Chemistry

electrical resistance is ohm denoted by the From Eq. (5.2) and Eq. (5.4), we write
symbol Ω (omega). Thus, Ω = VA-1. l 1 l
k=G a = R a (5.5)

The electrical conductance, G, of a Combination of Eq. (5.3) and Eq. (5.5)

solution is reciprocal of resistance. shows that k = 1/ρ.
1
G = R (5.2) Units of electrolytic conductivity

The SI unit of G is siemens, denoted by Quantity SI unit Common unit
Length m cm
S, which is equal to Ω-1. Therefore, we write Area m2 cm2
Resistance
S = Ω-1 = AV-1 = CV-1s-1 where C represents Conductivity Ω Ω
Ω-1 m-1 or Ω-1 cm-1
coulomb, the unit of electricity related to
S m-1
current strength in ampere and time in

seconds as C = A s.

The electrical resistance of a conductor 5.3.2 Molar conductivity (∧) : The electrolytic
conductivity is not suitable for comparing
is proportional to length l and inversely conductivities of different solutions. The
conductivity of a solution depends on number
proportional to cross sectional area a. Thus, of ions present in unit volume of solution.
l l The solution of higher concentration contains
R∝ a or R = ρ a (5.3) more ions and exhibits higher conductivity
than the solution of lower concentration. To
where ρ, the proportionality constant is compare conductivities of different solutions,
they must have the same concentration.
called resistivity of the conductor. It is the

resistance of conductor of unit length and unit

cross sectional area.

Can you recall ?

What is the SI unit of In 1880, the German physicist F.W.G.
resistivity ? Kohlrausch introduced the term molar
conductivity denoted by ∧ (lambda).
5.3.1 Conductivity (k) : We have seen that

G = 1/R and R is directly proportional to The molar conductivity of an electrolytic

length and inversely proportional to its cross solution is the electrolytic conductivity, k,

sectional area. It, therefore, follows that G divided by its molar concentration c.
k
is directly proportional to a and inversely ∧ = c (5.6)

proportional to the length l. Thus SI units of k are S m-1 and that of
a a
G ∝ l or G = k l (5.4) c are mol m-3. Hence SI units of ∧ are

The proportionality constant k is called S m2 mol-1. Common units employed for

conductivity. G = k if length and cross molar conductivity are Ω-1 cm2 mol-1.

sectional area of conductor are unity. Significance of molar conductivity : To
understand the significance of ∧, consider
Thus, conductivity is the electrical volume of a solution containing 1 mole of
conductance of a conductor of unit length dissolved electrolyte. Suppose the solution is
and unit area of cross section. In other words, placed between two parallel electrodes 1 cm
the conductivity is the electrical conductance apart and large enough to accommodate it.
of unit cube of material. Conductivity of The electrical conductance exhibited by this
solution of an electrolyte is called electrolytic solution is the molar conductivity. The molar
conductivity which refers to the electrical conductivity is the electrical conductance
conductance of unit volume (1 m3 or 1 cm3) generated by all the ions in 1 mole of the
of solution. electrolyte.

92

Remember... ∧ = 223 Ω-1 cm2 mol-1, c = 0.05 mol L-1

Conductivity is electrical Hence
conductance due to all the ions in 223 cm2 mol-1 × 0.05 mol L-1
1 cm3 of given solution. Molar conductivity k = Ω-1 1000 cm3L-1
is the electrical conductance due to the ions
obtained from 1 mole of an electrolyte in = 0.01115 Ω-1 cm-1
a given volume of solution.
5.3.4 Variation of conductivity with
concentration

5.3.3 Relation between k and ∧ : Conductivity i. The electrolytic conductivity is electrical
k is the electrical conductance of 1 cm3 of conductance of unit volume (1 cm3) of
solution. If V is volume of solution in cm3 solution. It depends on the number of
containing 1 mole of dissolved electrolyte, current carrying ions present in unit
its electrical conductance is ∧. Each 1 cm3 volume of solution.
portion in the volume V has conductance
k. Hence, total conductance of V cm3 is kV ii. On dilution total number of ions increase
which is molar conductivity. as a result of increased degree of
dissociation.
Thus, we have ∧ = k V (5.7)
iii. An increase in total number of ions is
Concentration of solution not in proportion of dilution. Therefore,
the number of ions per unit volume
= c mol L-1 of solution decreases. This results in
decrease of conductivity with decrease
= c mol L-1 = c mol cm-3 in concentration of solution.
1000 cm3L-1 1000
Suppose 100 cm3 of solution of an
Volume, V of solution in cm3 containing electrolyte contains 8 × 1020 ions. The
number of ions per cm3 is 8 × 1018.
1 mole of an electrolyte is reciprocal of
If the solution is diluted to 1000 cm3
concentration. Therefore, the total number of ions will increase but
not by a factor of 10. Assume that the
V= 1 = 1000 cm3 mol-1 number of ions increases from 8 × 1020
concentraion c to 64 × 1020 on dilution. After dilution the
number of ions per cm3 is 6.8 × 1018.
(5.8)
It is evident that the number of ions
Substitution for V in Eq. (5.7) yields per cm3 decreases from 8 × 1018 to 6.8 ×
1018 on dilution from 100 cm3 to 1000 cm3
∧= 1000k (5.9) and in turn, the conductivity decreases.
c
5.3.5 Variation of molar conductivity with
Try this... concentration

What must be the
concentration of a solution of silver
nitrate to have the molar conductivity of
121.4 Ω-1 cm2 mol-1 and the conductivity of
2.428 × 10-3 Ω-1 cm-1 at 25 0C ?

Problem 5.1 : The molar conductivity of i. The molar conductivity is the electrical
0.05 M BaCl2 solution at 250C is 223 Ω-1 conductance of 1 mole of an electrolyte
cm2 mol-1. What is its conductivity ? in a given volume of solution.

Solution : ii. The increasing number of ions produced
1000k ∧c in solution by 1 mole of the electrolyte
∧= c or k = 1000 lead to increased molar conductivity.

93

5.3.6 Variation of molar conductivity with Molar conductivity of strong electrolytes
concentration : The variation of molar
conductivity with concentration in case of at zero concentration can be determined by
strong and weak electrolytes is qualitatively
different. extrapolation of linear part of ∧ versus c

i. Strong electrolytes : The molar conductivity curve as shown in Fig. 5.1. This method
of solution of strong electrolyte increases
rapidly with dilution. It approaches the limiting cannot be used for weak electrolytes since ∧
value for 0.001 M or 0.0001 M solution. The
dilution has no effect on molar conductivity versus c curve does not approach linearity.
thereafter. The maximum limiting value of
molar conductivity is the molar conductivity Kohlrausch law is useful for calculating ∧0
at zero concentration or at infinite dilution.
It is denoted by ∧0. The zero concentration of weak electrolytes.
or infinite dilution means the solution is so
dilute that further dilution does not increase 5.3.7 Kohlrausch law of independent
the molar conductivity. migration of ions : The law states that at
infinite dilution each ion migrates independent
of co-ion and contributes to total molar
conductivity of an elctrolyte irrespective of the
nature of other ion to which it is associated.

During nineteenth century F. Kohlrausch Both cation and anion contribute to
with repeated experiments showed that the molar conductivity of the electrolyte at zero
molar conductivity of strong electrolytes varies
linearly with square root of concentration as : concentration and thus ∧0 is sum of molar

conductivity of cation and that of the anion
at zero concentration. Thus,

∧0 = n λ0⊕ + n λ0 (5.11)
⊕

∧ = ∧0 - a c (5.10) where λ⊕ and λ are molar conductivities of
cation and anion, respectively, and n and
where a is constant. For strong
⊕

n are the number of moles of cation and

electrolytes a plot of ∧ versus c is linear as anion, specified in the chemical formula of

shown in Fig. 5.1. the electrolyte.

Applications of Kohlrausch theory

Strong 1. The theory can be used to calculate the
electrolyte molar conductivity of an electrolyte at
the zero concentration. For example,

Weak ∧0 (KCl) = λ0K⊕+ λ0Cl
electrolyte
∧0 [Ba(OH)2] = λ0Ba2⊕+ 2 λ0OH
Fig. 5.1 : Variation of ∧ with c
Knowing the molar conductivites of ions

at infinite dilution, ∧0 values of electrolyte

can be obtained.

ii. Weak electrolytes : The molar conductivity 2. The theory is particularly useful in
of weak electrolytes increases rapidly on
dilution. For concentrations of 0.001M or calculating ∧0 values of weak electrolytes

0.0001 M, the ∧ value is lower than ∧0 the from those of strong electrolytes. For

molar conductivity at zero concentration. example, ∧0 of acetic acid can be

calculated by knowing those of HCl, NaCl

For weak electrolytes the variation of ∧ and CH3COONa as described below :

with c shown in Fig. 5.1 is not linear. ∧0 (HCl) + ∧0 (CH3COONa) - ∧0 (NaCl)

= λ0H⊕ + λ0Cl + λ0 + λ0 - λ0 - λ0Cl
CH3COO Na⊕ Na⊕

94

= λ0H⊕ + λ0 = ∧0 (CH3COOH) Problem 5.3 : Calculate molar conductivities
CH3COO at zero concentration for CaCl2 and Na2SO4.
Given : molar ionic conductivitis of Ca2⊕,
Thus, Cl , Na⊕ and SO42 ions are respectively,
104, 76.4, 50.1 and 159.6 Ω-1 cm2 mol-1.
∧0 (CH3COOH) = ∧0 (HCl) + ∧0 (CH3COONa)
- ∧0 (NaCl) Solution :

Because ∧0 values of strong electrolytes, According to Kohrausch law,

HCl, CH3COONa and NaCl, can be determined i. ∧0 (CaCl2) = λ0 + 2λ0Cl
Ca2⊕
by extrapolation method, the ∧0 of acetic acid

can be obtained.

Problem 5.2 : Calculate the molar = 104 Ω-1 cm2 mol-1 + 2 × 76.4 Ω-1 cm2 mol-1
conductivity of AgI at zero concentration
if the molar conductivities of NaI, AgNO3 = 256.8 Ω-1 cm2 mol-1
and NaNO3 at zero concentration are
respectively, 126.9, 133.4 and 121.5 Ω-1 ii. ∧0 (Na2SO4) = 2λ0Na⊕ + λ0SO42
cm2 mol-1.
= 2 × 50.1 Ω-1 cm2 mol-1
+ 159.6 Ω-1 cm2 mol-1

Solution : = 259.8 Ω-1 cm2 mol-1

According to Kohrausch law, Problem 5.4 : The molar conductivity of
0.01M acetic acid at 25 0C is 16.5 Ω-1 cm2
i. ∧0 (NaI) = λ0 + λ0I mol-1. Calculate its degree of dissociation in
Na⊕ 0.01 M solution and dissociation constant
if molar conductivity of acetic acid at zero
= 126.9 Ω-1 cm2 mol-1 concentration is 390.7 Ω-1 cm2 mol-1.

ii. ∧0 (AgNO3) = λ0 + λ0
Ag⊕ NO3

= 133.4 Ω-1 cm2 mol-1

iii. ∧0 (NaNO3) = λ0 + λ0 Solution :
Na⊕ NO3

= 121.5 Ω-1 cm2 mol-1 The degree of dissociation,

Eq. (i) + eq. (ii) - eq. (iii) gives ∝ = ∧c
∧0
∧0 (NaI) + ∧0 (AgNO3) - ∧0 (NaNO3)
16.6 Ω-1 cm2 mol-1
= λ0 + λ0I + λ0 + λ0 - λ0 -λ0NO3 = 390.7 Ω-1 cm2 mol-1 = 0.0422
Na⊕ Ag⊕ NO3 Na⊕

= λ0 + λ0I
Ag⊕

= ∧0 (AgI) Ka = ∝2c = (0.0422)2 × 0.01 = 1.85 × 10-5
1- ∝ (1 - 0.0422)
= 126.9 Ω-1 cm2 mol-1 + 133.4 Ω-1 cm2 mol-1

- 121.5 Ω-1 cm2 mol-1 5.3.8 Molar conductivity and degree of

= 138.8 Ω-1 cm2 mol-1 dissociation of weak electrolytes : The

degree of dissociation (∝) of weak electrolyte

Try this... is related to its molar conductivity at a given

Calculate ∧0 (CH2ClOOH) concentration c by the equation,
if ∧0 values for HCl, KCl and ∧c
∝ = ∧0 (5.12)
CH2ClCOOK are repectively, 4.261,
1.499 and 1.132 Ω-1 cm2 mol-1. where ∧c is the molar conductivity of
weak electrolyte at concentration c ; ∧0 is
molar conductivity at zero concentration.

95

Try this... 1. Determination of cell constant : The cell
constant is determined using the 1 M, 0.1 M
Obtain the expression for or 0.01 M KCl solutions. The conductivity
of KCl solution is well tabulated at various
dissociation constant in terms of ∧c temperatures. The resistance of KCl solution
and ∧0 using Ostwald's dilution law. is measured by Wheatstone bridge. (Refer to
standard XII Physics Textbook Chapter 9)
5.3.9 Measurement of conductivity : The
conductivity of a solution can be determined In Fig. 5.3 AB is the uniform wire. Rx is
from the resistance measurements by the variable known resistance placed in one
Wheatstone bridge. arm of Wheatstone bridge.

Conductivity Cell : The conductivity cell conductivity cell
consists of a glass tube with two platinum
plates coated with a thin layer of finely solution of unknown A.C.
divided platinium black. This is achieved by resistance
the electrolysis of solution of chloroplatinic
acid. The cell is dipped in a solution whose
resistance is to be measured as shown in Fig.
5.2.

Fig. 5.3 : Measurement of resistance

The conductivity cell containing KCl
solution of unknown resistance is placed in
the other arm of Wheatstone bridge. D is a
current detector. F is the sliding contact that
moves along AB. A.C. represents the source
of alternating current.

Fig. 5.2 : Conductivity cell The sliding contact is moved along AB
until no current flows. The detector D shows
Cell constant : The conductivity of an no deflection. The null point is, thus, obtained
at C.
electrolytic solution is given by Eq. (5.5),

k = 1 l According to Wheatstone bridge principle,
R a
Rsolution Rx
For a given cell, the ratio of separation l (AC) = l (BC)

(l) between the two electrodes divided by the l(AC)
Hence, Rsolution = l(BC) × Rx
area of cross section (a) of the electrode is (5.15)

called the cell constant. Thus,
l
Cell constant = a (5.13) By measuring lengths AC and BC and
knowing Rx, resistance of KCl solution can
SI unit of cell constant is m-1 which is be calculated. The cell constant is given by
Eq. (5.13).
conveniently expressed in cm-1. The Eq. (5.5)

then becomes Cell constant = kKCl × Rsolution
cell constant
k= R (5.14) The conductivity of KCl solution is
known. The cell constant, thus, can be
The determination of conductivity consists calculated.

of three steps :

96

2. Determination of conductivity of given 5.4.1 Electrochemical reactions :

solution : KCl solution in the conductivity Can you recall ?

cell in step (1) is replaced by the given What is the reaction involving
transfer of electrons from one
solution whose conductivity is to be measured. chemical species to another called?

Its resistance is measured by the process The chemical reaction occuring in
electrochemical cells involves transfer
described in step (1). The conductivity of of electrons from one species to the
other. It is a redox reaction, we learnt in
given solution is then calculated as : (Std. XI, Chapter 6).
Cell constant
k = Rsolution Electrochemical reactions, are made of
oxidation and reduction half reactions. The
3. Calculation of molar conductivity : The oxidation half reaction occurs at one electrode
and the reduction half reaction occurs at the
molar conductivity of the given solution is other electrode. The net cell reaction is the
sum of these half reactions.
then calculated using Eq. (5.9).
5.4.2 Electrodes : Electrodes are the surfaces
∧ = 1000 k on which oxidation and reduction half
c reactions take place. Electrodes may or may
not participate in the reactions. The electrodes
Problem 5.5 : Aconductivity cell containing which do not take part in reactions are inert
0.01M KCl gives at 250C the resistance electrodes.
of 604 ohms. The same cell containing
0.001M AgNO3 gives resistance of 6530 Cathode : It is an electrode at which the
ohms. Calculate the molar conductivity of reduction takes place. At this electrode the
0.001M AgNO3. [Conductivity of 0.01M species undergoing reduction gains electrons.
KCl at 25 0C is 0.00141 Ω-1 cm-1]
Anode : It is an electrode at which oxidation
Solution : takes place. At this electrode, the species
i. Calculation of cell constant undergoing oxidation loses electrons.

Cell constant = kKCl × RKCl 5.4.3 Types of electrochemical cells : There
are two types of electrochemical cells.
= 0.00141 Ω- cm-× 604 Ω
1. Electrolytic cell : In this type of cell,
= 0.852 cm-1 a nonspontaneous reaction, known as
electrolysis, is forced to occur by passing a
ii. Calculation of conductiviy of AgNO3' direct current from an external source into
the solution. In such cells electrical energy
k = Cell constant where R = 6530 Ω is converted into chemcial energy. The anode
R of electrolytic cell is positive and cathode is
negative.
= 0.852 cm-1
6530 Ω 2. Galvanic or voltaic cell : In galvanic
(voltaic) cell a spontaneous chemical reaction
= 1.3 × 10-4 Ω-1 cm-1 produces electricity. In these cells chemical
energy is converted into electrical energy.
iii. Calculation of molar conductivity of The anode of galvanic cell is negative and
cathode is positive.
AgNO3 1000k where c = 0.001 M
∧= c

= 1000 cm3 L-1 × 1.3 × 10-4 Ω-1 cm-1
0.001 mol L-1

= 130 Ω-1 cm2 mol-1

5.4 Electrochemical cells : An electrochemical
cell consists of two metal plates or carbon
(graphite) rods. These electronic conductors
are dipped into an electrolytic or ionic
conductor.

97

Use your brain power The carbon electrode connected to
terminal electrode of the battery is anode
Distinguish between and that connected to negative terminal of
the battery is cathode.
electrolytic and galvanic cells.

5.5 Electrolytic cell Remember...
In electrolysis the electrodes
Do you know ?
Michael Faraday was the are usually inert, Pt or graphite.

first person to explain electrolysis Reactions occuring in the cell : Fused NaCl
nearly 200 years ago. contains Na⊕ and Cl ions which are freely
mobile. When potential is applied, cathode
Electrolytic cell consists of a container attracts Na⊕ ions and anode attracts Cl ions.
in which electrolyte is placed. Two electrodes As these are charged particles, their migration
are immersed in the electrolyte and connected results in an electric current. When these
to a source of direct current. ions reach the respective electrodes, they
are discharged according to the following
At anode (+) a species oxidises with reactions.
the removal of electrons. These electrons are
pulled from anode and pushed to cathode Oxidation half reaction at anode :
through an external circuit. The electrons Cl ions migrate to anode. Each Cl
are supplied to species at cathode which are
reduced. ion, that reaches anode, gives one electron
to anode. It oxidises to neutral Cl atom in the
Remember... primary process. Two Cl atoms then combine
Electrolysis is the process to form chlorine gas in the secondary process.

of breaking down of an ionic 2 Cl (l) Cl (g) + Cl (g) + 2e
compound in molten state or in aqueous (primary process)
solution by the passage of electricity.
Cl (g) + Cl (g) Cl2 (g)
5.5.1 Electrolysis of molten NaCl (secondary process)

Construction of cell : The electrolytic cell 2Cl (l) Cl2 (g) + 2e
consists of a container in which fused NaCl is (overall oxidation)
placed. Two graphite electrodes are immersed
in it. They are connected by metallic wires to The battery sucks electrons so produced
a source of direct current that is battery. This at the anode and pushes them to cathode
is shown in Fig. 5.4. through a wire in an external circuit. The
battery thus serves as an electron pump. The
Battery electrons from the battery enter into solution
(D.C. source) through cathode and leave the solution
Carbon cathode through anode.

Carbon anode Reduction half reaction at cathode : The
electrons supplied by the battery are used
Cl2 gas Fused Na in cathodic reduction. Each Na⊕ ion, that
Fused NaCl reaches cathode accepts an electron from the
cathode and reduces to metallic sodium.

Na⊕ (l) +e Na (l)

Fig. 5.4 : Electrolysis of fused NaCl

98

Net cell reaction The other is the reduction of water to
hydrogen gas.
The net cell reaction is the sum of two
electrode reactions. ii. 2 H2O (l) + 2e H2 (g) + 2 OH (aq),
E0 = - 0.83 V
2 Cl (l) Cl2 (g) + 2e
(oxidation half reaction)
The standard potential (section 5.7.1) for
2 Na⊕ (l) + 2e 2 Na (l) the reduction of water is higher than that for
(reduction half reaction) reduction of Na⊕. This implies that water has
much greater tendency to get reduced than the
2 Na⊕ (l) + 2 Cl (l) 2 Na (l) + Cl2(g) Na⊕ ion. Hence reaction (ii), that is, reduction
of water is the cathode reaction when the
(overall cell reaction) aqueous NaCl is electrolysed.

Results of electrolysis of molten NaCl

i. A pale green Cl2 gas is released at anode. Oxidation half reaction at anode : At anode

ii. A molten silvery-white sodium is formed there will be competition between oxidation
at cathode.
of Cl ion to Cl2 gas as in case of molten
Decomposition of NaCl into metallic NaCl and the oxidation of water to O2 gas.
sodium and Cl2(g) is nonspontaneous. The
electrical energy supplied by the battery i. 2 Cl (aq) Cl2 (g) +2e ,E0oxi = - 1.36 V
forces the reaction to occur.
ii. 2H2O (l) O2 (g) + 4H⊕ (aq) + 2e

E0 = - 0.4 V
oxi

Remember... Standard electrode potential for the

When molten ionic compound oxidation of water is greater than that of Cl
is electrolysed, a metal is formed
at the negative electrode and a nonmetal ion or water has greater tendency to undergo
at the positive electrode.
oxidation. It is, therefore, expected that anode

half reaction would be oxidation of water.

The experiments have shown, however, that

5.5.2 Electrolysis of aqueous NaCl : the gas produced at the anode is Cl2 and
Electrolysis of an aqueous NaCl can be not O2. This suggests that anode reaction is
carried out in the cell used for the electrolysis oxidation of Cl to Cl2 gas. This is because
of molten NaCl using inert electrodes shown of the overvoltage, discussion of which is
in Fig. 5.4. The fused NaCl is replaced by
moderately concentrated aqueous solution of beyond the scope of the present book.
NaCl. The water involved in electrolysis of
aqueous NaCl, leads to electrode reactions It has been found experimentally
that differ from electrolysis of molten NaCl. that the actual voltage required for
electrolysis is greater than that calculated
Reduction half reaction at cathode : At using standard potentials. This additional
cathode, two reduction reactions compete. voltage required is the overpotential.
One is the reduction of Na⊕ ions as in case
of molten NaCl. Do you know ?

i. Na⊕ (aq) + e Na (s), E0 = -2.71 V Refining of metal and
electroplating are achieved by
electrolysis.

99

Overall cell reaction Q (C) = I (A) × t (s) (5.16)

It is the sum of electrode reactions. ii. Calculation of moles of electrons passed

2 Cl (aq) Cl2 (g) + 2e Total charge passed is Q(C). The charge
(oxidation at anode) of one mole electrons is 96500 coulombs (C).
It is referred to as one faraday (IF). Hence,
2 H2O (l) + 2e H2 (g) + 2 OH (aq)
(reduction at cathode) Moles of electrons actually passed
Q(C)
2 Cl (aq) + 2 H2O (l) Cl2 (g) + H2(g) = 96500 (C/mol e ) (5.17)
+ 2 OH (aq)
iii. Calculation of moles of product formed
(overall cell reaction)
The balanced equation for the half
Results of electrolysis of aqueous NaCl reaction occuring at the electrode is devised.
The stoichiometry of half reaction indicates
i. H2 gas is liberated at cathode. the moles of electrons passed and moles of
the product formed. For the reaction,
ii. Cl2 gas is released at anode.
Cu2⊕ (aq) + 2e Cu (s), two moles of
iii. Because Na⊕ ions remain unreacted and
OH ions are formed at cathode, NaCl electrons are required for the production of
solution is converted to NaOH solution.
one mole of Cu. So we can calculate the moles

Do you know ? of product formed. The moles of electrons

Sea water is the source of actually passed are given by Eq. (5.16).
300000 tones of Mg produced
every year by electrolysis. To simplify further we introduce the
entity mole ratio given by
Electrochemical art : Al, Cr and Sn
can be coloured by an electrochemical Mole ratio =
process called anodizing. In this process
metal anode oxidizes to give metal oxide moles of product formed in the half reaction
coat. When an organic dye is added to the
electrolyte, dye molecules soak forming moles of electrons required in the half reaction
spongy surface of coating and become 1
trapped with the hardening of the metal For the reaction of Cu, mole ratio = 2
oxide surface.
Therefore,

Moles of product formed

= moles of electrons actually passed × mole

Q(C) ratio
96500 (C/mol
5.5.3 Quantitative aspects of electrolysis : = e ) × mole ratio (5.18)

a. The mass of reactant consumed or =965I 0(0A)(C×/mt o(ls)e ) × mole ratio (5.19)
the mass of product formed at an electrode
during electrolysis can be calculated by iv. Calculation of mass of product :
knowing stoichiometry of the half reaction
at the electrode. Mass of product

i. Calculation of quantity of electricity W = moles of product × molar mass of product
passed : To calculate the quantity of
eletricity (Q) passed during electrolysis, the = I (A) × t (s) ) × mole ratio × molar mass
amount of current, I, passed through the cell 96500 (C/mol e of product
is measured. The time for which the current
is passed is noted. (5.20)

100

b. Suppose two cells containing different ii. Mass of Cu formed,
electrolytes are connected in series. The same
quantity of electricity is passed through them. W=
The masses of the substances liberated at the I (A) × t (s)
electrodes of the two cells are related as given 96500 (C/mol e ) ×mole ratio×molarmass
below : of Cu

The mass of the substance produced at = 5 A × 100 × 60 s × 1 mol
96500 (C/mol e-) 2 mol e-
the electrode of first cell is given by
Q(C) × 63.5 g mol-1
96500 (C/mol
W1 = e )× (mole ratio)1 × M1 = 9.87 g

Hence, Q(C) e ) Problem 5.7 : How long will it take to
96500 (C/mol produce 2.415 g of Ag metal from its salt
solution by passing a current of 3 ampere ?
= (mole W1 × M1 Molar mass of Ag is 107.9 g mol-1.
ratio)1

Similarly mass of substance liberated Solution :

at the electrode of second cell is W2 in the i. Stoichiometry :
equation,
Q(C) W2
96500 (C/mol e ) = (mole ratio)2 × M2 Ag⊕ (aq) + e Ag (s)

M1 and M2 are the molar masses of mole ratio = 1 mol
substances produced at the electrodes of cells 1 mol e

1 and 2. ii. W =

Because Q(C) is the same I (A) × t (s) ) × mole ratio × molar mass
for both, 96500 (C/mol e ) 96500 (C/mol e of Ag

We have 2.415 g = 3A×t ) × 1 mol
96500 (C/mol e 1 mol e

(mole W1 × M1 = (mole W2 × M2 × 107.9 g mol-1
ratio)1 ratio)2

(5.21) t = 2.415 × 96500 (C = As)
3 A × 107.9

Problem 5.6 : What is the mass of Cu = 720 s = 12 min.
metal produced at the cathode during
the passage of 5 ampere current through Do you know ?
CuSO4 solution for 100 minutes. Molar Names, galvanic or voltaic
mass of Cu is 63.5 g mol-1.
are given in honour of Italian
Solution : scientists L. Galvani and A. Volta for their
work in electrochemistry.
i. Stoichiometry for the formation of Cu is
Cu2⊕ (aq) + 2 e = Cu (s)

Hence, 1 mol
2 mol e
mole ratio =

101

Problem 5.8 : How many moles of Substitution of the quantities gives
electrons are required for reduction of 3
moles of Zn2⊕ to Zn ? How many Faradays 1 4.36g g mol-1 =
of electricity will be required ? mol/2mol e × 65.4

Solution : W2
1 mol/3mol e- × 27 g mol-1

i. The balanced equation for the reduction or 4.36 g × 2 = W2 × 3
of Zn2⊕ to Zn is 65.4 27

Zn2⊕ (aq) + 2e Zn (s)

The equation shows that 1 mole of Zn2⊕ Hence, W2 = 4.36 g × 2 × 27 = 1.2 g
65.4 × 3
is reduced to Zn by 2 moles of electrons.

For reduction of 3 moles of Zn2⊕ 6 moles

of electrons will be required. 5.6 Galvanic or voltaic cell : In galvanic or
voltaic cells, electricity is generated through
Faraday (96500 Coulombs) is the amount the use of spontaneous chemical reactions.

of charge on one mole of electrons. A galvanic (or voltaic) cell is made of
two half cells. Each half cell consists of a
Therefore, for 6 moles of electrons, 6 F metal strip immersed in the solution of its own
ions of known concentration. For example, a
electricity will be required. strip of zinc metal immersed in 1 M aqueous
solution of zinc ions forms an half cell.
Problem 5.9 : In a certain electrolysis
experiment 4.36 g of Zn are deposited It follows that two metal plates and
in one cell containing ZnSO4 solution. the solutions of their ions with known
Calculate the mass of Al deposited in concentrations are required for the construction
another cell containing AlCl3 solution of a galvanic (voltaic) cell. Two half cells
connected in series with ZnSO4 cell. Molar are constructed by immersing the two metal
masses of Zn and Al are 65.4 g mol-1 and plates in solutions of their respective ions
27 g mol-1, respectively. placed in separate containers. The two half
cells so constructed are combined together
Solution : to form the galvanic cell. The metal plates
called electrodes are connected through
Cell 1 : voltmeter by a conducting wire for transfer
of electrons between them. To complete the
Zn2⊕ (aq) + 2e Zn (s) circuit the two solutions are connected by
conducting medium through which cations
(mole ratio)1 = 1 mol and anions move from one compartment to
2 mol e the other. This requirement is fulfilled by a
salt bridge.
Cell 2 :

Al3⊕ (aq) + 3e Al (s)

(mole ratio)2 = 1 mol
3 mol e

(mole W1 × M1 = (mole W2 × M2 5.6.1 Salt bridge : In a galvanic cell, the two
ratio)1 ratio)2 solutions are connected by a salt bridge. It is
an U tube containing a saturated solution of
W1 = 4.36 g, M1 = 65.4 g mol-1, an inert electrolyte such as KCl or NH4NO3
M2 = 27 gmol-1 and 5 % agar solution. The ions of electrolyte
do not react with the ions of electrode
solutions or the electrodes.

102

Salt bridge is prepared by filling a U i. The metal electrodes or the inert electrodes
tube with hot saturated solution of the salt are denoted by vertical lines placed at the
and agar agar solution allowing it to cool. ends of the formula or the short notation.
The cooled solution sets into a gel which does The anode (-) is written at the extreme
not come out on inverting the tube. The salt left and cathode (+) at extreme right.
bridge is kept dipped in distilled water when
not in use as shown in Fig. 5.5. ii. The insoluble species if any or gases are
placed in the interior position adjacent to
U tube Saturated KCl the metal electrodes.
solution + agar gel
iii. The aqueous solutions of ions are placed
Beaker at the middle of the cell formula.

Distilled water iv. A single vertical line between two phases
indicates the phase boundary. It indicates
Glass the direct contact between them.
wool plugs
v. A double vertical line between two
solutions indicates that they are connected
by salt bridge.

Fig. 5.5 : Salt bridge vi. The additional information such as
concentration of solutions and gas
Try this... pressures is also given.

Salt bridge can be prepared vii. A single half cell is written in the order:
with a laminated long strip of good aqueous solution of ions first and then
quality filter paper. Cut the two ends of the solid electrode.
a laminated strip. Dip the two ends in a
saturated solution of KCl for 24 hours. For example Zn2⊕(1M) Zn (s). This
This strip can be used as salt bridge by order is reversed when the electrode
dipping the two ends in two solutions. acts as anode in the cell. The following
example illustrates these conventions.
Functions of salt bridge The cell composed of Mg (anode) and
Ag (cathode) consists of two half cells,
The salt bridge serves the following Mg2⊕ (1M) Mg (s) and Ag⊕ (1M)Ag(s).
functions : The cell is represented as :

i. It provides an electrical contact between Mg (s) Mg2⊕ (1M) Ag⊕(1M) Ag(s).
two solutions and thereby completes the
electrical circuit. Can you tell ?
You have learnt Daniel cell
ii. It prevents mixing of two solutions.
in XI th standard. Write notations
iii. It maintains electrical neutrality in both for anode and cathode. Write the cell
the solutions by transfer of ions. formula.

5.6.2 Formulation or short notation of 5.6.3 Writing of cell reaction : The cell
galvanic cells : A galvanic cell is represented reaction corresponding to the cell notation is
by a formula or short notation that includes written on the assumption that the right hand
electrodes, aqueous solutions of ions and other side electrode is cathode (+) and left hand
species which may or may not be involved in side electrode is anode (-).
the cell reaction. The following conventions
are used to write the cell notation.

103

As mentioned in section 5.4.2, oxidation Try this...
half reaction occurs at anode and reduction
half reaction at cathode. It, therefore, follows Write electrode reactions and
that in galvanic cell oxidation half reaction overall cell reaction for Daniel cell
takes place on the left hand side electrode you learnt in standard XI.

and reduction half reaction on the right hand 5.7 Electrode potential and cell potential
side electrode. : A galvanic cell is composed of two half
cells, each consisting of electronic (metal
The following steps are followed to write plates) and electrolytic (solution of ions)
the cell reaction. conductors in contact. At the surface of
i. Write oxidation half reaction for the left separation of solid metal and the solution,
there exists difference of electrical potential.
hand side electrode and reduction half
reaction for the right hand side electrode.

ii. Add two electrode half reactions to get This potential difference established due

the overall cell reaction. While adding to electrode half reaction occurring at the

the electrons must be cancelled. For this electrode surface, is the electrode potential.

purpose, it may be necessary to multiply The potential is associated with each of
one or both the half reactions by a suitable the half reaction, be it oxidation or reduction.
numerical factor (s). No electrons should The potential associated with oxidation
appear in the overall reaction. reaction is oxidation potential while that
associated with reduction gives the reduction
iii. It is important to note that the individual potential. The overall cell potential, also
called electromotive force (emf), is made of
half reactions may be written with one the contributions from each of the electrodes.
In other words, the cell potential is algebraic
or more electrons. For example, half sum of the electrode potentials,

reactions for H2 gas, whether written as Ecell = Eoxi (anode) + Ered (cathode)
2H⊕ (aq) + 2e H2 (g) or H⊕(aq) +

e 1/2 H2 (g) makes no difference.
In writing the overall cell reaction, the

electrons must be balanced.

Consider the cell, (5.22)

e where Eoxi is the oxidation potential of
Ni (s)Ni2⊕ (1M)Al3⊕ (1M)Al (s) anode (-) and Ered is the reduction potential
The oxidation at anode is of cathode (+).

Ni (s) Ni2⊕ (1M) + 2e When galvanic cell operates, electrons

The reduction half reaction at cathode is are generated at the anode. These electrons

Al3⊕ (1M) + 3 e Al (s). move through external circuit to the cathode.

To balance the electrons, oxidation The cell potential is the force that pushes

reaction is multiplied by 3 and reduction electrons away from anode (-) and pulls them

reaction by 2. The two half reactions so toward cathode where they are consumed.

obtained when added give the overall cell 5.7.1 Standard potentials : The electrode
reaction. Thus, potential and the cell potential depend on
concentrations of solutions, pressures of gases
3 Ni (s) 3 Ni2⊕ (1M) + 6 e and the temperature. To facitilate comparison
of different galvanic cells, it is necessary to
(oxidation half reaction) measure the cell voltage under given set of

2 Al3⊕ (1M) + 6 e 2 Al (s)

(reduction half reaction)

3 Ni (s) + 2Al3⊕ (1M) 3Ni2⊕ (1M) + 2 Al (s) standard conditions of concentration and
(overall cell reaction) temperature.

104

The standard conditions chosen are 1 5.7.2 Dependence of cell potential on
M concentration of solution, 1 atm pressure concentration (Nernst equation) : The
for gases, solids and liquids in pure form standard cell potential tells us whether or
and 250C. The voltage measured under not the reactants in their standard states
these conditions is called standard potential form the products in their standard states
designated as E0. spontaneously. To predict the spontaneity of
reactions for anything other than standard
The standard cell potential is the concentration conditions we need to know
algebraic sum of the standard electrode how voltage of galvanic cell varies with
potentials similar to Eq. (5.22). concentration.

E0 = E0 (anode) + E0 (cathode) Dependence of cell voltage on
cell oxi red concentrations is given by Nernst equation.
(5.23) For any general reaction,

Here E0 is standard oxidation potential
oxi
and E0 is the standard reduction potential.
red
aA + bB cC + dD
According to IUPAC convention,
standard potential of an electrode is taken The cell voltage is given by
as the standard reduction potential.

It must be realised that standard oxidation Ecell = E0 - RT 1n [C]c [D]d
potential of any electrode is numerically cell nF [A]a [B]b
equal to its standard reduction potential with
the reversal of sign. For example standard = E0 - 2.303RT log10 [C]c [D]d
oxidation potential of Zn2⊕ (1M) Zn electrode cell nF [A]a [B]b
is 0.76V. Its standard reduction potential will
be -0.76 V. Hereafter the standard reduction (5.25)
potential will be called standard potential, the
voltage associated with a reduction reaction. where n = moles of electrons used in

the reaction, F = Faraday = 96500 C, T =

It follows that the standard cell potential temperature in kelvin, R = gas constant =

(emf) is written in terms of the standard 8.314 J K-1mol-1

potentials of the electrodes. In Eq. (5.23), At 25 0C, 2.303RT = 0.0592 V
F
E0oxi(anode) is replaced by - E0 (anode).
red
We then write,

E0 = - E0 (anode) + E0 (cathode) Therefore at 25 0C, eq. (5.24) becomes
cell red red

Omitting the subscript red, we have 0.0592V [C]c [D]d
n [A]a [B]b
E0 = E0 (cathode, +) - E0 (anode, -) Ecell = E0 - log10
cell cell
(5.24)

Remember... (5.26)
• While constructing a galvanic
The Eq. (5.25) or Eq. (5.26) is the
cell from two electrodes, the Nernst equation. The first term on the right
electrode with higher standard hand of Nernst equation represents standard
potential is cathode (+) and that with state electrochemical conditions. The second
lower standard potential is anode (-). term is the correction for non standard state
conditions. The cell potential equals standard
• The difference in electrical potential potential if the concentrations of reactants
between anode and cathode is cell and products are 1 M each. Thus,
voltage.

105

if [A] = [B] = [C] = [D] = 1M, Problem 5.10 : Calculate the voltage

Ecell = E0 of the cell, Sn (s)Sn2⊕ (0.02M)Ag⊕
cell
(0.01M)Ag (s) at 25 0C.
If a gaseous substance is present in the
E0Sn = - 0.136 V, E0Ag = 0.800 V.
cell reaction its concentration term is replaced
Solution :
by the partial pressure of the gas.

The Nernst equation can be used to First we write the cell reaction.
calculate cell potential and electrode potential.
Sn (s) Sn2⊕ (0.02M) + 2 e

i. Calculation of cell potential : Consider the (oxidation at anode)
cell
[Ag⊕ (0.01M) + e Ag (s)] × 2
Cd(s)Cd2⊕(aq) Cu2⊕ (aq) Cu.
Let us first write the cell reaction (reduction at cathode)

Cd (s) Cd2⊕ (aq) + 2 e Sn (s) + 2 Ag⊕ (0.01M)
Sn2⊕ (0.02 M) + 2 Ag (s)

(oxidation at anode) (overall cell reaction)

Cu2⊕ (aq) + 2 e Cu (s) The cell potential is given by

(reduction at cathode) Ecell = E0 - 0.0592V log10 [Sn2⊕]
cell 2 [Ag⊕]2

Cd (s) + Cu2⊕ (aq) Cd2⊕ (aq) + Cu (s) E0 = E0Ag - E0Sn = 0.8 V + 0.136 V
cell

(overall cell reaction) = 0.936 V

Here n = 2 Hence, 0.0592V log10 0.02
Ecell = 0.936V - 2 (0.01)2
The potential of cell is given by Nernst
0.0592V
equation, [Cd2⊕] = 0.936V - 2 log10 200
[Cu2⊕]
Ecell = E0 - 0.0592 log10 0.0592V
cell 2 2
= 0.936V - × 2.301
at 25 0C.

(Concentration of solids and pure liquids = 0.936 V - 0.0681V = 0.8679 V
are taken to be unity.)
Problem 5.11 : The standard potential
ii. Calculation of electrode potential of the electrode, Zn2⊕ (0.02 M) Zn (s)
is - 0.76 V. Calculate its potential.
Consider Zn2⊕(aq)Zn(s)

The reduction reaction for the electrode is Solution :

Zn2⊕ (aq) + 2 e Zn (s) Electrode reaction :

Applying Nernst equation, electrode potential Zn2⊕ (0.02M) + 2 e Zn (s)

is given by 1 0.0592V 1
[Zn2⊕] n [Zn2 ⊕]
EZn = E0Zn - 0.0592 log10 EZn = E0Zn - log10
2

= E0Zn + 0.0592 log10 [Zn2⊕] at 25 0C = - 0.76 V + 0.0592V log10 (0.02)
2 2

= - 0.76 V + 0.0592V × (-1.6990)
2
= - 0.76 V - 0.0503V = - 0.81 V

106

5.8 Thermodynamics of galvanic cells Remember...

5.8.1 Gibbs energy of cell reactions and For chemical reaction to be
cell potential : The electrical work done in a spontaneous, ∆G must be negative.
galvanic cell is the electricity (charge) passed Because ∆G = - nFEcell, Ecell must be positive
multiplied by the cell potential. for a cell reaction if it is spontaneous.

Electrical work 5.8.2 Standard cell potential and equilibrium
= amount of charge passed × cell potential. constant : The relation between standard
Gibbs energy change of cell reaction and
Charge of one mole electrons is F standard cell potential is given by Eq. (5.27).
coulombs. For the cell reaction involving n
moles of electrons.

charge passed = nF coulombs - ∆G0 = nFE0cell

Hence, electrical work = nFEcell The relation between standard Gibbs
energy change of a chemical reaction
W. Gibbs in 1878 concluded that and its equilibrium constant as given in
electrical work done in galvanic cell is equal thermodynamics is :
to the decrease in Gibbs energy, - ∆G, of cell
reaction. It then follows that ∆G0 = - RT ln K (5.29)

Electrical work = - ∆G Combining Eq. (5.28) and Eq. (5.29), we
have
and thus - ∆G = nFEcell

or ∆G = -nFEcell (5.27) -nFE0cell = - RT ln K

Under standard state conditions, we write RT

∆G0 = -nFE0cell (5.28) or E0 = nF ln K
cell

The Eq. (5.28) explains why E0 is an 2.303 RT
cell nF
intensive property.
= log10K

We know that ∆G0 is an extensive

property since its value depends on the = 0.0592 log10K at 25 0C
n
amount of substance. If the stoichiometric

equation of redox reaction is multiplied by (5.29)

2 that is the amounts of substances oxidised

and reduced are doubled, ∆G0 doubles. The Try this...

moles of electrons transferred also doubles. Write expressions to calculate
equilibrium constant from
The ratio, ∆G0 then becomes
nF i. Concentration data
E0 = - ii. Thermochemical data
cell iii. Electrochemical data

E0 = - 2∆G0 = - ∆G0
cell 2 nF nF

Thus, E0 remains the same by
cell
multiplying the redox reaction by 2. It

means E0 is independent of the amount of
cell
substance and the intensive property.

107

Problem 5.12 : Calculate standard Gibbs What would happen if potential of one of
energy change and equilibrium constant at the electrodes in a galvanic cell is zero ? Can
250C for the cell reaction, we measure the potential of such a galvanic
cell ? There are two electrodes combined
Cd (s) + Sn2⊕ (aq) Cd2⊕ (aq) + Sn (s) together and a redox reaction results. The
measured cell potential is algebraic sum of two
Given : E0Cd = -0.403V and electrode potentials. One electrode potential is
zero. Therefore, the measured cell potential is
E0Sn = - 0.136 V. Write formula of the cell. equal to the potential of other electrode.

Solution : From foregoing arguments, it follows that
it is necessary to choose an arbitrary standard
i. The cell is made of two electrodes, electrode as a reference point. The chemists
Cd2⊕ (aq) Cd (s) and Sn2⊕ (aq) Sn (s). have chosen hydrogen gas electrode consisting
E0 value for Sn2⊕ (aq) Sn (s) electrode of H2 gas at 1 atm pressure in contact with
is higher than that of Cd2⊕(aq) Cd 1 M H⊕ ion solution as a primary reference
(s) electrode. Hence, Sn2⊕ (aq) Sn (s) electrode. The potential of this electrode has
electrode is cathode and Cd2⊕ (aq) Cd (s) arbitrarily been taken as zero. The electrode
anode. Cell formula is Cd(s) Cd2⊕ (aq) is called standard hydrogen electrode (SHE).
Sn2⊕ (aq) Sn (s)
We will see later that SHE is not the most
ii. ∆G0 = - nF E0 convenient electrode. Several other electrodes
cell namely calomel, silver-silver chloride and
glass electrodes with known potentials are
E0 = E0Sn - E0Cd = - 0.136 V - (-0.403V) used as secondary reference electrodes. The
cell potentials of these electrodes are determined
using SHE.
= 0.267 V.

n = 2 mol e

∆G0 = -2 mol e 96500 C/mol e × 0.267 V

= - 51531 V C = - 51531 J = - 51.53 kJ
0.0592 V
iii. E0 = 2 log10K
cell

0.267 V = 0.0592 V log10K A reference electrode is then defined as
2 an electrode whose potential is arbitrarily
taken as zero or is exactly known.

log10K = 0.267 × 2 = 9.0203 5.9.1 Standard hydrogen electrode (SHE) :
0.0592 Construction : SHE consists of a platinum
plate, coated with platinum black used as
K = antilog 9.0203 = 1.05 × 109 electrodes. This plate is connected to the

5.9 Reference electrodes : Every oxidation external circuit through sealed narrow glass
needs to be accompanied by reduction. The tube containing mercury. It is surrounded by
occurrence of only oxidation or only reduction an outer jacket.

is not possible. (Refer to Std. XI Chemistry The platinum electrode is immersed in

Textbook Chapter 6) 1 M H⊕ ion solution. The solution is kept

In a galvanic cell oxidation and reduction saturated with dissolved H2 by bubbling
occur simultaneously. The potential associated hydrogen gas under 1 atm pressure through
with the redox reaction can be experimentally the side tube of the jacket as shown in Fig.5.6.

measured. For the measurement of potential Platinum does not take part in the electrode

two electrodes need to be combined together reaction. It is inert electrode and serves as the

where the redox reaction occurs. site for electron transfer.

108

Cu wire Pure and ee
dry Hat2mgas
Vessel at 1 Salt bridge
Glass jacket
Pt wire Zn (Hg2, 1atm)
Solution Mercury anode SHE
H⊕ ions Platinised 1M
(1M) platinum sZonuStiOo4n 1M
plate H⊕ ion
solution

Fig. 5.6 : Standard hydrogen electrode Fig. 5.7 : Determination of standard potential
using SHE
Formulation : Standard hydrogen electrode
is represented as iii. The concentration of H⊕ ion solution
cannot be exactly maintained at 1 M.
H⊕ (1M) H2 (g, 1atm)Pt Due to bubbling of gas into the solution,
evaporation of water may take place.
Electrode reaction : The platinum black This results in changing the concentration
capable of adsorbing large quantities of H2 of solution.
gas, allows the change from gaseous to ionic
form and the reverse process to occur. Hydrogen gas electrode

The reduction half reaction at the For hydrogen gas electrode,
electrode is Hhy⊕d(raoqg)enH2g(ags,PdHi2f)ferPftr,om[H⊕u]niatyn.d pressure of

2H⊕ (1M) + 2e H2 (g, 1atm) Electrode reaction :
Application of SHE E0H2 = 0.000 V

SHE is used as a primary reference 2H⊕ (aq) + 2e H2 (g, PH2)
electrode to determine the standard potentials
of other electrodes. From the Nernst equation

To determine the standard potential of EH2 = E0H2- 0.0592 log10 PH2
Zn2⊕(1M)Zn (s), it is combined with SHE 2 [H⊕]2
to form the cell,
= - 0.0592 log10 PH2
2 [H⊕]2

ZnZn2⊕(1M)H⊕ (1M)H2 (g, 1atm)Pt Because E0H2 = 0
This is shown in Fig. 5.7.
5.10 Galvanic cells useful in day-to-day life
The standard cell potential, E0cell, is measured. Voltaic (or galvanic) cells in common use can
be classified as primary and secondary cells.
E0 = E0H2 - E0Zn = - E0Zn , because E0H2 is zero. i. Primary voltaic cells : When a galvanic
cell
cell discharges during current generation,
Thus, the measured emf of the cell is equal the chemicals are consumed. In primary
voltaic cell, once the chemicals are
to standard potential of Zn2⊕(1M)Zn (s) completely consumed, cell reaction stops.
The cell reaction cannot be reversed even
electrode. after reversing the direction of current
flow or these cells cannot be recharged.
Difficulties in setting SHE The most familiar example is dry cell.
i. It is difficult to obtain pure and dry

hydrogen gas.

ii. The pressure of hydrogen gas cannot be
maintained exactly at 1 atm throughout
the measurement.

109

ii. Secondary voltaic cells : In secondary Brass cap
voltaic cell, the chemicals consumed
during current generation can be Paper spacer
regenerated. For this purpose an external
potential slightly greater than the cell Zn container
potential is applied across the electrodes.
This results in reversal of the direction of Paste of MnO2 +
current flow causing the reversal of cell carbon
reaction This is recharging of cell. The Graphite rod
voltaic cells which can be recharged are
called secondary voltaic cells. Paste of NH4Cl
+ ZnCl2
It is amazing to see that secondary
cells are galvanic cells during discharge and Bottom of Zn container
electrolytic cells during recharging. Examples Fig. 5.8 : Dry cell (Leclanche' cell)
of secondary cells are lead storage battery,
mercury cell and nickel-cadmium cell. Cell reactions:

5.10.1 Dry cell (Leclanche' cell) : It is a cell i. Oxidation at anode : When the cell
without liquid component, but the electrolyte operates the current is drawn from the
is not completely dry. It is a viscous aqueous cell and metallic zinc is oxidised to zinc
paste. ions.

Construction : The container of the cell is Zn (s) Zn2⊕ (aq) + 2e
made of zinc which serves as anode (-). It
is lined from inside with a porous paper to ii. Reduction at cathode : The electrons
separate it from the other material of the cell.
liberated in oxidation at anode flow along
An inert graphite rod in the centre of the
cell immersed in the electrolyte paste serves the container and migrate to cathode. At
as cathode. It is surrounded by a paste of
manganese dioxide (MnO2) and carbon black. cathode NH4⊕ ions are reduced.

The rest of the cell is filled with an 2NH4⊕ (aq) + 2e 2NH3 (aq) + H2 (g)
electrolyte. It is a moist paste of ammonium
chloride (NH4Cl) and zinc chloride (ZnCl2). Hydrogen gas produced in reduction
Some starch is added to the paste to make it
thick so that it cannot be leaked out. reaction is oxidised by MnO2 and
prevents its collection on cathode.
The cell is sealed at the top to prevent
drying of the paste by evaporation of moisture. H2(g)+2MnO2(s) Mn2O3(s)+H2O(l)
See Fig. 5.8.
The net reduction reaction at cathode is

combination of these two reactions.

2NH4⊕(aq) + 2 MnO2(s) + 2e
Mn2O3 (s) + 2 NH3 (aq) + H2O (l)

iii. Net cell reaction : The net cell reaction is
sum of oxidation at anode and reduction
at cathode.

Zn (s) + 2 NH4⊕ (aq) + 2 MnO2(s)

Zn2⊕ (aq) + Mn2O3 (s) + 2 NH3 (aq) + H2O(l)

The ammonia produced combines with

Zn2⊕ to form soluble compound containing

complex ion.

Zn2⊕ (aq) + 4 NH3 (aq) [Zn (NH3)4]2⊕(aq)

110

Do you know ? The electrodes are immersed in
an electrolytic aqueous solution of
Alkaline dry cell : The Leclanche' 38 % (by mass) of sulphuric acid of density
dry cell works under acidic 1.2 g/mL.
conditions due to the presence of NH4Cl.
The difficulty with this dry cell is that Zn Notation of the cell : The cell is formulated
anode corrodes due to its actions with H⊕ as Pb(s)PbSO4(s)38%H2SO4(aq)PbSO4(s)
ions from NH4⊕ ions. PbO2(s)  Pb(s)

This results in shortening the life of a. Cell reactions during discharge
dry cell. To avoid this a modified or of
the dry cell called alkaline dry cell has i. Oxidation at anode (-) : When the
been proposed. In alkaline dry cell NaOH cell provides current, spongy lead is
or KOH is used as electrolyte in place of oxidised to Pb2⊕ ions and negative charge
NH4Cl. accumulates on lead plates. The Pb2⊕
ions so formed combine with SO42 ions
The alkaline dry cell has longer life from H2SO4 to form insoluble PbSO4. The
than acidic dry cell since the Zn corrodes net oxidation is the sum of these two
more slowly. processes.

Uses of dry cell : Dry cell is used as a source Pb (s) Pb2⊕ (aq) + 2 e
of power in flashlights, portable radios, tape
recorders, clocks and so forth. (oxidation)

5.10.2 Lead storage battery (Lead Pb2⊕ (aq) + SO42 (aq) PbSO4 (s)
accumulator) : Lead accumulator stores (precipitation)
electrical energy due to regeneration of original
reactants during recharging. It functions as Pb (s)+SO42 (aq) PbSO4 (s) + 2 e ...(i)
galvanic cell and as electrolytic cell, as well. (overall oxidation)

Construction : A group of lead plates packed ii. Reduction at cathode (+) : The electrons
with spongy lead serves as anode (-). Another produced at anode travel through external
group of lead plates bearing lead dioxide circuit and re-enter the cell at cathode. At
(PbO2) serves as cathode (+). cathode PbO2 is reduced to Pb2⊕ ions in
presence of H⊕ ions. Subsequently Pb2⊕
Anode (-) Cathode (+) ions so formed combine with SO42 ions
from H2SO4 to form insoluble PbSO4 that
gets coated on the electrode.

PbO2 (s) + 4H⊕ (aq) + 2 e Pb2⊕ (aq)
+2H2O(l)
38 % (reduction)
H2SO4
Pb (s) + SO42 (aq) PbSO4 (s)
(precipitation)

Pb Pb plates with PbO2 (s) + 4H⊕ (aq) + SO42 (aq)+ 2 e
plates PbO2 PbSO4 (s) + 2H2O (l) ...(ii)

Fig. 5.9 : Lead storage cell (overall reduction)

To provide large reacting surface, the iii. Net cell reaction during discharge: The
cell contains several plates of each type. The net cell reaction is the sum of overall
two types of plates are alternately arranged oxidation at anode and overall reduction
as shown in Fig. 5.9. at cathode.

111

Pb (s) + PbO2 (s) + 4H⊕ (aq) + 2SO42 (aq) ii. A 12 V lead storage battery constructed
2PbSO4 (s) + 2H2O (l) by connecting six 2 V cells in series is
used in automobiles and inverters.
or
Pb (s) + PbO2 (s) + 2H2SO4 (aq) 5.10.3 Nickel-Cadmium or NICAD storage
cell : Nickel-cadmium cell is a secondary dry
2PbSO4(s) + 2H2O (l) ...(iii) cell. In other words it is a dry cell that can
As the cell operates to generate current, be recharged.
H2SO4 is consumed. Its concentration (density)
decreases and the cell potential is decreased. Anode of the NICAD storage cell is
The cell potential thus depends on sulphuric cadmium metal. The cathode is nickel (IV)
acid concentration (density). oxide, NiO2 supported on Ni. The electrolyte
solution is basic.
b. Cell reactions during recharging :
The potential of lead accumulator is 2V. It The electrode reactions and overall cell
must be recharged with the falling of the reaction are as follows :
cell potential to 1.8 V. To recharge the cell
external potential slightly greater than 2 V Cd (s) + 2OH (aq) Cd(OH)2 (s) + 2 e
needs to be applied across the electrodes. (anodic oxidation)

During recharging the cell functions as NiO2 (s) + 2 H2O (l) + 2 e
electrolytic cell. The anode and cathode are Ni(OH)2 (s) + 2OH (aq)
interchanged with PbO2 electrode being anode (cathodic reduction)
(+) and lead electrode cathode (-).
iv. Oxidation at anode (+) : It is reverse Cd (s) + NiO2 (s) + 2 H2O (l)
Cd(OH)2 (s) + Ni(OH)2 (s)
of reduction reaction (ii) at cathode that
occurs during discharge. (overall cell reaction)

PbSO4 (s) + 2H2O (l) The reaction product at each electrode
is solid that adheres to electrode surface.
PbO2 (s) + 4H⊕(aq) + SO42 (aq) + 2 e ...(iv) Therefore the cell can be recharged. The
potential of the cell is about 1.4 V. The cell
v. Reduction at cathode (-) : It is reverse has longer life than other dry cells. It can
of oxidation reaction (i) at anode that occurs be used in electronic watches, calculators,
during discharge. photographic equipments, etc.

PbSO4 (s)+2 e Pb (s)+SO42 (aq) ...(v)

vi. Net cell reaction : It is the sum of 5.10.4 Mercury battery : Mercury battery is
a secondary dry cell and can be recharged.
reaction (iv) and (v) or the reverse of The mercury battery consists of zinc anode,
amalgamated with mercury. The cathode is
net cell reaction (iii) that occurs during a paste of Hg and carbon. The electrolyte
is strongly alkaline and made of a paste of
discharge KOH and ZnO. The electrode ractions and
net cell reaction are :
2PbSO4 (s) + 2H2O (l) Pb (s) +

PbO2 (s) + 2 H2SO4 (aq)

The above reaction shows that H2SO4 is Zn(Hg)+2OH (aq) ZnO(s) +H2O(l) + 2 e
regenerated. Its concentration (density) HgO(s)+ H2O(l)+2e (anode oxidation)
Zn (Hg) + HgO(s) Hg(l) + 2 OH (aq)
and in turn, the cell potential increases.
(cathode reduction)
Applications of lead accumulator
ZnO(s) + Hg(l)
i. It is used as a source of direct current in
the laboratory. (overall cell reaction)

112

The overall reaction involves only solid Hydrogen gas is continuously bubbled,
substances. There is no change in electrolyte through anode and oxygen gas through
composition during operation. The mercury cathode into the electrolyte.
battery, therefore, provides more constant
voltage (1.35V) than the Leclanche' dry cell. Cell reactions
It also has considerably higher capacity and
longer life than dry cell. i. Oxidation at anode (-) : At anode

hydrogen gas is oxidised to H2O.

The mercury dry cell finds use in hearing 2H2 (g) + 4OH (aq) 4H2O (l) + 4 e
aids, electric watches, pacemakers, etc.
ii. Reduction at cathode (+) : The electrons

5.11 Fuel cells : The functioning of fuel cells released at anode travel, through external
is based on the fact that combustion reactions
are of redox type and can be used to generate circuit to cathode. Here O2 is reduced to
electricity. OH-.

O2 (g) + 2H2O (aq)+ 4 e 4OH (aq)

The fuel cells differ from ordinary iii. Net cell reaction : The overall cell
galvanic cells in that the reactants are not
placed within the cell. They are continuously reaction is the sum of electrode reactions
supplied to electrodes from a reservoir.
(i) and (ii).

In these cells one of the reactants is a 2H2 (g) + O2 (g) 2H2O (l)
fuel such as hydrogen gas or methanol. The
other reactant such as oxygen, is oxidant. The overall cell reaction is combustion

The simplest fuel cell is hydrogen-oxygen of H2 to form liquid water. Interestingly, the
fuel cell. fuel H2 gas and the oxidant O2 do not react
directly.

5.11.1 Hydrogen-oxygen fuel cell : In H2 - The chemical energy released during the
O2 fuel cell, the fuel is hydrogen gas. Oxygen formation of O-H bond is directly converted
gas is an oxidising agent. The energy of the into electrical energy accompanying in above
combustion of hydrogen is converted into combustion reaction. The cell continues
electrical energy. to operate as long as H2 and O2 gases are
supplied to electrodes.

H2O The cell potential is given by

(anode) (cathode) E0 = E0 - E0 = 0.4V - (-0.83V)
cell cathode anode

= 1.23 V.

Advantages of fuel cells

H2 (g) O2 (g) i. The reacting substances are continuously
Aqueous supplied to the electrodes. Unlike
conventional galvanic cells, fuel cells do
KOH not have to be discarded on consuming
of chemicals.
Fig. 5.10 : H2 - O2 fuel cell
ii. They are nonpolluting as the only reaction
Construction : The anode and cathode are product is water.
porous carbon rods containing small amount
of finely divided platinum metal that acts iii. Fuel cells provide electricity with an
as a catalyst. The electrolyte is hot aqueous efficiency of about 70 % which is twice
solution of KOH. The carbon rods immersed as large when compared with efficiency
into electrolyte are shown in Fig. 5.10. of thermal plants (only 40 %).

113

Drawbacks of fuel cell ii. Below hydrogen electrode the negative
H2 gas is hazardous to handle and the standard potential increases and above
hydrogen electrode the positive standard
cost of preparing H2 is high. potential increases.

Internet my friend iii. E0 values apply to the reduction half
Fuel cells are also used in cell reactions that occur in the forward
direction as written.
phones and laptop computers. The
cell proposed for use in these products iv. Higher (more positive) E0 value for a half
is direct methanol fuel cell (DMFC). reaction indicates its greater tendency to
Collect information of this cell. occur in the forward direction and in
turn greater tendency for the substance
Applications of fuel cells to reduce. Conversely, the low (more
negative) E0 value of a half reaction
i. The fuel cells are used on experimental corresponds to its greater tendency to
basis in automobiles. occur in the reverse direction or for the
substance to oxidise.
ii. The fuel cell are used for electrical power
in the space programme.

iii. In space crafts the fuel cell is operated The half reactions are listed in order of
at such a high temperature that the their decreasing tendency in the forward
water evaporates at the same rate as it direction.
is formed. The vapour is condensed and
pure water formed is used for drinking Remember...
by astronauts. The left side of half reaction

iv. In future, fuel cells can possibly be has cations of metals or non-
explored as power generators in hospitals, metallic molecules (oxidants). There are
hotels and homes. free metals or anions of non metals on the
right side (reductants).

Can you tell ? Applications of electrochemical series
i. Relative strength of oxidising agents:
In what ways are fuel cells
and galvanic cells similar and in The species on the left side of half
what ways are they different ? reactions are oxidizing agents. E0 value

5.12 Electrochemical series (Electromotive is a measure of the tendency of the species
series) : The standard potentials of a number to accept electrons and get reduced.
of electrodes have been determined using In other words, E0 value measures the
standard hydrogen electrode. These electrodes strength of the substances as oxidising
with their half reactions are arranged according agents. Larger the E0 value greater is
to their decreasing standard potentials as the oxidising strength. The species in the
shown in Table 5.1. This arrangement is top left side of half reactions are strong
called electrochemical series. oxidising agents. As we move down the
Key points of electrochemical series table, E0 value and strength of oxidising
agents decreases from top to bottom.
i. The half reactions are written as

reductions. The oxidizing agents and Use your brain power
electrons appear on the left side of half
Indentify the strongest and the
reactions while the reducing agents weakest oxidizing agents from the
are shown on the right side in the half electrochemical series.
reaction.

114

ii. Relative strength of reducing agents: From Table 5.1 of electrochemical series we
The species on the right side of half have
reactions are reducing agents.
E0Mg = -2.37 V and E0Ag = 0.8 V. For the
The half reactions at the bottom of the cell having Mg as anode and Ag cathode.
table with large negative E0 values have
a little or no tendency to occur in the E0 = E0Ag - E0Mg = 0.8V - (-2.37V)
forward direction as written. They tend Cell
to favour the reverse direction. It follows,
that the species appearing at the bottom = 3.17 V.
right side of half reactions associated
with large negative E0 values are the EMF being positive the cell reaction
effective electron donors. They serve as is spontaneous. Thus Ag⊕ ions oxidise to
strong reducing agents. The strength of metallic Mg.
reducing agents increases from top to
bottom as E0 values decrease. General rules

Use your brain power i. An oxidizing agent can oxidize any
Identify the strongest and the reducing agent that appears below it,
and cannot oxidize the reducing agent
weakest reducing agents from the appearing above it in the electrochemical
electrochemical series. series.

ii. An reducing agent can reduce the
oxidising agent located above it in the
electrochemical series.

iii. Spontaneity of redox reactions : A redox Use your brain power
reaction in galvanic cell is spontaneous
only if the species with higher E0 value From E0 values given in
is reduced (accepts electrons) and that Table 5.1, predict whether Sn can
with lower E0 value is oxidised (donates reduce I2 or Ni2⊕.
electrons).

The standard cell potential must be positive Do you know ?
for a cell reaction to be spontaneous under
the standard conditions. Noteworthy The fuel cells for power
application of electromotive series is electric vehicles incorporate the
predicting spontaneity of redox reactions proton conducting plastic membrane.
from the knowledge of standard potentials. These are proton exchange membranes
(PEM) fuel cells.
Suppose, we ask a question : At standard
conditions would Ag⊕ ions oxidise
metallic magnesium ? To answer this
question, first we write oxidation of Mg
by Ag⊕.

Mg (s) Mg2⊕ (aq) + 2 e (oxidation)

2Ag2⊕ (aq) + 2 e 2Ag (s)(reduction)

Mg (s) +2Ag2⊕ (aq) Mg2⊕ (aq) + 2Ag (s)
(overall reaction)

115

Table 5.1 : The standard aqueous electrode potentials at 298 K

(Electrochemical series)

Electrode Half reaction E0 V

Left side species Right side species
(oxidizing agents) (oxidizing agents)

F F2Pt F2+ 2 e F +2.870
Au⊕Au Au⊕ + e Au +1.680

Ce4⊕, Ce3⊕Pt Ce4⊕ + e Ce3⊕ +1.610

Au3⊕ Au Au3⊕ + 3e Au +1.500

Cl Cl2Pt Cl2 + 2e 2Cl +1.360
Pt2⊕Pt Pt2⊕ + 2e Pt +1.200

Br Br2Pt Br2 + 2e 2Br +1.080
Hg2⊕Hg Hg2⊕ + 2e- Hg +0.854

Ag⊕ Ag Ag⊕ + e Ag +0.799
Increasing strength as oxidising agent
Hg22⊕Hg Increasing strength as reducing agentHg22⊕ + 2e2Hg+0.79
Fe3⊕,Fe2⊕Pt Fe3⊕ + e Fe2⊕ +0.771

I I2(s) Pt I2 + 2e 2I +0.535
Cu2⊕  Cu Cu2⊕ + 2e Cu +0.337

AgAgCl(s)Cl AgCl (s) + e Ag + Cl- +0.222

Cu2⊕,Cu⊕Pt Cu2⊕ + e Cu⊕ +0.153

Sn4⊕, Sn2⊕Pt Sn4⊕ + 2e Sn2⊕ +0.15

H⊕H2Pt 2H⊕ + 2e H2 0.00
Pb2⊕Pb Pb2⊕ + 2e Pb -0.126

Sn2⊕Sn Sn2⊕ + 2e Sn -0.136

Ni2⊕Ni Ni2⊕ + 2e Ni -0.257

Co2⊕Co Co2⊕ + 2e- Co -0.280

Cd2⊕Cd Cd2⊕ + 2e Cd -0.403

Fe2⊕Fe Fe2⊕ + 2e Fe -0.440

Cr3⊕Cr Cr3⊕ + 3e Cr -0.740

Zn2⊕Zn Zn2⊕ + 2e Zn -0.763

Al3⊕Al Al3⊕ + 3e Al -1.66

Mg2⊕Mg Mg2⊕ + 2e Mg -2.37
Na⊕Na Na⊕ + e Na -2.714

Ca2⊕Ca Ca2⊕ + 2e Ca -2.866

K⊕K K⊕ + e K -2.925

Li⊕Li Li⊕ + e Li -3.045

Note : (i) all ions are at 1 M concentration in water.
(ii) all gases are at 1 atm pressure.

116

Exercises

1. Choose the most correct option.

i. Two solutions have the ratio of their ii. I2 (s) + 2e 2I (aq)
concentrations 0.4 and ratio of their
conductivities 0.216. The ratio of E0 = 0.53V
their molar conductivities will be
iii. Pb2⊕ (aq) + 2e Pb (s)

a. 0.54 b. 11.574 E0 = -0.13V

c. 0.0864 d. 1.852 iv. Fe2⊕ (aq) + 2e Fe (s)

ii. On diluting the solution of an E0 = - 0.44V
electrolyte
The strongest oxidising and reducing
a. both ∧ and k increase agents respectively are

b. both ∧ and k decrease a. Ag and Fe2⊕ b. Ag⊕ and Fe

c. ∧ increases and k decreases c. Pb2⊕ and I d. I2 and Fe2⊕

d. ∧ decreases and k increases vii. For the reaction Ni(s) + Cu2⊕ (1M)
iii. 1 S m2 mol-1 is eual to
Ni2⊕ (1M) + Cu (s), E0 =
cell
0.57V ∆G0 of the reaction is

a. 10-4 S m2 mol-1 a. 110 kJ b. -110 kJ

b. 104 Ω-1 cm2 mol-1 c. 55 kJ d. -55 kJ

c. 10-2 S cm2 mol-1 viii. Which of the following is not correct?

d. 102 Ω-1 cm2 mol-1 a. Gibbs energy is an extensive
property
iv. The standard potential of the cell in
which the following reaction occurs b. Electrode potential or cell potential
is an intensive property.
H2 (g,1atm) + Cu2⊕(1M)
2H⊕(1M) + Cu (s), (E0Cu = 0.34V) is c. Electrical work = - ∆G

a. -0.34 V b. 0.34 V d. If half reaction is multiplied by a
numerical factor, the corresponding
c. 0.17 V d. -0.17 V E0 value is also multiplied by the
same factor.
v. For the cell, Pb
(s)Pb2⊕(1M)Ag⊕(1M) Ag (s), if ix. The oxidation reaction that takes
concentraton of an ion in the anode place in lead storage battery during
compartment is increased by a factor discharge is
of 10, the emf of the cell will

a. increase by 10 V a. Pb2⊕ (aq) + SO42 (aq) PbSO4(s)

b. increase by 0.0296 V b. PbSO4 (s) + 2H2O (l) PbO2 (s)
+ 4H⊕(aq) + SO42 (aq) + 2e
c. decrease by 10 V
c.Pb(s)+SO42 (aq) PbSO4 (s) +
d. decrease by 0.0296 V 2e

vi. Consider the half reactions with d. PbSO4(s) + 2e Pb(s)
standard potentials + SO42 (aq)

i. Ag⊕ (aq) + e Ag (s)

E0 = 0.8V

117

x. Which of the following expressions ii. What is a salt bridge ?

represent molar conductivity of iii. Write electrode reactions for the
electrolysis of aqueous NaCl.
Al2(SO4)3?

a. 3 λ0 + 2 λ0 iv. How many moles of electrons are
Al3⊕ SO42 passed when 0.8 ampere current is
passed for 1 hour through molten
b. 2 λ0 + 3 λ0 CaCl2 ?
Al3⊕ SO42
v. Construct a galvanic cell from
c. 1/3 λ0 + 1/2 λ0 the electrodes Co3⊕Co and
Al3⊕ SO42 Mn2⊕ Mn. E0Co = 1.82 V,

d. λ0 + λ0 E0Mn = - 1.18V. Calculate E0cell.
Al3⊕ SO42
vi. Using the relationsip between ∆G0
2. Answer the following in one or two of cell reaction and the standard
potential associated with it, how will
sentences. you show that the electrical potential
is an intensive property ?
i. What is a cell constant ?

ii. Write the relationship between
conductivity and molar conductivity
and hence unit of molar conductivity.

iii. Write the electrode reactions during
electrolysis of molten KCl.

iv. Write any two functions of salt viii. Derive the relationship between
bridge. standard cell potential and
equilibrium constant of cell reaction.
v. What is standard cell potential for
the reaction ix. It is impossible to measure the
potential of a single electrode.
3Ni (s) + 2Al3⊕(1M) Comment.
3Ni2⊕(1M)
x. Why do the cell potential of lead
+ 2Al(s) if E0Ni = - 0.25 V and E0Al = accumulators decrease when it
-1.66V? generates electricity ? How the cell
potential can be increased ?
vi. Write Nerst equation. What part of
it represents the correction factor for xi. Write the electrode reactions and net
nonstandard state conditions ? cell reaction in NICAD battery.

vii. Under what conditions the cell 4. Answer the following :
potential is called standard cell
potential ? i. What is Kohrausch law of
independent migration of ions?
viii. Formulate a cell from the following How is it useful in obtaining molar
electrode reactions : conductivity at zero concentration of
a weak electrolyte ? Explain with an
Au3⊕(aq) + 3e Au(s) example.

Mg(s) Mg2⊕(aq) + 2e ii. Explain electrolysis of molten NaCl.

ix. How many electrons would have a iii. What current strength in amperes
total charge of 1 coulomb ? will be required to produce 2.4
g of Cu from CuSO4 solution in
x. What is the significance of the single 1 hour ? Molar mass of Cu = 63.5 g
vertical line and double vertical line mol-1. (2.03 A)
in the formulation galvanic cell.

3. Answer the following in brief iv. Equilibrium constant of the reaction,

i. Explain the effect of dilution of 2Cu⊕(aq) Cu2⊕(aq) + Cu(s)
solution on conductivity ?
is 1.2 × 106. What is the standard

118

potential of the cell in which the order of increasing strength under
reaction takes place ? (0.36 V) standard state conditions. The
standard potentials are given in
v. Calculate emf of the cell parenthesis.

Zn(s)Zn2⊕(0.2M)H⊕(1.6M) K (-2.93V), Br2(1.09V), Mg(-2.36V),
H2(g, 1.8 atm) Pt at 250C. Ce3⊕(1.61V), Ti2⊕(-0.37V), Ag⊕(0.8
(0.785V) V), Ni (-0.23V).

vi. Calculate emf of the following cell at xiv. Predict whether the following
250C. reactions would occur spontaneously
under standard state conditions.
Zn (s) Zn2⊕(0.08M)Cr3⊕(0.1M)Cr
a. Ca (s) + Cd2⊕ (aq)
E0Zn = - 0.76 V, E0Cr = - 0.74 V
Ca2⊕(aq) + Cd(s)
(0.0327 V)
b. 2 Br (s) + Sn2⊕ (aq)
vii. What is a cell constant ? What are
its units? How is it determined Br2(l) + Sn(s)
experimentally? c. 2Ag(s) + Ni2⊕ (aq)

viii. How will you calculate the moles 2 Ag⊕ (aq) + Ni (s)
of electrons passed and mass of
the substance produced during (use information of Table 5.1)
electrolysis of a salt solution using
reaction stoichiometry. Activity :

ix. Write the electrode reactions when 1. Write electrode reactions net
lead storage cell generates electricity. cell reaction in the electrolysis
What are the anode and cathode and of molten barium chloride.
the electrode reactions during its
recharging? 2. Prepare the salt bridge and set up
the Daniel cell in your laboratory.
x. What are anode and cathode of H2- Measure its emf using voltmeter and
O2 fuel cell ? Name the electrolyte compare it with the value calculated
used in it. Write electrode reactions from the information in Table 5.1
and net cell reaction taking place in
the fuel cell. 3. k1 and k2 are conductivities of two
solutions and c1 and c2 are their
xi. What are anode and cathode for concentrations. Establish the
Leclanche' dry cell ? Write electrode relationship between k1, k2, c1, c2 and
reactions and overall cell reaction molar conductivities ∧1 and ∧2of the
when it generates electricity. two solutions.

xii. Identify oxidising agents and arrange 4. Find and search working of power
them in order of increasing strength inverters in day-to-day life.
under standard state conditions.
The standard potentials are given in 5. Collect information of pollution free
parenthesis. battery.

Al (-1.66V), Al3⊕(-1.66V),Cl2
(1.36V), Cd2⊕(-0.4V), Fe(-0.44V),
I2(0.54V), Br (1.09V).

xiii. Which of the following species are
reducing agents? Arrange them in

119

6. CHEMICAL KINETICS

Can you recall ? Average rate = change in concentration of a species
change in time
• What is the influence of particle
size of reacting solid on rate of a = ∆c
chemical reaction ? ∆t

• Why is finely divided nickel used in Consider the reaction A B in which A is
hydrogenation of oil ?
consumed and B is produced.
• What is effect of change of temperature
on the rate of a chemical reaction ? average rate of consumption of A = - ∆[A]
∆t

Average rate of formation of B = + ∆[B]
∆t

6.1 Introduction : Three important Therefore, average rate of reaction = - ∆[A]
characteristics of chemical reactions include ∆t
: extent of reaction, feasibility and its rate.
In standard XI, we learnt how equibrium = + ∆[B]
constants predict the extent of reaction. In unit ∆t
3 of this text, we learnt how thermodynamic
properties such as change in entropy or The rate of reaction represents a
enthalpy tell us whether under the given set
of conditions chemical reaction represented decrease in concentration of the reactant
by chemical equation occurs or not. Chemical
kinetics is a branch of chemistry which deals per unit time or increase in concentration of
with the rate of chemical reactions and the
factors those affect them. product per unit time. The dimensions of rate

A chemist wants to know the rates of are concentration divided by time, that is,
reactions for different reasons. One the study
of reaction rates help us to predict how rapidly mol dm-3 sec-1.
the reaction approaches equilibrium. Secondly
it gives information on the mechanism of 6.2.2 Instantaneous rate of : To determine the
chemical reactions. instantaneous rate of a reaction the progress
of a reaction is followed by measuring
the concentrations of reactant or product
for different time intervals. The changes
in concentration are relatively fast in the

A number of reactions occur as a sequence product concerntration reactant concerntration
of elementary steps constituting the mechanism
of reaction.

6.2 Rate of reactions : The rate of reaction
describes how rapidly the reactants are
consumed or the products are formed.

6.2.1 Average rate of chemical reaction : The
average rate of a reaction can be described by
knowing change in concentration of reactant
or product divided by time interval over which
the change occurs. Thus,

Fig. 6.1 : Determination of instantaneous rate

120

beginning which later become slow. The In general, For
concentration of a reactant or a product plotted
against time are shown in Fig. 6.1 (a) and aA + bB cC + dD,
6.1 (b). A tangent drawn to the curve at time
t1 gives the rate of the reaction. The slope thus rate = - 1 d[A] =- 1 d[B]
obtained gives the instantaneous rate of the a dt b dt
reaction at time t1. The instantaneous rate dc/
dt, is represented by replacing ∆ by derivative = 1 d[C] = 1 d[D]
dc/dt in the expression of average rate. In c dt d dt
chemical kinetics we are concerned with
instantaneous rates. Problem 6.1 : For the reaction

2 N2O5(g) 4 NO2(g) + O2(g) in liquid
bromine, N2O5 disappears at a rate of 0.02
moles dm-3 sec-1. At what rate NO2 and O2
are formed? What would be the rate of
For the reaction, A B,
reaction?
Rate of consumption of A at any time t = - d[A]
dt Solution :
d[B]
Rate of formation of B at any time t = ∆t Given : d[N2O5] = 0.02 1 d[NdOt 2]
dt 4
Rate of reaction at time t = - d[A] = d[B]
dt dt = d[O2]
For the reaction involving one mole dt

of A and B each, the rate of consumption of Rate of reaction = - 1 d[N2O5]
2 dt
A equals the rate of formation of B. This is

not true for the reactions involving different = 1 - d[Nd2tO5]
2
stoichiometries. Consider, for example, a

reaction : Rate of formation of O2 = d[dOt2]

A + 3B 2C

When one mole of A is consumed three = 1 × 0.02
moles of B are consumed and two moles of 2
C are formed. The stoichiometric coefficients d[Nd2tO5]
of the three species are different. Thus the = 1 - = 1 × 0.02 moles dm-3 sec -1
rate of consumption of B is three times the 2 2
rate of consumption of A. Likewise the rate of
formation of C is twice the rate of consumption = 0.01 moles dm-3 sec -1 d[NO5]
of A. We write, dt
Rate of formation of NO2 =

= 4 - d[N2O5]
2 dt
d[A] d[C] d[A]
- d[B] = -3 dt and dt = -2 dt = 2 × moles dm3 sec -1
dt

With this = 0.04 moles dm3 sec -1

- d[A] =- 1 d[B] = 1 d[C] Try this...
dt 3 dt 2 dt
For the reaction,
or rate of reaction = - d[A] =- 1 d[B] 3I (aq)+S2O82 (aq) I3 (aq) + 2 SO42 (aq)
dt 3 dt
Calculate the rate of formation of
= 1 d[C] I3 , the rates of consumption of I- and
2 dt S2O82 and the overall rate of reaction
if the rate of formation of SO42 is
Write the expression for: 0.022 moles dm-3 sec -1

2 N2O(g) 4 NO2(g) + O2(g) ?

121

6.3 Rate of reaction and reactant (iii) If x = 0, the rate is independent of
concentration : The rate of a reaction at a concentration of A.
given temperature for a given time instant (iv) If x < 0 the rate decreases as [A] increases.
depends on the concentration of reactant. Such 6.3.2 Writing the rate law
rate-concentration relation is the rate law. Consider the reaction,

6.3.1 Rate law : Consider the general reaction, 2H2O2(g) 2 H2O(l) + O2(g).

aA + bB cC + dD (6.1) If the rate of the reaction is proportional

The rate of reaction at a given time is to concentration of H2O2. The rate law is given
proportional to its molar concentration at that by
time raised to simple powers or
rate = k[H2O2]

Rate of reaction ∝ [A]x [B]y or Try this...
For the reaction,
rate = k [A]x[B]y (6.2)

where k the proportionality constant is called NO2(g) + CO(g) NO(g) + CO2(g),
the rate constant, which is independent of the rate of reaction is experimetally found
concentration and varies with temperature. to be proportional to the squre of the
For unit concentrations of A and B, k is equal concentration of NO2 and independent
to the rate of reaction. Equation (6.2) is called that of CO. Write the rate law.
differential rate law.

The powers x and y of the concentration Problem 6.2 : Write the rate law for the
terms A and B in the rate law not necessarily reaction, A + B P from the following
equal to stoichiometric coefficients (a and data :
b) appearing in Eq. (6.1). Thus x and y may
be simple whole numbers, zero or fraction. [A] moles [B] moles Initial rate/
Realize that x and y are experimentally dm-3 sec -1 dm-3 sec -1 moles dm-3 sc -1
determined. The rate law in Eq. (6.2) is (Initial)
determined experimentally and expresses (Initial) 4.0 × 10-5
the rate of a chemical reaction in terms of (i) 0.4 6.0 × 10-5
molar concentrations of the reactants and 0.2 3.2 × 10-4
not predicted from the stoichiometries of the (ii) 0.6
reactants. 0.2
(iii) 0.8
The exponents x and y appearing in the rate 0.4
law tell us how the concentration change
affects the rate of the reaction. Solution :
(i) For x = y = 1, Eq. (6.2) gives
a. From above data (i) and (ii), when [A]
rate = k[A][B] increases by a factor 1.5 keeping [B] as
constant, the rate increases by a factor
The equation implies that the rate of a 1.5. It means rate ∝ [A] and x = 1
reaction depends linearly on concentrations of
A and B. If either of concentration of A or B is b. From observations (i) and (iii), it can be
doubled, the rate would be doubled. seen that when concentrations of A and
B are doubled, the rate increases by a
(ii) For x = 2 and y = 1. The Eq. (6.2) then factor 8. Due to doubling of [A] the rate
leads to rate = k[A]2[B]. If concentration of A is doubled (because x = 1) that is rate
is doubled keeping that of B constant, the rate increases by a factor 2.
of reaction will increase by a factor of 4. This implies that doubling [B], the

rate increases by a factor 4. or rate ∝ [B]2
and y = 2. Therefore, rate = k[A] [B]2

contd....

122

Problem 6.2 contd.... Try this...
Alternatively • For the reaction

The rate law gives rate = k [A]x[B]y. 2A + 2B 2C +D, if

a. From above observations (i) and (ii) concentration of A is doubled at constant

(i) 4 × 10-5 = (0.4)x(0.2)y [B] the rate increases by a factor of 4. If

(ii) 6 × 10-5 = (0.6)x(0.2)y the concentration of B is doubled with [A]

being constant the rate is doubled. Write

Dividing (ii) by (i), we have the rate law of the reaction.

(6

4
× 10-5 = 1.5 = ((0.6)x(0.2)y = 0.6 x • The rate law for the reaction
× 10-5 ((0.4)x(0.2)y0.4
A+B C is found to be
= (1.5)x
rate = k[A]2[B].
Hence x = 1
The rate constant of the reaction at 25 0C
b. From observations (i) and (iii) separately is 6.25 M-2s-1. What is the rate of reaction
in the rate law gives when [A] = 1.0 moles dm-3 sec -1 and
[B] = 0.2 moles dm-3 sec -1?
iii) 4 × 10-5 = (0.4) × (0.2)y since x = 1

iv) 3.2 × 10-4 = 0.8 × (0.4)y 6.3.3 Order of the reaction : For the reaction,

Dividing (iv) by (iii) we write aA + bB cC + dD is

3.2 × 10-4 = 0.8 (0.4) y If the rate of the reaction is given as
4 × 10-5 0.4 (0.2)y

(y rate = k[A]x[B]y.

or 8 = 2 × 0.6 Then the sum x + y gives overall
0.2 order of the reaction. Thus overall order of
the chemical reaction is given as the sum of
or 4 = 22 = 2y powers of the concentration terms in the rate
law expression. For example :
Therefore y = 2.

The rate law is then rate = k[A][B]2.

i. For the reaction,

Problem 6.3 : For the reaction, 2H2O2(g) 2H2O(l) + O2(g)

2 NOBr(g) 2 NO2(g) + Br2(g), experimentally determined rate law is

the rate law is rate = k[NOBr]2. If the rate rate = k[H2O2].
of the reaction is 6.5 × 10-6 mol L-1 s-1 when The reaction is of first order.
the concentration of NOBr is 2 × 10-3 mol
L-1. What would be the rate constant for the ii. If the experimentally determined rate law
reaction? for the reaction

Solution : rate

rate = k[NOBr]2 or k = [NOBr]2 H2(g) + I2(g) 2 HI(g) is

= 6.5 × 10-6 mol L-1s-1 rate = k[H2][I2].
(2 × 10-3mol L-1)2
The reaction is of first order in H2 and I2 each
= 1.625 mol-1 L1 s-1 and hence overall of second order.

123

Key points about the order of reaction Solution : The reaction is first order in A
and second order in B. Hence, the rate law
a. The order of chemical reaction is gives
experimentally determined.

b. The order can be integer or fractional. rate = k[A][B]2
Look at the reaction,
or k = rate
CH3CHO(g) CH4(g) + CO(g). [A][B]2

The rate law for the reaction was found to be rate = 3.6 × 10-2 mol/s,
[A] = 0.2 mol dm-3 and [B] = 0.1 mol dm-3
rate = k[CH3CHO]3/2.
Here the order of the reaction is 3/2. Substitution gives
3.6 × 10-2 mol dm-3s-1
c. The order of the reaction, can be zero for : k = 0.2 mol dm-3 × (0.1 mol dm-3)2

NO2(g) + CO(g) NO(g) + CO2(g) 3.6 × 10-2 s-1
= 0.2 × 0.01 mol2 dm-6 sec-1
The rate expression for this is : rate = k[NO2]2.
This shows that order of reaction with respect = 18 mol-2 dm-6 sec-1

to NO2 is 2 and with CO is zero or the rate Use your brain power
is independent of concentration of CO. The
The rate of the reaction
overall order of reaction is 2. 2A + B 2C + D is 6 × 10-4 mol
dm-3 s-1 when [A] = [B] = 0.3 mol dm-3. If
d. Only a few reactions of third order are the reaction is of first order in A and zeroth
known. Reactions with the orders higher than order in B, what is the rate constant?
three are scanty.

Problem 6.4 : For the reaction

2NO(g) + 2H2(g) N2(g) + 2 H2O(g), Problem 6.6 : Consider,
the rate law is rate = k[NO]2 [H2]. What is
the order with respect to NO and H2 ? What A+B P. If the concentration of A is

is the overall order of the reaction ? doubled with [B] being constant, the rate of

Solution : In the rate law expression, the the reaction doubles. If the concentration
exponent of [NO] is 2 and that of [H2] is 1.
Hence, reaction is second order in NO, first of A is tripled and that of B is doubled,
order in H2 and the reaction is third order.
the rate increases by a factor 6. What is

order of the reaction with respect to each

reactant ? Determine the overall order of

the reaction.

Try this... Solution :
The reaction, CHCl3(g) + Cl2(g)
The rate law of the reaction :
CCl4(g) + HCl(g) is first
order in CHCl3 and 1/2 order in Cl2. rate = k[A]x[B]y (i)
Write the rate law and overall order of
reaction. If [A] is doubled, the rate doubles.

Hence 2 × rate = k [2A]x[B]y

= k 2x [A]x[B]y (ii)

Problem 6.5 : The rate of the reaction, 6 × rate = k [3A]x[2B]y (iii)

(iii) gives 6 × rate = 3 k[A]2y [B]y
(i) rate k [A][B]y
A+B P is 3.6 × 10-2 mol dm-3 s-1

when [A] = 0.2 moles dm-3 and [B] = 0.1 or 6 = 3 × 2y or 2 = 2y and y =1
The reaction is first order in A and first
moles dm-3. Calculate the rate constant if order in B. The overall reaction is of the
second order.
the reaction is first order in A and second

order in B.

124

6.4 Molecularity of elementary reactions : these steps is slower than others. The slowest
step is the rate determining step.
Complex reactions are those which
constitute a series of elementary reactions. The slowest step determines the rate of overall
reaction.
6.4.1 Elementary reaction

Consider, Consider, 2NO2Cl(g) 2NO2(g) + Cl2(g).

O3(g) O2(g) + O(g) The reaction takes place in two steps:

C2H5I(g) C2H4(g) + HI(g) i. NO2Cl(g) k1 NO2(g) + Cl(g) (slow)
ii. NO2Cl(g) + Cl (g) k2 NO2(g) + Cl2(g)(fast)
These reactions occur in a single step and

cannot be broken down further into simpler

reactions. These are elementary reactions.

6.4.2 Molecularity of reaction : The Overall 2NO2Cl(g) 2NO2(g) + Cl2(g)
molecularity refers to how many reactant
molecules are involved in reactions. In the The first step being slower than the second it is
above reactions there is only one reactant the rate determining step.
molecule. These are unimolecular reactions or
their molecularity is one. The rate law is

O3(g) + O(g) 2 O2(g) rate = k[NO2Cl]

2 NO2(g) 2 NO(g) + O2(g) This also represents the rate law of the overall
reaction. The reaction thus is of the first order.

The elementary reactions involving two Reaction intermediate:

reactant molecules are bimolecular reactions In the above reaction Cl is formed in the
first step and consumed in the second. Such
or they have molecularity as two. species represents the reaction intermediate.
The concentration of reaction intermediate
The molecularity of an elementary does not appear in the rate law.
reaction is the number of reactant molecules
taking part in it.

6.4.3 Order and molecularity of elementary Distinction between order and molecularity
reactions: of a reaction :

The rate law for the elementary reaction Order Molecularity

2NO2(g) 2NO2(g) + O2(g) is found to 1. It is experimentally i. It is theoretical
be rate = k[NO2]2. The reaction is second order
and bimolecular. The order of reaction is 2 and determined property. entity.

its molecularity is also 2. 2. It is the sum ii. It is the number of
of powers of the reactant molecules
For the elementary reaction, concerntration terms taking part in an
of reactants those elementary reaction.
C2H5 I(g) C2H4(g) + HI(g) appear in the rate
equation.
rate = k[C2H5I]

It is unimolecular and first order. However the

order and molecularity of the reaction may or 3. It may be an iii. It is integer.
integer, fraction or
may not be the same. zero.

6.4.4 Rate determining step : A number of
chemical reactions are complex. They take
place as a series of elementary steps. One of

125

Problem 6.7 : A reaction occurs in the The differential rate law is given by
following steps
rate = - d[A] = k [A] (6.4)
dt
where [A] is the concentration of reactant at
i. NO2(g) + F2(g) NO2F(g) + F(g) (slow)
time t.
ii. F(g) + NO2(g) NO2F(g) (fast)
Rearranging Eq. (6.4),
a. Write the equation of overall reaction.
d[A]
b. Write down rate law. [A] = -k dt (6.5)

c. Identify the reaction intermediate. Let [A]0 be the initial concentration of the
reactant A at time t = 0. Suppose [A]t is the
Solution : concentration of A at time = t.

a. The addition of two steps gives the overall The equation (6.5) is integrated between limits
reaction as
[A] = [A]0 at t = 0 and [A] = [A]t at t = t
2NO2(g) + F2(g) 2 NO2F(g)
[A]t
b. Step (i) is slow. The rate law of the reaction ∫ ∫ [A]0 d[A] t dt
[A]
is predicted from its stoichiometry. Thus, = -k

rate = k[NO2] [F2] 0
c. F is produced in step (i) and consumed in
step (ii) or F is the reaction intermediate. On integration,

[ ] [A]t = -k (t)0t
ln[A] [A]0

Substitution of limits gives

Try this... ln [A]t - ln [A]0 = -k t

A complex reaction takes place in or ln [A]t = -kt (6.6)
two steps: [A]0 (6.7)

i) NO(g) + O3(g) NO3(g) + O(g) 1 [A]0
t [A]t
ii) NO3(g) + O(g) NO2(g) + O2(g) or k = ln

The predicted rate law is rate = k[NO][O3]. Converting ln to log10, we write

Identify the rate determining step. Write k = 2.303 log10 [A]0
t [A]t
the overall reaction. Which is the reaction

intermediate? Why? Eq. (6.7) gives the integrated rate law for the
first order reactions.
6.5 Integrated rate law : We introduced the
differential rate law earlier. It describes how The rate law can be written in the following
rate of a reaction depends on the concentration
of reactants in terms of derivatives. forms [A]t
[A]0
i. Eq. (6.6) is ln = -kt

The differential rate laws are converted By taking antilog of both sides we get
into integrated rate laws. These tell us the
concentrations of reactants for different times. [A]t = e-kt or [A]t = [A]0e-kt (6.8)
[A]0
6.5.1 Integrated rate law for the first order
reactions in solution : Consider first order ii. Let ‘a’ mol dm-3 be the initial concentration
reaction,
of A at t = 0

A product (6.3) Let x mol dm-3 be the concentration of
A that decreases (reacts) during time t. The

126

concentration of A that remains unreacted at k = 0.693
time t would be (a - x) mol/dm3 t1/2

Substitution of [A]0 and [A]t = (a - x) t1/2 = 0.693 (6.10)
k1/2
2.303 a
k= t log10 (a-x) (6.9) Eq. (6.10) shows that half life of the

first order reaction is independent of initial

Equations (6.7), (6.8) and (6.9) represent the reactant concentration. This is shown in Fig
integrated rate law of first order reactions.
(6.2) as a plot of [A]t versus t.

6.5.2 Units of rate constant for the first order
reaction:

The integrated rate law is
2.303 [A]0
k = t log10 [A]t

Because log10 [A]0 is unitless quantity, the [A]
[A]t
dimensions of k will be (time)-1. The units of k rate

will be s-1, min-1or (hour)-1 time

6.5.3 Half life of the first order reactions (t1/2) Fig. 6.2 : Half life period of first order reaction

Radioactive processes follow the first 6.5.5 Graphical representation of the first
order kinetics. The half life of reaction is time order reactions
required for the reactant concentration to fall
to one half of its initial value. i. The differential rate law for the first order

6.5.4 Half life and rate constant of the first reaction A P is
order reaction :
d[A]
The integrated rate law for the first order rate = - dt = k [A]t + 0

reaction is y mx c

k = 2.303 log10 [A]0 The equation is of the form y = mx + c. A plot
t [A]t of rate versus [A]t is a straight line passing
through origin. This is shown in Fig. 6.3. The
where [A]0 is the initial concentration of slope of straight line = k.
reactant at t = 0. It falls to [A]t at time t after
the start of the raction. The time required for

[A]0 to become [A]0/2 is denoted as t1/2

or

[A]t = [A]0/2 at t = t1/2

Putting this condition in the integrated rate

law we write
2.303 [A]t
k = t1/2 log10 [A]0/2

= 2.303 log10 2 intitial Concerntration
t1/2 Fig. 6.3 : Variation of rate with [A]

= 2.303 × 0.3010
t1/2

127

ii. From Eq. (6.7) the integrated rate law is 6.5.6 Examples of first order reactions

k = 2.303 log10 [A]0 Some examples of reactions of first order are :
t [A]t
i. 2 H2O2(l) 2 H2O(l) + O2(g),
On rearrangement, the equation becomes

kt = log10 [A]0 - log10 [A]t rate = k[H2O2]
2.303
k ii. 2 N2O5(g) 4 NO2(g) + O2(g),
2.303
Hence, log10 [A]t = - t + log10 [A]0 rate = k [N2O5]

6.5.7 Integrated rate law for gas phase f

y mx c reactions

The equation is of the straight line. A graph of For the gas phase reaction,

log10 [A]0 versus t yields a stright line with A(g) B(g) + C(g)
[A]t
slope -k/2.303 and y-axis intercept as log10[A]0 Let initial pressure of A be Pi that decreases by
x within time t.
This is shown in Fig. 6.4
Pressure of reactant A at time t

PA = Pi - x

[A]0 (6.11)
[A]t
log The pressures of the products B and C at time
t are

time PB = PC = x
The total pressure at time t is then
Fig. 6.4 : A plot showing log [At]t/[A]0 vs time

iii. Eq. (6.7) gives P = Pi - x + x + x = Pi + x

log10 [A]0 = k t Hence, x = P - Pi (6.12)
[A]t 2.303
Pressure ofAat time t is obtained by substitution

of Eq. (6.12) into Eq. (6.11). Thus

v mx PA = Pi - (P - Pi) = Pi - P + Pi = 2Pi - P
The integrated rate law turns out to be
The equation has a straight line form y = mx .

Hence, the graph of log10 [A]0 versus t is k = 2.303 log10 [A]0
[A]t t [A]t
straight line passing through origin as shown
The concentration now expressed in terms of
in Fig. 6.5.
pressures.

Thus, [A]0 = Pi and [A]t = PA = 2 Pi - P

Substitution gives in above

log [A]0 k = 2.303 log10 2 Pi P (6.13)
[A]t t Pi -

P is the total pressure of the reaction mixture

at time t.

conc
Fig. 6.5 :

128

Problem 6.8 : The half life of first order Problem 6.10 : Following data were

reaction is 990 s. If the initial concentration obtained during the first order decomposition

of the reactant is 0.08 mol dm-3, what of SO2Cl2 at the constant volume.

concentration would remain after 35 SO2Cl2(g) SO2(g) + Cl2(g)

minutes?

Solution : 0.693 0.693 Times/s Toatl pressure/bar
t1/2 990 s
k = = = 7 × 10-4 s-1 0 0.5

k = 2.303 log10 [A]0 100 0.6
t [A]t
Calculate the rate constant of the reaction.

[A]0 = 0.08 mol dm-3, t = 35 min or 2100 s, Solution : 2.303 Pi
[A]t = ? t Pi -
k= log10 2 P
log10 [[AA]]t0
= kt = 7 × 10-4 s-1 × 2100 s Pi = 0.5 bar, P = 0.6 bar, t = 100 s
2.303 2.303
( )k
[A]0 = 0.6383 = 2.303 log10 0.5 bar
[A]t 100 2 × 0.5 bar - 0.6 bar
= antilog 0.6383 = 4.35
Hence, [A]t ( )= 0.5
= [A]0 = 0.08 2.303 log10 0.4 = 2.23 × 10-3 s-1
4.35 4.35 100

= 0.0184 mol dm-3 6.5.8 Zero order reactions: The rate of zero
order reaction is independent of the reactant
Problem 6.9 : In a first order reaction 60% concentration.
of the reactant decomposes in 45 minutes.
Calculate the half life for the reaction Integrated rate law for zero order reactions
: For zero order reaction,
Solution : [A]0
[A]t
k = 2.303 log10 AP
t
the differntial rate law is given by
[A]0 = 100, [A]t = 100 - 60 = 40, t = 45 min d[A]
Substitution of these in above rate = - [A] = k [A]0 = k .....(6.14)

2.303 100 Rearrangement of Eq. (6.14) gives
t 40
k = log10 d[A] = -k dt

= 2.303 log10 2.5 Integration between the limits
45
[A] = [A]0 at t = 0 and [A] = [A]t at t = t gives
2.303
= 45 × 0.3979 = 0.0204 min-1 [A]t t

t1/2 = 0.693 = 0.693 = 34 min ∫ d[A] = -k ∫ dt
k 0.0204 min-1 [A]0 0

or [A]t - [A]0 = - kt

Hence, k t = [A]0 - [A]t (6.15)

Try this... Units of rate constant of zero order reactions

The half life of a first order k= [A]0 - [A]t = mol L-1 = mol dm-3 t-1
reaction is 0.5 min. Calculate time t t
needed for the reactant to reduce to 20% and The units of rate constant of zero order reaction
the amount decomposed in 55 s.
are the same as the rate.

129

Half life of zero order reactions : The rate The metals surface gets completely
covered by a layer of NH3 molecules. A number
constant of zero order reaction is given by eqn of NH3 molecules attached on platinum
surface is small compared to ammonia. A large
(6.15) [A]0 - [A]t number of the NH3 molecules tend to remain as
t gas which do not react. The molecules present
k= on the metal surface only react. The rate of
a reaction was thus independent of the total
Using the conditions t = t1/2, [A]t = [A]1/2, concentration of NH3 and remains constant.

Eq. (6.15) becomes ii. Decomposition of nitrous oxide in the
presence of Pt catalyst.
k= [A]0 - [A]0/2 = [A]0
t1/2 2 t1/2

Hence, t1/2 = [A]0 6.16
2k

The half life of zero order reactions is 2 N2O(g) Pt 2 N2(g) + O2(g)
iii. The catalytic decomposition of PH3 on hot
proportional to the initial concentration of tungsten at high pressure.

reactant.

Graphical representation of zero oder 6.5.9 Pseudo-first order reactions : Certain
reactions : The rate law in Eq. (6.15) gives reactions which are expected to be of higher
order follow the first order kinetics. Consider
[A]t = -k t + [A]0 6.17 hydrolysis of methyl acetate.

CH3COOCH3(aq) + H2O(l)
y mx c

Which is straight line given by CH3COOH(aq) + CH3OH(aq)
y = mx + c.
The rate law is
A plot of [A]t versus t is a straight line as
shown in Fig 6.6. rate = k' [CH3COOCH3] [H2O]

The reaction was expected to follow
the second order kinetics, however, obeys the
first order.

[A]t The reason is that solvent water is
present in such large excess that the change
time in its concentration is negligible compared
Fig. 6.6 : [A]t vs t for zero order reaction to initial one or its concentration remains
constant.

Thus [H2O] = constant = k''. The rate law
becomes

The slope of straight line is -k and its intercept rate = k' [CH3COOCH3] k''
on y-axis is [A]0. = k [CH3COO CH3]

The t1/2 of zero order reaction is where k = k'k''
directly proportional to the initial concentration.
The reaction is thus of first order.
Examples of zero order reactions :
The reaction C12H22O11(aq) + H2O (l) (excess)
Here follow some examples
C6H12O6(aq) + C6H12O6(aq)
Decomposition of NH3 on platinum metal glucose fructose

2 NH3(g) N2(g) + 3 H2(g) Can it be of pseudo-first order type ?

130

6.6 Collision theory of bimolecular reactions 6.6.4 Potential energy barrier

6.6.1 Collision between reactant molecules Consider again the reaction

Chemical reactions occur as a result A+C-B A-B+C
of collisions between the reactant species. It
may be expected that the rate of the reaction is During a course of collision, new bond
equal to the rate of collision. For the gas-phase A - B developes. At the same time bond C - B
reactions the number of collisions is far more breaks. A configuration in which all the three
and typically many powers of tens compared atoms are weakly connected together is called
to the observed rate. activated complex.

6.6.2 Activation : For the reaction to occur A+B-C ABC A-B+C
the colliding reactant molecules must possess
the minimum kinetic energy. This minimum To attain the configuration A B C
kinetic energy is the activation energy. atoms need to gain energy, which comes from
The reaction would occur only if colliding the kinetic energy of colliding molecules.
molecules possess kinetic energies equal to or
greater than the activation energy. The energy barrier between reactants
and products is shown in Fig. 6.7. The reactant
6.6.3 Orientation of reactant molecules molecules need to climb up and overcome this
before they get converted to products. The
The requirement for successful collision height of the barrier is called as activation
described above is sufficient for reactions energy (Ea). Thus the reactant molecules
involving simple molecules (or ions) however transform to products only if they possess
not for those involving complex molecules. energy equal to or greater than such activation
energy. A fraction of molecules those possess
Besides the above considerations energy greater than Ea is given by f = e-Ea/RT.
the colliding molecules must have proper
orientation. The molecules need to be so
oriented relative to each other that the reacting
groups approach closely.

Consider, A + C - B A-B+C Ea

i. The collision of A with C approaching toward ∆HEnergy
A would not lead to reaction.
Fig. 6.7 : Potential energy barrier
+ + As a result only a few collisions lead to
A CB A CB products. The number of successful collisions
are further reduced by the orientation
No reaction will takes place. The reactant requirement already discussed.
molecules would collide and separate owing
to the improper orientation of C - B.

ii. + +
A B C AB C

The reaction is successful as a result of
proper orientation of C - B. A fraction of such
collisions bring forth conversion of reactants
to products.

131

Do you know ? 6.7.2 Graphical determination of activation

For a gaseous reaction at energy : Taking logarithm of both sides of eqn
298 K, Ea = 75 kJ/mol. The fraction
of successful collisions is given by f = e-Ea/RT (6.18) we obtain Ea
RT
= e-75000/8.314 × 298 = 7 × 10-14 or only 7 collisions ln k = - + ln A (6.19)
in 1014 collisions are sufficiently energetic
to lead to the reaction. Converting natural base to base 10 we write
Ea 1
log10k = - 2.303 R T + log10 A (6.20)

y mx c

Remember... This equation is of the form of straight
line y = mx + c.
All collisions of reactant
molecules do not lead to a chemical The Arrhenius plot of log10 k versus
reaction. The colliding molecules need to 1/T giving a straight line is shown in fig.
possess certain energy which is greater (6.8). A slope of the line is -Ea /2.303R with its
than the activation energy Ea and proper intercept being log10 A.
orientation.

6.7 Temperature dependence of reaction
rates

Do you know ? Fig. 6.8 : Variation of log10k with 1/T

It has been observed that From a slope of the line the activation
the rates of most of the chemical energy can be determined. Eq. (6.18) shows
reactions usually increase with temperature. that with an increase of temperature, -Ea/RT
In everyday life we see that the fuels such as and in turn, the rate of reaction would increase.
oil, coal are inert at room temperature but
burn rapidly at higher temperatures. Many
foods spoil rapidly at room temperature and
lasts longer in freezer.

The concentrations change only a little 6.7.3 Determination of activation energy :
with temperature. The rate constant shows a
strong dependence on the temperature. For two different temperatures T1 amd T2
Ea
6.7.1 Arrhenius equation: Arrhenius log10 k1 = log10A - 2.303 RT1 (6.21)
suggested that the rate of reactions varies with
temperature as

k = A e-Ea /RT 6.18 Ea
2.303 RT2
where k is the rate constant, Ea is the activation log10k2 = log10A - (6.22)
energy, R molar gas constant, T temperature in
kelvin, and A is the pre-exponential factor. Eq. where k1 and k2 are the rate constants
(6.18) is called as the Arrhenius equation. at temperatures T1 and T2 respectively.
Subtracting Eq. (6.21) from Eq. (6.22),
The pre exponential factor A and the
rate constant have same unit in case of the log10k2 - log10 k1 = - Ea R 1 + Ea 1
first order reactions. Besides A is found to be 2.303 T2 2.303 R T1
related to frequency of collisions.

132

(Hence, increase with temperature. The rate of reaction
increases accordingly.
log10 k2 = Ea R [1-1
k1 2.303 T1 T2
(
( = Ea T2 - T1 ( ..........(6.23) Problem 6.12 : The rate constants for a first
2.303 R T1T2 order reaction are 0.6 s-1 at 313 K and 0.045
( s-1 at 293 K. What is the activation energy?

6.7.4 Graphical description of effect of Solution - Ea
temperature : It has been realized that average (log10 k2 2.303 T2 - T1
kinetic energy of molecules is proportional to k1 = R T1T2
temperature. The collision theory suggested a
bimolecular reaction occurs only if the reacting k1 = 0.045 s-1, k2 = 0.6 s-1, T1 = 293 K,
molecules have sufficient kinetic energies (at T2 = 313 K, R = 8.314 J K-1mol-1
least Ea) and proper orientation when they
collide. Substituting

At a given temperature, the fraction log10 0.6 = Ea ×
of molecules with their kinetic energy equal 0.045 2.303 × 8.314
to or greater than Ea may lead to the product.
With an increase of temperature the fraction [313 - 293
of molecules having their energies (Ea) would
increases. The rate of the reaction thus would 293 × 313
increase. This is depicted by plotting a fraction Ea
of molecules with given kinetic energy versus log10 13.33 = 2.303 × 8.314 ×
kinetic energy for two different temperatures 20
T1 and T2 (T2 being > T1) in Fig. 6.9. 293 × 313
Ea
1.1248 = 19.15 × 2.18 × 10-4

Ea = 1.1248 × 19.15 J mol-1/ 2.18 × 10-4

= 98810 J/mol-1 = 98.8 kJ/mol-1

Problem 6.13 : A first order gas phase
reaction has activation energy of 240 kJ
mol-1. If the pre-exponential factor is 1.6
× 1013 s-1. what is the rate constant of the
reaction at 600 K?

Solution : Arrhenius equation

k = A e-Ea/RT is written as

A = Ea
log10 k 2.303 RT
Fig. 6.9 : Comparison of fraction of molecules
activated at T1 and T2 Ea = 240 kJ mol-1 = 240 × 103 J mol-1,

The shaded areas is proportional to T = 600 K, A = 1.6 × 1013 s-1
total number of molecules. The total area
is the same at T1 and T2. The area (a + b) Hence log10 A =
represents a fraction of molecules with kinetic k
energy exceeding Ea is larger at T2 than at T1
(since T2 > T1). This indicates that a fraction of 240 × 103 J mol-1 = 20.89
molecules possessing energies larger than Ea
2.303 × 8.314 J mol-1 K-1× 600 K

contd....

133

A = antilog 20.89 6.8 Effect of a catalyst on the rate of reaction
k
= 7.78 × 1020 A catalyst is a substance added to the
reactants that increases the rate of the reaction
A without itself being consumed in the reaction.
and k = 7.78 × 1020
Consider
= 1.6 × 1013 s-1
7.78 × 1020 2 KClO3(s)+ 3O2(g) MnO2 2 KCl(s)

= 2.055 × 10-8 s-1 Here MnO2 is the catalyst. It has been observed
that the decomposition rate increases with
Problem 6.14 : The half life of a first order the addition of catalyst. A catalyst provides
reaction is 900 min at 820 K. Estimate its alternative pathway associated with lower
half life at 720 K if the activation energy is activation energy.
250 kJ mol-1.
Fig. 6.10 compares the potential energy
Solution : 0.693 barriers for the catalysed and uncatalysed
k reactions. The barrier for uncatalysed reaction
t1/2 = (Ea) is larger than that for the same reaction in
the presence of a catalyst Ea.
Rate constants at two different

temperatures, T1 and T2 are k1 and k2
respectively, and the corresponding half

lives (t1/2)1 and (t1/2)2. 0.693
0.693 k
(t1/2)1 = k and (t1/2)2 =

[ THheenceeq ,ua((ttt11i//o22 ))n12, l=og kk1012 kk12 =Ea×
2.303 RT
T2 - T1
[T1T2 Fig. 6.10 : Potential energy barriers for
catalyzed and uncatalyzed reactions
[
(t1/2)1 Ea T2 - T1 [ Consider the decomposition of H2O2 in
(t1/2)2 2.303 RT T1T2 aqueous solution catalysed by I- ions.
[log10 =

Ea = 250 kJ mol-1, T1 = 720 K, 2H2O2(l) I- 2 H2O(l) + O2(g)

T2 = 820 K, (t1/2)2 = 900 min At room temperature the rate of reaction is

(t1/2)1 slower in the absence of catalyst with its
(t1/2)2
Thus, log10 = activation energy being 76 kJ mol-1. In the

presence of iodide ion/ the catalyst I- the

[250 × 103 J mol-1 820 K - 720 K reaction is faster since the activation energy

2.303 × 8.314 J K-1mol-1 820 K × 720 K decreases to 57 kJ mol-1.

= 2.212

(t1/2)1 = antilog 2.212 = 162.7
(t1/2)2

(t1/2)1 = (t1/2)2 × 162.7 = 900 × 162.7
= 1.464 × 105 min

134

Fig 6.11 shows a plot of fraction of molecules
as a function of energy. A catalyst lowers
the the threshold energy. Consequently more
molecules acquire the minimum amount of
energy and tend to cross the energy barrier. A
fraction of activated molecules is greater for
the catalyzed reaction. The rate of catalyzed
reaction thus is larger than the reaction with
no catalyst.

Fig. 6.11 : Comparison of fraction of molecules
for catalyzed and uncatalyzed reactions

Exercises

1. Choose the most correct option.

i. The rate law for the reaction v. Slope of the graph ln[A]t versus t for
aA + bB P is rate = k[A] [B]. first order reaction is

The rate of reaction doubles if a. -k b. k

a. concentrations of A and B are both c. k/2.303 d. -k/2.303
doubled.
vi. What is the half life of a first order
b. [A] is doubled and [B] is kept reaction if time required to decrease
constant concentration of reactant from 0.8 M to
0.2 M is 12 h?
c. [B] is doubled and [A] is halved

d. [A] is kept constant and [B] is a. 12 h b. 3 h
halved.
c. 1.5 h d. 6 h

ii. The order of the reaction for which the vii. The reaction, 3 ClO ClO3 +2 Cl
units of rate constant are mol dm-3 s-1 occurs in two steps,
is
(i) 2 ClO- ClO2

a. 1 b. 3 (ii) ClO2 + ClO ClO3 + Cl

c. 0 d. 2 The reaction intermediate is

iii. The rate constant for the reaction a. Cl b. ClO2

2 N2O5(g) 2 N2O4(g) + O2(g) is c. ClO3 d. ClO

4.98 × 10-4 s-1. The order of reaction is viii. The elementary reaction

a. 2 b. 1 O3(g) + O(g) 2 O2(g) is

c. 0 d. 3 a. unimolecular and second order

iv. Time required for 90 % completion of b. bimolecular and first order
a certain first order reaction is t. The c. bimolecular and second order
time required for 99.9 % completion d. unimolecular and first order
will be
ix. Rate law for the reaction,

a. t b. 2t 2 NO + Cl2 2 NOCl is
c. t/2 d. 3t
rate = k[NO2]2[Cl2]. Thus of k would
increase with

135

a. increase of temperature if time is expressed in seconds and
concentration of reactants in mol/L?
b. increase of concentration of NO
vii. Write Arrhenius equation and explain
c. increase of concentration of Cl2 the terms involved in it.
d. increase of concentrations of
viii. What is the rate determining step?
both Cl2 and NO
ix. Write the relationships between rate
x. For an endothermic reaction, X Y. constant and half life of first order and
zeroth order reactions.
If Ef is activation energy of the forward
reaction and Er that for reverse reaction, x. How do half lives of the first order
which of the following is correct? and zero order reactions change with
initial concentration of reactants?
a. Ef = Er b. Ef < Er
c. Ef > Er

d. ∆H = Ef - Er is negative 3. Answer the following in brief.

2. Answer the following in one or two i. How instantaneous rate of reaction is
sentences. determined?

i. For the reaction, ii. Distinguish between order and
molecularity of a reaction.
N2(g) + 3 H2(g) 2 NH3(g),
iii. A reaction takes place in two steps,
what is the relationship among d[N2] ,
dt 1. NO(g) + Cl2(g) NOCl2(g)
d[H2] d[NH3]
dt and dt ? 2. NOCl2(g) + NO(g) 2
NOCl(g)
ii. For the reaction,

CH3Br(aq) + OH-(aq) a. Write the overall reaction. b. Identify
reaction intermediate. c. What is the
CH3OH (aq) +Br (aq), rate law is molecularity of each step?
rate = k[CH3Br][OH ]
iv. Obtain the relationship between the
a. How does reaction rate changes if rate constant and half life of a first
[OH ] is decreased by a factor of 5 ? order reaction.

b. What is change in rate if v. How will you represent zeroth order
concentrations of both reactants are reaction graphically?
doubled?
vi. What are pseudo-first order reactions?
iii. What is the relationship between Give one example and explain why it
coefficients of reactants in a balanced is pseudo-first order.
equation for an overall reaction and
exponents in rate law. In what case the vii. What are requirements for the colliding
coefficients are the exponents? reactant molecules to lead to products?

iv. Why all collisions between reactant viii. How catalyst increases the rate of
molecules do not lead to a chemical reaction? Explain with the help of
reaction? potential energy diagram for catalyzed
and uncatalyzed reactions.

v. What is the activation energy of a ix. Explain with the help of Arrhenius
reaction? equation, how does the rate of reaction
changes with (a) temperature and (b)
vi. What are the units for rate constants for activation energy.
zero order and second order reactions

136

x. Derive the integrated rate law for first v. The rate constant of a reaction at 5000C
order reaction. is 1.6 × 103 M-1s-1. What is the frequency
factor of the reaction if its activation
xi. How will you represent first order energy is 56 kJ/mol. (9.72 × 106 M-1s-1)
reactions graphically.
vi. Show that time required for 99.9%
xii. Derive the integrated rate law for the first completion of a first order reaction is
order reaction, A(g) B(g) + C(g) three times the time required for 90%
completion.
in terms of pressure.
vii. A first order reaction takes 40 minutes
xiii. What is zeroth order reaction? Derive its for 30% decomposition. Calculate its half
integrated rate law. What are the units of life. (77.66 min)
rate constant?
viii. The rate constant for the first order reaction
xiv. How will you determine activation is given by log10 k = 14.34 - 1.25 × 104
energy: (a) graphically using Arrhenius T. Calculate activation energy of the
equation (b) from rate constants at two reaction. (239.3 kJ/mol)
different temperatures?
ix. What fraction of molecules in a gas at
xv. Explain graphically the effect of 300 K collide with an energy equal to
temperature on the rate of reaction. activation energy of 50 kJ/mol ? (2 × 10-9)

xvi. Explain graphically the effect of catalyst
on the rate of reaction.

xvii. For the reaction 2A + B products, Activity :

find the rate law from the following data. 1. You wish to determine
the reaction order and rate
[A]/M [B]/M rate/M s-1 constant for the reaction, 2AB2
0.3 0.05 0.15 A2 + 2B2. a) What data would
0.6 0.05 0.30 you collect? b) How would you use
0.6 0.2 1.20 these data to determine whether the
reaction is zeroth or first order?
4. Solve
2. The activation energy for two
i. In a first order reaction, the concentration reactions are Ea and E'a with Ea > E'a.
of reactant decreases from 20 mmol dm-3 If the temperature of reacting system
to 8 mmol dm-3 in 38 minutes. What is increase from T1 to T2, predict which
the half life of reaction? (28.7 min) of the following is correct?

ii. The half life of a first order reaction is 1.7 a. k'1 = k'2 b. k'1 > k'2
hours. How long will it take for 20% of k1 k2 k1 k2
the reactant to react? (32.9 min)
k'2 k'1 k'2
iii. The energy of activation for a first order k2 k1 k2
reaction is 104 kJ/mol. The rate constant c. k'1 < d. <2
at 25 0C is 3.7 × 10-5 s-1. What is the rate k1
constant at 300C? (R = 8.314 J/K mol) ks are rate constants at lower
(7.4 × 10-5) temperature and k's at higher
temperature.
iv. What is the energy of activation of a
reaction whose rate constant doubles
when the temperature changes from 303
K to 313 K? (54.66 kJ/mol)

137

7. ELEMENTS OF GROUPS 16, 17 AND 18

Can you recall ? Fluorine (9F), chlorine (17Cl), bromine
(35Br), iodine (53I) and astatine (85At) constitute
• How does the valence shell Group 17. These are collectively known as
electronic configuration of the halogens (Greek halo means salt, gene means
elements vary in the p-block of periodic born), that is, salt producing element.
table ?
Halogens are very reactive due to high
• Name the first element of groups 16, electronegativities and hence they are not
17 and 18. found in free sate. They occur in the form
of compounds.
7.1 Introduction : You have learnt in
Std. XI that in the p-block elements the Fluorine occurs mainly as insoluble
differentiating electron (the last filling fluorides (fluorspar CaF2, cryolite Na3AlF6,
electron) enters the p-orbital of the outermost fluorapatite 3Ca3(PO4)2.CaF2) and small
shell. Since maximum of six electrons can be quantities are present in soil, fresh water
accommodated in a p-subshell it gives rise to plants, and bones and teeth of animals. Sea
groups 13 to 18, in the p-block. In this chapter water contains chlorides, bromides and iodides
we shall study the properties of elements of of Na, K, Mg and Ca. However it mainly
groups 16, 17 and 18. contains NaCl (2.5 % by mass). The deposits
of dried up sea beds contain sodium chloride
7.2 Occurence : The elements oxygen (8O), and carnallite, KCl.MgCl2.6H2O. Marine life
sulfur (16S), selenium (34Se), tellurium (52Te) also contains iodine in their systems. For
and polonium (84Po) constitute Group 16, example, sea weed contains upto 0.5 % iodine
called the oxygen family. Large number of and chile saltpetre contains upto 0.2 % of
metal ores are oxides or sulfides. Group 16 sodium iodate. Astatine, the last member of
elements are also called chalcogens or ore halogen family is radioactive and has a half
forming elements. life of 8.1 hours.

Oxygen is the most abundant of all the The elements helium (2He), neon (10Ne),
elements on earth. Oxygen forms 20.95 % argon (18Ar), krypton (36Kr), xenon (54Xe) and
by volume of air and about 46.6 % by mass radon (86Rn) constitute the Group 18.
of earth's crust. Sulfur forms 0.034% by
mass of the earths crust. It occurs mainly in All the noble gases except radon occur
combined forms as sulfates such as gypsum in the atmosphere. Their abundance in dry air
(CaSO4.2H2O), epsom salt (MgSO4.7H2O), is ∼ 1% (by volume) with argon as the major
baryte (BaSO4) and sulfides such as galena constituent. The main commercial source of
(PbS), zinc blende (ZnS), copper pyrites helium is natural gas. Helium and neon are
(CuFeS2). found in minerals of radioactive origin e.g.
pitchblende, monazite, cleveite. Xenon and
Selenium and tellurium are also found as radon are the rarest elements of the group.
metal selenides and tellurides in sulfide ores. Radon is a decay product of 226Ra.
Polonium which is radioactive is a decay
product of thorium and uranium.

138

Table 7.1 : Condensed electronic configuration of elements of group 16, 17 and 18

Group 16 (Oxygen family) Group 17 (Halogen family) Group 18 (Noble gases)

Element Condensed Element Condensed Element Condensed
Electronic Electronic Electronic
Configuration Configuration Configuration

2He 1s2

8O [He]2s22p4 9F [He]2s22p5 10Ne [He]2s22p6

16S [Ne]3s23p4 17Cl [Ne]3s23p5 18Ar [Ne]3s23p6

34Se [Ar]3d104s24p4 35Br [Ar]3d104s24p5 36Kr [Ar]3d104s24p6

52Te [Kr]4d105s25p4 53I [Kr]4d105s25p5 54Xe [Kr]4d105s25p6

84Po [Xe]4f145d106s26p4 85At [Xe]4f145d106s26p5 86Rn [Xe]4f145d106s26p6

7.3 Electronic configuration of elements of down the group, as a result of increase in the
group 16, 17 and 18: : The general electronic number of quantum shells.
configuration of the group 16 elements is
ns2np4 while that of group 17 elements is Across a period atomic or ionic radii decrease
ns2np5. The group 18 elements are shown by with increasing atomic number, consequent
ns2np6 configuration. to increase in (Zeff) effective nuclear charge.
Group 17 elements (Halogens) have the
The elements of groups 16 and 17 smallest atomic radii in their respective
repectively have two and one electrons less periods.
than the stable electronic configuration of the
nearest noble gas. ii. Ionisation enthalpy : The group 16, 17 and
18 elements have high ionisation enthalpy.
Table 7.1 shows the condensed electronic The ionisation enthalpy decreases down the
configuration of the elements of group 16, 17 group due to increase in the atomic size.
and 18.
Across a period ionisation enthalpy increases
7.4 Atomic and physical properties of with increase of atomic number. This is due
elements of group 16, 17 and 18. to addition of electrons in the same shell.

7.4.1 Atomic properties of Group 16, 17 However the elements of group 16 have lower
and 18 elements : These properties are given ionisation enthalpy values compared to those
in Tables 7.2, 7.3 and 7.4. of group 15 in the corresponding periods,
owing to extra stable half filled electronic
i. Atomic and Ionic radii : In group 16, configuration of p-orbitals in elements of
17 and 18 atomic and ionic radii increase group 15.

Table 7.2 : Atomic and physical properties of group 16 elements.

Element Atomic Atomic Atomic Ionic Ionization Electro- Electron Density M.P. B.P.
number mass radius radius enthalpy negativity gain g /cm3 (K) (K)
g/mol (pm)
E2 (∆iH1) 3.50 enthalpy
16.00 66 (pm) kJ/mol 2.44 kJ/mol
32.06 104 2.48
O8 78.96 117 140 1314 2.01 -141 1.32 55 90
S 16 127.60 137 1.76
Se 34 210.00 146 184 1000 -200 2.06 393 718
Te 52
Po 84 198 941 -195 4.19 490 958

221 869 -190 6.25 725 1260

230 813 -174 - 520 1235

139

Table 7.3 : Atomic and physical properties of group 17 elements

Element Atomic Atomic Atomic Ionic Ionization Electro Electron Density M.P. B.P.
number mass radius radius enthalpy negativity gain g/cm3 (K) (K)
F g/mol (pm) E (pm)
Cl 9 (∆iH1) 4.0 enthalpy 1.5 54.4 84.9
Br 17 19.00 64 133 kJ/mol 3.2 kJ/mol 1.66 172.0 239.0
I 35 35.45 99 184 3.0 3.19 265.8 332.5
At 53 79.90 114 196 1680 2.7 -333 4.94 386.6 458.2
85 126.90 133 220 2.2
210 - 1256 -349 - - -
-
1142 -325

1008 -296

- -

Table 7.4 : Atomic and physical properties of group 18 elements.

Element Atomic Atomic Atomic Ionization Electron Density M.P. B.P. Atmospheric
number mass radius enthalpy gain g/cm3 (K) (K) content
g/mol (pm)
(∆iH1) enthalpy 1.8 × 10-4 - 4.2 (% by volume)
kJ/mol kJ/mol 9.0 × 10-4 24.6 27.1
1.8 × 10-3 83.8 87.2 5.24 × 10-4
He 2 4.00 120 2372 48 3.7 × 10-3 115.8 119.7
5.9 × 10-3 161.3 165.0 -
Ne 10 20.18 160 2080 116 9.7 × 10-3 202 211 1.82 × 10-3

Ar 18 39.95 190 1520 96 0.934
1.14 × 10-4
Kr 36 83.80 200 1351 96 8.7 × 10-6

Xe 54 131.30 220 1170 77

Rn 86 222.00 - 1037 68

iii. Electronegativity : In a group (16, 17 and Try this...
18) the electronegativity decreases down the
group. • Observe Table no 7.3 and
explain the trend in following
* Oxygen has the highest electronegativity atomic properties of group 17 elements.
next to fluorine amongst all the elements.
i. Atomic size, ii. Ionisation enthalpy,
* Halogens have very high electronegativity. iii. electronegativity, iv. electron gain
Fluorine is the most electronegative element enthalpy
in the periodic table.
• Oxygen has less negative electron gain
iv. Electron gain enthalpy : In the groups 16 enthalpy than sulfur. Why ?
and 17 electron gain enthalpy becomes less
negative down the group. Problem 7.1 : Elements of group 16
generally show lower values of first
However in group 16, oxygen has less ionisation enthalpy compared to the
negative electron gain enthalpy than sulfur elements of corresponding period of group
due to its small atomic size. 15. Why ?

* In group 17, fluorine has less negative Solution : Group 15 elements have extra
electron gain enthalpy than that of chlorine. stable, half filled p-orbitals with electronic
This is due to small size of fluorine atom. configuration (ns2np3). Therefore more
amount of energy is required to remove an
* Group 18 elements (noble gases) have electron compared to that of the partially
no tendency to accept electrons because of filled orbitals (ns2np4) of group 16 elements
their stable electronic configuration (ns2np6) of the corresponding period.
and thus have large positive electron gain
enthalpy.

140

Problem 7.2 : The values of first ionisation b. Group 17 elements (Halogen family) :
enthalpy of S and Cl are 1000 and 1256 kJ
mol-1, respectively. Explain the observed Fluorine, chlorine are gases, bromine is a liquid
trend. and iodine is a solid at room temperature. F2
is yellow, Cl2 greenish yellow, Br2 red and I2
Solution : The elements S and Cl belong is violet, in colour.
to second period of the periodic table.
Fluorine and chlorine react with water.
Across a period effective nuclear charge Bromine and iodine are only sparingly soluble
increases and atomic size decreases with in water and are soluble in various organic
increase in atomic number. solvents such as chloroform, carbon disulfide,
carbon tetrachloride, hydrocarbons which
Therefore the energy required for the give coloured solutions. Bond dissociation
removal of electron from the valence enthalpies of halogen molecules follow the
shell (I.E.) increases in the order order : Cl - Cl > Br - Br > F - F > I - I.
S < Cl.
Problem 7.4 : Fluorine has less negative
7.4.2 Physical properties of group 16, 17 electron gain affinity than chlorine. Why ?
and 18 elements :
Solution : The size of fluorine atom is
a. Group 16 elements (Oxygen family or smaller than chlorine atom. As a result, there
chalcogens) : Oxygen is a gas while other are strong inter electronic repulsions in the
elements are solids at room temperature. relatively small 2p orbitals of fluorine and
therefore, the incoming electron does not
Oxygen and sulfur are nonmetals, selenium experience much attraction. Thus fluorine
and tellurium are metalloids, while polonium has less negative electron gain affinity than
is a metal. Polonium is radioactive with its chlorine.
half life of 13.8 days.
Problem 7.5 : Bond dissociation enthalpy
Melting and boiling points increase with of F2 (158.8 KJ mol-1) is lower than that
increasing atomic number. of Cl2 (242.6 KJ mol-1) Why ?
Solution : Fluorine has small atomic size
All the elements of group 16 exhibit allotropy. than chlorine. The lone pairs on each F
atoms in F2 molecule are so close together
Problem 7.3 : Why is there a large that they strongly repel each other, and
difference between the melting and boiling make the F - F bond weak (fig. 7.1) Thus
points of oxygen and sulfur ? it requires less amount of energy to break
the F - F bond. In Cl2 molecule the lone
Solution : Oxygen exists as diatomic pairs on each Cl atom are at a larger
molecule (O2) where as sulfur exists as distance and the repulsion is negligible.
polyatomic molecule (S8). Thus Cl - Cl bond is comparitively stronger.
The van der Waals forces of attraction Therefore bond dissociation enthalpy of F2
between O2 molecules are relatively weak is lower than that of Cl2.
owing to its much smaller size.
lone pair - lone pair
The large van der Waals attractive forces in repulsion
the S8 molecules can be noticed because of
large molecular size. Therefore oxygen has F-F
low m.p. and b.p. as compared to sulfur.
Fig. 7.1

141