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Electrochemistry
Electrochemistry Electrochemical Cell

The area of chemistry

The devices which
concerned with
convert electrical
the interconversion of energy into chemical
chemical energy energy or vice versa.
and electrical energy.
 Example
Electrochemical cell

A battery is an electrochemical cell

Electrolytic
Galvanic cell
cell

It takes the energy released by a

Also known as spontaneous chemical reaction and
voltaic cell. uses it to produce electricity.
 Galvanic Cell

A galvanic cell is an electrochemical

cell that produces electricity as a
result of the spontaneous reaction Anode Cathode
occurring inside it. (oxidation) (Reduction)

− +
ΔrG < 0

Example Daniel cell

Chemical energy is
converted
into electrical energy. Electron flow
 Electrolytic Cell

An electrolytic cell is an electrochemical

cell in which a non-spontaneous reaction
is driven by an external source of current.

+ −
ΔrG > 0

Electrical energy is converted

into chemical energy.
Current Battery
Electrons
 Important
Redox Couple
Points

Galvanic and electrolytic cells Any redox reaction may be expressed

1 as the difference of two half-reactions,
are reverse of each other.
which are conceptual reactions showing
gain and loss of electrons, i.e., reduction
and oxidation, respectively.

Redox reactions are

2
involved in both cells. The reduced and oxidised species in
a half-reaction form a redox couple.
 Reduction Oxidation

Redox Redox
Half-reaction Half-reaction
couple couple

M+/M M+ (aq) + e− M (s) M/M+ M (s) M+ (aq) + e−

Electrochemical Cell

An electrochemical cell consists of An electrode and its electrolyte

two electrodes, or metallic comprise an electrode
conductors, in contact with an compartment (half cell).The two
electrolyte, an ionic conductor (which electrodes may share
may be a solution, a liquid, or a solid). the same compartment.
 Half Cell

Example
Whenever a metal strip is put
in an electrolyte, the process of
oxidation or reduction takes
place within the system.

Some reactive metals have a

tendency to go into the solution
phase, when placed in contact with
their ions or their salt solution, it is
called oxidation.

Zn rod in ZnSO4 solution

 Electrode Potential

Whenever a metal strip is

put in an electrolyte
Electrode potential can be
either oxidation potential or
reduction potential.

A potential difference is
developed between the metal
electrode and its ions in solution.
 Electrode Potential

Cu rod in CuSO4 solution

Some metals (Cu, Ag, Au, etc.)

when in contact with their ions
in the solution have a
tendency to get deposited on
the metal rod.

Reduction
 Electrode Potential

Reduction Potential (R.P.)

Cu2+
(aq) + 2e‒ Cu (s) ECu2+/Cu

Greater the R.P., greater will

be the tendency to get
reduced.
 Electrode Potential

Oxidation Potential (O.P.)

Zn (s) Zn2+ (aq) + 2e‒ E Zn/Zn2+

Greater the O.P., greater will

be the tendency to get
oxidised.
 Properties of
Electrode Potential
 Electrode potential or cell
Electrode potential depends on:
potential are intensive properties
and does not depend on:

1 Concentration of solution
1 Amount of solution

Thickness of metal
Nature of metal 2
2 rod/plate/wire, etc.

Stoichiometric variation
3
Nature of the electrolyte of a given cell reaction
3

A2+ (aq) + 2e‒ A (s), E = x volt

4
Pressure, temperature
conditions
2A2+ (aq) + 4e‒ 2A (s), E = x volt
 Electrode Potential

An electrode is chosen as a
The potential of a single
reference with respect to which
electrode
all other electrodes are valued.
cannot be determined. But
the potential difference
between two electrodes can
Standard Hydrogen Electrode
be measured accurately.
(SHE) is taken as standard
reference electrode.

The potential of an electrode Its electrode potential

is calculated using a is arbitrarily assumed
reference electrode. to be 0.00 volt.
 Standard Hydrogen Electrode (SHE)

At 298 K
Pure H2 gets adsorbed
at Pt surface

A platinum
Pure
foil (platinum
Dipped in hydrogen Which is in contact with
electrode
an acidic gas is H+ ions in the solution
coated with
solution bubbled at
finely
of 1 bar into the
powdered
1 M HCl. solution.
Pt black).
Reduction or oxidation
occurs on Pt foil.
 Standard Hydrogen Electrode (SHE)

If it acts as anode, If it acts as cathode,

2H+(aq)
H2(g) 2H+(aq) + 2e‒ SOP = 0 + 2e− H2 (g) SRP=0

Cell representation Cell representation

Pt (s) | H2 (g, 1 bar) | H+ (aq, 1 M) H+ (aq, 1 M) l H2 (g, 1 bar) l Pt (s)

 Reference Potential

Electrode potential of SHE is

taken to be zero at all temperature.

S.O.P. = -S.R.P. = 0
 Note

On reversing the chemical equation,

the effect on E is as follows:

A (s) An+ (aq) + ne‒ EA/A

0
n+ = SOP

An+ (aq) + ne‒ A (s) EA0 n+/A = SRP

E0 n+
A /A = - EA/A
0
n+
 Standard Electrode Potential

Electrode reaction in
Potential difference Representation
standard conditions
developed between metal
electrodes and the
solution of its ions at 1 M
Reduction: Zn2+ (aq) + 2e‒ Zn (s) E0 (SRP)
concentration at 1 bar Zn2+/Zn
pressure and at a
particular temperature is
known as standard Oxidation: Zn (s) Zn2+ (aq) + 2e‒ E0 (SOP)
Zn/Zn2+
electrode potential (E0).
 Cell Potential

The driving force that pushes the Electromotive force

electrons away from the anode and of a cell is equal to the
pulls them towards the cathode is an potential difference
electrical potential called electromotive between its terminals
force, also known as cell potential when no current is
or the cell voltage. passing through
the circuit.

Unit: Volt
 Cell Potential

0
SRP of SRP of
Ecell = substance ‒ substance
reduced oxidised

The sign of the cell

SRP stands for standard reduction potential tells us the
potential which measures the direction in which the
tendency for a given chemical reaction must shift to
species to be reduced. reach equilibrium.

0
Ecell is a positive number
 Cell Potential

Reactions for which E° cell is The magnitude of the cell

positive. potential

Have equilibrium constants that is a measure of the driving force

favor the formation of the products behind a reaction. The larger the
of the reaction. value of the cell potential.

Occurs naturally and referred The farther is the reaction

to as spontaneous reaction. from equilibrium.
 Relation between ΔrG and Ecell

From thermodynamics
Maximum non-expansion
(electrical) work that a system
dG = dWnon-exp + VdP - SdT (the cell) can do is equal to
decrease in Gibbs free energy
At constant T & P

∫dG = ∫dWnon-exp We, max = |ΔG|

| ΔG | = Wnon-exp, max. ΔG identified with the Gibbs

energy of the cell reaction, ΔrG
 Relation between ΔrG and Ecell

Maximum electrical work Charge present on

obtained from a cell = |ΔrG| 1F = 1 mole of electron

ΔrG = ‒q×V = ‒ nFE

= 96,485 C/mol

where, F is Faraday’s constant

n = Number of moles of electron
≈ 96,500 C/mol
E = Electrode potential or cell EMF
ΔrG = Change in Gibbs energy for
half-cell or cell
 Relation between ΔrG and Ecell

By knowing the Δ r G at a specified

composition, the cell emf at that
composition can be stated.

Spontaneous
Δ rG < 0 Ecell > 0
cell reaction

Non-spontaneous
Δ G>0 Ecell < 0
cell reaction r
 Calculation of Ecell

Electrode reaction Δ rG

1 Anode: A (s) An+ (aq) + ne‒ ΔrG1 = -nFE

A/An+

2 Cathode: Bm+ (aq) + me‒ B (s) ΔrG2 = -mFE Bm+/B

Multiply eqn (1) by m, & eqn (2) by n and then add both the equations

mA (s) + nBm+ (aq) mAn+ (aq) + nB (s) ΔrG3

Cell reaction Δ rG

ΔrG3 = m ΔrG1 + n ΔrG2

 Calculation of Ecell

When reduction potential of both

electrodes are taken into account:

Reduction Reduction
Ecell = potential ‒ potential
ΔrG3 = m ΔrG1 + n ΔrG2 of cathode of anode

⇒ -nmFEcell= -nmFEA/An+ ‒ nmFEBm+/B Ered Ered

Ecell = ‒
(cathode) (anode)

Ecell = E + E The difference in electrode

A/An+ Bm+/B
potentials of the two half-cell
reactions (oxidation half-cell and
reduction half-cell) is known
as emf of the cell or cell
potential.
 Ecell of Daniell Cell

At Anode

Zn (s) Zn2+(aq) + 2e‒

ΔrG1 = -2 × F × EZn/Zn2+
At Cathode

Cu2+(aq) + 2e‒ Cu (s)

ΔrG2 = -2 × F × ECu2+/Cu
 Ecell of Daniell Cell

Cell reaction

Zn Cu2+ Cu Zn2+
(s) + (aq) (s) + (aq)

ΔrG3 = ΔrG1 + ΔrG2

-2 × F × E cell = -2 × F × E Zn/Zn2+ + -2 × F × E Cu2+/Cu

 Ecell of Daniell Cell

E cell = E
Cu2+/Cu + E Zn/Zn2+

E cell = E
Cu2+/Cu ‒ E
Zn2+/Zn

E0cell = E0Cu2+/Cu ‒ E0Zn2+/Zn

E0cell = 0.34 ‒ ‒ 0.76

E0cell = 1.1 V
 Calculation of Standard Electrode Potential

(at 298 K experimentally)

Example
To calculate standard potential
of any other electrode, it
Construct a cell
is coupled with SHE

Zinc
Anode Cathode SHE
electrode
Its potential is measured and that
gives the value of standard electrode 0
potential of that electrode. Ecell = 0.76 V
 Example

0
=
0 0
Ecell E ‒ E
H+/H2 Zn2+/Zn

0.76 = 0 ‒ E
0
Zn2+/Zn

E0
Zn2+/Zn = -0.76 V

W.r.t. H2, Zn has greater

=
0
E 0.76 V tendency to get oxidised.
Zn/Zn2+
 Half-Cell Potential

E0
Zn2+/Zn
= - 0.76 V

0
E Zn/Zn2+ = 0.76 V

With respect to hydrogen,

zinc has greater tendency
to get oxidised.
 Measuring SRP of Zinc Electrode

Standard
Zinc Hydrogen
Electrode Electrode
Standard Electrode Potential

Similarly, Standard
Reduction Potential (SRP)
(at 298 K) for many other
electrodes are calculated.

They are arranged in a series in

decreasing order of electrode
potential known as an
electrochemical series.
 Electrochemical Series

SRP for various half-cells

are measured at 25℃
with respect to SHE.

Electrochemical
series
These SRPs are
arranged in a
decreasing order.
 Electrochemical Series

SRP, E0
Half-reaction
(at 298 K)

F2 (g) + 2e− 2F− (aq) 2.87 V

Co3+ (aq) + e− Co2+ (aq) 1.81 V

H2O2 (l ) + 2H+ (aq) + 2e− 2H2O (l) 1.78 V

−
MnO4 (aq) + 8H+ (aq) + 5e− Mn2+ (aq) + 4H2O (l) 1.51 V
 Electrochemical Series
SRP, E0
Half-reaction
(at 298 K)

Au3+ (aq) + 3e− Au (s) 1.40 V

Cl2 (g) + 2e− 2Cl− (aq) 1.36 V

2−
Cr2O7 (aq) + 14H+ (aq) + 6e− 2Cr3+ (aq) + 7H2O (l) 1.33 V

O2 (g) + 4H+ (aq) + 4e− 2H2O (l ) 1.23 V

MnO2 (s) + 4H+ (aq) + 2e− Mn2+ (aq) + 2H2O (l ) 1.23 V

 Electrochemical Series

SRP, E0
Half-reaction
(at 298 K)

Br2 (l ) + 2e− 2Br− (aq) 1.09 V

−
NO3 (aq) + 4H+ (aq) + 3e− NO (g) + 2H2O (l) 0.97 V

2+
2Hg2+ (aq) + 2e− Hg2 (aq) 0.92 V

Ag+ (aq) + e− Ag (s) 0.80 V

Fe3+ (aq) + e− Fe2+ (aq) 0.77 V

 Electrochemical Series

SRP, E0
Half-reaction
(at 298 K)

O2 (g) + 2H+ (aq) + 2e− H2O2 0.68 V

I2 (s) + 2e− 2I− (aq) 0.54 V

Cu+ (aq) + e− Cu (s) 0.52 V

Cu2+ (aq) + 2e− Cu (s) 0.34 V

AgCl (s) + e− Ag (s) + Cl− (aq) 0.22 V

 Electrochemical Series

SRP, E0
Half-reaction
(at 298 K)

AgBr (s) + e− Ag (s) + Br− (aq) 0.10 V

2H+ (aq) + 2e− H2 (g) 0.00 V

Pb2+ (aq) + 2e− Pb (s) −0.13 V

Sn2+ (aq) + 2e− Sn (s) −0.14 V

Ni2+ (aq) + 2e− Ni (s) −0.25 V

 Electrochemical Series

SRP, E0
Half-reaction
(at 298 K)

Fe2+ (aq) + 2e− Fe (s) −0.44 V

Cr3+ (aq) + 3e− Cr (s) −0.74 V

Zn2+ (aq) + 2e− Zn (s) −0.76 V

2H2O (l ) + 2e− H2 (g) + 2OH− (aq) −0.83 V

Al3+ (aq) + 3e− Al (s) −1.66 V

 Electrochemical Series

SRP, E0
Half-reaction
(at 298 K)

Mg2+ (aq) + 2e− Mg (s) −2.36 V

Na+ (aq) + e− Na (s) −2.71 V

Ca2+ (aq) + 2e− Ca (s) −2.87 V

K+ (aq) + e− K (s) −2.93 V

Li+ (aq) + e− Li (s) −3.05 V

 Application of Electrochemical Series

Species (particularly metals)

having highly negative SRP
(i)
are powerful reducing agents.

Examples Li, Na, Ca etc.

Species having highly +ve SRP

(ii) are powerful oxidising agents.

‒
Examples F2, MnO4 etc.
 Application of Electrochemical Series

More active metals can

(iii) displace less active metals
from their salt solution

CuSO4 (aq) + Zn (s) Cu (s) + ZnSO4 (aq)

ZnSO4 (aq) + Cu (s) No reaction Oxides of active metals

(iv) are thermally stable.

Examples Na2O, Al2O3, CaO

Application of Electrochemical Series

Oxides of less active metals

(v) are thermally less stable.

Δ
1 O (g)
Ag2O (s) 2Ag (s) + 2
2

Δ 1 O (g)
HgO (s) Hg (l) + 2
2
Characteristics of Anode in Daniell Cell

Oxidation takes place at anode.

It acts as a source of electrons.

It has negative polarity.

The electrode potential

is represented by E
Zn /Zn2+
Characteristics of Cathode in Daniell Cell

Reduction takes place at cathode.

It acts as a sink of electrons.

It has positive polarity.

The electrode potential

E
is represented by Cu2+/Cu
 Daniell Cell

On joining the metal strips through

a wire (of negligible resistance)

The current flows as long as the

potential difference exists between
the metal phase and the liquid phase
 Why?

The current is
generated,
but it is not
sustainable
 Anode

Zn atoms will move in the Accumulation of extra

solution to form Zn2+ ions positive charge in the solution

After some time, equilibrium Will not allow extra zinc ions
is established to move in the solution.

Zn (s) ⇌ Zn2+ (aq) + 2e−

The solution will be saturated
with Zn2+ ions.
 Cathode

Cu2+ ions will get deposited

on electrode to form Cu Accumulation of sufficient
atoms positive charge on the rod will
not allow extra copper
ions to get deposited.

After some time, equilibrium

is established

Cu2+ (aq) + 2e− ⇌ Cu (s)

 Cathode

Dynamic equilibrium
 Daniell Cell
Metallic wire

Salt bridge
Zinc rod Copper rod
(Anode) (Cathode)

ZnSO4 CuSO4
solution solution
Two solutions
are connected
Electron by
flow Current a salt bridge.
 Salt Bridge

U-shaped inverted tube that contains

a gel permeated with an inert electrolyte
(E.g.: potassium chloride in agar jelly)

Functions of Salt Bridge

It connects the solution

1 of two half-cells to
complete the circuit.

It minimises the
2 liquid junction
potential.
 Functions of Salt Bridge

Ensures that the two

It maintains the electrical
electrolytic solutions do not
neutrality of the solution in
3 order to give continuous 4 mix but slight diffusion of
ions from one electrode to
flow or generation of current.
another is possible.

Salt bridge is not required for

The simultaneous electrical neutrality galvanic cell in which a
of the anodic oxidation chamber and common electrolyte of anode
cathodic reduction chamber is due to half and cathode half is
the same mobility or velocity of K+ and present.
NO3- ions taken into salt bridge. Example: Concentration cell
 Selection of Electrolyte for Salt Bridge

The electrolyte in salt bridge

Generally tube is filled with a should be such that, speed of its
paste of agar-agar powder cation equals to the speed of its
with a natural electrolyte anion in electrical field.

Generally, not common to

The ions of The ions
anodic/cathodic compartment with
the inert are not
porous plugs at each mouth of tube.
electrolyte do oxidised
not react or reduced
with other ion at the
in the electrodes.
solution.
 Selection of Electrolyte for Salt Bridge

If Ag+, Hg22+, Pb2+, Tl+ ions are present in

a cell, then in salt bridge, KCl is not used.

KCl is generally
preferred but KNO3
or NH4NO3 can Because there can be formation of
also be used. precipitate of AgCl, Hg2Cl2, PbCl2
or TlCl at the mouth of the tube which
will prevent the migration
of ions and its functioning will stop.
 Cell Representation

Oxidation Reduction
half cell + half cell

Electrochemical Cell

A cell can be represented using some

universal conventions and notations.
 Cell Representation

Rule 1

On the left side Anode half cell

On the right
Cathode half cell
side

Rule 2

The separation of two phases shown

by single vertical line (|)
 Cell Representation

Rule 3 Example

Sn(s) | Sn2+(aq), Sn4+(aq)

Various materials present
in the same phase are shown
together with comma (,)

Anode half cell

 Cell Representation

Example

The significant features Zn (s) | Zn2+ (aq, 1 M)

of the substance, like
pressure of a gas,
concentration of ions, etc.
is indicated in brackets
immediately after
writing the substance. Anode half-cell
 Cell Representation

Rule 4 Example

Anode half cell Cathode half cell

The salt bridge is represented by
double slash (||)

Zn (s) | Zn2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu (s)

Salt bridge
 Cell Representation

Rule 5 Example

For gas electrode

Pt (s) | H2 (g) | H+ (aq) H+ (aq) | H2 (g) | Pt (s)
Gas indicated after electrode

For anode
For anode For cathode
Gas indicated before electrode

For cathode
 Complete Cell Representation

Pt (s) | H2 (g, 1 bar) | H+ (aq, 1 M) || Cu2+ (aq, 1 M) l Cu (s)

Salt bridge

Cathode half cell

Anode half cell

Phase boundary

Electrode
 Complete Cell Representation

Example

Cu (s) | Cu2+ (aq, 1 M) || Cl2 (g, 1 atm) | Cl− (aq, 1 M) | C (s)

Factor Affecting Cell Potential

(1) Temperature

(2) Composition of the reaction mixtures

(3) The partial pressure of the gas (if any)

 Nernst Equation
For any reaction,

ΔrG = ΔrG0 + RT ln Q
The dependence
of the concentration and
pressure of the gas
on the cell potential −nFEcell = −nFE0
cell
+ 2.303 RT log Q
can be derived from
thermodynamics.

Q - Reaction quotient
 Nernst Equation
For any reaction,
Where,
R = Universal gas constant
aA + bB cC + dD T = Temperature
n = Number of transferred
electrons
[C]c [D]d F = Faraday’s constant
Q = [A]a [B]b
...(1) Q = Reaction quotient

2.303 RT T = 298 K
Ecell = E0cell − nF
log Q ...(2)
Take R = 8.314 J/mol K
F = 96500 C/mol

Nernst equation
 Nernst Equation
From thermodynamics,

Ecell =
0
Ecell −
0.059
log Q
ΔrG = ΔrG0 + RT ln Q
n

At chemical equilibrium,

0.059
log [C] [D] =
c d
= −
0
Ecell ΔrG 0
n [A]a [B]b

Ecell = 0

Cell will be of no use

 Equilibrium in Electrochemical Cell

ΔrGO = -RT ln Keq log Keq =

nF
E0cell
2.303 RT

=
0
-nFEcell -2.303 RT log (Keq) T = 298 K,
Take R = 8.314 J/mol K,
F = 96500 C

2.303 RT
E0cell = nF
log Keq
n
log Keq = 0.059
E0cell
Nernst Equation for Different Types of Half-Cells

Based on the constituents,

electrodes are classified as:

Metal-metal
Gas
soluble salt
electrode
electrode
 Metal–Metal Soluble Salt Electrode

Considering a half-cell reaction,

M
Mn+ (aq)
n+
(aq) + ne
ne‒
‒ M (s) Cell representation

Mn+ (aq) | M (s)

Nernst equation for the half-cell,

RT [M(s)]
EMn+/M = 0
EM n+/M − ln
nF [Mn+(aq)]

2.303RT 1
= 0
EM n+/M − log
nF [Mn+]
 Metal–Metal Soluble Salt Electrode

Similarly for oxidation

B(s) Bm+ (aq) + me‒

2.303RT
= −
0
E B/Bm+ E B/Bm+ log[Bm+ (aq)]
mF

Cell representation: B (s) | Bm+ (aq)

 Metal–Metal Soluble Salt Electrode

Example Zinc half-cell,

Zn2+ (aq) + 2e− Zn (s)

Zinc Rod

Electrode potential

0 0.059 1
EZn2+/Zn = EZn 2+/Zn − 2
log
[Zn2+]

ZnSO4 Solution Cell

Zn2+ (aq) | Zn (s)
representation:
 Gas Electrode

Cl2 gas
Example
electrode

If it
acts as Pt (s) | Cl2 (g) | Cl‒ (aq)
cathode
If it
Cl‒ (aq) | Cl2 (g) | Pt (s) acts as
anode
 Concentration Cells

A concentration cell consists of

two electrodes of the same
material, each electrode dipping in
a solution of its own ions.
Concentration Cells

Current flows in the outer

circuit due to difference in
concentration of solution or
partial pressure of the gases
involved.
 Concentration Cells

Example Cu (s) | Cu2+ (aq., C1) || Cu2+ (aq., C2) | Cu (s)

Two copper electrodes having Cu2+

ions at different concentrations.

C1 & C2 are concentrations

of Cu2+ ions of each half-cell.
 Concentration Cells

Same metal and

its solution
As cathode and anode
consist of same electrode.

0
ECathode = 0
EAnode
 Concentration Cells

0
Ecell for all
concentration cells

0
ECell = E0cathode ─ 0
EAnode

0
ECell = 0.00 Volt
 Concentration Cells

0 0.059 C1 …(1)
Ecell = Ecell − 2
log
C2

Putting values in eq. (1)

0.059 C1
ECell = 0 ─
2
log
C2

0.059 C1
ECell = 2
log
C2
 Electrolytic Cell

Cell reaction:

An electrolytic cell is an 2H2O (l) 2H2 (g) + O2 (g) Ecell = -1.23 V

electrochemical
cell in which a non-
spontaneous reaction
is driven by an external The spontaneous process is actually
source of current. the reverse reaction (i.e., formation
of H2O from H2 and O2)

Converts electrical energy

into chemical energy.
For this at pH = 7, Ecell = 1.23 V
Construction of Electrolytic Cell

Source of current

External Battery

Cathode is attached to negative

terminal of the battery

Reduction occurs at cathode

So, cathode acts as a

negative electrode
 Electrolysis

Electrolysis is a process of
oxidation and reduction due to
current in the electrolytic solution

Electrolyte is a combination
of cations and anions which
in fused state or in aqueous
solution can conduct electricity.
 Construction of Electrolytic Cell

Electrode Half-Reaction
Anode is attached to positive
terminal of the battery Cathode Na+ (l) + e− Na (l)

Anode 2Cl− (l) Cl2 (g) + 2e−

Oxidation occurs at anode

Overall reaction:

2Na+ (l) + 2Cl− (l) 2Na (l) + Cl2 (g)

So, anode acts as a
positive electrode

<
0
Ecell 0V
 Electrolysis of Molten NaCl

+ −

Anode Cathode

Molten NaCl

Battery Electrons
Current
 Electrolysis of Molten NaCl

Product obtained during electrolysis

depends on following factors:

Sodium metal and

chlorine gas are called 1. Amount of charge passed
product of electrolysis into the electrolyte
of molten NaCl.
2. Nature of the electrolyte

3. Concentration the of
electrolyte

4. Nature of the electrode

Faraday’s Laws of Electrolysis

These laws are about the amount

of product of electrolysis (or the
mass of substance produced
or deposited on electrodes).
Faraday’s First Law

The mass deposited/

released/produced of any
substance during electrolysis
is proportional to the
amount of charge passed
into the electrolyte.
 Faraday’s First Law

W ∝ Q If Q = 1C

W = ZQ W = Z

Mass deposited or liberated

W: Mass deposited or liberated Z when 1 C of charge is
Q: Amount of charge passed passed into the solution
Z: Electrochemical equivalent
of the substance

Unit: kg/C or g/C

Electrochemical Equivalent of the Substance (Z)

If Q = 1C

W = Z

Mass deposited or liberated

Z when 1 C of charge is
passed into the solution

Unit: kg/C or g/C

 Equivalent Mass (E)

Mass of any Molar mass

substance produced
when 1 mole of
E = No. of e– involved in oxidation/reduction
electrons are passed
through the solution
during electrolysis. M
= n-factor
 Calculating Equivalent Mass (E)

Al3+
Ag+ (aq) e− Ag (s) 3 3e− Al (s)
1
+ (aq) +

M
E = 1 M
E = 3

2 Cu2+ (aq) + 2e− Cu (s)

M
E = 2
 Faraday’s First Law

We know,
1 C charge E
= g metal deposited
1 mole of e– = 1 F of Charge flow 96500

1F = 96500 C
So, Z =
E
96500

E gram metal
96500 C
charge flow = deposited or Z is the mass deposited or
liberated liberated when 1 C of charge
is passed into the solution.
 Faraday’s First Law

From Faraday’s first law EQ

W = 96500

W = ZQ
M
E= Q=i×t
n-factor

Substituting the value of Z

Molar mass i×t
E
W = n-factor
×
96500
W = 96500
×Q
 Faraday’s Second Law

If equal charge (Q) is passed

through two electrolytic cells and
cells are connected in the series

The mass deposited at electrode will

be in the ratio of their electrochemical
equivalents or in the ratio of the
equivalents masses.
 Faraday’s Second Law

Considering two Since charge passed is

electrolytic cells, 1 & 2 same for both the cells,
connected in series

Applying 1st law,

Q1 = Q2

W1 = Z1 × Q1 W2 = Z2 × Q2

W1 Z1
So,
W2
=
Z2
For two cells
 Faraday’s Second Law

E
As, Z = 96500

E1 E2
Z1 = 96500
& Z2 = 96500

W1 E1
W2 = E2

W1 Z1 E1
W2 = Z2 = E2
 Faraday’s Second Law

W1 Z1 E1
W2 = Z2 = E2
 Nature of the electrolyte

Product of Electrolysis

Electrolyte

Molten form Aqueous form

 Aqueous Electrolyte

When electrolysis of aqueous

solution of electrolyte is carried out,
there may be two or more species
present in the solution which may
compete for reduction at cathode
or oxidation at anode.
 Aqueous Electrolyte

For species competing for For species competing for

reduction at cathode, oxidation at anode,

Higher is the SRP, greater Lower is the SRP (or higher in

will be its ease to undergo SOP) greater will be its ease to
reduction at cathode. undergo oxidation at anode.
 Molten Electrolyte

Electrolysis of molten NaF solution

Electrode Half-reaction E0

Cathode Na+ (l) + e− Na (l) E0 = -2.71 V

Anode 2F− F2 + 2e− E0 = -2.87 V

Product of electrolysis:

Anode F2 Cathode Na
 Aqueous Electrolyte

Electrolysis of aq. NaF solution

Possibilities at cathode:

Na+ (aq) + e− Na (s) E0 = -2.71 V

2H2O (l) + 2e− H2 (g) + 2OH‒ (aq) E0 = -0.83 V

Product of
electrolysis at H2
cathode
 Aqueous Electrolyte

Electrolysis of aq. NaF solution

Possibilities at anode:

2F− (aq) F2 (g) + 2e− Eox

0 = -2.87 V

2H2O (l) O2 + 4H+ + 4e− E0ox = -1.23 V

Product of
O2
electrolysis at anode
 Aqueous Electrolyte

Electrolysis of aq. NaCl solution

Possibilities at cathode:

Na+ (aq) + e− Na (s) E0 = -2.71 V

2H2O (l) + 2e− H2 (g) + 2OH− (aq) E0 = -0.83 V

Product of
electrolysis at H2
cathode
 Aqueous Electrolyte

Electrolysis of aq. NaCl solution

Possibilities at anode:

2Cl− (aq) Cl2 (g) + 2e− E0ox = -1.36 V

2H2O (l) O2 (g) + 4H+ (aq) + 4e− E0ox = -1.23 V

If the oxidation potential of

Product of H2O is more than that of Cl‒,
Cl2
electrolysis at anode
then why Cl2 is liberated?
 Aqueous Electrolyte

According to thermodynamics, To increase its rate, the greater

oxidation of H2O to produce potential difference is applied known
O2 should take place on anode. as over voltage or over potential

However, experimentally the But because of this oxidation

rate of oxidation of water is of Cl– ions also become feasible
found to be very slow. and this takes place on anode.
Concentration of electrolyte

Concentration
of electrolyte

Dilute Concentrated
 Dilute Electrolyte

Electrolysis of aq. H2SO4 solution

Possibilities at cathode:

2H+ (aq) + 2e− H2 (g) E0 = 0.00 V

2H2O (l) + 2e− H2 (g) + 2OH− (aq) E0 = -0.83 V

Product of
electrolysis at H2
cathode
 Dilute Electrolyte

Electrolysis of aq. H2SO4 solution

Possibilities at anode:

2SO42− (aq) S2O82− (aq) + 2e− E0ox = -2.05 V

2H2O (l) O2 (g) + 4H+ (aq) + 4e− E0ox = -1.23 V

Product of
O2
electrolysis at anode
 Concentrated Electrolyte

Electrolysis of highly conc. aq H2SO4 solution

Possibilities at cathode:

2H+ (aq) + 2e− H2 (g) E0 = 0.00 V

2H2O (l) + 2e− H2 (g) + 2OH− (aq) E0 = -0.83 V

Product of
electrolysis at H2
cathode
 Concentrated Electrolyte

Electrolysis of highly conc. aq H2SO4 solution

Possibilities at anode:

2SO42− (aq) S2O82− (aq) + 2e− Eox

0 = -2.05 V

2H2O (l) O2 (g) + 4H+ (aq)+ 4e− Eox

0 = -1.23 V

Product of 2−
S2O8
electrolysis at anode
 Nature of the electrode

Electrode

Inert/ inactive Active

electrode electrode
 Inert Electrode Active Electrode

Electrode that The metal electrodes

conduct electrons in the cell that are
into or out of the cell active, because the
metals themselves are
but cannot take part
components of the
in the half-reactions. half reactions.

e.g. Pt, graphite, etc. E.g. Cu, Ag etc.

Zn – SHE Cell

Inert
electrode
(Pt)
 Conductors

Metallic Electrolytic Metallic Electrolytic

conductors conductors conductors conductors

No transfer of
Charge carriers are Charge carriers Transfer of mass
mass
electrons are ions
Resistance is
because of
Decomposition of Resistance is
No chemical collision of ions
electrolyte takes because of
changes with solvent
place collision of
molecules &
electrons with
because of
fixed metal atom.
interionic force of
attraction.
 Conductors

Metallic Electrolytic Metallic Electrolytic

conductors conductors conductors conductors

Metallic Electrolytic T Electrolytic T Electrolytic

conduction conduction conduction conduction
depends on the depends on the
nature nature of
of metal (e.g. no. electrolyte
of valence e- per (strong/ weak),
metal atom, nature of
crystal structure solvent
etc.), and its
temperature, viscosity,
length & area of temperature
cross section. etc.
 Electrical Resistance (R)

The electrical resistance

of a conductor

Directly Inversely
proportional to
proportional
its area of cross
to its length (l)
section (A).

Cross- sectional
area (A)

Length (l)
 Electrical Resistance (R)
Similarly for electrolytic conductors

l 𝜌 l
R ∝
A
R = A

l l Cell
R = 𝜌
A A constant

l Separation between the

Known as electrodes
resistivity/specific
resistance Overlapping area of
A cross-section of electrodes
dipped in the solution
 Resistivity

Platinised Pt Cross
electrodes sectional A
area (A) 𝜌 = R
l
Length (l)

When l = 1 cm,
A = 1 cm2

𝜌 = R

Connectin
g wires
 Resistivity
Ohm
Unit centimetre
(Ω cm)

C.G.S.
Resistivity of a solution is Ohm meter
defined as the resistance (Ω m)
of the solution between
S.I.
the two electrodes of 1 cm2
area of cross-section
and 1 cm apart.

1Ωm = 100 Ω cm

1 Ω cm = 0.01 Ω m
Interconversion of Units of Resistivity

1Ωm = 100 Ω cm

1 Ω cm = 0.01 Ω m
 Conductance

The inverse of the resistance 1

is known as conductance.It is G = R
represented by G.
l
R= 𝜌 A

A
G = l𝜌
Unit mho or Ω-1 or S (siemens)

S.I.
 Factors Affecting Conductance & Resistance

Solute-solute interactions Solute-solvent interaction

(i) (Inter-Ionic force of attraction) (ii) (Hydration of ions)

Greater the force of attraction, Greater the solvation,

greater will be the resistance. greater will be resistance

1
Force ∝ Charge Solvation ∝ Charge ∝
Size
 Factors Affecting Conductance & Resistance

Solvent-solvent interaction
Hydrated largest (iii)
Li+ (Viscosity):

Hydrated smallest Cs+ Greater the viscosity,

greater will be resistance

Resistance Resistance
of LiCl > of CsCl
Temperature , Resistance
(iv)
Factors Affecting Conductance & Resistance

Nature of electrolyte
(v)

Electrolyte Conductance

Weak Low

Strong High
 Conductivity (𝞳)

The inverse of resistivity is

known as conductivity/
specific conductance (𝞳).

1
𝞳
= 𝜌 G
= 𝞳 × A
l
1 A Gl
G
= 𝜌 × l
𝞳 = A
 Units of Conductivity (𝞳)

Unit S cm-1 / Ω-1 cm-1

Gl Gl l G l2
𝞳 = A
= A
× =
l V
C.G.S.
S m-1 / Ω-1 m-1

When l = 1 cm, V = 1 cm3

S.I.

𝞳 = G
1 S cm-1 = 100 S m-1
 Conductivity (𝞳)

Conductivity is the conductance

of unit volume of solution filled
between two parallel electrodes
at unit distance.

We know,
Accurate measurement
of resistance (R) performed on a
1 l Wheatstone bridge
𝞳 = R × A

Conductivity is computed by
measuring the resistance (R)
of the electrolyte and cell constant.
 Cell Constant (G*)
Cross-sectional
Area (A)

Length (l)
l Using G* and R of the
solution, conductivity of
the cell is determined.

G*
𝞳 = R
Value of cell constant is written
in the description of cell.
 Note

Conductance is Conductivity is additive

1 extensive property. 4 property when l/A is same.

Conductivity is
2 intensive property. 𝞳total = 𝞳1 + 𝞳2 + 𝞳3 + . . . . + 𝞳water

3 𝞳 ∝ Number of ions
G1 × l +
G2 ×
l
+
l A A
Gtotal ×
A = l l
Conductivity or resistivity G3 × + .... + Gwater ×
of the solution is dependent A A
on its concentration.
 Note

Gtotal = G1 + G2 + G3 + .... + Gwater

1 1 1 1 1
Rtotal = R1
+
R2
+
R3
+....+
Rwater

All electrolytes in solution

are connected in parallel.
 Molar Conductivity (∧m)

l = 1 , and solution contains

Conductance of a solution containing
1 mole of an electrolyte between two 1 mol of electrolyte
electrodes which are unit length
apart
∧m = G

Gl
𝞳 = A
∧m = 𝞳 ×V

𝞳A l 𝞳V Volume of solution which

G = l
×
l
= l
2
contain 1 mol of electrolyte
 In C.G.S. Unit

Molarity of
a solution
C mol/L ∧m = 𝞳 ×V

C mol
electrolyte 103 cm3 solution 𝞳 (S cm-1 ) × 1000
dissolves in ∧m (S cm2 mol−1 ) = Molarity

1 mol
electrolyte 1000
cm3 solution
will dissolve C
in
 In S.I. Unit

Molarity of
C mol/L ∧m = 𝞳×V
a solution

C mol 𝞳 (S m-1 )
electrolyte 10‒3 m3 solution ∧m (S m2 mol−1 ) = 1000 × Molarity
dissolves in

1 mol
electrolyte 1000 m3 1 S m2 mol-1 = 104 S cm2 mol-1
will dissolve C solution
in
1 S cm2 mol-1 = 10-4 S m2 mol-1
 Equivalent Conductivity (∧eq)

Normality of
N eq./L
a solution

The conductance of
a solution containing N eq.
1 g-equivalent of electrolyte 103 cm3 solution
the electrolyte dissolves in

1 mol
electrolyte 1000
cm3 solution
will dissolve N
in
 Equivalent Conductivity (∧eq)

∧eq = 𝞳× V
Normality
of N eq./L
𝞳 (S cm-1 ) × 1000 a solution
∧eq (S cm2 eq−1) =
Normality
N eq. 10-3 m3
electrolyte
solution
∧eq = 𝞳 ×V dissolves in

1 eq.
𝞳 (S m-1 ) electrolyte 1 m3
∧eq (S m2 eq−1) = 1000 × Normality
will 1000 N solution
dissolve in
 Relation between 𝝠eq and 𝝠m

Dividing (eq. 2) by (eq. 1)

1S m2 eq-1 = 104 S cm2 eq-1

Λeq M
Λm = N
1S cm2 eq-1 = 10-4 S m2 eq-1

N = M × n-factor
𝞳 (S cm-1) × 1000 …..(eq. 1)
∧m (S cm2 mol-1) =
M Λm
Λeq = n-factor
𝞳 (S cm-1) × 1000
∧eq (S cm2 eq-1) = …..(eq. 2)
N
 Variation of Conductivity and Molar
Conductivity with Concentration

Effect of concentration
Both conductivity change on
& molar conductivity
change with the
concentration of
the electrolyte.
Molar
Conductivity
conductivity
 Variation of Conductivity (𝞳) with Concentration

Both for weak and

strong electrolytes. On dilution, number of ions
per unit volume decreases.

Concentration Conductivity
Also, conductivity depends
upon the number of ions
per unit volume.

Dilution

Conductivity decreases
Since, conductivity on dilution.
is the conductance
per unit volume of
electrolytic solution.
 Variation of Molar Conductivity (∧m) with Concentration

Molar conductance
increases upon dilution
The variation in Λm
with concentration for
Or decreases on increasing
the concentration.

Weak Strong
electrolytes electrolytes
Reason is different for strong
and weak electrolytes
 The Variation in Λm for Weak Electrolyte
At higher
concentration,
Weak electrolytes like acetic
Increase in the
acid have lower degree of
number of ions in
dissociation
solution

Near lower
With dilution of concentration
an ionic solution

At infinite Steep increase in

dilution, α 1 molar conductance
The degree of
dissociation (α) of
the electrolyte
increases
 Plot of Λm vs. √C

For CH3COOH solution

400

Λm (S cm2 mol-1 )
200

0 0.2 0.4
√C (mol/L)1/2
 The Variation in Λm for Strong Electrolytes

Λm increases slowly with In concentrated solution

dilution.

As strong electrolytes Strong attraction

are mostly in ionised state between oppositely
charged ions

No abrupt change in the Slows down the

number of ions with movement of ions
dilution.
 The Variation in Λm for Strong Electrolytes

With dilution,
Decrease in When concentration
distance
interionic approaches zero
between the
attractions
ions increases

The molar conductivity is

Gradual Influence of known as limiting molar
increase in the other ions on conductivity
molar the movement
conductivity of of the ion
the solution decreases
Molar conductivity
at infinite dilution
 Plot of Λm vs. √C
0
The variation in Λm for strong
Λm
= Λm − A√C

electrolyte is given by

400
For KCl solution

Λm (S cm2 mol-1 )
Λm
= Λ 0m − A√C
200 Slope = - A

∧m : Molar conductivity
KCl
0
Λm
∧0m : Limiting molar conductivity
C : Concentration of electrolyte
A : Constant
0 0.2 0.4
√C (mol/L)1/2
 Plot of Λm vs. √C

Value of the constant 'A' for a given

solvent and temperature depends
on the type of electrolyte i.e., the
charges on the cations and anion
produced on the dissociation of
the electrolyte in the solution.
 Types of Electrolytes

Examples

Electrolyte Type

All electrolytes of
a particular type
have the same NaCl 1-1
value of ‘A’.

CaCl2 2-1

MgSO4 2-2
Kohlrausch Law

0
Λm for a strong electrolyte can
be determined by extrapolation
(as variation is linear).
 Kohlrausch Law

0
Λm for weak electrolyte can’t be
determined by extrapolation
(as variation is non-linear).
Λm

0
√C Λm for weak electrolytes is obtained
using Kohlrausch law of
independent migration of ions.
 Kohlrausch’s Law

At 298 K,
Kohlrausch examined ∧0m values
0 0 0 0
for ∧m ‒ ∧m ∧m ‒ ∧m
a number of strong electrolytes
(KCl) (NaCl) = (KBr) (NaBr)

and observed certain regularities.

= ∧0m (KI) ‒ ∧0m(NaI) ≃ 23.4 S cm2 mol‒1

Similarly, it was found that

He noted that the difference in
∧m0 of the electrolytes NaX and ∧0m (NaBr) ‒ ∧0m(NaCl) = ∧0m (KBr) ‒ ∧0m (KCl)
KX for any ‘X’ is nearly constant.

≃ 1.8 S cm2 mol‒1

 Kohlrausch’s Law of Independent Migration of Ions

AxBy solution

At infinite dilution or near zero AxBy (aq) xAy+ (aq) + yBx− (aq)
concentration when dissociation is 100%,
each ion makes a definite contribution
towards molar conductivity of electrolyte 0
irrespective of the nature of the other Λ0m =
0
x λ Ay+ + y λ Bx−
ion.

x and y are the number of

cations and anions, respectively,
after dissociation of the
electrolyte
 Examples

NaCl solution

Limiting molar conductivity of NaCl

Λ0m (NaCl) = Λ0m (Na+ ) + Λ0m (Cl - )

 Examples

CaCl2 solution

CaCl2 (aq) Ca2+ (aq) + 2Cl─ (aq)

0
∧m (CaCl
2)
= ∧0m (Ca2+) + 0
2∧m (Cl- )
 General Case:

y 0 x 0
AxBy (aq) xAy+ (aq) + yBx- (aq) Λ0eq = y
λeq (Ay+) + λeq(Bx-)
x

Λ0m = x λm0 (Ay+) + y λm0 (Bx-)

Λ0eq(A B )
x y
= λ0eq(Ay+) + λ0eq(Bx-)

0
Λm
Λ 0eq = Al2(SO4)3
n-factor
solution

0
1 0 1 0
=
0
+
0
Λ eq = y
λm (Ay+) + x
λm (Bx-) ∧0eq(Al
2(SO4)3)
λeq(Al3+) λ eq(SO 2–)
4
 Applications of Kohlrausch's Law

1
Kohlrausch law is used to evaluate
Λ0 for weak electrolytes.
m
To calculate limiting
molar conductivity
(Λ0m) for weak
electrolytes.
From the individual
values of Λ0m of the ions.
CH3COOH,
NH4OH etc.
Limiting Molar Conductivity of Weak Electrolytes

CH3COOH solution

At infinite dilution,

CH3COOH (aq) CH3COO− (aq) + H+ (aq)

Limiting molar conductivity of CH3COOH,

0 0
∧0m [CH3COOH] = ∧m [CH3COO−] + ∧m [H+]
…(1)
 Limiting Molar Conductivity of Weak Electrolytes

For different electrolytes having

The Λ0m values of strong electrolytes common ion, limiting molar
can be graphically obtained. conductivity of the common ion remains
same for all such electrolytes.

The Λ0m values of weak electrolytes 0

are calculated indirectly by evaluating E.g:- λm (H + ) is same for both the
Λ0m values of selected strong electrolytes, CH3COOH and HCl.
electrolytes.
Limiting Molar Conductivity of Weak Electrolytes

Let’s consider three strong

electrolytes HCl, NaCl, CH3COONa.

0 0
∧m [HCl] = 0
∧m [H+] + ∧0m [Cl-] …(2)

0
∧m [NaCl] = ∧0m[Na+] + ∧0m [Cl-] …(3)

0
∧m [CH3COONa] = ∧0m[CH3COO-] + ∧0m [Na+] …(4)
Limiting Molar Conductivity of Weak Electrolytes

Adding eq. (2) and eq. (4) and subtracting eq. (3)

∧0m (HCl) + 0
∧m (CH3COONa) − ∧0m (NaCl) =

∧0m [H+] + ∧0m [Cl-] + ∧0m[CH3COO-] +

…(5)
∧0m [Na+] − ∧0m[Na+]
− ∧0m [Cl-]

Comparing eq. (1) and eq. (5)

∧0m [CH3COOH] = 0 (HCl)

∧m + ∧0m(CH3COONa) ‒ ∧0m(NaCl)
 Remember!

For different electrolytes having a common

ion, the limiting molar conductivity of the ion
remains the same for all such electrolytes.

0 +
E.g.: λm(H ) is same for both the
electrolytes, CH3COOH and HCl.
Applications of Kohlrausch’s Law

To determine the degree

of dissociation (𝛼) and
dissociation constant ‘Ka’
of a
weak electrolyte
 Degree of Dissociation

Dissociation of weak
electrolyte ‘HA’
∧m ∝ Degree of ionisation
HA ⇌ H+ A‒
+
c
∧m1 𝛼1
t=0 C 0 0 ∧m2
c = 𝛼2

If C2 0 , 𝛼2 1
t = teq C − C𝛼 C𝛼 C𝛼

∧m
where, C: Initial concentration of HA
𝛼 = …(1)
∧m0
 Dissociation Constant (Ka)
Dissociation of weak ‒
electrolyte ‘HA’ [H+] [A ]
So, Ka = [HA]
HA ⇌ H+ + A‒
(C𝛼)(C𝛼)
Ka =
(C − C𝛼)
t=0 C 0 0
C𝛼2
Ka = …(2)
1−𝛼
t = teq C − C𝛼 C𝛼 C𝛼

where,
C: Initial concentration of undissociated electrolyte
𝛼: Degree of dissociation
Dissociation Constant of Weak Electrolyte

∧m
C𝛼2 …(2)
𝛼= …(1) Ka =
∧0m 1−𝛼

Put eq. (1) in eq. (2)

2
∧m
C 2
∧0m ∧m
Ka = ≃ C 0
∧m ∧m
1−
∧0m
 Batteries

Types of batteries

Cell refers to a single

galvanic cell,,Whereas
battery is formed when
two or more cells are Primary Secondary
connected in series or in batteries batteries
series-parallel.

Non-
rechargeable
 Primary Batteries

Primary cells can be used +

only so long the active
materials are present.
Carbon rod
(cathode)
Once these are consumed, the
cell cannot be recharged by MnO2 + carbon
passage of current through it
black + NH4Cl
paste
Hence, it has to be discarded.
Zinc cup
(anode)
Examples: 1. Dry cell or Leclanche cell
̶
2. Mercury Cell
 Dry Cell

Anode Cathode

Cathode:
Carbon (graphite)
rod surrounded by +
Zinc container powdered MnO2 + NH4 + e− MnO(OH) + NH3
manganese dioxide
and carbon.
Anode:
The space between the Zn (s) Zn2+ + 2e−
electrodes is filled by a moist
paste of ammonium chloride
(NH4Cl) and zinc chloride
(ZnCl2).
 Dry Cell

Ammonia produced in
the reaction forms a
complex with Zn2+

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

Cell It decreases
1.5 V
potential over use

Used in flashlights,
portable radios, etc.
 Mercury Cell

Anode cap

Zinc - mercury
Anode
amalgam

Paste of HgO
Cathode
and carbon

Paste of KOH Anode Cathode

Electrolyte
and ZnO
Gasket Separator Cell can
 Mercury Cell

At the
anode: ECell = 1.35 V

Zn(Hg) + 2OH− ZnO (s) + H2O + 2e−

Overall reaction does not

At the cathode:
involve any ion in the solution
whose concentration can
HgO + H2O + 2e− Hg (l) + 2OH− change during its life time.

Overall cell reaction

Used in watches, hearing
aids, calculators, etc.
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)
 Secondary Batteries

Examples

A secondary cell after

Lead storage cell
use can be recharged
by passing current
Nickel-cadmium cell
through it in the
opposite direction so
that it can
be used again.
 Secondary Batteries

Lead storage cell

Examples

Nickel-cadmium
cell
 Lead Storage Cell

Cathode
Lead Anode Anode
̶

Grid of lead
+ Negative
plates: lead
packed with Cathode grids filled
lead dioxide with spongy
(PbO2) lead
Positive plates:
38% solution of lead grids
Electrolyte filled with
sulphuric acid
PbO2
38% sulphuric
acid solution
 Lead Storage Cell

When battery is in use

At the anode:

2−
Pb (s) + SO4 (aq) PbSO4 (s) + 2e−

At the cathode:

2−
PbO2 (s) + SO4 (aq) + 4H+ (aq) + 2e− PbSO4 (s) + 2H2O (l)
 Lead Storage Cell

Overall reaction
Pb (s) + PbO2 (s) + 2H2SO4 (aq) 2PbSO4 (s) + 2H2O (l)
(when in use)

On charging the battery While discharging lead

the reaction is reversed. storage battery

PbSO4 (s) on anode and

If 1 F of electricity is
cathode is converted into
withdrawn from battery,
Pb and PbO2, respectively.
1 mol of H2SO4 is consumed
 Lead Storage Cell

During working of cell

H2SO4
[H2SO4] in Density
will be of solution
solution
consumed
Used in automobiles
(cars/bikes) and
inverters.
During charging of cell

PbSO4 will get converted into Pb (s)

and PbO2 (s) and H2SO4 will be
produced.
 Nickel‒Cadmium Cell

Cadmium Anode

Nickel
dioxide Cathode Negative
(NiO2) plate

Separator

KOH Electrolyte Positive

plate
 Nickel‒Cadmium Cell

Anode:

Cd (s) + 2OH‒ (aq) Cd(OH)2 (s) + 2e‒

Cathode:

NiO2 (s) + 2H2O (l) + 2e‒ Ni(OH)2 (s) + 2OH‒ (aq)

 Nickel‒Cadmium Cell

Overall reaction: Cd (s) + NiO2 (s) + 2H2O (l) Cd(OH)2 (s) + Ni(OH)2 (s)

ECell = 1.4 V

Used in cordless razors,

portable electronics, etc. This cell has comparatively
longer life than a lead storage cell.
 Fuel Cell

Here, the cathode and anode

constituents are continually supplied.
Galvanic cell
that are designed
to convert the energy
of combustion of fuels
like H2, CH4, CH3OH, etc, Thus, energy can be withdrawn
directly into electrical indefinitely from a fuel cell as
energy. long as the outside supply
of fuel is maintained.
 H2‒O2 Fuel Cell

Electrodes Electrolyte H2 gas is fed into one

compartment and the O2 gas
is fed into another
compartment.

Made of a conducting
material, such as Aqueous solution
graphite, with a of a base
sprinkling of platinum These gases diffuse slowly
through the electrodes and
react with an electrolyte that
Acts as is in the central
compartment.
catalyst
H2–O2 Fuel Cell
 H2‒O2 Fuel Cell

At the cathode:

O2 (g) + 2H2O (l) + 4e− 4OH− (aq)

At the anode:

2H2 (g) + 4OH− (aq) 4H2O (l) + 4e−

Overall reaction:

2H2 (g) + O2 (g) 2H2O (l)

 H2‒O2 Fuel Cell Advantages of Fuel Cells

Cell used for providing They offer high energy

electrical power in the
Apollo space program
1 conversions (almost 70%).
Compared to thermal plants
whose efficiency is about 40%.

The water vapour produced during

the reaction were condensed and
added to the drinking water supply
for the astronauts.
2 These cells are pollution-free.
 Corrosion

1 Rusting of iron

Slowly coating
of the surfaces of
metallic objects with
2 Tarnishing of silver
oxides or other
salts of the metal.

Development of green coating

3 on copper and bronze
Tarnishing of Silver Green Coating on Copper
 Rusting of Iron

Formation of porous coating

of iron oxide (Fe2O3.xH2O)
on iron surface in moist air

May be considered as an
electrochemical phenomenon
 Water
Iron surface O2 drop

Fe2+ (aq)

Fe Fe2+ + 2e−

O2 + 4H+ + 4e− 2H2O

 Rusted iron
(Fe2O3.xH2O)
 Anode reaction Cathode reaction

Behaves
Electrons released at
as
anodic spot move
cathodic
through the metal
Fe (s) Fe2+ (aq) 2e− region
+ … (1)

Go to another
spot on the metal
Eox
0
= 0.44 V

Reduce oxygen (of air)

in the presence of H+
 Cathode reaction

O2 (g) + 4H+ (aq) + 4e− 2H2O (l) .… (2)

E0 = 1.23 V H+ available can be

from

H2CO3 formed Dissolution of

due to other acidic
dissolution of oxides from
CO2 from air the
into water atmosphere
 Overall Reaction

Fe (s) Fe2+ (aq) + 2e− …1

O2 (g) + 4H+ (aq) + 4e− 2H2O (l) …2

Overall reaction

2Fe (s) + O2 (g) + 4H+ (aq) 2Fe2+ (aq) + 2H2O (l)

= −
0
Ecell
0 0
Ecathode Eanode

Ecell
0
= 1.23 − − 0.44

0
Ecell = 1.67 V
 Rusting of Iron

Fe2+ is further oxidised by

atmospheric oxygen to Fe3+.
Due to porous coating of rust

This comes out as rust in the

form of hydrated ferric oxide Further attack of
Strength
with moist air occur
of metal
on iron metal
further production of H+ ions.

Fe2O3. xH2O
 Prevention of Corrosion

Prevention

Using paint
Protective Cathodic
or some
coating protection
chemicals
 Prevention of Corrosion

1. Using Paint or Some Chemicals

Covering the surface with paint or

by some chemicals (e.g. bisphenol)
prevent the surface of the metallic
object to come in contact with the
atmosphere. 2. Protective Coating

Cover the surface by other

metals that are inert or
react to save the object.
Galvanisation: Coating of Zn metal
 Atmosphere

Zinc layer

Iron slab (to be protected)

Partial corrosion
of zinc layer
 Complete corrosion of
zinc layer and appearance
of rust on Fe surface
 Prevention of Corrosion

3. Cathodic Protection

In cathodic protection, the object is

connected to a metal with a more
negative standard reduction potential
(E.g.: Mg, E0 = −2.36 V). A block of magnesium
replaced occasionally is much
cheaper than the ship,
Mg acts as a sacrificial anode, building, or pipeline for which
supplying it is being sacrificed.
its own electrons to the iron and
becoming oxidised to Mg2+ in the
process.
Cathodic Protection

Other active
Iron
metal

Cathode Anode