5.

Electrochemistry

Introduction : Electrochemistry is the area of chemistry which is concerned with interconversion of

chemical and electrical energy.

Electric conduction :
i) The electric current represents a charge transfer.
ii) A charge transfer or flow of electricity occurs through substances called conductors.
iii) There are two types of conductors which give rise to two types of conduction of electricity.

Metallic conduction :
i) Electrical conduction through metals involves a direct flow of electrons from one point to the other.
ii) Metallic conductors are, thus, electronic conductors.
iii) Conduction through electronic conductors involves no transfer of matter from one part of the
conductor to the other.

Electrolytic or ionic conduction :

i. Electrolytic conduction involves conduction of electric current by the movement of ions of the
electrolytes.
ii. Electrolytes dissociate into ions when dissolved in polar solvents such as water. Ionic solids
dissociate into ions in molten state as well.
iii. Conduction through electrolytic conductors involves transfer of matter from one part of the
conductor to the other.

Information provided by measurement of conductivities of solutions :

i. The conducting and nonconducting nature of solutions can be identified by measurement of their
conductivity.
ii. Sucrose and urea do not dissociate in their aqueous solutions. The conductivities of these solutions
are nearly the same as that of water. These substances are called nonelectrolytes.
iii. Substances like potassium chloride, acetic acid, sodium hydroxide, HCl dissociate in their aqueous
solutions. The conductivities of their aqueous solutions are higher than that of water. These are called
electrolytes.
iv. Electrolytes conduct electricity in molten state or when dissolved in water.
v. On the basis of high or low electrical conductivity electrolytes are classified into strong and weak
electrolytes.
vi. The substances such as ionic salts, strong acids or bases are almost completely dissociated in
aqueous solutions. These are strong electrolytes. The solutions of strong electrolytes exhibit high
conductivities
vii. The weak acids and weak bases are weak electrolytes. They dissociate to a very small extent in
aqueous solutions and show lower conductivities than those of strong electrolytes.

Electrical conductance of solution :

i) According to Ohm's law, the electrical resistance R of a conductor is equal to the electric potential
difference V divided by the electric current, I
𝑉
R=
𝐼
ii) The SI unit of potential is volt (V) and that of current is ampere (A). The unit of electrical
resistance is ohm denoted by the symbol Ω (omega). Thus, Ω = VA-1.

73 | P a g e
iii) The electrical resistance of a conductor is proportional to length l and inversely
proportional to cross sectional area a. Thus,
𝑙
R ∝
𝑎
𝑙
R=𝜌
𝑎
Where 𝜌 the proportionality constant is called resistivity of the conductor.
𝑎
iv) 𝜌 = R If area is unit cross-sectional area and unit length, then 𝜌 = R
𝑙
v) Resistivity is the resistance of conductor of unit length and unit cross sectional area.
vi) The unit of resistivity :
𝑎
𝜌=R
𝑙
SI Unit of resistivity : Ω m
Common unit of resisitivity : Ω cm

Electrical conductance: The electrical conductance, ‘G ` of a solution is reciprocal of resistance.

1
G=
𝑅
The SI unit of electrical conductance (G) is = S (Siemens), which is equal to Ω-1.
(1S = 1 Ω-1)

Conductivity:
i) The electrical conductance is directly proportional to its cross sectional area and inversely
proportional to the length ,
𝑎
G∝
𝑙
𝑎
G=k
𝑙
Where k is the proportionality constant is called as conductivity
𝑙
ii) k = G , If length and cross sectional area are unity, then
𝑎
k=G
iii) Thus, conductivity is the electrical conductance of a conductor of unit length and unit area of
cross section. OR The conductivity is the electrical conductance of unit cube of material.
iv) Conductivity of solution of an electrolyte is called electrolytic conductivity which refers to the
electrical conductance of unit volume (1 m3 or 1 cm3) of solution.
𝑙
v) k = G
𝑎
1 𝑙 𝑙 1 1 𝑎
k= ( but, R = 𝜌 , Therefore = X )
𝑅𝑎 𝑎 𝑅 𝜌 𝑙
1
Therefore, k =
𝜌
Thus conductivity and resistivity are inversely proportional.
Units of electrolytic conductivity

Quantity SI unit Common unit

Length m cm
Area m2 Cm2
Resistance Ω Ω
Conductivity Ω-1 m-1 or S m-1 Ω-1 cm-1

Molar conductivity (∧ ) :
i) The molar conductivity of an electrolytic solution is the electrolytic conductivity, k, divided by its
molar concentration c.
𝑘
Ʌ=
𝐶

74 | P a g e
The SI unit if conductivity are S m-1 and that of concentration are mol m-3 .
Hence the SI unit of molar conductivity is the S m2 mol-1 .
Common unit is: S cm2 mol-1 or Ω-1 cm2 mol-1
Significance of molar conductivity :
i) To understand the significance of molar conductivity ( Ʌ) consider volume of a solution
containing 1 mole of dissolved electrolyte.
ii) Suppose the solution is placed between two parallel electrodes 1 cm apart and large enough to
accommodate it.
iii) The electrical conductance exhibited by this solution is the molar conductivity.
iv) The molar conductivity is the electrical conductance generated by all the ions in 1 mole
of the electrolyte.

Relation between conductivity (k) and molar conductivity ( Ʌ) :

i) Conductivity k is the electrical conductance of 1 cm3 of solution.
ii) If V is volume of solution in cm3 containing 1 mole of dissolved electrolyte its electrical conductance
is Ʌ.
iii) Each 1 cm3 portion in the volume V has conductance k. Hence, total conductance of V cm3 is
kV which is molar conductivity.
Thus, Molar conductivity = k V
Ʌ = K V ------------- (i)
Let the Concentration of solution = C mol L-1
𝐶 𝑚𝑜𝑙
=
𝐿
𝐶 𝑚𝑜𝑙
=
1000 𝑐𝑚3
𝐶
= mol cm-3
1000
𝑚𝑜𝑙𝑒
But, Concentration =
𝑉𝑜𝑙𝑢𝑚𝑒
𝑚𝑜𝑙𝑒𝑠
∴ Volume =
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
1
Volume =
𝐶/1000
1000
V = cm mol-1 3
𝐶
Subustitute the value of V in equation (i)

Ʌ =K XV
1000
Ʌ =K X
𝐶

Variation of conductivity with concentration:

i) The electrolytic conductivity is electrical conductance of unit volume (1 cm3) of solution.
ii) It depends on the number of current carrying ions present in unit volume of solution.
iii) On dilution total number of ions increase as a result of increased degree of dissociation. But increase
in total number of ions is not in proportion of dilution.
iv) Therefore, the number of ions per unit volume of solution decreases.
v) This results in decrease of conductivity with decrease in concentration of solution.

Variation of molar conductivity with concentration:

i) The molar conductivity is the electrical conductance of 1 mole of an electrolyte in a given volume of
solution.
ii) The increasing number of ions produced in solution by 1 mole of the electrolyte lead to increased
molar conductivity
75 | P a g e
 Variation of molar conductivity with concentration : The variation of molar conductivity with
concentration in case of strong and weak electrolytes is qualitatively different.
A) Strong electrolytes :
i) The molar conductivity of solution of strong electrolyte increases rapidly with dilution.
ii) It approaches the maximum limiting value at 0.001 M or 0.0001 M solution.
iii) The maximum limiting value of molar conductivity is the molar conductivity at zero concentration
or at infinite dilution. It is denoted by Ʌ 0 or Ʌ ꝏ .
iv) The zero concentration or infinite dilution means the solution is so dilute that further dilution does
not increase the molar conductivity.
v) F. Kohlrausch with repeated experiments showed that the molar conductivity of strong
electrolytes varies linearly with square root of concentration as :
Ʌ = Ʌ 0 + a √𝑐
Where a is constant. For strong electrolytes a plot of Ʌ versus √𝑐 is linear.

vi) Molar conductivity of strong electrolytes at zero concentration can be determined by

extrapolation of linear part of Ʌ versus √𝑐 curve as shown in the figure.

B) Weak electrolytes :
i) The molar conductivity of weak electrolytes increases rapidly on dilution.
ii) But at the concentrations of 0.001 M or 0.0001 M, the molar conductivity is much lower than its
maximum molar conductivity at zero concentration.
iii) For weak electrolyte the variation of molar conductivity with concentration is not linear.
iv) This method cannot be used for weak electrolytes since Ʌ versus √𝑐 curve does not approach
linearity.
v) Therefore Kohlrausch law is useful for calculating maximum limiting molar conductivity (Ʌ0 )
of weak electrolytes.

Kohlrausch law of independent migration of ions :

i) The law states that at infinite dilution each ion migrates independent of co-ion and contributes to
total molar conductivity of an electrolyte irrespective of the nature of other ion to which it is associated.
ii) Both cation and anion contribute to molar conductivity of the electrolyte at zero concentration and
thus Ʌ 0 is sum of molar conductivity of cation and that of the anion at zero concentration.
Thus,
Ʌ 0 = n+ 𝜆0+ + n- 𝜆0−
Where 𝜆0+ and 𝜆0− are molar conductivities of cation and anion, respectively, and n+ and n- are the
number of moles of cation and anion, specified in the chemical formula of the electrolyte.

Applications of Kohlrausch theory:

i) The theory can be used to calculate the molar conductivity of an electrolyte at the zero concentration.
For example,
Ʌ 0 (KCl) = 𝜆0𝐾+ + 𝜆0𝑐𝑙−
Ʌ 0 [ Ba (OH)2 ] = 𝜆𝐵𝑎2+ + 2 𝜆0𝑂𝐻−
0

76 | P a g e
 Knowing the molar conductivities of ions at infinite dilution, Ʌ 0 values of electrolyte can be
obtained.
ii) The theory is particularly useful in calculating Ʌ 0 values of weak electrolytes from those of strong
electrolytes.
For example, Ʌ 0 of acetic acid can be calculated by knowing the molar conductivities of HCl, NaCl and
CH3COONa as given below.
Ʌ 0 (CH3COOH) = Ʌ 0 (HCl) + Ʌ 0 (CH3COONa) - Ʌ 0 (NaCl)
= 𝜆0𝐻+ + 𝜆0𝑐𝑙− + 𝜆0𝐶𝐻3𝐶𝑂𝑂− + 𝜆0𝑁𝑎+ - 𝜆0𝑁𝑎+ - 𝜆0𝑐𝑙−
= 𝜆0𝐶𝐻3𝐶𝑂𝑂− + 𝜆0𝐻+
Because Ʌ 0 Values of strong electrolyte, HCl, CH3COONa and NaCl can be determined by
extrapolation method, and from this Ʌ 0 of acetic acid can be obtained.

Molar conductivity and degree of dissociation of weak electrolytes :

i) The degree of dissociation ( ∝) of weak electrolyte is related to its molar conductivity at a given
concentration C by the equation,
Ʌ
∝= 𝐶
Ʌꝏ
where Ʌ C is the molar conductivity of weak electrolyte at concentration c and Ʌ ꝏ is molar
conductivity at zero concentration.
∝2 𝐶
Ka =
1−∝
Where Ka = dissociation constant for acid,
∝ = Degree of Dissociation of weak electrolyte
C = Molar concentration.

Measurement of conductivity : The conductivity of a solution can be determined by measuring the

resistance of solution by using Wheatstone bridge.

Conductivity Cell :
i) The conductivity cell consists of a glass tube with two platinum plates coated with a thin layer of
finely divided platinium black. This is achieved by the electrolysis of solution of chloroplatinic
acid.
ii) The cell is dipped in a solution whose resistance is to be measured

Cell constant:
i) The conductivity of an electrolytic solution is given by equation.
1 𝑙
k=
𝑅𝑎
ii) For a given cell, the ratio of separation (l) between the two electrodes is divided by the area of cross
section (a) of the electrode is called the cell constant.
𝑙
Cell constant =
𝑎

77 | P a g e
 iii) SI unit of cell constant is m-1 which is conveniently expressed in cm-1.
𝑐𝑒𝑙𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
iv) k=
𝑅

The Determination of conductivity consists of three steps:

1) Determination of cell constant :
i) The cell constant is determined using the 1 M, 0.1 M or 0.01 M KCl solutions.
ii) The conductivity of KCl solution is well tabulated at various temperatures.
iii) The resistance of KCl solution is measured by Wheatstone bridge.

Measurement of resistance.
iv) AB is the uniform wire.
v) Rx is the variable known resistance placed in one arm of Wheatstone bridge.
vi) The conductivity cell containing KCl solution of unknown resistance is placed in the other
arm of Wheatstone bridge.
vii) D is a current detector. F is the sliding contact that moves along AB and A.C. represents the
source of alternating current.
viii) The sliding contact is moved along AB until no current flows. Then detector D shows no
deflection. The null point is, thus, obtained at C.
ix) According to Wheatstone bridge principle,

𝑅 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑅𝑥
=
𝑙(𝐴𝐶) 𝑙(𝐵𝐶)
𝑙(𝐴𝐶)
Hence , 𝑅𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = x 𝑅𝑥
𝑙(𝐵𝐶)
x) By measuring lengths AC and BC and knowing Rx , resistance of KCl solution can be
calculated.
xi) The cell constant is given by equation,
Cell constant = 𝑘𝐾𝐶𝑙 X 𝑅𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Thus by knowing the conductivity of KCl, cell constant can be determined.

2) Determination of conductivity of given solution :

i) KCl solution in the conductivity cell is replaced by the given solution whose conductivity is to be
measured.
ii) Its resistance is measured by the process described in step (1).
iii) The conductivity of given solution is then calculated as
𝐶𝑒𝑙𝑙 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
k=
𝑅𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛

3) Calculation of molar conductivity :

i) The molar conductivity of the given solution is then calculated using equation,
1000 𝑋 𝑘
Ʌ=
𝐶
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.
78 | P a g e
 Electrochemical reactions :
i) The chemical reaction occuring in electrochemical cells involves transfer of electrons from one
species to the other.
ii) Electrochemical reactions, are made of oxidation and reduction half reactions. The oxidation half
reaction occurs at one electrode and the reduction half reaction occurs at the other electrode. The net
cell reaction is the sum of these half reactions.

Electrodes :
i) Electrodes are the surfaces on which oxidation and reduction half reactions take place.
ii) Electrodes may or may not participate in the reactions.
iii)The electrodes which do not take part in reactions are inert electrodes.
iv) Cathode : It is an electrode at which the reduction takes place. At this electrode the species
undergoing reduction gains electrons.
v)Anode : It is an electrode at which oxidation takes place. At this electrode, the species undergoing
oxidation loses electrons.

Types of electrochemical cells : There are two types of electrochemical cells.

1) Electrolytic cell :
i) In this type of cell, a nonspontaneous reaction, known as electrolysis, is forced to occur by passing a
direct current from an external source into the solution.
ii) In such cells electrical energy is converted into chemical energy.
iii)The anode of electrolytic cell is positive and cathode is negative.
2) Galvanic or voltaic cell :
i) In galvanic (voltaic) cell a spontaneous chemical reaction produces electricity.
ii) In these cells chemical energy is converted into electrical energy.
iii)The anode of galvanic cell is negative and cathode is positive.

1) Electrolytic cell
i) Electrolytic cell consists of a container in which electrolyte is placed.
ii) Two electrodes are immersed in the electrolyte and connected to a source of direct current.
iii) At anode (+) a species oxidises with the removal of electrons. These electrons are pulled from
anode and pushed to cathode through an external circuit. The electrons are supplied to species at
cathode which are reduced.

Electrolysis of molten NaCl

a) Construction of cell :
i) The electrolytic cell consists of a container in which fused NaCl is placed.
ii) Two graphite electrodes are immersed in it. They are connected by metallic wires to a source of
direct current that is battery.

79 | P a g e
iii) The carbon electrode connected to positive terminal electrode of the battery is anode and that
connected to negative terminal of the battery is cathode.

Reactions occuring in the cell : Fused NaCl contains Na⊕ and Cl ions which are freely mobile.
When potential is applied, cathode attracts Na⊕ ions and anode attracts Cl- ions.
Oxidation half reaction at anode :
Cl ions migrate to anode. Each Cl ion, that reaches anode, gives one electron to anode. It
oxidises to neutral Cl atom in the primary process. Two Cl atoms then combine to form chlorine gas
in the secondary process.
2 Cl (l)  Cl (g) + Cl (g) + 2e
(primary process)
Cl (g) + Cl (g)  Cl2 (g)
(secondary process)
2Cl (l) Cl2 (g) + 2e
(overall oxidation)

Reduction half reaction at cathode : The electrons supplied by the battery are used in cathodic
reduction. Each Na⊕ ion, that reaches cathode accepts an electron from the cathode and reduces to
metallic sodium.
Na⊕ (l) + e-  Na (l)

Net cell reaction

The net cell reaction is the sum of two electrode reactions.
2 Cl (l)  Cl2 (g) + 2e
(oxidation half reaction)
2 Na (l) + 2e 
⊕
 2 Na (l)
(reduction half reaction)
2 Na (l) + 2 Cl (l) 
⊕
 2 Na (l) + Cl2(g)
(overall cell reaction)

Results of electrolysis of molten NaCl

i. A pale green Cl2 gas is released at anode.
ii. A molten silvery-white sodium is formed at cathode.
Decomposition of NaCl into metallic sodium and Cl2(g) is nonspontaneous. The electrical energy
supplied by the battery forces the reaction to occur

Electrolysis of aqueous NaCl :

i) Electrolysis of an aqueous NaCl can be carried out in the cell used for the electrolysis of molten
NaCl using inert electrodes.
ii) The fused NaCl is replaced by moderately concentrated aqueous solution of NaCl.
iii) The water involved in electrolysis of aqueous NaCl, leads to electrode reactions that differ from
electrolysis of molten NaCl.

Reduction half reaction at cathode : At cathode, two reduction reactions compete. One is the
reduction of Na⊕ ions as in case of molten NaCl.
i. Na⊕ (aq) + e  Na (s), E0 = -2.71 V
The other is the reduction of water to hydrogen gas.
ii. 2 H2O (l) + 2e   H2 (g) + 2 OH (aq), E0 = - 0.83 V

80 | P a g e
 The standard potential for the reduction of water is higher than that for reduction of Na⊕.
This implies that water has much greater tendency to get reduced than the Na⊕ ion. Hence reaction
(ii), that is, reduction of water is the cathode reaction when the aqueous NaCl is electrolysed.

Oxidation half reaction at anode : At anode there will be competition between oxidation of Cl ion
to Cl2 gas as in case of molten NaCl and the oxidation of water to O2 gas.
i. 2 Cl (aq)   Cl2 (g) + 2e ,E0oxi = - 1.36 V
ii. 2H2O (l)  O2 (g) + 4H⊕ (aq) + 2e E0oxi = - 0.4 V

Standard electrode potential for the oxidation of water is greater than that of Cl ion or water
has greater tendency to undergo oxidation. It is, therefore, expected that anode half reaction would be
oxidation of water. The experiments have shown, however, that the gas produced at the anode is Cl2
and not O2. This suggests that anode reaction is oxidation of Cl to Cl2 gas. This is because of the
overvoltage.
Overall cell reaction
It is the sum of electrode reactions.
2 Cl (aq) Cl2 (g) + 2e
(oxidation at anode)
2 H2O (l) + 2e H2 (g) + 2 OH (aq)
(reduction at cathode)
2 Cl (aq) + 2 H2O (l) Cl2 (g) + H2(g) + 2 OH (aq)
(overall cell reaction)

Results of electrolysis of aqueous NaCl

i. H2 gas is liberated at cathode.
ii. Cl2 gas is released at anode.
iii. Because Na⊕ ions remain unreacted and OH ions are formed at cathode, NaCl solution is
converted to NaOH solution.

Quantitative aspects of electrolysis :

A) The mass of reactant consumed or the mass of product formed at an electrode during electrolysis can
be calculated by knowing stoichiometry of the half reaction at the electrode.
i. Calculation of quantity of electricity passed : To calculate the quantity of electricity (Q) passed
during electrolysis, the amount of current, I, passed through the cell is measured. The time for which
the current is passed is noted.
Q (C) = I (A) × t (s)
ii. Calculation of moles of electrons passed : Total charge passed is Q(C). The charge of one mole
electrons is 96500 coulombs (C). It is referred to as one faraday (1 F). Hence,
Q(C )
Moles of electrons actually passed 
96500(C/mol e )
iii. Calculation of moles of product formed : The balanced equation for the half reaction occuring at
the electrode is written.
mole of productformed in the half reaction
Mole ratio =
mole of electrons required in the half reaction
Therefore,

81 | P a g e
 Moles of product formed = moles of electrons actually passed × mole ratio
Q(C )
 × mole ratio
96500 (C / mol e )
I ( A)  t (s)
 × mole ratio
96500 (C / mol e )

iv. Calculation of mass of product :

Mass of product W = moles of product × molar mass of product
I ( A)  t (s)
= × mole ratio × molar mass of product
96500(C / mol e )

B) Suppose two cells containing different electrolytes are connected in series. The same quantity of
electricity is passed through them. The masses of the substances liberated at the electrodes of the two
cells are related as given below :
The mass of the substance produced at the electrode of first cell is given by
Q(C)
W1   (mole ratio)1  M1
96500(C / mol e )

Q(C ) W1
Hence, =
96500(C / mol e ) (mole ratio)1  M1

Similarly mass of substance liberated at the electrode of second cell is W2 in the equation,
Q(C) W2

96500(C / mol e ) (mole ratio)2  M 2

M1 and M2 are the molar masses of substances produced at the electrodes of cells 1 and 2.
Q(C )
Because is the same for both,
96500(C / mol e )

W1 W2
We have 
(mole ratio)1  M1 (mole ratio) 2  M 2

Galvanic or voltaic cell :

i) In galvanic or voltaic cells, electricity is generated through the use of spontaneous chemical
reactions.
ii) A galvanic (or voltaic) cell is made of two half cells
iii) Each half cell consists of a metal strip immersed in the solution of its own ions of known
concentration. For example, a strip of zinc metal immersed in 1 M aqueous solution of zinc ions forms
an half cell.

Salt bridge :
i) In a galvanic cell, the two solutions are connected by a salt bridge.
ii) It is an U tube containing a saturated solution of an inert electrolyte such as KCl or NH4NO3 and
5% agar solution.

82 | P a g e
Functions of salt bridge
i. It provides an electrical contact between two solutions and thereby completes the electrical circuit.
ii. It prevents mixing of two solutions.
iii. It maintains electrical neutrality in both the solutions by transfer of ions.

Formulation or short notation of galvanic cells : The following conventions are used to write the
cell notation or cell formula
i. The metal electrodes or the inert electrodes are placed at the ends of the formula or the short
notation. The anode (-) is written at the extreme left and cathode (+) at extreme right.
ii. The insoluble species if any or gases are placed in the interior position adjacent to the metal
electrodes.
iii. The aqueous solutions of ions are placed at the middle of the cell formula.
iv. A single vertical line between two phases indicates the phase boundary. It indicates the direct
contact between them.
v. A double vertical line between two solutions indicates that they are connected by salt bridge.
vi. The additional information such as concentration of solutions and gas pressures is also given.
vii. A single half cell is written in the order: aqueous solution of ions first and then the solid electrode.
For example Zn2⊕(1M) Zn (s) . This order is reversed when the electrode acts as anode in the cell.
The following example illustrates these conventions. The cell composed of Mg (anode) and Ag
(cathode) consists of two half cells, Mg2⊕ (1M) | Mg (s) and Ag⊕ (1M) | Ag(s).
The cell is represented as : Mg (s) | Mg2⊕ (1M) | Ag⊕(1M) | Ag(s).

Writing of cell reaction : The following steps are followed to write the cell reaction.
i. Write oxidation half reaction for the left hand side electrode (anode) and reduction half reaction for
the right hand side electrode, (cathode).
ii. Add two electrode half reactions to get the overall cell reaction. While adding the electrons must be
balanced. For this purpose, it may be necessary to multiply one or both the half reactions by a suitable
numerical factor (s). No electrons should appear in the overall reaction.
iii. It is important to note that the individual half reactions may be written with one or more electrons.
For example, half reactions for H2 gas, whether written as
2H⊕ (aq) + 2e   H2 (g) or H⊕(aq) + e   1/2 H2 (g) makes no difference.

Example

The oxidation at anode is

Al (s)  Al3⊕ (1M) + 3e
83 | P a g e
 The reduction half reaction at cathode is
Ni2⊕ (1M) + 2e   Ni (s).
To balance the electrons, oxidation reaction is multiplied by 2 and reduction reaction by 3. The
two half reactions so obtained when added give the overall cell reaction. Thus,
2 Al (s)  2 Al3⊕ (1M) + 6e
(oxidation half reaction)
3 Ni (1M) + 6 e 
2⊕
 3 Ni (s)
(reduction half reaction)
2 Al (s) + 3Ni (1M) 
2⊕
 2Al3⊕ (1M) + 3 Ni (s)
(overall cell reaction)

Electrode potential and cell potential :

i) At the surface of separation of solid metal and the solution, there exists difference of electrical
potential. This potential difference established due to electrode half reaction occurring at the
electrode surface, is the electrode potential.
ii) The potential associated with oxidation reaction is oxidation potential while that associated with
reduction gives the reduction potential.
iii) The overall cell potential, also called electromotive force (emf), it is algebraic sum of the
electrode potentials,
Ecell = Eoxi (anode) + Ered (cathode)
where Eoxi is the oxidation potential of anode (-) and Ered is the reduction potential of cathode (+).
Standard potentials :
The standard conditions chosen are 1 M concentration of solution, 1 atm pressure for gases, solids and
liquids in pure form and 250C. The voltage measured under these conditions is called standard
potential designated as E0.
i) The standard cell potential is the algebraic sum of the standard electrode potentials
E0cell = E0oxi (anode) + E0red (cathode) ……(1)
Here E0oxi is standard oxidation potential and E0red is the standard reduction potential.
ii) According to IUPAC convention, standard potential of an electrode is taken as the standard
reduction potential.
iii) It must be realised that standard oxidation potential of any electrode is numerically equal to its
standard reduction potential with the reversal of sign.
iv) For example standard oxidation potential of Zn2⊕ (1M) | Zn electrode is 0.76V. Its standard
reduction potential will be -0.76 V.
v) Hereafter the standard reduction potential will be called standard potential, the voltage associated
with a reduction reaction.
vi) It follows that the standard cell potential (emf) is written in terms of the standard potentials of the
electrodes. E0oxi (anode) in equation (1) is replaced by - E0red (anode).
We then write,
E0cell = - E0red (anode) + E0red (cathode)
Omitting the subscript red, we have
E0cell = E0(cathode, +) - E0 (anode, -)

Note:
• While constructing a galvanic cell from two electrodes, the electrode with higher standard
potential is cathode (+) and that with lower standard potential is anode (-).

Dependence of cell potential on concentration (Nernst equation) :

Dependence of cell voltage on concentrations is given by Nernst equation. For any general
reaction,

84 | P a g e
 aA + bB   cC + dD
The cell voltage is given by
RT [C ]c [ D] d
Ecell  E 0 cell  ln
nF [ A] a [ B]b
2.303RT [C ]c [ D] d
 E 0 cell  log 10 ....(1)
nF [ A] a [ B]b

where n = moles of electrons used in the reaction, F = Faraday = 96500 C,

T = temperature in kelvin, R = gas constant = 8.314 J K-1mol-1
2.303RT
At 25 0C,  0.0592V
F
Therefore at 25 0C, eq. (5.24) becomes
0.0592V [C]c [ D]d
0
Ecell = E cell - log10 ……(2)
n [ A]a [ B]b
The Eq. (1) or Eq. (2) is the Nernst equation.
The first term on the right hand of Nernst equation represents standard state electrochemical
conditions. The second term is the correction for non standard state conditions. The cell potential
equals standard potential if the concentrations of reactants and products are 1 M each. Thus,
if [A] = [B] = [C] = [D] = 1M,
Ecell = E0cell
If a gaseous substance is present in the cell reaction its concentration term is replaced by the
partial pressure of the gas.
Example
The Nernst equation can be used to calculate cell potential and electrode potential.
i. Calculation of cell potential : Consider the cell
Cd(s) |Cd2⊕(aq) || Cu2⊕ (aq) | Cu.
Let us first write the cell reaction
Cd (s)   Cd2⊕ (aq) + 2 e
(oxidation at anode)
Cu2⊕ (aq) + 2 e   Cu (s)
(reduction at cathode)
Cd (s) + Cu (aq) 
2⊕
 Cd2⊕ (aq) + Cu (s)
(overall cell reaction)
Here n = 2
The potential of cell is given by Nernst equation,
0.0592 [Cd 2 ]
Ecell = E0cell log10 2
at 25 0C.
2 [Cu ]
(Concentration of solids and pure liquids are taken to be unity.)

ii. Calculation of electrode potential : Consider Zn2⊕(aq) | Zn(s)

The reduction reaction for the electrode is
Zn2⊕ (aq) + 2 e   Zn (s)
Applying Nernst equation, electrode potential is given by
0.0592 1
EZn  E 0 Zn  log10
2 [ Zn2 ]
0.0592
 E 0 Zn  log10[ Zn2 ] at 250 C
2
85 | P a g e
Thermodynamics of galvanic cells
Gibbs energy of cell reactions and cell potential :
i) The electrical work done in a galvanic cell is the electricity (charge) passed multiplied by the cell
potential.
Electrical work = amount of charge passed × cell potential.
Charge of one mole electrons is F coulombs. For the cell reaction involving n moles of electrons.
charge passed = nF coulombs
Hence, electrical work = nFEcell
The electrical work done in galvanic cell is equal to the decrease in Gibbs energy, - ΔG, of cell
reaction. It then follows that
Electrical work = - ΔG
and thus - ΔG = nFEcell
or ΔG = -nFEcell
Under standard state conditions, we write ΔG0 = -nFE0cell

Explains why E0cell is an intensive property.

We know that ΔG0 is an extensive property since its value depends on the amount of substance. If the
stoichiometric equation of redox reaction is multiplied by 2 that is the amounts of substances oxidised
and reduced are doubled, ΔG0 doubles. The moles of electrons transferred also doubles.
G 0 2G 0 G0
The ratio, E 0 cell   then becomes E 0cell   
nF 2nF nF
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.

Standard cell potential and equilibrium constant : The relation between standard Gibbs energy
change of cell reaction and standard cell potential is given by equation
- ΔG0 = nFE0cell …… (1)
The relation between standard Gibbs energy change of a chemical reaction and its equilibrium
constant as given in thermodynamics is :
ΔG0 = - RT ln K …..(2)
Combining Eq. (1) and Eq. (2), we have
-nFE0cell = - RT ln K
RT
or E0cell  ln K
nF
2.303 RT
 log 10 K
nF
0.0592
 log 10 K at 25 0 C ..
n

Reference electrodes : A reference electrode is then defined as an electrode whose potential is

arbitrarily taken as zero or is exactly known.
The chemists have chosen hydrogen gas electrode consisting of H2 gas at 1 atm pressure in contact
with 1 M H⊕ ion solution as a primary reference electrode. The potential of this electrode has
arbitrarily been taken as zero. The electrode is called standard hydrogen electrode (SHE).

Standard hydrogen electrode (SHE) :

Construction : SHE consists of a platinum plate, coated with platinum black used as electrodes. This
plate is connected to the external circuit through sealed narrow glass tube containing mercury. It is
surrounded by an outer jacket.

86 | P a g e
 The platinum electrode is immersed in 1 M H⊕ ion solution. The solution is kept saturated with
dissolved H2 by bubbling hydrogen gas under 1 atm pressure through the side tube of the jacket.
Platinum does not take part in the electrode reaction. It is inert electrode and serves as the site for
electron transfer.

Formulation : Standard hydrogen electrode is represented as

H⊕ (1M) | H2 (g, 1atm) | Pt
Electrode reaction : The platinum black capable of adsorbing large quantities of H2 gas, allows the
change from gaseous to ionic form and the reverse process to occur.
The reduction half reaction at the electrode is
2H⊕ (1M) + 2e   H2 (g, 1atm) E0H2 = 0.000 V
Application of SHE
SHE is used as a primary reference electrode to determine the standard potentials of other
electrodes.To determine the standard potential of Zn2⊕(1M) | Zn (s), it is combined with SHE to form
the cell,
Zn | Zn2⊕(1M) || H⊕ (1M) | H2 (g, 1atm) | Pt
The standard cell potential, E0cell, is measured.
E0cell = E 0 H2 - E0Zn = - E0Zn , because E 0 H2 is zero.
Thus, the measured emf of the cell is equal to standard potential of Zn2⊕(1M) | Zn (s) electrode.

Difficulties in setting SHE

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.
iii. The concentration of H+ ion solution cannot be exactly maintained at 1 M. Due to bubbling of gas
into the solution, evaporation of water may take place. This results in changing the concentration
of solution.

87 | P a g e
 Galvanic cells useful in day-to-day life
Voltaic (or galvanic) cells in common use can be classified as primary and secondary cells.
i. Primary voltaic cells : In primary voltaic cell, once the chemicals are completely consumed, cell
reaction stops. these cells cannot be recharged. example is dry cell.
ii. Secondary voltaic cells : In secondary voltaic cell, the chemicals consumed during current
generation can be regenerated. The voltaic cells which can be recharged are called secondary voltaic
cells.
.Examples of secondary cells are lead storage battery, mercury cell and nickel-cadmium cell.

Dry cell (Leclanche' cell) : It is a cell without liquid component, but the electrolyte is not completely
dry. It is a viscous aqueous paste.
Construction :
i) The container of the cell is made of zinc which serves as anode (-). It is lined from inside with a
porous paper to separate it from the other material of the cell.
ii) An inert graphite rod in the centre of the cell serves as cathode. It is surrounded by a paste of
manganese dioxide (MnO2) and carbon black.
iii) The rest of the cell is filled with an electrolyte. It is a moist paste of ammonium chloride (NH4Cl)
and zinc chloride (ZnCl2). Some starch is added to the paste to make it thick so that it cannot be leaked
out.
iv) The cell is sealed at the top to prevent drying of the paste by evaporation of moisture.

Cell reactions:
i. Oxidation at anode : When the cell operates the current is drawn from the cell and metallic zinc is
oxidised to zinc ions.
Zn | (s) 
 Zn2⊕ (aq) + 2e
ii. Reduction at cathode : The electrons liberated in oxidation at anode flow along the container and
migrate to cathode. At cathode NH4⊕ ions are reduced.
2NH4⊕ (aq) + 2e   2NH3 (aq) + H2 (g)
Hydrogen gas produced in reduction reaction is oxidised by MnO2 and prevents its collection
on cathode.
H2(g)+2MnO2(s)   Mn2O3(s)+H2O(l)
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)
88 | P a g e
Uses of dry cell : Dry cell is used as a source of power in flashlights, portable radios, tape recorders,
clocks and so forth.

Lead storage battery (Lead accumulator) : Lead accumulator stores electrical energy during
recharging. It functions as galvanic cell and as electrolytic cell, as well.
Construction : A group of lead plates packed with spongy lead serves as anode (-). Another group of
lead plates bearing lead dioxide (PbO2) serves as cathode (+).

To provide large reacting surface, the cell contains several plates of each type. The two types
of plates are alternately arranged.
The electrodes are immersed in an electrolytic aqueous solution of 38 % (by mass) of sulphuric
acid of density 1.2 g/mL.
Notation of the cell : The cell is formulated as
Pb(s) | PbSO4(s) | 38%H2SO4(aq) | PbSO4(s) | PbO2(s) | Pb(s)

a. Cell reactions during discharge

i. Oxidation at anode (-) : lead is oxidised to Pb2⊕ ions and the Pb2⊕ions so formed combine with
SO4 2 ions from H2SO4 to form insoluble PbSO4. The net oxidation is the sum of these two processes.
Pb (s)  Pb2⊕ (aq) + 2 e (oxidation)
Pb (aq) + SO4 2 (aq) PbSO4 (s) (precipitation)
2⊕

Pb (s) + SO4 2 (aq) PbSO4 (s) + 2 e ...(i) (overall oxidation)

ii. Reduction at cathode (+) : The electrons produced at anode travel through external circuit and re-
enter the cell at cathode. At cathode PbO2 is reduced to Pb2⊕ ions in presence of H⊕ ions.
Subsequently Pb2⊕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) (reduction)
Pb (s) + SO42 (aq)  PbSO4 (s) (precipitation)
PbO2 (s) + 4H (aq) + SO42 (aq) + 2 e 
⊕

PbSO4 (s) + 2H2O (l) ...(ii) (overall reduction)

iii. Net cell reaction during discharge: The net cell reaction is the sum of overall oxidation at anode
and overall reduction at cathode.
Pb (s) + PbO2 (s) + 4H⊕ (aq) + 2SO42 (aq)   2PbSO4 (s) + 2H2O (l) or
Pb (s) + PbO2 (s) + 2H2SO4 (aq)   2PbSO4(s) + 2H2O (l) ...(iii)
As the cell operates to generate current, H2SO4 is consumed. Its concentration (density)
decreases and the cell potential is decreased. The cell potential thus depends on sulphuric acid
concentration (density).
89 | P a g e
b. Cell reactions during recharging :
The potential of lead accumulator is 2V. It must be recharged with the falling of the cell
potential to 1.8 V. To recharge the cell external potential slightly greater than 2 V needs to be applied
across the electrodes.
During recharging the cell functions as electrolytic cell. The anode and cathode are
interchanged with PbO2 electrode being anode (+) and lead electrode cathode (-).
iv. Oxidation at anode (+) : It is reverse of reduction reaction (ii) at cathode that occurs during
discharge.
PbSO4 (s) + 2H2O (l)   PbO2 (s) + 4H⊕(aq) + SO42 (aq) + 2 e ...(iv)
v. Reduction at cathode (-) : It is reverse of oxidation reaction (i) at anode that occurs during
discharge.
PbSO4(s) + 2 e   Pb (s) + SO42 (aq) .....(v)

vi. Net cell reaction : It is the sum of reaction (iv) and (v) or the reverse of net cell reaction (iii) that
occurs during discharge
2PbSO4 (s) + 2H2O (l)   Pb (s) + PbO2 (s) + 2 H2SO4 (aq)
The above reaction shows that H2SO4 is regenerated. Its concentration (density) and in turn, the
cell potential increases.

Applications of lead accumulator

i. It is used as a source of direct current in the laboratory.
ii. A 12 V lead storage battery constructed by connecting six 2 V cells in series is used in automobiles
and inverters.

Nickel-Cadmium or NICAD storage cell : Nickel-cadmium cell is a secondary dry cell. In other
words it is a dry cell that can be recharged.
Anode of the NICAD storage cell is cadmium metal. The cathode is nickel (IV) oxide, NiO2
supported on Ni. The electrolyte solution is basic.
The electrode reactions and overall cell reaction are as follows :
Cd (s) + 2OH (aq)   Cd (OH)2 (s) + 2 e
(anodic oxidation)
NiO2 (s) + 2 H2O (l) + 2 e   Ni(OH)2 (s) + 2OH (aq)
(cathodic reduction)
Cd (s) + NiO2 (s) + 2 H2O (l)  Cd(OH)2 (s) + Ni(OH)2 (s)
(overall cell reaction)

The reaction product at each electrode is solid that adheres to electrode surface. Therefore the
cell can be recharged. The potential of the cell is about 1.4 V. The cell has longer life than other dry
cells. It can be used in electronic watches, calculators, photographic equipments, etc.

Mercury battery : Mercury battery is a secondary dry cell and can be recharged. The mercury battery
consists of zinc anode, amalgamated with mercury. The cathode is a paste of Hg and carbon. The
electrolyte is strongly alkaline and made of a paste of KOH and ZnO. The electrode reactions and
net cell reaction are :
Zn(Hg)+2OH (aq)   ZnO(s) +H2O(l) + 2 e (anode oxidation)
HgO(s)+ H2O(l)+2e   Hg(l) + 2 OH (aq) (cathode reduction)
Zn (Hg) + HgO(s)   ZnO(s) + Hg(l) (overall cell reaction)

90 | P a g e
 The overall reaction involves only solid substances. There is no change in electrolyte
composition during operation.
The mercury dry cell finds use in hearing aids, electric watches, pacemakers, etc.

Fuel cells : In these cells one of the reactants is a fuel such as hydrogen gas or methanol. The other
reactant such as oxygen, is oxidant. The simplest fuel cell is hydrogen-oxygen fuel cell.

Hydrogen-oxygen fuel cell : In H2 - O2 fuel cell, the fuel is hydrogen gas. Oxygen gas is an oxidising
agent. The energy of the combustion of hydrogen is converted into electrical energy.

Construction : The anode and cathode are porous carbon rods containing small amount of finely
divided platinum metal that acts as a catalyst. The electrolyte is hot aqueous solution of KOH. The
carbon rods immersed into electrolyte. Hydrogen gas is continuously bubbled, through anode and
oxygen gas through cathode into the electrolyte.

Cell reactions
i. Oxidation at anode (-) : At anode hydrogen gas is oxidised to H2O.
2H2 (g) + 4OH (aq)   4H2O (l) + 4 e

ii. Reduction at cathode (+) : The electrons released at anode travel, through external circuit to
cathode. Here O2 is reduced to OH-.
O2 (g) + 2H2O (aq)+ 4 e   4OH (aq)

iii. Net cell reaction : The overall cell reaction is the sum of electrode reactions(i) and (ii).
2H2 (g) + O2 (g)   2H2O (l)
The overall cell reaction is combustion of H2 to form liquid water.
The cell continues to operate as long as H2 and O2 gases are supplied to electrodes.
The cell potential is given by
E 0cell  E 0cathode  E 0 anode  0.4V  (0.83V )
 1.23V
Advantages of fuel cells
i. The reacting substances are continuously supplied to the electrodes.
ii. They are nonpolluting as the only reaction product is water.
iii. Fuel cells provide electricity with an efficiency of about 70 % which is twice as large when
compared with efficiency of thermal plants (only 40 %).

91 | P a g e
Drawbacks of fuel cell
H2 gas is hazardous to handle and the cost of preparing H2 is high.

Applications of fuel cells

i. The fuel cells are used on experimental basis in automobiles.
ii. The fuel cell are used for electrical power in the space programme.
iii. In space crafts the fuel cell is operated at such a high temperature that the water evaporates at the
same rate as it is formed. The vapour is condensed and pure water formed is used for drinking by
astronauts.
iv. In future, fuel cells can possibly be explored as power generators in hospitals, hotels and homes.

Electrochemical series (Electromotive series) : The electrodes with their half reactions are
arranged according to their decreasing standard potentials. This arrangement is called electrochemical
series.

Key points of electrochemical series

i. The half reactions are written as reductions. The oxidizing agents and electrons appear on the left side
of half reactions while the reducing agents are shown on the right side in the half reaction.
ii. Below hydrogen electrode the negative standard potential increases and above hydrogen electrode the
positive standard potential increases.
iii. E0 values apply to the reduction half reactions that occur in the forward direction as written.
iv. Higher (more positive) E0 value for a half reaction indicates its greater tendency to occur in the
forward direction and in turn greater tendency for the substance to reduce. Conversely, the low (more
negative) E0 value of a half reaction corresponds to its greater tendency to occur in the reverse direction
or for the substance to oxidise.
The half reactions are listed in order of their decreasing tendency in the forward direction.

Applications of electrochemical series

i. Relative strength of oxidising agents : The species on the left side of half reactions are oxidizing
agents. E0 value is a measure of the tendency of the species to accept electrons and get reduced.
In other words, E0 value measures the strength of the substances as oxidising agents. Larger the E0
value greater is the oxidising strength. The species in the top left side of half reactions are strong
oxidising agents. As we move down the table, E0 value and strength of oxidising agents decreases
from top to bottom.

ii. Relative strength of reducing agents : The species on the right side of half reactions are reducing
agents. The half reactions at the bottom of the table with large negative E0 values have a little or no
tendency to occur in the forward direction as written. They tend to favour the reverse direction. It
follows, that the species appearing at the bottom right side of half reactions associated with large
negative E0 values are the effective electron donors. They serve as strong reducing agents. The
strength of reducing agents increases from top to bottom as E0 values decrease.

iii. Spontaneity of redox reactions : A redox reaction in galvanic cell is spontaneous only if the

92 | P a g e
species with higher E0 value is reduced (accepts electrons) and that with lower E0 value is oxidised
(donates electrons).
The standard cell potential must be positive for a cell reaction to be spontaneous under the standard
conditions.

Example – Can 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) +2Ag (aq) 
2⊕
 Mg2⊕ (aq) + 2Ag (s) (overall reaction)

E0Mg = -2.37 V and E0Ag = 0.8 V. For the cell having Mg as anode and Ag cathode.
E0Cell = E0Ag - E0Mg = 0.8V - (-2.37V) = 3.17 V.
EMF being positive the cell reaction is spontaneous. Thus Ag⊕ ions oxidise to metallic Mg.

General rules
i. An oxidizing agent can oxidize any reducing agent that appears below it, and cannot oxidize the
reducing agent appearing above it in the electrochemical series.
ii. An reducing agent can reduce the oxidising agent located above it in the electrochemical series.

93 | P a g e
94 | P a g e