5.

ELECTROCHEMISTRY
makes possible the manufacture of essential
Can you recall ? chemicals. You have learnt preparation of
• What is a redox reaction ? NaOH, widely used in the manufacture of
• Which form of energy is soaps, detergents and paper, by electrolysis of
converted into electrical energy in dry NaCl. Electrolysis is possibly the only means
cells ? to produce fluorine. The processes such as
electro-refining (for purification of metals),
electroplating (for coating one metal is on the
surface of another) are also electrochemical
processes.
In standard XI, you learnt redox reactions.
Redox reaction forms the basis for the
• How is NaOH manufactured from generation of electricity by chemical reactions
NaCl ? and also for chemical reactions brought out
by means of electricity. These processes
5.1 Introduction : Dry cell is used to power are carried out in an electrochemical cell.
our electrical and electronic equipments Electrochemistry deals with the design and
because it generates electricity. Do you operation of such cells.
know how does a dry cell generate electricity
? A chemical reaction occurs in it which The current research in electrochemistry
generates electricity. Thus in a dry cell is focused on the design of fuel cells. The
chemical energy is converted into electrical fuel cells are being explored as convenient
energy. and compact source of electricity.

You are familiar with the electrolysis 5.2 Electric conduction : We know that the
of solutions of ions. Electrolysis is breaking electric current represents a charge transfer.
down of an ionic compound by the passage A charge transfer or flow of electricity occurs
of eletricity. Breaking down of an electrolyte through substances called conductors. There
during electrolysis is a chemical reaction are two types of conductors which give rise
that takes place by the passage of electricity. to two types of conduction of electricity.
Electrical energy is, thus, converted into 5.2.1 Metallic conduction :
chemical energy.
Can you recall ?
Electrochemistry is the area of chemistry
• What is the origin of electrical
which is concerned with interconversion of
conductivity of metals ?
chemical and electrical energy.
It also deals with the resistance and Electrical conduction through metals
conductance of aqueous electrolytic solutions. involves a direct flow of electrons from one
The determination of conductivities of aqueous point to the other. The outermost electrons of
electrolytic solutions provide an information metals form conduction bond. The electrons in
on the extent of ionization of electrolytes in conduction band are free to move and hence
water. (Refer to Chapter 3). flow under the influence of applied electrical
potential (Chapter 1). Metallic conductors
The study of electrochemical cells is
are, thus, electronic conductors.
important in science and technology. It
90
5.2.2 Electrolytic or ionic conduction : i. The conducting and nonconducting
Electrolytic conduction involves conduction nature of solutions can be identified
of electric current by the movement of ions by measurement of their conductivity.
of the electrolytes. In this type of conduction Sucrose and urea do not dissociate in their
the charge transfer occurs in the form of aqueous solutions. The conductivities of
movement of ions through molten electrolytes these solutions are nearly the same as
or the aqueous solutions of electrolytes. that of water. These substances are called
Substances such as ionic salts, strong or weak nonelectrolytes.
acids and bases are the electrolytes. These On the other hand, substances like
dissociate into ions when dissolved in polar potassium chloride, acetic acid, sodium
solvents such as water. Ionic solids dissociate hydroxide, HCl dissociate in their aqueous
into ions in molten state as well. solutions. The conductivities of their
Conduction through electrolytic aqueous solutions are higher than that
conductors involves transfer of matter from of water. These are called electrolytes.
one part of the conductor to the other. It Electrolytes conduct electricity in molten
means that the current flowing through an state or when dissolved in water.
electrolytic conductor is accompanied by a ii. On the basis of high or low electrical
chemical change. conductivity electrolytes are classified
5.2.3 Information provided by measurement into strong and weak electrolytes. The
of conductivities of solutions : substances such as ionic salts, strong
acids or bases are almost completely
Try this... dissociated in aqueous solutions. These
are strong electrolytes. The solutions
of strong electrolytes exhibit high
Switch Battery
conductivities.
The weak acids and weak bases are
Non conductive weak electrolytes. They dissociate to a
frame Lamp very small extent in aqueous solutions
and show lower conductivities than those
Electrodes of strong electrolytes.
Solution Remember...
• Arrange a simple set up as shown in Electrolyte is a compound that
the diagram above. conducts electricity when molten or
in aqueous solution and breaks down into
• The lamp will glow when circuit is ions during electrolysis.
complete.
5.3 Electrical conductance of solution :
• Prepare 5 % (mass/volume) solutions
According to Ohm's law, the electrical
of cane sugar, acetic acid, sodium
resistance R of a conductor is equal to the
chloride and urea in distilled water.
electric potential difference V divided by the
• Check the electrical conductivity electric current, I :
of these solutions using the above V
R= (5.1)
assembly. Compare these with that I
observed with distilled water. The SI unit of potential is volt (V) and
that of current is ampere (A). The unit of

91
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 = (5.5)
a R a
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= (5.2) Units of electrolytic conductivity
R
The SI unit of G is siemens, denoted by
Quantity SI unit Common unit
S, which is equal to Ω-1. Therefore, we write
Length m cm
S = Ω-1 = AV-1 = CV-1s-1 where C represents
coulomb, the unit of electricity related to Area m2 cm2
current strength in ampere and time in Resistance Ω Ω
seconds as C = A s. Conductivity Ω m-1 or
-1
Ω-1 cm-1
S m-1
The electrical resistance of a conductor
is proportional to length l and inversely 5.3.2 Molar conductivity (∧) : The electrolytic
proportional to cross sectional area a. Thus, conductivity is not suitable for comparing
l l conductivities of different solutions. The
R∝ or R = ρ (5.3)
a a conductivity of a solution depends on number
where ρ, the proportionality constant is of ions present in unit volume of solution.
called resistivity of the conductor. It is the The solution of higher concentration contains
resistance of conductor of unit length and unit more ions and exhibits higher conductivity
cross sectional area. than the solution of lower concentration. To
Can you recall ? compare conductivities of different solutions,
they must have the same concentration.
What is the SI unit of In 1880, the German physicist F.W.G.
resistivity ? Kohlrausch introduced the term molar
5.3.1 Conductivity (k) : We have seen that conductivity denoted by ∧ (lambda).
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.
is directly proportional to a and inversely k
∧= (5.6)
proportional to the length l. Thus c
a a SI units of k are S m-1 and that of
G∝ or G = k (5.4) c are mol m-3. Hence SI units of ∧ are
l l
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
Thus, conductivity is the electrical understand the significance of ∧, consider
conductance of a conductor of unit length volume of a solution containing 1 mole of
and unit area of cross section. In other words, dissolved electrolyte. Suppose the solution is
the conductivity is the electrical conductance placed between two parallel electrodes 1 cm
of unit cube of material. Conductivity of apart and large enough to accommodate it.
solution of an electrolyte is called electrolytic The electrical conductance exhibited by this
conductivity which refers to the electrical solution is the molar conductivity. The molar
conductance of unit volume (1 m3 or 1 cm3) conductivity is the electrical conductance
of solution. generated by all the ions in 1 mole of the
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 Ω-1 cm2 mol-1 × 0.05 mol L-1
k=
1 cm3 of given solution. Molar conductivity 1000 cm3L-1
is the electrical conductance due to the ions = 0.01115 Ω cm-1
-1

obtained from 1 mole of an electrolyte in 5.3.4 Variation of conductivity with

a given volume of solution. 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 ii. On dilution total number of ions increase
k. Hence, total conductance of V cm3 is kV as a result of increased degree of
which is molar conductivity. 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,
= c mol L-1 the number of ions per unit volume
of solution decreases. This results in
c mol L-1 c decrease of conductivity with decrease
= 3 -1 = mol cm-3
1000 cm L 1000 in concentration of solution.
Volume, V of solution in cm3 containing
1 mole of an electrolyte is reciprocal of Suppose 100 cm3 of solution of an
concentration. Therefore, electrolyte contains 8 × 1020 ions. The
1 1000 number of ions per cm3 is 8 × 1018.
V= = cm3 mol-1
concentraion c If the solution is diluted to 1000 cm3
(5.8) the total number of ions will increase but
Substitution for V in Eq. (5.7) yields not by a factor of 10. Assume that the
1000k number of ions increases from 8 × 1020
∧= (5.9) to 64 × 1020 on dilution. After dilution the
c
number of ions per cm3 is 6.8 × 1018.
Try this...
It is evident that the number of ions
What must be the per cm3 decreases from 8 × 1018 to 6.8 ×
concentration of a solution of silver 1018 on dilution from 100 cm3 to 1000 cm3
nitrate to have the molar conductivity of and in turn, the conductivity decreases.
121.4 Ω-1 cm2 mol-1 and the conductivity of
2.428 × 10-3 Ω-1 cm-1 at 25 0C ? 5.3.5 Variation of molar conductivity with
concentration
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
∧= or k =
c 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 at zero concentration can be determined by
conductivity with concentration in case of extrapolation of linear part of ∧ versus c
strong and weak electrolytes is qualitatively curve as shown in Fig. 5.1. This method
different. cannot be used for weak electrolytes since ∧
i. Strong electrolytes : The molar conductivity versus c curve does not approach linearity.
of solution of strong electrolyte increases Kohlrausch law is useful for calculating ∧0
rapidly with dilution. It approaches the limiting of weak electrolytes.
value for 0.001 M or 0.0001 M solution. The 5.3.7 Kohlrausch law of independent
dilution has no effect on molar conductivity migration of ions : The law states that at
thereafter. The maximum limiting value of infinite dilution each ion migrates independent
molar conductivity is the molar conductivity of co-ion and contributes to total molar
at zero concentration or at infinite dilution. conductivity of an elctrolyte irrespective of the
It is denoted by ∧0. The zero concentration nature of other ion to which it is associated.
or infinite dilution means the solution is so Both cation and anion contribute to
dilute that further dilution does not increase molar conductivity of the electrolyte at zero
the molar conductivity. concentration and thus ∧0 is sum of molar
During nineteenth century F. Kohlrausch conductivity of cation and that of the anion
with repeated experiments showed that the at zero concentration. Thus,
molar conductivity of strong electrolytes varies ∧0 = n⊕ λ0⊕ + n λ0 (5.11)
linearly with square root of concentration as :
where λ⊕ and λ are molar conductivities of
∧ = ∧0 - a c (5.10) 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
1. The theory can be used to calculate the
Strong molar conductivity of an electrolyte at
electrolyte
the zero concentration. For example,
∧0 (KCl) = λ0K⊕+ λ0Cl
Weak
electrolyte
∧0 [Ba(OH)2] = λ0Ba2⊕+ 2 λ0OH
Knowing the molar conductivites of ions
at infinite dilution, ∧0 values of electrolyte
Fig. 5.1 : Variation of ∧ with c can be obtained.
ii. Weak electrolytes : The molar conductivity 2. The theory is particularly useful in
of weak electrolytes increases rapidly on calculating ∧0 values of weak electrolytes
dilution. For concentrations of 0.001M or from those of strong electrolytes. For
0.0001 M, the ∧ value is lower than ∧0 the example, ∧0 of acetic acid can be
molar conductivity at zero concentration. 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 + λ0CH3COO + λ0Na⊕ - λ0Na⊕ - λ0Cl
94
= λ0H⊕ + λ0CH3COO = ∧0 (CH3COOH) Problem 5.3 : Calculate molar conductivities
Thus, at zero concentration for CaCl2 and Na2SO4.
Given : molar ionic conductivitis of Ca2⊕,
∧0 (CH3COOH) = ∧0 (HCl) + ∧0 (CH3COONa) Cl , Na⊕ and SO42 ions are respectively,
- ∧0 (NaCl)
104, 76.4, 50.1 and 159.6 Ω-1 cm2 mol-1.
Because ∧0 values of strong electrolytes,
Solution :
HCl, CH3COONa and NaCl, can be determined
by extrapolation method, the ∧0 of acetic acid According to Kohrausch law,
can be obtained. i. ∧0 (CaCl2) = λ0Ca2⊕ + 2λ0Cl
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 = 256.8 Ω-1 cm2 mol-1
if the molar conductivities of NaI, AgNO3
ii. ∧0 (Na2SO4) = 2λ0Na⊕ + λ0SO42
and NaNO3 at zero concentration are
respectively, 126.9, 133.4 and 121.5 Ω-1 = 2 × 50.1 Ω-1 cm2 mol-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
i. ∧0 (NaI) = λ0Na⊕ + λ0I 0.01M acetic acid at 25 0C is 16.5 Ω-1 cm2
= 126.9 Ω-1 cm2 mol-1 mol-1. Calculate its degree of dissociation in
0.01 M solution and dissociation constant
ii. ∧0 (AgNO3) = λ0Ag⊕ + λ0NO3
if molar conductivity of acetic acid at zero
= 133.4 Ω-1 cm2 mol-1 concentration is 390.7 Ω-1 cm2 mol-1.
iii. ∧0 (NaNO3) = λ0Na⊕ + λ0NO3 Solution :
= 121.5 Ω cm mol
-1 2 -1
The degree of dissociation,
∧c
Eq. (i) + eq. (ii) - eq. (iii) gives ∝= ∧
0
∧0 (NaI) + ∧0 (AgNO3) - ∧0 (NaNO3)
= λ0Na⊕ + λ0I + λ0Ag⊕ + λ0NO3 - λ0Na⊕ -λ0NO3 16.6 Ω-1 cm2 mol-1
= = 0.0422
390.7 Ω-1 cm2 mol-1
= λ0Ag⊕ + λ0I
= ∧0 (AgI) ∝2c (0.0422)2 × 0.01
Ka = = = 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
is related to its molar conductivity at a given
Try this... concentration c by the equation,
Calculate ∧0 (CH2ClOOH) ∧c
if ∧0 values for HCl, KCl and ∝ = ∧ (5.12)
0
CH2ClCOOK are repectively, 4.261, where ∧c is the molar conductivity of
1.499 and 1.132 Ω-1 cm2 mol-1. weak electrolyte at concentration c ; ∧0 is
molar conductivity at zero concentration.

95
 1. Determination of cell constant : The cell
Try this...
constant is determined using the 1 M, 0.1 M
Obtain the expression for or 0.01 M KCl solutions. The conductivity
dissociation constant in terms of ∧c of KCl solution is well tabulated at various
and ∧0 using Ostwald's dilution law. temperatures. The resistance of KCl solution
is measured by Wheatstone bridge. (Refer to
5.3.9 Measurement of conductivity : The standard XII Physics Textbook Chapter 9)
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
consists of a glass tube with two platinum
conductivity cell
plates coated with a thin layer of finely
divided platinium black. This is achieved by
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. A.C.
solution of unknown
resistance
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.
The sliding contact is moved along AB
Fig. 5.2 : Conductivity cell
until no current flows. The detector D shows
Cell constant : The conductivity of an no deflection. The null point is, thus, obtained
electrolytic solution is given by Eq. (5.5), at C.
1 l
k= 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
area of cross section (a) of the electrode is l(AC)
Hence, Rsolution = × Rx (5.15)
called the cell constant. Thus, l(BC)
l By measuring lengths AC and BC and
Cell constant = (5.13)
a 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
conveniently expressed in cm-1. The Eq. (5.5) Eq. (5.13).
then becomes
cell constant Cell constant = kKCl × Rsolution
k= (5.14)
R The conductivity of KCl solution is
The determination of conductivity consists known. The cell constant, thus, can be
of three steps : calculated.

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
solution whose conductivity is to be measured. What is the reaction involving
Its resistance is measured by the process transfer of electrons from one
described in step (1). The conductivity of chemical species to another called?
given solution is then calculated as : The chemical reaction occuring in
Cell constant
k= electrochemical cells involves transfer
Rsolution
of electrons from one species to the
3. Calculation of molar conductivity : The
other. It is a redox reaction, we learnt in
molar conductivity of the given solution is
(Std. XI, Chapter 6).
then calculated using Eq. (5.9).
Electrochemical reactions, are made of
1000 k
∧= c oxidation and reduction half reactions. The
oxidation half reaction occurs at one electrode
Problem 5.5 : A conductivity cell containing and the reduction half reaction occurs at the
0.01M KCl gives at 250C the resistance other electrode. The net cell reaction is the
of 604 ohms. The same cell containing sum of these half reactions.
0.001M AgNO3 gives resistance of 6530 5.4.2 Electrodes : Electrodes are the surfaces
ohms. Calculate the molar conductivity of on which oxidation and reduction half
0.001M AgNO3. [Conductivity of 0.01M reactions take place. Electrodes may or may
KCl at 25 0C is 0.00141 Ω-1 cm-1] not participate in the reactions. The electrodes
Solution : which do not take part in reactions are inert
i. Calculation of cell constant electrodes.
Cell constant = kKCl × RKCl Cathode : It is an electrode at which the
= 0.00141 Ω- cm-× 604 Ω reduction takes place. At this electrode the
= 0.852 cm-1 species undergoing reduction gains electrons.
ii. Calculation of conductiviy of AgNO3' Anode : It is an electrode at which oxidation
Cell constant takes place. At this electrode, the species
k= where R = 6530 Ω undergoing oxidation loses electrons.
R
5.4.3 Types of electrochemical cells : There
0.852 cm-1
= are two types of electrochemical cells.
6530 Ω
1. Electrolytic cell : In this type of cell,
= 1.3 × 10-4 Ω-1 cm-1
a nonspontaneous reaction, known as
iii. Calculation of molar conductivity of electrolysis, is forced to occur by passing a
AgNO3 direct current from an external source into
1000k
∧= where c = 0.001 M the solution. In such cells electrical energy
c
is converted into chemcial energy. The anode
1000 cm3 L-1 × 1.3 × 10-4 Ω-1 cm-1
= of electrolytic cell is positive and cathode is
0.001 mol L-1
negative.
= 130 Ω-1 cm2 mol-1
2. Galvanic or voltaic cell : In galvanic
5.4 Electrochemical cells : An electrochemical (voltaic) cell a spontaneous chemical reaction
cell consists of two metal plates or carbon produces electricity. In these cells chemical
(graphite) rods. These electronic conductors energy is converted into electrical energy.
are dipped into an electrolytic or ionic The anode of galvanic cell is negative and
conductor. cathode is positive.
97
 The carbon electrode connected to
Use your brain power terminal electrode of the battery is anode
Distinguish between
and that connected to negative terminal of
electrolytic and galvanic cells.
the battery is cathode.
5.5 Electrolytic cell
Remember...
Do you know ? In electrolysis the electrodes
Michael Faraday was the are usually inert, Pt or graphite.
first person to explain electrolysis
nearly 200 years ago. Reactions occuring in the cell : Fused NaCl
contains Na⊕ and Cl ions which are freely
Electrolytic cell consists of a container mobile. When potential is applied, cathode
in which electrolyte is placed. Two electrodes attracts Na⊕ ions and anode attracts Cl ions.
are immersed in the electrolyte and connected As these are charged particles, their migration
to a source of direct current. results in an electric current. When these
ions reach the respective electrodes, they
At anode (+) a species oxidises with
are discharged according to the following
the removal of electrons. These electrons are
reactions.
pulled from anode and pushed to cathode
through an external circuit. The electrons Oxidation half reaction at anode :
are supplied to species at cathode which are Cl ions migrate to anode. Each Cl
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
The battery sucks electrons so produced
in it. They are connected by metallic wires to
at the anode and pushes them to cathode
a source of direct current that is battery. This
through a wire in an external circuit. The
is shown in Fig. 5.4.
battery thus serves as an electron pump. The
electrons from the battery enter into solution
through cathode and leave the solution
Battery through anode.
(D.C. source)
Reduction half reaction at cathode : The
Carbon anode Carbon cathode
electrons supplied by the battery are used
in cathodic reduction. Each Na⊕ ion, that
reaches cathode accepts an electron from the
Cl2 gas Fused Na
cathode and reduces to metallic sodium.
Fused NaCl
Na⊕ (l) +e Na (l)
Fig. 5.4 : Electrolysis of fused NaCl
98
Net cell reaction The other is the reduction of water to
The net cell reaction is the sum of two hydrogen gas.
electrode reactions. ii. 2 H2O (l) + 2e H2 (g) + 2 OH (aq),
2 Cl (l) Cl2 (g) + 2e E0 = - 0.83 V
(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
(overall cell reaction) of water is the cathode reaction when the
Results of electrolysis of molten NaCl aqueous NaCl is electrolysed.
i. A pale green Cl2 gas is released at anode. Oxidation half reaction at anode : At anode
there will be competition between oxidation
ii. A molten silvery-white sodium is formed
of Cl ion to Cl2 gas as in case of molten
at cathode.
NaCl and the oxidation of water to O2 gas.
Decomposition of NaCl into metallic
i. 2 Cl (aq) Cl2 (g) +2e ,E0oxi = - 1.36 V
sodium and Cl2(g) is nonspontaneous. The
electrical energy supplied by the battery ii. 2H2O (l) O2 (g) + 4H⊕ (aq) + 2e
forces the reaction to occur. E0oxi = - 0.4 V
Standard electrode potential for the
Remember...
oxidation of water is greater than that of Cl
When molten ionic compound ion or water has greater tendency to undergo
is electrolysed, a metal is formed oxidation. It is, therefore, expected that anode
at the negative electrode and a nonmetal half reaction would be oxidation of water.
at the positive electrode. The experiments have shown, however, that
the gas produced at the anode is Cl2 and
5.5.2 Electrolysis of aqueous NaCl : not O2. This suggests that anode reaction is
Electrolysis of an aqueous NaCl can be oxidation of Cl to Cl2 gas. This is because
carried out in the cell used for the electrolysis of the overvoltage, discussion of which is
of molten NaCl using inert electrodes shown beyond the scope of the present book.
in Fig. 5.4. The fused NaCl is replaced by
moderately concentrated aqueous solution of It has been found experimentally
NaCl. The water involved in electrolysis of that the actual voltage required for
aqueous NaCl, leads to electrode reactions electrolysis is greater than that calculated
that differ from electrolysis of molten NaCl. using standard potentials. This additional
Reduction half reaction at cathode : At voltage required is the overpotential.
cathode, two reduction reactions compete.
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).
2 H2O (l) + 2e H2 (g) + 2 OH (aq) It is referred to as one faraday (IF). Hence,
(reduction at cathode) Moles of electrons actually passed
Q(C)
2 Cl (aq) + 2 H2O (l) Cl2 (g) + H2(g) = (5.17)
96500 (C/mol e )
+ 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.
i. H2 gas is liberated at cathode. The stoichiometry of half reaction indicates
ii. Cl2 gas is released at anode. the moles of electrons passed and moles of
the product formed. For the reaction,
iii. Because Na⊕ ions remain unreacted and
OH ions are formed at cathode, NaCl Cu2⊕ (aq) + 2e Cu (s), two moles of
solution is converted to NaOH solution. electrons are required for the production of
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 To simplify further we introduce the
every year by electrolysis. entity mole ratio given by
Electrochemical art : Al, Cr and Sn Mole ratio =
can be coloured by an electrochemical moles of product formed in the half reaction
process called anodizing. In this process moles of electrons required in the half reaction
metal anode oxidizes to give metal oxide 1
For the reaction of Cu, mole ratio =
coat. When an organic dye is added to the 2
electrolyte, dye molecules soak forming Therefore,
spongy surface of coating and become Moles of product formed
trapped with the hardening of the metal
oxide surface. = moles of electrons actually passed × mole
ratio
Q(C)
5.5.3 Quantitative aspects of electrolysis : = × mole ratio (5.18)
96500 (C/mol e )
a. The mass of reactant consumed or
the mass of product formed at an electrode I (A) × t (s)
= × mole ratio (5.19)
during electrolysis can be calculated by 96500 (C/mol e )
knowing stoichiometry of the half reaction iv. Calculation of mass of product :
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
I (A) × t (s)
eletricity (Q) passed during electrolysis, the = × mole ratio × molar mass
96500 (C/mol e ) of product
amount of current, I, passed through the cell
is measured. The time for which the current (5.20)
is passed is noted.
100
 b. Suppose two cells containing different
electrolytes are connected in series. The same ii. Mass of Cu formed,
quantity of electricity is passed through them. W=
The masses of the substances liberated at the I (A) × t (s)
× mole ratio × molar mass
electrodes of the two cells are related as given 96500 (C/mol e ) of Cu
below :
5 A × 100 × 60 s 1 mol
The mass of the substance produced at = ×
96500 (C/mol e )
- 2 mol e-
the electrode of first cell is given by
× 63.5 g mol-1
Q(C)
W1 = × (mole ratio)1 × M1 = 9.87 g
96500 (C/mol e )
Q(C)
Hence,
96500 (C/mol e ) Problem 5.7 : How long will it take to
W1 produce 2.415 g of Ag metal from its salt
= solution by passing a current of 3 ampere ?
(mole ratio)1 × M1
Molar mass of Ag is 107.9 g mol-1.
Similarly mass of substance liberated
Solution :
at the electrode of second cell is W2 in the
equation, i. Stoichiometry :
Q(C) W2
= (mole ratio) Ag⊕ (aq) + e Ag (s)
96500 (C/mol e ) 2
× M2
1 mol
mole ratio =
M1 and M2 are the molar masses of 1 mol e
substances produced at the electrodes of cells ii. W =
1 and 2.
Q(C) I (A) × t (s)
Because is the same × mole ratio × molar mass
96500 (C/mol e ) 96500 (C/mol e ) of Ag
for both, 3A×t 1 mol
2.415 g = ×
We have 96500 (C/mol e ) 1 mol e
W1 W2 × 107.9 g mol-1
=
(mole ratio)1 × M1 (mole ratio)2 × M2
2.415 × 96500 (C = As)
(5.21) t=
3 A × 107.9

= 720 s = 12 min.
Problem 5.6 : What is the mass of Cu
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
mole ratio =
2 mol e

101
Problem 5.8 : How many moles of Substitution of the quantities gives
electrons are required for reduction of 3 4.36g
moles of Zn2⊕ to Zn ? How many Faradays =
1 mol/2mol e × 65.4 g mol-1
of electricity will be required ?
W2
Solution : 1 mol/3mol e- × 27 g mol-1
i. The balanced equation for the reduction
of Zn2⊕ to Zn is 4.36 g × 2 W2 × 3
or =
65.4 27
Zn2⊕ (aq) + 2e Zn (s)
The equation shows that 1 mole of Zn2⊕ 4.36 g × 2 × 27
is reduced to Zn by 2 moles of electrons. Hence, W2 = = 1.2 g
65.4 × 3
For reduction of 3 moles of Zn2⊕ 6 moles
of electrons will be required. 5.6 Galvanic or voltaic cell : In galvanic or
Faraday (96500 Coulombs) is the amount voltaic cells, electricity is generated through
of charge on one mole of electrons. the use of spontaneous chemical reactions.
Therefore, for 6 moles of electrons, 6 F A galvanic (or voltaic) cell is made of
electricity will be required. two half cells. Each half cell consists of a
metal strip immersed in the solution of its own
ions of known concentration. For example, a
Problem 5.9 : In a certain electrolysis
strip of zinc metal immersed in 1 M aqueous
experiment 4.36 g of Zn are deposited
solution of zinc ions forms an half cell.
in one cell containing ZnSO4 solution.
Calculate the mass of Al deposited in It follows that two metal plates and
another cell containing AlCl3 solution the solutions of their ions with known
connected in series with ZnSO4 cell. Molar concentrations are required for the construction
masses of Zn and Al are 65.4 g mol-1 and of a galvanic (voltaic) cell. Two half cells
27 g mol-1, respectively. are constructed by immersing the two metal
plates in solutions of their respective ions
Solution :
placed in separate containers. The two half
Cell 1 : cells so constructed are combined together
Zn2⊕ (aq) + 2e Zn (s) to form the galvanic cell. The metal plates
called electrodes are connected through
1 mol voltmeter by a conducting wire for transfer
(mole ratio)1 =
2 mol e of electrons between them. To complete the
Cell 2 : circuit the two solutions are connected by
Al3⊕ (aq) + 3e Al (s) conducting medium through which cations
and anions move from one compartment to
1 mol the other. This requirement is fulfilled by a
(mole ratio)2 =
3 mol e salt bridge.
W1 W2 5.6.1 Salt bridge : In a galvanic cell, the two
= solutions are connected by a salt bridge. It is
(mole ratio)1 × M1 (mole ratio)2 × M2
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
and 5 % agar solution. The ions of electrolyte
M2 = 27 gmol-1 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 ii. The insoluble species if any or gases are
not in use as shown in Fig. 5.5. placed in the interior position adjacent to
Saturated KCl
the metal electrodes.
solution + agar gel
iii. The aqueous solutions of ions are placed
U tube at the middle of the cell formula.

Beaker iv. A single vertical line between two phases

indicates the phase boundary. It indicates
the direct contact between them.
Distilled water
v. A double vertical line between two
Glass solutions indicates that they are connected
wool plugs
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
For example Zn2⊕(1M) Zn (s). This
saturated solution of KCl for 24 hours.
order is reversed when the electrode
This strip can be used as salt bridge by
acts as anode in the cell. The following
dipping the two ends in two solutions.
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 ?
ii. It prevents mixing of two solutions. You have learnt Daniel cell
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
galvanic cells : A galvanic cell is represented 5.6.3 Writing of cell reaction : The cell
by a formula or short notation that includes reaction corresponding to the cell notation is
electrodes, aqueous solutions of ions and other written on the assumption that the right hand
species which may or may not be involved in side electrode is cathode (+) and left hand
the cell reaction. The following conventions side electrode is anode (-).
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
The following steps are followed to write cells, each consisting of electronic (metal
the cell reaction. plates) and electrolytic (solution of ions)
i. Write oxidation half reaction for the left conductors in contact. At the surface of
hand side electrode and reduction half separation of solid metal and the solution,
reaction for the right hand side electrode. there exists difference of electrical potential.
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
iii. It is important to note that the individual associated with reduction gives the reduction
half reactions may be written with one potential. The overall cell potential, also
or more electrons. For example, half called electromotive force (emf), is made of
reactions for H2 gas, whether written as the contributions from each of the electrodes.
2H⊕ (aq) + 2e H2 (g) or H⊕(aq) + In other words, the cell potential is algebraic
e 1/2 H2 (g) makes no difference. sum of the electrode potentials,
In writing the overall cell reaction, the
Ecell = Eoxi (anode) + Ered (cathode)
electrons must be balanced.
Consider the cell, (5.22)
e where Eoxi is the oxidation potential of
Ni (s)Ni (1M)Al3⊕ (1M)Al (s)
2⊕ anode (-) and Ered is the reduction potential
of cathode (+).
The oxidation at anode is
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
3 Ni (s) 3 Ni2⊕ (1M) + 6 e concentrations of solutions, pressures of gases
(oxidation half reaction) and the temperature. To facitilate comparison
2 Al (1M) + 6 e
3⊕
2 Al (s) of different galvanic cells, it is necessary to
(reduction half reaction) measure the cell voltage under given set of
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
The standard cell potential is the reactions for anything other than standard
algebraic sum of the standard electrode concentration conditions we need to know
potentials similar to Eq. (5.22). how voltage of galvanic cell varies with
concentration.
E0cell = E0oxi (anode) + E0red (cathode)
(5.23) Dependence of cell voltage on
Here E0oxi is standard oxidation potential concentrations is given by Nernst equation.
and E0red is the standard reduction potential. For any general reaction,

According to IUPAC convention, aA + bB cC + dD

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 RT [C]c [D]d
Ecell = E0cell - 1n
potential of any electrode is numerically nF [A]a [B]b
equal to its standard reduction potential with
the reversal of sign. For example standard 2.303RT [C]c [D]d
= E0cell - log10
oxidation potential of Zn2⊕ (1M) Zn electrode 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
where n = moles of electrons used in
voltage associated with a reduction reaction.
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), 2.303RT
At 25 0C, = 0.0592 V
E0oxi(anode) is replaced by - E0red (anode). F
We then write,
E0cell = - E0red (anode) + E0red (cathode) Therefore at 25 0C, eq. (5.24) becomes
Omitting the subscript red, we have
0.0592V [C]c [D]d
E0cell = E0 (cathode, +) - E0 (anode, -) Ecell = E0cell - log10
n [A]a [B]b
(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
• The difference in electrical potential conditions. The cell potential equals standard
between anode and cathode is cell potential if the concentrations of reactants
voltage. and products are 1 M each. Thus,

105
 if [A] = [B] = [C] = [D] = 1M,
Problem 5.10 : Calculate the voltage
Ecell = E cell
0
of the cell, Sn (s)Sn2⊕ (0.02M)Ag⊕
If a gaseous substance is present in the (0.01M)Ag (s) at 25 0C.
cell reaction its concentration term is replaced E0Sn = - 0.136 V, E0Ag = 0.800 V.
by the partial pressure of the gas. Solution :
First we write the cell reaction.
The Nernst equation can be used to
calculate cell potential and electrode potential. Sn (s) Sn2⊕ (0.02M) + 2 e
(oxidation at anode)
i. Calculation of cell potential : Consider the
cell [Ag⊕ (0.01M) + e Ag (s)] × 2
Cd(s)Cd2⊕(aq) Cu2⊕ (aq) Cu. (reduction at cathode)
Let us first write the cell reaction Sn (s) + 2 Ag⊕ (0.01M)
Cd (s) Cd2⊕ (aq) + 2 e Sn2⊕ (0.02 M) + 2 Ag (s)
(oxidation at anode) (overall cell reaction)
The cell potential is given by
Cu2⊕ (aq) + 2 e Cu (s)
0.0592V [Sn2⊕]
(reduction at cathode) Ecell = E cell -
0
log10
2 [Ag⊕]2
Cd (s) + Cu2⊕ (aq) Cd2⊕ (aq) + Cu (s) E0cell = E0Ag - E0Sn = 0.8 V + 0.136 V
(overall cell reaction) = 0.936 V
Here n = 2 Hence,
0.0592V 0.02
Ecell = 0.936V - log10
The potential of cell is given by Nernst 2 (0.01)2
equation, 0.0592V
= 0.936V - log10 200
0.0592 [Cd2⊕] 2
Ecell = E0cell - log10
2 [Cu2⊕] 0.0592V
= 0.936V - × 2.301
at 25 0C. 2
(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)
Consider Zn2⊕(aq)Zn(s) is - 0.76 V. Calculate its potential.
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
0.0592 1 0.0592V 1
EZn = E0Zn - log10 EZn = E0Zn - log10 2⊕
2 [Zn2⊕
] n [Zn ]

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

0.0592V
= - 0.76 V + × (-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
cell potential : The electrical work done in a For chemical reaction to be
galvanic cell is the electricity (charge) passed spontaneous, ∆G must be negative.
multiplied by the cell potential. Because ∆G = - nFEcell, Ecell must be positive
for a cell reaction if it is spontaneous.
Electrical work
= amount of charge passed × cell potential.
5.8.2 Standard cell potential and equilibrium
Charge of one mole electrons is F constant : The relation between standard
coulombs. For the cell reaction involving n Gibbs energy change of cell reaction and
moles of electrons. standard cell potential is given by Eq. (5.27).
charge passed = nF coulombs - ∆G0 = nFE0cell
Hence, electrical work = nFEcell The relation between standard Gibbs
W. Gibbs in 1878 concluded that energy change of a chemical reaction
electrical work done in galvanic cell is equal and its equilibrium constant as given in
to the decrease in Gibbs energy, - ∆G, of cell thermodynamics is :
reaction. It then follows that ∆G0 = - RT ln K (5.29)
Electrical work = - ∆G Combining Eq. (5.28) and Eq. (5.29), we
and thus - ∆G = nFEcell have
or ∆G = -nFEcell (5.27) -nFE0cell = - RT ln K
Under standard state conditions, we write
RT
∆G0 = -nFE0cell (5.28) or E0cell = ln K
nF
The Eq. (5.28) explains why E0cell is an
intensive property. 2.303 RT
= log10K
nF
We know that ∆G0 is an extensive
property since its value depends on the 0.0592
amount of substance. If the stoichiometric = log10K at 25 0C
n
equation of redox reaction is multiplied by (5.29)
2 that is the amounts of substances oxidised
and reduced are doubled, ∆G0 doubles. The
moles of electrons transferred also doubles. Try this...
The ratio, Write expressions to calculate
∆G0 equilibrium constant from
E0cell = - then becomes
nF
i. Concentration data
2∆G 0
∆G 0
ii. Thermochemical data
E0cell = - =-
2 nF nF
iii. Electrochemical data
Thus, E0cell remains the same by
multiplying the redox reaction by 2. It
means E0cell is independent of the amount of
substance and the intensive property.

107
 What would happen if potential of one of
Problem 5.12 : Calculate standard Gibbs
the electrodes in a galvanic cell is zero ? Can
energy change and equilibrium constant at
we measure the potential of such a galvanic
250C for the cell reaction,
cell ? There are two electrodes combined
Cd (s) + Sn2⊕ (aq) Cd2⊕ (aq) + Sn (s) together and a redox reaction results. The
Given : E0Cd = -0.403V and measured cell potential is algebraic sum of two
E0Sn = - 0.136 V. Write formula of the cell. electrode potentials. One electrode potential is
Solution : zero. Therefore, the measured cell potential is
equal to the potential of other electrode.
i. The cell is made of two electrodes,
Cd2⊕ (aq) Cd (s) and Sn2⊕ (aq) Sn (s). From foregoing arguments, it follows that
E0 value for Sn2⊕ (aq) Sn (s) electrode it is necessary to choose an arbitrary standard
is higher than that of Cd2⊕(aq) Cd electrode as a reference point. The chemists
(s) electrode. Hence, Sn2⊕ (aq) Sn (s) have chosen hydrogen gas electrode consisting
electrode is cathode and Cd2⊕ (aq) Cd (s) of H2 gas at 1 atm pressure in contact with
anode. Cell formula is Cd(s) Cd2⊕ (aq) 1 M H⊕ ion solution as a primary reference
Sn2⊕ (aq) Sn (s) electrode. The potential of this electrode has
ii. ∆G0 = - nF E0cell arbitrarily been taken as zero. The electrode
is called standard hydrogen electrode (SHE).
E0cell = E0Sn - E0Cd = - 0.136 V - (-0.403V)
We will see later that SHE is not the most
= 0.267 V.
convenient electrode. Several other electrodes
n = 2 mol e namely calomel, silver-silver chloride and
∆G0 = -2 mol e 96500 C/mol e × 0.267 V glass electrodes with known potentials are
= - 51531 V C = - 51531 J = - 51.53 kJ used as secondary reference electrodes. The
0.0592 V potentials of these electrodes are determined
iii. E0cell = log10K
2 using SHE.
A reference electrode is then defined as
0.0592 V an electrode whose potential is arbitrarily
0.267 V = log10K
2 taken as zero or is exactly known.
5.9.1 Standard hydrogen electrode (SHE) :
0.267 × 2
log10K = = 9.0203 Construction : SHE consists of a platinum
0.0592
plate, coated with platinum black used as
K = antilog 9.0203 = 1.05 × 109
electrodes. This plate is connected to the
external circuit through sealed narrow glass
5.9 Reference electrodes : Every oxidation
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
 Pure and e e
Cu wire dry H2 gas
at 1 atm Salt bridge
Glass jacket
Zn H2
Vessel anode (g, 1atm)
Mercury SHE
Pt wire
1M
Solution Platinised ZnSO4 1M
H⊕ ions platinum soution
plate H⊕ ion
(1M) solution

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

2H⊕ (1M) + 2e H2 (g, 1atm) 2

hydrogen gas differ from unity.
E0H2 = 0.000 V
Electrode reaction :
Application of SHE
2H⊕ (aq) + 2e H2 (g, PH2)
SHE is used as a primary reference
electrode to determine the standard potentials From the Nernst equation
of other electrodes.
0.0592 P
To determine the standard potential of EH = E0H2- log10 H⊕2 2
2 2 [H ]
Zn (1M)Zn (s), it is combined with SHE
2⊕
0.0592 P
to form the cell, = - log10 H⊕2 2
2 [H ]
ZnZn2⊕(1M)H⊕ (1M)H2 (g, 1atm)Pt
Because E0H2 = 0
This is shown in Fig. 5.7.
The standard cell potential, E0cell, is measured. 5.10 Galvanic cells useful in day-to-day life
Voltaic (or galvanic) cells in common use can
E0cell = E0H2 - E0Zn = - E0Zn , because E0H2 is zero. be classified as primary and secondary cells.
Thus, the measured emf of the cell is equal i. Primary voltaic cells : When a galvanic
to standard potential of Zn2⊕(1M)Zn (s) cell discharges during current generation,
electrode. the chemicals are consumed. In primary
Difficulties in setting SHE voltaic cell, once the chemicals are
i. It is difficult to obtain pure and dry completely consumed, cell reaction stops.
hydrogen gas. The cell reaction cannot be reversed even
ii. The pressure of hydrogen gas cannot be after reversing the direction of current
maintained exactly at 1 atm throughout flow or these cells cannot be recharged.
the measurement. The most familiar example is dry cell.
109
ii. Secondary voltaic cells : In secondary
voltaic cell, the chemicals consumed Brass cap
during current generation can be Paper spacer
Zn container
regenerated. For this purpose an external Paste of MnO2 +
potential slightly greater than the cell carbon
Graphite rod
potential is applied across the electrodes.
Paste of NH4Cl
This results in reversal of the direction of + ZnCl2
current flow causing the reversal of cell
reaction This is recharging of cell. The Bottom of Zn container

voltaic cells which can be recharged are Fig. 5.8 : Dry cell (Leclanche' cell)
called secondary voltaic cells. Cell reactions:
It is amazing to see that secondary i. Oxidation at anode : When the cell
cells are galvanic cells during discharge and operates the current is drawn from the
electrolytic cells during recharging. Examples cell and metallic zinc is oxidised to zinc
of secondary cells are lead storage battery, ions.
mercury cell and nickel-cadmium cell. Zn (s) Zn2⊕ (aq) + 2e
5.10.1 Dry cell (Leclanche' cell) : It is a cell
ii. Reduction at cathode : The electrons
without liquid component, but the electrolyte liberated in oxidation at anode flow along
is not completely dry. It is a viscous aqueous the container and migrate to cathode. At
paste. cathode NH4⊕ ions are reduced.
Construction : The container of the cell is 2NH4⊕ (aq) + 2e 2NH3 (aq) + H2 (g)
made of zinc which serves as anode (-). It
is lined from inside with a porous paper to Hydrogen gas produced in reduction
reaction is oxidised by MnO2 and
separate it from the other material of the cell.
prevents its collection on cathode.
An inert graphite rod in the centre of the
cell immersed in the electrolyte paste serves H2(g)+2MnO2(s) Mn2O3(s)+H2O(l)
as cathode. It is surrounded by a paste of The net reduction reaction at cathode is
manganese dioxide (MnO2) and carbon black. combination of these two reactions.
The rest of the cell is filled with an 2NH4⊕(aq) + 2 MnO2(s) + 2e
electrolyte. It is a moist paste of ammonium Mn2O3 (s) + 2 NH3 (aq) + H2O (l)
chloride (NH4Cl) and zinc chloride (ZnCl2). iii. Net cell reaction : The net cell reaction is
Some starch is added to the paste to make it sum of oxidation at anode and reduction
thick so that it cannot be leaked out. at cathode.
The cell is sealed at the top to prevent Zn (s) + 2 NH4⊕ (aq) + 2 MnO2(s)
drying of the paste by evaporation of moisture.
Zn2⊕ (aq) + Mn2O3 (s) + 2 NH3 (aq) + H2O(l)
See Fig. 5.8.
The ammonia produced combines with
Zn2⊕ to form soluble compound containing
complex ion.
Zn2⊕ (aq) + 4 NH3 (aq) [Zn (NH3)4]2⊕(aq)

110
 The electrodes are immersed in
Do you know ?
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 i. Oxidation at anode (-) : When the
the dry cell called alkaline dry cell has cell provides current, spongy lead is
been proposed. In alkaline dry cell NaOH oxidised to Pb2⊕ ions and negative charge
or KOH is used as electrolyte in place of accumulates on lead plates. The Pb2⊕
NH4Cl. 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 (oxidation)
recorders, clocks and so forth. Pb2⊕ (aq) + SO42 (aq) PbSO4 (s)
5.10.2 Lead storage battery (Lead (precipitation)
accumulator) : Lead accumulator stores Pb (s)+SO42 (aq) PbSO4 (s) + 2 e ...(i)
electrical energy due to regeneration of original (overall oxidation)
reactants during recharging. It functions as
ii. Reduction at cathode (+) : The electrons
galvanic cell and as electrolytic cell, as well.
produced at anode travel through external
Construction : A group of lead plates packed circuit and re-enter the cell at cathode. At
with spongy lead serves as anode (-). Another cathode PbO2 is reduced to Pb2⊕ ions in
group of lead plates bearing lead dioxide presence of H⊕ ions. Subsequently Pb2⊕
(PbO2) serves as cathode (+). ions so formed combine with SO42 ions
from H2SO4 to form insoluble PbSO4 that
Anode (-) gets coated on the electrode.
Cathode (+)
PbO2 (s) + 4H⊕ (aq) + 2 e Pb2⊕ (aq)
+2H2O(l)
(reduction)
38 %
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
or used in automobiles and inverters.
Pb (s) + PbO2 (s) + 2H2SO4 (aq) 5.10.3 Nickel-Cadmium or NICAD storage
2PbSO4(s) + 2H2O (l) ...(iii) cell : Nickel-cadmium cell is a secondary dry
As the cell operates to generate current, cell. In other words it is a dry cell that can
H2SO4 is consumed. Its concentration (density) be recharged.
decreases and the cell potential is decreased.
The cell potential thus depends on sulphuric Anode of the NICAD storage cell is
acid concentration (density). cadmium metal. The cathode is nickel (IV)
oxide, NiO2 supported on Ni. The electrolyte
b. Cell reactions during recharging : solution is basic.
The potential of lead accumulator is 2V. It
must be recharged with the falling of the The electrode reactions and overall cell
cell potential to 1.8 V. To recharge the cell reaction are as follows :
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 (-).
Cd (s) + NiO2 (s) + 2 H2O (l)
iv. Oxidation at anode (+) : It is reverse
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
v. Reduction at cathode (-) : It is reverse potential of the cell is about 1.4 V. The cell
of oxidation reaction (i) at anode that occurs has longer life than other dry cells. It can
during discharge. be used in electronic watches, calculators,
PbSO4 (s)+2 e Pb (s)+SO42 (aq) ...(v) photographic equipments, etc.

vi. Net cell reaction : It is the sum of 5.10.4 Mercury battery : Mercury battery is
reaction (iv) and (v) or the reverse of a secondary dry cell and can be recharged.
net cell reaction (iii) that occurs during The mercury battery consists of zinc anode,
discharge amalgamated with mercury. The cathode is
a paste of Hg and carbon. The electrolyte
2PbSO4 (s) + 2H2O (l) Pb (s) + is strongly alkaline and made of a paste of
PbO2 (s) + 2 H2SO4 (aq) KOH and ZnO. The electrode ractions and
The above reaction shows that H2SO4 is net cell reaction are :
regenerated. Its concentration (density) Zn(Hg)+2OH (aq) ZnO(s) +H2O(l) + 2 e
and in turn, the cell potential increases. (anode oxidation)
Applications of lead accumulator HgO(s)+ H2O(l)+2e Hg(l) + 2 OH (aq)
i. It is used as a source of direct current in (cathode reduction)
the laboratory. Zn (Hg) + HgO(s) ZnO(s) + Hg(l)
(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 Cell reactions
voltage (1.35V) than the Leclanche' dry cell.
It also has considerably higher capacity and i. Oxidation at anode (-) : At anode
longer life than dry cell. 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 circuit to cathode. Here O2 is reduced to
are of redox type and can be used to generate OH-.
electricity. 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 reaction is the sum of electrode reactions
placed within the cell. They are continuously (i) and (ii).
supplied to electrodes from a reservoir.
2H2 (g) + O2 (g) 2H2O (l)
In these cells one of the reactants is a
The overall cell reaction is combustion
fuel such as hydrogen gas or methanol. The
of H2 to form liquid water. Interestingly, the
other reactant such as oxygen, is oxidant.
fuel H2 gas and the oxidant O2 do not react
The simplest fuel cell is hydrogen-oxygen directly.
fuel cell.
The chemical energy released during the
5.11.1 Hydrogen-oxygen fuel cell : In H2 - formation of O-H bond is directly converted
O2 fuel cell, the fuel is hydrogen gas. Oxygen into electrical energy accompanying in above
gas is an oxidising agent. The energy of the combustion reaction. The cell continues
combustion of hydrogen is converted into to operate as long as H2 and O2 gases are
electrical energy. supplied to electrodes.
H2O The cell potential is given by
(anode) (cathode) E0cell = E0cathode - E0anode = 0.4V - (-0.83V)
= 1.23 V.
Advantages of fuel cells
H2 (g) O2 (g) i. The reacting substances are continuously
Aqueous supplied to the electrodes. Unlike
KOH conventional galvanic cells, fuel cells do
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
iii. Fuel cells provide electricity with an
of finely divided platinum metal that acts
efficiency of about 70 % which is twice
as a catalyst. The electrolyte is hot aqueous
as large when compared with efficiency
solution of KOH. The carbon rods immersed
of thermal plants (only 40 %).
into electrolyte are shown in Fig. 5.10.

113
Drawbacks of fuel cell ii. Below hydrogen electrode the negative
H2 gas is hazardous to handle and the standard potential increases and above
cost of preparing H2 is high. hydrogen electrode the positive standard
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
phones and laptop computers. The direction as written.
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
Applications of fuel cells turn greater tendency for the substance
to reduce. Conversely, the low (more
i. The fuel cells are used on experimental
negative) E0 value of a half reaction
basis in automobiles.
corresponds to its greater tendency to
ii. The fuel cell are used for electrical power occur in the reverse direction or for the
in the space programme. substance to oxidise.
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
has cations of metals or non-
iv. In future, fuel cells can possibly be metallic molecules (oxidants). There are
explored as power generators in hospitals, free metals or anions of non metals on the
hotels and homes. right side (reductants).
Can you tell ?
Applications of electrochemical series
In what ways are fuel cells
i. Relative strength of oxidising agents:
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 E0Cell = E0Ag - E0Mg = 0.8V - (-2.37V)
a little or no tendency to occur in the
forward direction as written. They tend = 3.17 V.
to favour the reverse direction. It follows, EMF being positive the cell reaction
that the species appearing at the bottom is spontaneous. Thus Ag⊕ ions oxidise to
right side of half reactions associated metallic Mg.
with large negative E0 values are the
General rules
effective electron donors. They serve as
strong reducing agents. The strength of i. An oxidizing agent can oxidize any
reducing agents increases from top to reducing agent that appears below it,
bottom as E0 values decrease. and cannot oxidize the reducing agent
appearing above it in the electrochemical
Use your brain power series.
Identify the strongest and the
ii. An reducing agent can reduce the
weakest reducing agents from the
oxidising agent located above it in the
electrochemical series.
electrochemical series.
iii. Spontaneity of redox reactions : A redox
reaction in galvanic cell is spontaneous Use your brain power
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
for a cell reaction to be spontaneous under Do you know ?
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
Suppose, we ask a question : At standard (PEM) fuel cells.
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
Ce , Ce Pt
4⊕ 3⊕
Ce + e
4⊕
Ce 3⊕
+1.610
Au3⊕ Au Au3⊕ + 3e Au +1.500
Cl Cl2Pt Cl2 + 2e 2Cl +1.360
Pt Pt
2⊕
Pt + 2e
2⊕
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

Increasing strength as reducing agent

Hg22⊕Hg Hg22⊕ + 2e 2Hg +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
Cu ,Cu Pt
2⊕ ⊕
Cu + e
2⊕
Cu ⊕
+0.153
Sn4⊕, Sn2⊕Pt Sn4⊕ + 2e Sn2⊕ +0.15
H⊕H2Pt 2H⊕ + 2e H2 0.00
Pb Pb2⊕
Pb + 2e
2⊕
Pb -0.126
Sn2⊕Sn Sn2⊕ + 2e Sn -0.136
Ni2⊕Ni Ni2⊕ + 2e Ni -0.257
Co Co2⊕
Co + 2e
2⊕ -
Co -0.280
Cd2⊕Cd Cd2⊕ + 2e Cd -0.403
Fe2⊕Fe Fe2⊕ + 2e Fe -0.440
Cr Cr3⊕
Cr + 3e
3⊕
Cr -0.740
Zn2⊕Zn Zn2⊕ + 2e Zn -0.763
Al Al3⊕
Al + 3e
3⊕
Al -1.66
Mg2⊕Mg Mg2⊕ + 2e Mg -2.37
Na⊕Na Na⊕ + e Na -2.714
Ca Ca2⊕
Ca + 2e
2⊕
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 E0 = 0.53V
conductivities 0.216. The ratio of iii. Pb2⊕ (aq) + 2e Pb (s)
their molar conductivities will be
E0 = -0.13V
a. 0.54 b. 11.574
iv. Fe2⊕ (aq) + 2e Fe (s)
c. 0.0864 d. 1.852
E0 = - 0.44V
ii. On diluting the solution of an
The strongest oxidising and reducing
electrolyte
agents respectively are
a. both ∧ and k increase
a. Ag and Fe2⊕ b. Ag⊕ and Fe
b. both ∧ and k decrease
c. Pb2⊕ and I d. I2 and Fe2⊕
c. ∧ increases and k decreases
vii. For the reaction Ni(s) + Cu2⊕ (1M)

d. ∧ decreases and k increases Ni2⊕ (1M) + Cu (s), E0cell =
iii. 1 S m2 mol-1 is eual to 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
iv. The standard potential of the cell in property
which the following reaction occurs b. Electrode potential or cell potential
H2 (g,1atm) + Cu2⊕(1M) is an intensive property.
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
c. 0.17 V d. -0.17 V numerical factor, the corresponding
E0 value is also multiplied by the
v.
For the cell, Pb same factor.
(s)Pb2⊕(1M)Ag⊕(1M) Ag (s), if
concentraton of an ion in the anode ix. The oxidation reaction that takes
compartment is increased by a factor place in lead storage battery during
of 10, the emf of the cell will discharge is
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) + SO4 (aq) + 2e
2
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)
E = 0.8V
0

117
 x. Which of the following expressions ii. What is a salt bridge ?
represent molar conductivity of iii. Write electrode reactions for the
Al2(SO4)3? electrolysis of aqueous NaCl.
a. 3 λ0Al3⊕ + 2 λ0SO42 iv. How many moles of electrons are
b. 2 λ Al3⊕ + 3 λ SO42
0 0
passed when 0.8 ampere current is
c. 1/3 λ0Al3⊕ + 1/2 λ0SO42 passed for 1 hour through molten

CaCl2 ?
d. λ0Al3⊕ + λ0SO42
v. Construct a galvanic cell from
2. Answer the following in one or two the electrodes Co3⊕Co and
sentences. Mn2⊕ Mn. E0Co = 1.82 V,
i. What is a cell constant ? E0Mn = - 1.18V. Calculate E0cell.
ii. Write the relationship between vi. Using the relationsip between ∆G0
conductivity and molar conductivity of cell reaction and the standard
and hence unit of molar conductivity. potential associated with it, how will
iii. Write the electrode reactions during you show that the electrical potential
electrolysis of molten KCl. is an intensive property ?
iv. Write any two functions of salt viii. Derive the relationship between
bridge. standard cell potential and
v. What is standard cell potential for equilibrium constant of cell reaction.
the reaction ix. It is impossible to measure the
3Ni (s) + 2Al3⊕(1M) potential of a single electrode.
3Ni2⊕(1M) Comment.

+ 2Al(s) if E0Ni = - 0.25 V and E0Al = x. Why do the cell potential of lead
-1.66V? accumulators decrease when it
generates electricity ? How the cell
vi. Write Nerst equation. What part of potential can be increased ?
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
i. What is Kohrausch law of
potential ?
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
Au3⊕(aq) + 3e Au(s) a weak electrolyte ? Explain with an
Mg(s) Mg2⊕(aq) + 2e example.
ii. Explain electrolysis of molten NaCl.
ix. How many electrons would have a
total charge of 1 coulomb ? iii. What current strength in amperes
will be required to produce 2.4
x. What is the significance of the single
g of Cu from CuSO4 solution in
vertical line and double vertical line
1 hour ? Molar mass of Cu = 63.5 g
in the formulation galvanic cell.
mol-1. (2.03 A)
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 × 10 . What is the standard
6

118
 potential of the cell in which the order of increasing strength under
reaction takes place ? (0.36 V) standard state conditions. The
v. Calculate emf of the cell standard potentials are given in
parenthesis.
Zn(s)Zn 2⊕ (0.2M)H ⊕ (1.6M)
H2(g, 1.8 atm) Pt at 250C. K (-2.93V), Br2(1.09V), Mg(-2.36V),
(0.785V) Ce3⊕(1.61V), Ti2⊕(-0.37V), Ag⊕(0.8
V), Ni (-0.23V).
vi. Calculate emf of the following cell at
250C. xiv. Predict whether the following
reactions would occur spontaneously
Zn (s) Zn2⊕(0.08M)Cr3⊕(0.1M)Cr under standard state conditions.
E0Zn = - 0.76 V, E0Cr = - 0.74 V a. Ca (s) + Cd2⊕ (aq)
(0.0327 V) Ca2⊕(aq) + Cd(s)
vii. What is a cell constant ? What are b. 2 Br (s) + Sn2⊕ (aq)
its units? How is it determined
experimentally? Br2(l) + Sn(s)

viii. How will you calculate the moles c. 2Ag(s) + Ni2⊕ (aq)
of electrons passed and mass of 2 Ag⊕ (aq) + Ni (s)
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
lead storage cell generates electricity. 1. Write electrode reactions net
What are the anode and cathode and cell reaction in the electrolysis
the electrode reactions during its of molten barium chloride.
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.
5. Collect information of pollution free
The standard potentials are given in
battery.
parenthesis.
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