73.5 Br
–
78.1
Ca
2+
119.0 CH
3
COO
–
40.9
Mg
2+
106.0 SO
4
2
−
160.0
2020-21
84Chemistry
then it can be approximated to the ratio of molar conductivity
Λ
m
at the
concentration c to limiting molar conductivity,
Λ
0
m
. Thus we have:
°
=
m
m
Λ
α
Λ
(3.26)
But we know that for a weak electrolyte like acetic acid (Class XI,
Unit 7),
( )
( )
2 2
2
2
= = =
a
1
1
m m
m m mm
m
m
c c
c
K
Λ Λ
Λ Λ Λ
Λ
Λ
Λ
ο ο
ο
ο
α
− α
−
−
(3.27)
Applications of Kohlrausch law
Using Kohlrausch law of independent migration of ions, it is possible to
calculate
Λ
0
m
for any electrolyte from the
λ
o
of individual ions. Moreover,
for weak electrolytes like acetic acid it is possible to determine the value
of its dissociation constant once we know the
Λ
0
m
and
Λ
m
at a given
concentration c.
Calculate
Λ
0
m
for CaCl
2
and MgSO
4
from the data given in Table 3.4.
We know from Kohlrausch law that
( )
2
CaCl
m
Λ
ο
=
2+ –
Ca Cl
2
ο ο
λ + λ
= 119.0 S cm
2
mol
–1
+ 2(76.3) S cm
2
mol
–1
= (119.0 + 152.6) S cm
2
mol
–1
= 271.6 S cm
2
mol
–1
( )
4
MgSO
m
Λ
ο
=
2–
2+
4
Mg SO
ο ο
λ + λ
= 106.0 S cm
2
mol
–1
+ 160.0 S cm
2
mol
–1
= 266 S cm
2
mol
–1
.
Λ
0
m
for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm
2
mol
–1
respectively. Calculate
Λ
0
for HAc.
( )
HAc
m
Λ
ο
=
+ –
H Ac
ο ο
λ + λ
+ – – + – +
H Cl Ac Na Cl Na
ο ο ο ο ο ο
= λ + λ + λ + λ − λ − λ
=
( ) ( ) ( )
HCl NaAc NaCl
m m m
Λ Λ Λ
ο ο ο
+ −
= (425.9 + 91.0 – 126.4 ) S cm
2
mol
–1
= 390.5 S cm
2
mol
–1
.
The conductivity of 0.001028 mol L
–1
acetic acid is 4.95 × 10
–5
S cm
–1
.
Calculate its dissociation constant if
Λ
0
m
for acetic acid is
390.5 S cm
2
mol
–1
.
m
Λ
=
.
.
5 1 3
1
4 95 10 S cm 1000 cm
0 001028 mol L Lc
κ
− −
−
×
= ×
= 48.15 S cm
3
mol
–1
α =
Λ
Λ
−
ο −
=
2 1
2 1
48.15 Scm mol
390.5 Scm mol
m
m
= 0.1233
k =
( )
. ( . )
c
1 .
–1 2
2
0 001028 mol L 0 1233
1 0 1233
α
α
×
=
− −
= 1.78 × 10
–5
mol L
–1
Example 3.8Example 3.8
Example 3.8Example 3.8
Example 3.8
SolutionSolution
SolutionSolution
Solution
Example 3.9Example 3.9
Example 3.9Example 3.9
Example 3.9
SolutionSolution
SolutionSolution
Solution
Example 3.7Example 3.7
Example 3.7Example 3.7
Example 3.7
SolutionSolution
SolutionSolution
Solution
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85
Electrochemistry
In an electrolytic cell external source of voltage is used to bring about
a chemical reaction. The electrochemical processes are of great importance
in the laboratory and the chemical industry. One of the simplest electrolytic
cell consists of two copper strips dipping in an aqueous solution of
copper sulphate. If a DC voltage is applied to the two electrodes, then
Cu
2+
ions discharge at the cathode (negatively charged) and the following
reaction takes place:
Cu
2+
(aq) + 2e
–
→ Cu (s) (3.28)
Copper metal is deposited on the cathode. At the anode, copper is
converted into Cu
2+
ions by the reaction:
Cu(s) → Cu
2+
(s) + 2e
–
(3.29)
Thus copper is dissolved (oxidised) at anode and deposited
(reduced) at cathode. This is the basis for an industrial process in
which impure copper is converted into copper of high purity. The
impure copper is made an anode that dissolves on passing current
and pure copper is deposited at the cathode. Many metals like Na, Mg,
Al, etc. are produced on large scale by electrochemical reduction of
their respective cations where no suitable chemical reducing agents
are available for this purpose.
Sodium and magnesium metals are produced by the electrolysis of
their fused chlorides and aluminium is produced (Class XII, Unit 6) by
electrolysis of aluminium oxide in presence of cryolite.
Quantitative Aspects of Electrolysis
Michael Faraday was the first scientist who described the quantitative
aspects of electrolysis. Now Faraday’s laws also flow from what has
been discussed earlier.
Faraday’s Laws of Electrolysis
After his extensive investigations on electrolysis of solutions and melts
of electrolytes, Faraday published his results during 1833-34 in the
form of the following well known Faraday’s two laws of electrolysis:
(i) First Law: The amount of chemical reaction which occurs at any
electrode during electrolysis by a current is proportional to the
quantity of electricity passed through the electrolyte (solution or
melt).
(ii) Second Law: The amounts of different substances liberated by the
same quantity of electricity passing through the electrolytic solution
are proportional to their chemical equivalent weights (Atomic Mass
of Metal ÷ Number of electrons required to reduce the cation).
3.53.5
3.5
3.5
3.5
Electrolytic
Electrolytic
Electrolytic
Electrolytic
Electrolytic
Cells and
Cells and
Cells and
Cells and
Cells and
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
3.7 Why does the conductivity of a solution decrease with dilution?
3.8 Suggest a way to determine the
Λ
°
m
value of water.
3.9 The molar conductivity of 0.025 mol L
–1
methanoic acid is 46.1 S cm
2
mol
–1
.
Calculate its degree of dissociation and dissociation constant. Given λ
0
(H
+
)
= 349.6 S cm
2
mol
–1
and λ
0
(HCOO
–
) = 54.6 S cm
2
mol
–1
.
2020-21
86Chemistry
There were no constant current sources available during Faraday’s
times. The general practice was to put a coulometer (a standard electrolytic
cell) for determining the quantity of electricity passed from the amount
of metal (generally silver or copper) deposited or consumed. However,
coulometers are now obsolete and we now have constant current (I)
sources available and the quantity of electricity Q, passed is given by
Q = It
Q is in coloumbs when I is in ampere and t is in second.
The amount of electricity (or charge) required for oxidation or
reduction depends on the stoichiometry of the electrode reaction. For
example, in the reaction:
Ag
+
(aq) + e
–
→ Ag(s) (3.30)
One mole of the electron is required for the reduction of one mole
of silver ions.
We know that charge on one electron is equal to 1.6021× 10
–19
C.
Therefore, the charge on one mole of electrons is equal to:
N
A
× 1.6021 × 10
–19
C = 6.02 × 10
23
mol
–1
× 1.6021 × 10
–19
C = 96487 C mol
–1
This quantity of electricity is called Faraday and is represented by
the symbol F.
For approximate calculations we use 1F Y 96500 C mol
–1
.
For the electrode reactions:
Mg
2+
(l) + 2e
–
→ Mg(s) (3.31)
Al
3+
(l) + 3e
–
→ Al(s) (3.32)
It is obvious that one mole of Mg
2+
and Al
3+
require 2 mol of
electrons (2F) and 3 mol of electrons (3F) respectively. The charge passed
through the electrolytic cell during electrolysis is equal to the product
of current in amperes and time in seconds. In commercial production
of metals, current as high as 50,000 amperes are used that amounts
to about 0.518 F per second.
A solution of CuSO
4
is electrolysed for 10 minutes with a current of
1.5 amperes. What is the mass of copper deposited at the cathode?
t = 600 s charge = current × time = 1.5 A × 600 s = 900 C
According to the reaction:
Cu
2+
(aq) + 2e
–
= Cu(s)
We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu.
For 900 C, the mass of Cu deposited
= (63 g mol
–1
× 900 C)/(2 × 96487 C mol
–1
) = 0.2938 g.
Example 3.10Example 3.10
Example 3.10Example 3.10
Example 3.10
SolutionSolution
SolutionSolution
Solution
Products of electrolysis depend on the nature of material being
electrolysed and the type of electrodes being used. If the electrode is
inert (e.g., platinum or gold), it does not participate in the chemical
reaction and acts only as source or sink for electrons. On the other
hand, if the electrode is reactive, it participates in the electrode reaction.
Thus, the products of electrolysis may be different for reactive and inert
3.5.1 Products of
Electrolysis
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87
Electrochemistry
electrodes.The products of electrolysis depend on the different oxidising
and reducing species present in the electrolytic cell and their standard
electrode potentials. Moreover, some of the electrochemical processes
although feasible, are so slow kinetically that at lower voltages these do
not seem to take place and extra potential (called overpotential) has to
be applied, which makes such process more difficult to occur.
For example, if we use molten NaCl, the products of electrolysis are
sodium metal and Cl
2
gas. Here we have only one cation (Na
+
) which is
reduced at the cathode (Na
+
+ e
–
→ Na) and one anion (Cl
–
) which is
oxidised at the anode (Cl
–
→ ½Cl
2
+ e
–
) . During the electrolysis of aqueous
sodium chloride solution, the products are NaOH, Cl
2
and H
2
. In this
case besides Na
+
and Cl
–
ions we also have H
+
and OH
–
ions along with
the solvent molecules, H
2
O.
At the cathode there is competition between the following reduction
reactions:
Na
+
(aq) + e
–
→ Na (s)
( )
cell
V
E
= – 2.71 V
H
+
(aq) + e
–
→ ½ H
2
(g)
( )
cell
V
E
= 0.00 V
The reaction with higher value of E
J
is preferred and therefore, the
reaction at the cathode during electrolysis is:
H
+
(aq) + e
–
→ ½ H
2
(g) (3.33)
but H
+
(aq) is produced by the dissociation of H
2
O, i.e.,
H
2
O (l) → H
+
(aq) + OH
–
(aq) (3.34)
Therefore, the net reaction at the cathode may be written as the sum
of (3.33) and (3.34) and we have
H
2
O (l) + e
–
→ ½H
2
(g) + OH
–
(3.35)
At the anode the following oxidation reactions are possible:
Cl
–
(aq) → ½ Cl
2
(g) + e
–
( )
cell
V
E
= 1.36 V (3.36)
2H
2
O (l) → O
2
(g) + 4H
+
(aq) + 4e
–
( )
cell
V
E
= 1.23 V (3.37)
The reaction at anode with lower value of E
J
is preferred and
therefore, water should get oxidised in preference to Cl
–
(aq). However,
on account of overpotential of oxygen, reaction (3.36) is preferred. Thus,
the net reactions may be summarised as:
NaCl (aq)
H O
2
→
Na
+
(aq) + Cl
–
(aq)
Cathode: H
2
O(l) + e
–
→ ½ H
2
(g) + OH
–
(aq)
Anode: Cl
–
(aq) → ½ Cl
2
(g) + e
–
Net reaction:
NaCl(aq) + H
2
O(l) → Na
+
(aq) + OH
–
(aq) + ½H
2
(g) + ½Cl
2
(g)
The standard electrode potentials are replaced by electrode potentials
given by Nernst equation (Eq. 3.8) to take into account the concentration
effects. During the electrolysis of sulphuric acid, the following processes
are possible at the anode:
2H
2
O(l) → O
2
(g) + 4H
+
(aq) + 4e
–
( )
cell
E
V
= +1.23 V (3.38)
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88Chemistry
2SO
4
2–
(aq) → S
2
O
8
2–
(aq) + 2e
–
( )
cell
E
V
= 1.96 V (3.39)
For dilute sulphuric acid, reaction (3.38) is preferred but at higher
concentrations of H
2
SO
4
, reaction (3.39) is preferred.
Any battery (actually it may have one or more than one cell connected
in series) or cell that we use as a source of electrical energy is basically
a galvanic cell where the chemical energy of the redox reaction is
converted into electrical energy. However, for a battery to be of practical
use it should be reasonably light, compact and its voltage should not
vary appreciably during its use. There are mainly two types of batteries.
In the primary batteries, the reaction occurs only once and after use
over a period of time battery becomes dead and cannot be reused
again. The most familiar example of this type is the dry
cell (known as Leclanche cell after its discoverer) which is
used commonly in our transistors and clocks. The cell
consists of a zinc container that also acts as anode and
the cathode is a carbon (graphite) rod surrounded by
powdered manganese dioxide and carbon (Fig.3.8). The
space between the electrodes is filled by a moist paste of
ammonium chloride (NH
4
Cl) and zinc chloride (ZnCl
2
). The
electrode reactions are complex, but they can be written
approximately as follows :
Anode: Zn(s) → Zn
2+
+ 2e
–
Cathode: MnO
2
+ NH
4
+
+ e
–
→ MnO(OH) + NH
3
In the reaction at cathode, manganese is reduced
from the + 4 oxidation state to the +3 state. Ammonia
produced in the reaction forms a complex with Zn
2+
to give
[Zn (NH
3
)
4
]
2+
. The cell has a potential of nearly 1.5 V.
Mercury cell, (Fig. 3.9) suitable for low current devices
like hearing aids, watches, etc. consists of zinc – mercury
amalgam as anode and a paste of HgO and carbon as the
cathode. The electrolyte is a paste of KOH and ZnO. The
electrode reactions for the cell are given below:
Anode: Zn(Hg) + 2OH
–
→ ZnO(s) + H
2
O + 2e
–
Cathode: HgO + H
2
O + 2e
–
→ Hg(l) + 2OH
–
3.6 Batteries3.6 Batteries
3.6 Batteries3.6 Batteries
3.6 Batteries
3.6.1 Primary
Batteries
Fig. 3.8: A commercial dry cell
consists of a graphite
(carbon) cathode in a
zinc container; the latter
acts as the anode.
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
3.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours,
then how many electrons would flow through the wire?
3.11 Suggest a list of metals that are extracted electrolytically.
3.12 Consider the reaction: Cr
2
O
7
2–
+ 14H
+
+ 6e
–
→ 2Cr
3+
+ 7H
2
O
What is the quantity of electricity in coulombs needed to reduce 1 mol
of Cr
2
O
7
2–
?
2020-21
89
Electrochemistry
Fig. 3.10: The Lead storage battery.
The overall reaction is represented by
Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
The cell potential is approximately
1.35 V and remains constant during its
life as the overall reaction does not
involve any ion in solution whose
concentration can change during its life
time.
A secondary cell after use can be recharged by passing current
through it in the opposite direction so that it can be used again. A
good secondary cell can undergo a large number of discharging
and charging cycles. The most important secondary cell is the lead
storage battery (Fig. 3.10) commonly used in automobiles and
invertors. It consists of a lead anode and a grid of lead packed with
lead dioxide (PbO
2
) as cathode. A 38% solution of sulphuric acid
is used as an electrolyte.
The cell reactions when the battery is in use are given below:
Anode: Pb(s) + SO
4
2–
(aq) → PbSO
4
(s) + 2e
–
Cathode: PbO
2
(s) + SO
4
2–
(aq) + 4H
+
(aq) + 2e
–
→ PbSO
4
(s) + 2H
2
O (l)
i.e., overall cell reaction consisting of cathode and anode reactions is:
Pb(s) + PbO
2
(s) + 2H
2
SO
4
(aq) → 2PbSO
4
(s) + 2H
2
O(l)
On charging the battery the reaction is reversed and PbSO
4
(s) on
anode and cathode is converted into Pb and PbO
2
, respectively.
Fig. 3.9
Commonly used
mercury cell. The
reducing agent is
zinc and the
oxidising agent is
mercury (II) oxide.
3.6.2 Secondary
Batteries
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90Chemistry
Positive plate
Separator
Negative plate
Another important secondary
cell is the nickel-cadmium cell
(Fig. 3.11) which has longer life
than the lead storage cell but
more expensive to manufacture.
We shall not go into details of
working of the cell and the
electrode reactions during
charging and discharging.
The overall reaction during
discharge is:
Cd (s) + 2Ni(OH)
3
(s) → CdO (s) + 2Ni(OH)
2
(s) + H
2
O (l)
Production of electricity by thermal plants is not a very efficient method
and is a major source of pollution. In such plants, the chemical energy
(heat of combustion) of fossil fuels (coal, gas or oil) is first used for
converting water into high pressure steam. This is then used to run
a turbine to produce electricity. We know that a galvanic cell directly
converts chemical energy into electricity and is highly efficient. It is
now possible to make such cells in which reactants are fed continuously
to the electrodes and products are removed continuously from the
electrolyte compartment. Galvanic cells that are designed to convert
the energy of combustion of fuels like hydrogen, methane, methanol,
etc. directly into electrical energy are called fuel cells.
One of the most successful fuel cells
uses the reaction of hydrogen with
oxygen to form water (Fig. 3.12). The
cell was used for providing electrical
power in the Apollo space programme.
The water vapours produced during the
reaction were condensed and added to
the drinking water supply for the
astronauts. In the cell, hydrogen and
oxygen are bubbled through porous
carbon electrodes into concentrated
aqueous sodium hydroxide solution.
Catalysts like finely divided platinum or
palladium metal are incorporated into
the electrodes for increasing the rate of
electrode reactions. The electrode
reactions are given below:
Cathode: O
2
(g) + 2H
2
O(l) + 4e
–
→ 4OH
–
(aq)
Anode: 2H
2
(g) + 4OH
–
(aq) → 4H
2
O(l) + 4e
–
Overall reaction being:
2H
2
(g) + O
2
(g) → 2H
2
O(l)
The cell runs continuously as long as the reactants are supplied.
Fuel cells produce electricity with an efficiency of about 70 % compared
Fig. 3.11
A rechargeable
nickel-cadmium cell
in a jelly roll
arrangement and
separated by a layer
soaked in moist
sodium or potassium
hydroxide.
3.7 Fuel Cells3.7 Fuel Cells
3.7 Fuel Cells3.7 Fuel Cells
3.7 Fuel Cells
Fig. 3.12: Fuel cell using H
2
and O
2
produces electricity.
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91
Electrochemistry
Fig. 3.13: Corrosion of iron in atmosphere
Oxidation: Fe (s)→ Fe
2+
(aq) +2e
–
Reduction: O
2
(g) + 4H
+
(aq) +4e
–
→ 2H
2
O(l)
Atomospheric
oxidation: 2Fe
2+
(aq)
+ 2H
2
O(l) + ½O
2
(g) → Fe
2
O
3
(s) + 4H
+
(aq)
to thermal plants whose efficiency is about 40%. There has been
tremendous progress in the development of new electrode materials,
better catalysts and electrolytes for increasing the efficiency of fuel cells.
These have been used in automobiles on an experimental basis. Fuel
cells are pollution free and in view of their future importance, a variety
of fuel cells have been fabricated and tried.
Corrosion slowly coats the surfaces of metallic objects with oxides or
other salts of the metal. The rusting of iron, tarnishing of silver,
development of green coating on copper and bronze are some of the
examples of corrosion. It causes enormous damage
to buildings, bridges, ships and to all objects made
of metals especially that of iron. We lose crores of
rupees every year on account of corrosion.
In corrosion, a metal is oxidised by loss of electrons
to oxygen and formation of oxides. Corrosion of iron
(commonly known as rusting) occurs in presence of
water and air. The chemistry of corrosion is quite
complex but it may be considered
essentially as an electrochemical
phenomenon. At a particular spot
(Fig. 3.13) of an object made of
iron, oxidation takes place and
that spot behaves as anode and
we can write the reaction
Anode: 2 Fe (s) → 2 Fe
2+
+ 4 e
–
2+
(Fe /Fe)
V
E
= – 0.44 V
Electrons released at anodic spot move through the metal and go
to another spot on the metal and reduce oxygen in the presence of H
+
(which is believed to be available from H
2
CO
3
formed due to dissolution
of carbon dioxide from air into water. Hydrogen ion in water may also
be available due to dissolution of other acidic oxides from the
atmosphere). This spot behaves as cathode with the reaction
Cathode: O
2
(g) + 4 H
+
(aq) + 4 e
–
→ 2 H
2
O (l)
+
2 2
H O H O
=1.23 V
| |
E
V
The overall reaction being:
2Fe(s) + O
2
(g) + 4H
+
(aq) → 2Fe
2 +
(aq) + 2 H
2
O (l)
(cell)
V
E
=1.67 V
The ferrous ions are further oxidised by atmospheric oxygen to
ferric ions which come out as rust in the form of hydrated ferric oxide
(Fe
2
O
3
. x H
2
O) and with further production of hydrogen ions.
Prevention of corrosion is of prime importance. It not only saves
money but also helps in preventing accidents such as a bridge collapse
or failure of a key component due to corrosion. One of the simplest
methods of preventing corrosion is to prevent the surface of the metallic
object to come in contact with atmosphere. This can be done by covering
the surface with paint or by some chemicals (e.g. bisphenol). Another
simple method is to cover the surface by other metals (Sn, Zn, etc.) that
are inert or react to save the object. An electrochemical method is to
provide a sacrificial electrode of another metal (like Mg, Zn, etc.) which
corrodes itself but saves the object.
3.83.8
3.83.8
3.8
CorrosionCorrosion
CorrosionCorrosion
Corrosion
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92Chemistry
The Hydrogen EconomyThe Hydrogen Economy
The Hydrogen EconomyThe Hydrogen Economy
The Hydrogen Economy
At present the main source of energy that is driving our economy is fossil fuels
such as coal, oil and gas. As more people on the planet aspire to improve their
standard of living, their energy requirement will increase. In fact, the per
capita consumption of energy used is a measure of development. Of course, it
is assumed that energy is used for productive purpose and not merely wasted.
We are already aware that carbon dioxide produced by the combustion of fossil
fuels is resulting in the ‘Greenhouse Effect’. This is leading to a rise in the
temperature of the Earth’s surface, causing polar ice to melt and ocean levels
to rise. This will flood low-lying areas along the coast and some island nations
such as Maldives face total submergence. In order to avoid such a catastrope,
we need to limit our use of carbonaceous fuels. Hydrogen provides an ideal
alternative as its combustion results in water only. Hydrogen production must
come from splitting water using solar energy. Therefore, hydrogen can be used
as a renewable and non polluting source of energy. This is the vision of the
Hydrogen Economy. Both the production of hydrogen by electrolysis of water
and hydrogen combustion in a fuel cell will be important in the future. And
both these technologies are based on electrochemical principles.
Intext QuestionsIntext Questions
Intext QuestionsIntext Questions
Intext Questions
3.13 Write the chemistry of recharging the lead storage battery, highlighting
all the materials that are involved during recharging.
3.14 Suggest two materials other than hydrogen that can be used as fuels in
fuel cells.
3.15 Explain how rusting of iron is envisaged as setting up of an
electrochemical cell.
SummarySummary
SummarySummary
Summary
An electrochemical cell consists of two metallic electrodes dipping in electrolytic
solution(s). Thus an important component of the electrochemical cell is the ionic
conductor or electrolyte. Electrochemical cells are of two types. In galvanic cell,
the chemical energy of a spontaneous redox reaction is converted into electrical
work, whereas in an electrolytic cell, electrical energy is used to carry out a non-
spontaneous redox reaction. The standard electrode potential for any electrode
dipping in an appropriate solution is defined with respect to standard electrode
potential of hydrogen electrode taken as zero. The standard potential of the cell
can be obtained by taking the difference of the standard potentials of cathode and
anode (
( )
cell
V
E
= E
V
cathode
– E
V
anode
). The standard potential of the cells are
related to standard Gibbs energy (∆
r
G
V
= –nF
( )
cell
V
E
) and equilibrium constant
(∆
r
G
V
= – RT ln K) of the reaction taking place in the cell. Concentration dependence
of the potentials of the electrodes and the cells are given by Nernst equation.
The conductivity,
κ
, of an electrolytic solution depends on the concentration
of the electrolyte, nature of solvent and temperature. Molar conductivity,
Λ
m
, is
defined by =
κ
/c where c is the concentration. Conductivity decreases but molar
conductivity increases with decrease in concentration. It increases slowly with
decrease in concentration for strong electrolytes while the increase is very steep
for weak electrolytes in very dilute solutions. Kohlrausch found that molar
conductivity at infinite dilution, for an electrolyte is sum of the contribution of the
2020-21
93
Electrochemistry
molar conductivity of the ions in which it dissociates. It is known as law of
independent migration of ions and has many applications. Ions conduct electricity
through the solution but oxidation and reduction of the ions take place at the
electrodes in an electrochemical cell. Batteries and fuel cells are very useful
forms of galvanic cell. Corrosion of metals is essentially an electrochemical
phenomenon. Electrochemical principles are relevant to the Hydrogen Economy.
3.1 Arrange the following metals in the order in which they displace each other
from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
3.2 Given the standard electrode potentials,
K
+
/K = –2.93V, Ag
+
/Ag = 0.80V,
Hg
2+
/Hg = 0.79V
Mg
2+
/Mg = –2.37 V, Cr
3+
/Cr = – 0.74V
Arrange these metals in their increasing order of reducing power.
3.3 Depict the galvanic cell in which the reaction
Zn(s)+2Ag
+
(aq) →Zn
2+
(aq)+2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
3.4 Calculate the standard cell potentials of galvanic cell in which the following
reactions take place:
(i) 2Cr(s) + 3Cd
2+
(aq) → 2Cr
3+
(aq) + 3Cd
(ii) Fe
2+
(aq) + Ag
+
(aq) → Fe
3+
(aq) + Ag(s)
Calculate the ∆
r
G
J
and equilibrium constant of the reactions.
3.5 Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s)|Mg
2+
(0.001M)||Cu
2+
(0.0001 M)|Cu(s)
(ii) Fe(s)|Fe
2+
(0.001M)||H
+
(1M)|H
2
(g)(1bar)| Pt(s)
(iii) Sn(s)|Sn
2+
(0.050 M)||H
+
(0.020 M)|H
2
(g) (1 bar)|Pt(s)
(iv) Pt(s)|Br
–
(0.010 M)|Br
2
(l )||H
+
(0.030 M)| H
2
(g) (1 bar)|Pt(s).
3.6 In the button cells widely used in watches and other devices the following
reaction takes place:
Zn(s) + Ag
2
O(s) + H
2
O(l) → Zn
2+
(aq) + 2Ag(s) + 2OH
–
(aq)
Determine ∆
r
G
J
and E
J
for the reaction.
3.7 Define conductivity and molar conductivity for the solution of an electrolyte.
Discuss their variation with concentration.
3.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm
–1
. Calculate
its molar conductivity.
3.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298
K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution
at 298 K is 0.146 × 10
–3
S cm
–1
.
ExercisesExercises
Exercises
Exercises
Exercises
2020-21
94Chemistry
Answers to Some Intext Questions
3.5 E
(cell)
= 0.91 V
3.6
1
r
G 45.54 kJ mol
−
∆ = −
V
, K
c
= 9.62 ×10
7
3.9 0.114, 3.67 ×10
–4
mol L
–1
3.10 The conductivity of sodium chloride at 298 K has been determined at different
concentrations and the results are given below:
Concentration/M 0.001
0.010 0.020 0.050 0.100
10
2
×
κ
/S m
–1
1.237 11.85 23.15 55.53 106.74
Calculate
Λ
m
for all concentrations and draw a plot between
Λ
m
and c
½
.
Find the value of
0
m
Λ
.
3.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10
–5
S cm
–1
. Calculate its
molar conductivity. If
0
m
Λ
for acetic acid is 390.5 S cm
2
mol
–1
, what is its
dissociation constant?
3.12 How much charge is required for the following reductions:
(i) 1 mol of Al
3+
to Al?
(ii) 1 mol of Cu
2+
to Cu?
(iii) 1 mol of MnO
4
–
to Mn
2+
?
3.13 How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl
2
?
(ii) 40.0 g of Al from molten Al
2
O
3
?
3.14 How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H
2
O to O
2
?
(ii) 1 mol of FeO to Fe
2
O
3
?
3.15 A solution of Ni(NO
3
)
2
is electrolysed between platinum electrodes using a
current of 5 amperes for 20 minutes. What mass of Ni is deposited at the
cathode?
3.16 Three electrolytic cells A,B,C containing solutions of ZnSO
4
,
AgNO
3
and CuSO
4
,
respectively are connected in series. A steady current of 1.5 amperes was
passed through them until 1.45 g of silver deposited at the cathode of cell B.
How long did the current flow? What mass of copper and zinc were deposited?
3.17 Using the standard electrode potentials given in Table 3.1, predict if the
reaction between the following is feasible:
(i) Fe
3+
(aq)
and I
–
(aq)
(ii) Ag
+
(aq) and Cu(s)
(iii) Fe
3+
(aq) and Br
–
(aq)
(iv) Ag(s) and Fe
3+
(aq)
(v) Br
2
(aq)
and
Fe
2+
(aq).
3.18 Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO
3
with silver electrodes.
(ii) An aqueous solution of AgNO
3
with platinum electrodes.
(iii) A dilute solution of H
2
SO
4
with platinum electrodes.
(iv) An aqueous solution of CuCl
2
with platinum electrodes.
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