Electrochemistry :

Electrochemistry is a branch of science

which deals with the production of

electricity from energy released during

spontaneous chemical reactions and the

use of electric energy to bring about non

spontaneous chemical transformations.

Conductors :

A conductor is a substance which allows the

passage of electric current.

The two types of conductors are

i) metallic conductors or electronic

conductors

ii) Electrolytic conductors or ionic

conductors

In metallic conductors the conduction of

electricity is due to the flow of electrons. So

they are called electronic conductors. The

electronic conductance depends on

(i) The nature and structure of the metal

(ii) The number of valence electrons per atom.

(iii)Temperature (it decreases with increase in

the temperature)

The composition of the metallic conductors

remain unchanged by the passage of

electric current.

In electrolytic conductors the

conduction is due to the movement of ions,

hence they are called ionic conductors. The

conductivity of electrolytic solution

depends upon

(i) The nature of the electrolyte

(ii) Size of the ions produced and their

solvation.

(iii)The nature of the solvent and its

viscosity.

(iv) Concentration of the electrolyte and

(v) Temperature (increases with increase in

temperature)

On passing a direct current through a

solution of an electrolytic conductor

change in the composition takes place.

Resistance :

Resistance is the opposition to the flow of

current. It is denoted by R and is measured

in ohm ()

lR ∝

Al

R = ρA

Where l is the length and

A is area of cross section

is resistivity

Conductance (G)

Conductance is inverse of resistance

But where k is conductivity or

specific conductance

1 AG = =

R ρl

1= k

ρ

AG = k

l

Conductivity

Conductivity of a solution of an electrolyte

is the conductance of a solution placed

between two electrodes each of one square

meter area kept at a distance of 1 meter

apart.

SI unit for conductivity is Sm-1.

(1 S cm-1 = 100 Sm-1) and S = ohm-1

conductivity of a solution decreases with

decrease in concentration of the solution

due to decrease in the number of ions per

unit volume of the solution.

Cell constant :

For measuring the conductivity of a

solution of an electrolyte conductivity

cell is used. It has two platinum

electrodes each of an area of cross

section of A and kept at a distance of l.

* lG = = Rk

A

cell constantor k =

R

Cell constant

Molar conductivity :

Molar conductivity of a solution at a

given concentration is the conductance

of the volume V of a solution containing

one mole of electrolyte kept between two

electrodes with area of cross section A

and distance of unit length.

It is represented by m

m =kv

SI unit for m is Sm2mol-1

If m is in Sm2mol-1 and k in Sm-1

where C is conc. in mol L-1

When m is in S cm2mol-1 and k is in

Scm-1

m

kλ =

1000C

m

1000kλ =

C

1. Conductivity of 0.01 M NaCl solution is

0.12 Sm-1. Calculate its molar

conductivity

-2 2

m

k 0.12λ = = = 1.2 ×10 Sm / mol

1000C 1000 × 0.01

2. The molar conductivity of 0.1M nitric

acid is 630 S cm2 /mol. Calculate its

conductivity.

m

-1

1000kλ =

C

1000k630 =

0.1

630×0.1∴ k = = 0.063 Scm

1000

Limiting molar conductivity :

Limiting molar conductivity is the molar

conductivity when concentration of the

solution approaches zero or at infinite

dilution. (Represented by om)

Variation of molar conductivity with ion

concentration :

1) For a strong electrolyte : For a strong

electrolyte m increases slowly with

dilution as

m is molar conductivity.

1o 2

m mλ = λ - AC

om is limiting molar conductivity, C

concentration in mol L-1 A is constant

for a given solvent and temperature and

depend on type of the electrolytes(1-1

NaCl, 2-1 CaCl2 and 2-2 MgSO4 etc). A

plot of m verses is a straight line 1

2C

The intercept of the line on the Y axis

gives om for a strong electrolyte. This is

called extrapolation method.

For a weak electrolyte :

Weak electrolytes have low degree of

dissociation at moderate concentrations.

Upon dilution the increase in m is very

steep due to increase in the degree of

dissociation of a weak electrolyte.

The increase in m with dilution is

more when concentration approaches

zero. Hence om can not be obtained by

extrapolating m to zero concentration.

Kohlrausch’s law of independent

migration of ions:

The limiting molar conductivity of an

electrolyte can be represented as the

sum of the individual contributions of

the anion and cation of the electrolyte.

Eg

In general for an electrolyte

+ -

o o o

Na Clλ m(NaCl)= λ + λ

+ -

y+ x -

2+ -

y x

o o ox y A B

o o o2 Ca Cl

AxBy xA + yB

λ m(A B )= xλ + yλ

λ mCaCl = λ + 2λ

The relationship between degree of

dissociation() and limiting molar

conductivity (om) for a weak electrolyte

mo

m

λα =

λ

Where m molar conductivity

at a given concentration

om is the limiting molar

conductivity.

2

2m

o om m m

CαKa =

1 - α

cλKa =

λ (λ - λ )

Calculate the molar conductivity of a

solution of MgCl2 at infinite dilution given

that the molar ionic conductivities of

2+ -

2+ -2

o 2 -1 o 2 -1

(Mg ) (Cl )

o o oMgCl Mg Cl

2 -1

λ = 106.1 Scm mol and λ = 76.3 Scm mol

λ = λ + 2λ

= 106.1 + 2(76.3)

= 258.7 Scm mol

Problem :

The values of limiting molar conductivities

(om) for NH4Cl, NaOH and NaCl are

respectively 149.74; 248.1 and 126.4

Scm2mol-1. Calculate the limiting molar

conductivity of NH4OH

4 4

o o o oNH OH NH Cl NaOH NaClλ = λ + λ - λ

= 149.74+248.1-126.4

= 271.44 Scm2 mol-1

2 faradys first law Faraday’s first law of electrolysis:

The amount of chemical reaction which

occurs at any electrode during electrolysis

by a current is proportional to the quantity

of electricity passed through the electrolyte

either through its aqueous solution or

molten state.

2 faradys first law If w is the mass of the substance deposited

and Q is the current passed in coulombs

w Q

But Q = I t where I is the current strength

in ampere and t is time in seconds.

2 faradys first law Faraday:

A Faraday is the charge of one mole of

electrons in coloumbs. It is represented

by F.

1 Faraday = Avogadro No. × Charge of one

electron in Coloumbs

2 faradys first law

= 6.02×1023×1.6021×10-19C

= 96487C 96500 C

1 Faraday of electric current is required to

reduce 1 mole of cations with unit positive

charge eg to reduce 1 mole of silver ions

to get 1 mole of silver i.e, 108g of Ag

96500 C of electricity is required

2 faradys first law

+ -Ag(aq)+ e Ag

1mol. 96500C 108g

To reduce 1 mole of cations with two unit

positive charge 2 Faraday of current is

required.

Eg to reduce 1 mole of Mg2+ ions to get 1

mole i.e, 24g of Mg 2F i.e, 2×96500C of

2 faradys first law

Electricity is required.

In general to reduce 1 mole of Mn+ cation nF

of electric current is required

2+ -(aq)Mg + 2e Mg

1 mole 2 × 96500C 24g

n+ -(aq)M + ne M

nF

2 faradys first law Faraday’s second law of electrolysis:

The amounts of different substances

liberated by the same quantity of electricity

passing through the electrolytic solution

are proportional to their chemical

equivalent weights.

2 faradys first law

Chemical equivalent weight

Eg

Atomic mass=

Number of electrons required to reduce the cation

+ -

(aq)Ag + e Ag

At mass∴ Eq weight of Ag =

1

2 faradys first law 2+ -

(aq)Mg + 2e Mg

At wtEq weight of Mg =

2

3+ -(aq)Al + 3e Al

At wtEq weight of Al =

3

2 faradys first law

Problems:

1) A solution of Ni(NO3)2 is electrolysed

between platinum electrodes using a

current of 5 amperes for 20 minutes.

What mass of nickel is deposited at the

cathode? (Mol mass of Ni = 58.7)

2 faradys first law

Solution:

Q = I t

= 5×20×60 = 6000C

2+ -Ni + 2e Ni

2 × 96500C

193000C 58.7g

2 faradys first law

For 193000C of electricity mass of nickel

obtained = 58.7g

For 6000C of electricity 6000 × 58.7

= 1.812g193000

2 faradys first law 2) How long it will take for the deposition of

0.2g of silver when silver nitrate solution

is electrolysed using 0.5 ampere of

current (Mol mass of Ag = 108)

2 faradys first law

+ -Ag + e Ag

96500C 108g

For 108g of silver to be deposited current

required is 96500C.

For 0.2g of Ag

But Q = I t

0.2 × 96500= 178.7C = Q

108

Q 178.7t = = = 357.4 sec

I 0.5

It is a device in which chemical energy of

spontaneous redox reaction is converted

into electrical energy.

An electrochemical cell consists of

two half cells or electrodes

Electrochemical Cell or Galvanic cell

Electrode – An electrode is a metal in contact

with a solution containing its own ions

e.g Zn (s) / Zn2+ ( aq) or

Cu(s)/Cu2+ (aq)

Non metallic electrodes may also be

prepared in presence of a inert metal like

Platinum which provides surface for the

the conduction of electrons.

Eg Pt (s) H2 (g) H+ (aq) or

Pt (s) Br2 (aq) Br- (aq)

oxidation or reduction and for

Electrode Potential

The potential difference developed between

the electrode (metal) and the electrolyte

(solution containing its own ions) when both

the metal and the solution are in equilibrium

is called electrode potential. It is expressed in

terms of volt.

When coupled with SHE, if reduction

takes place at the given electrode the

electrode

Nernst Equation :

Oxidised state reduced state

reductionn+ -(aq) (s)M + ne M

n+ n+

o

(M / M) (M / M)

RT [reduced state]E = E - ln

nF [oxidised state]

n+

o

n+(M / M)

RT [M]= E - ln

nF [M ]

n+

o10 n+(M / M)

2.303RT [M]= E - log

nF [M ]

Substituting the values R= 8.314JK-1mol-1

T =298K

F = 96500C

and [M] = 1

n+ n+

o10 n+(M / M) (M / M)

0.059 1E = E - log

n M

where E is the electrode potential

(reduction)

Eo standard electrode potential or

standard reduction potential (SRP)

n=number of electrons exchanged.

Standard Electrode potential :

Standard electrode potential is the

electrode potential when the

concentrations of all the species

involved is unity (1M) and if a gas is

involved its pressure should be 1 bar.

Standard Hydrogen Electrode SHE

It consists of a platinum electrode coated

with platinum black. The electrode is

dipped in 1M HCl. Pure hydrogen gas is

bubbled through it under a pressure of 1

bar. S.H.E is represented as

Pt(s) H2 (g)(1bar) H+

(aq)(1M)

The reduction reaction taking place is

S.H.E is assigned an electrode potential

of 0.0 V at all temperatures.

2

1( ) ( )

2H aq e H g

Electrochemical cell

Daniel cell

To prepare Daniel cell get a zinc electrode

by dipping zinc rod in 1M ZnSO4 solution.

Get a copper electrode by dipping a copper

plate in 1 M CuSO4 solution. Couple these

two electrodes using a salt bridge to get

Daniel cell

Reactions taking place

2

2

2 2

( ) ( ) ( ) ( )

t anode 2

cathode 2

oxdn

redn

s aq aq s

A Zn Zn e

At Cu e Cu

Net cell reaction Zn Cu Zn Cu

Zinc rod is the –ve pole and copper plate is

the +ve pole. Flow of electrons is from Zinc

to copper but the flow of current is from

copper to zinc.

Daniel cell can be represented as

Zn/ Zn2+ (aq) Cu2+

(aq) /Cu

EMF of Daniel cell Eo cell = EoR - E

oL

= EoCu - E

oZn = 0.34-(-0.76)

= 1.10V

If an external opposite potential is

applied in the Daniel cell and E ext is

increased slow

i) If E ext < 1.10 V

Zinc rod is the –ve pole, copper is the

positive pole, electrons flow from zinc to

copper and the current flows from copper

to zinc.

2) If E ext = 1.10V: No reaction takes place

and there is no flow of electrons or current.

3) If E ext > 1.10V

All the reactions taking place are

reversed. Zinc is deposited at the zinc

electrode and copper dissolves at

copper electrode. Electrons flow from

copper to zinc and current flows from

zinc to copper. Electrochemical cell

becomes an electrolytic cell.

Cell potential :

Cell potential is the potential difference

between the two electrodes of the galvanic

cell.

Cell electromotive force(emf) :

It is the difference between the electrode

potential of the cathode and anode when

no current is drawn through the cell.

E cell = E right – E left

= E - E

Eocell = Eo

R – EoL

e.g Eo Daniel cell = EoR – Eo

L

Of the electrode

undergoing

reduction

Of the electrode

undergoing

oxidation

2 + 2+/Cu /Cu

o o

Cu ZnE - E = 0.34 -(-0.76)=1.10V

Use SHE in the determination of SRP of a

given electrode

Construct a standard electrode of

the given metal by dipping the pure metal

in 1M solution of its own ion at 25o C

Couple this standard electrode with SHE

using a salt bridge to get galvanic cell.

Measure the emf of the cell using suitable

instrument like potentiometer.

Eo = EoR – Eo

L

One of the electrodes of the cell is SHE

and its electrode potential is 0.0V. So the

electrode potential of the given electrode

will be the emf of the cell in magnitude.

If reduction takes place at the given

electrode its Eo will be +ve but if oxidation

takes place at the given electrode is Eo will

be –ve.

Calculation of the emf of the cell for any

given concentration of the electrolytes

of the electrodes:

Consider Daniel cell with given

concentrations of the electrolyte

2 2 //

cell ZnCu Cu ZnEMF or E E E

2 22 2/ /

2.303 1 2.303 1log logo o

Cu Cu Zn Zn

RT RTE E

nF nFCu Zn

2 2

2

10 2/ /

2.303logo o

Cu Cu Zn Zn

ZnRTE E

nF Cu

Substituting R=8.314JK-1mol, T=298K

F=96500C;n=2

2+ 2+

2+

o o

2+Cu / Cu Zn / Zn

Zn0.059= E - E - log

2 Cu

2

2

0.059log

2

o

cell

ZnE

Cu

2

2

0.0591.10 log

2

Zn

Cu

In Daniel cell when [Zn2+] = [Cu2+]

Ecell =1.10V

if [Zn2+]>[Cu2+] Ecell<1.10v

if [Zn2+]<[Cu2+] Ecell<1.10v

For a general electrochemical reaction

of the type

neaA bB cC dD

ln

c d

o

cell cell a b

C DRTE E

nF A B

When Daniel cell is discharged

completely E cell =0 and

the equilibrium constant

2

2 e

ZnK

Cu

2+

ocell cell 2+

Zn2.303RTE = 0 = E - log

nF Cu

2

2

2.303logo

cell

ZnRTE

nF Cu

2.303logo

cell c

RTE K

nF

0.059logo

cell cE Kn

Problem 1. Calculate the emf of the cell

in which the following reaction takes

place.

2

( ) ( )2 (0.002 ) (0.160 ) 2s sNi Ag M Ni M Ag

Given that Eocell = 1.05V

2 2

( )

10 2

( )

[ ][ ]0.059log

2 [ ][ ]

so

cell cell

s

Ni AgE E

Ni Ag

But [M] for any element is taken as unity

= 0.914V

2+

ocell cell 10 2+

Ni0.059E = E - log

2 Ag

2

0.059 0.160= 1.05 - log

2 0.002

2. Calculate the equilibrium constant for

the reaction at 298K

Given that Eo Ag+/Ag = 0.80V and

Eo(Cu2+/Cu) = 0.34V

Solution :

2

( ) ( ) ( )2 ( ) 2s aq sCu Ag aq Cu Ag

log0.059

o

c

nE cellK

0.059logo

cell cE Kn

2( / ) ( / )

o o o

cell Ag Ag Cu CuE E E

=0.80-0.34=0.46V 2 0.46

log 15.590.059

cK

Taking the antilog Kc =3.92×1015

Relationship between cell potential and

Gibbs Energy:

o o

r cellG nFE

o

rG Is the Gibb’s energy

n = no. of electrons exchanged

in the net cell reaction

F= 96500C

Eo cell is standard emf of the cell.

But = -RT lnK

-nFEo = - RT lnK

o

rG

lno RTE K

nF

2.303log

RTK

nF

0.059logoE K

n

Problems: The cell in which the following

reaction occurs

Has Eocell = 0.236V at 298K. Calculate the

standard Gibb’s energy and the

equilibrium constant for the cell reaction.

3 2

( ) ( ) ( ) 2( )2 2 2aq aq aq sFe I Fe I

Solution : n = 2

.Go = -nFEo

= - 2×96500×0.236

= - 45548 J

0.059logcellE K

n

0.0590.236 log

2K

2 0.236log 8

0.059K

Taking the antilog K = 108

Electrolysis:

Electrolysis is the chemical reactions

taking place when a direct current is

passed through the aqueous solution or

molten state of an electrolyte.

The product of electrolysis

depends on

(i) Nature of the electrolyte (Aqueous

solution or molten state)

(ii) Types of the electrodes

(iii)Concentration of the solution

1) eg When aqueous solution of copper

sulphate is electrolysed using copper

electrodes

2

( ) ( )

2

( ) ( )

t anode 2

t cathode 2

oxdn

s aq

redn

aq s

A Cu Cu e

A Cu e Cu

Thus copper from anode dissolves and an

equivalent amount of pure copper is

deposited on cathode. This technique is

used in electrolytic refining of crude

copper.

2) When molten sodium chloride is

electrolysed using inert electrodes

2t anode 2 2

oxdn

redn

A Cl Cl e

At cathode Na e Na

Thus chlorine gas is liberated at anode

and Sodium metal is formed at cathode.

3) When aqueous solution of NaCl is

electrolysed.

2

NaCl Na Cl

H O H OH

At cathode there is a competition

between the following reduction

reactions

At cathode the reaction with higher value

of Eo is preferred. Therefore the reaction

taking place at cathode is

( )

2( )

2.71

1 0.02

o

aq s cell

o

aq g cell

Na e Na E V

H e H E V

2( )

1 2

aq gH e H

But H+ ions are produced by dissociation

of H2O. So the net reaction taking place

at cathode is

2 ( ) 2( )

1 2

l gH O e H OH

The possible reactions at anode are

- - oaq 2 aq cell

+ - o2 (l) 2 cell

1Cl Cl + e E = 1.36V

2

2H O O + 4H + 4e E = 1.23V

At anode the reaction with lower Eo is

preferred. But on account of over potential

of oxygen oxidation of Cl- ions takes place.

Thus when aqueous solution of NaCl

is electrolysed hydrogen is liberated at

cathode, chlorine at anode and NaOH

solution is formed

Batteries

A battery may have one or more than one

galvanic cells connected in series. A

battery should be of practical use,

reasonably light and its voltage should

not change very much during the use.

Batteries are divided into two types;

(i) Primary battery in which reaction occurs

only once and cannot be recharged. Eg

Dry cell or Leclanche cell and Mercury

cell

(ii) Secondary battery which can be

recharged by passing current through it

in opposite direction,so that it can be

Reused.

Eg Lead storage battery and Nickel

cadmium cell.

Dry cell or Leclanche cell

It consists of a zinc container as an

anode. A graphite rod surrounded by a

mixture of manganese dioxide and

carbon powder is cathode.

The space between the electrodes is

filled with electrolyte a moist paste of

ammonium chloride and zinc chloride

Reaction taking place

NH3 produced in the reaction forms a

2+ -(s)

+ -2 4 3

At anode Zn Zn + 2e

At cathode MnO + NH + e MnO(OH)+ NH

complex with Zn2+ to form [Zn(NH3)4]2+.

Zinc container is the –ve pole and the

graphite rod in the middle is the +ve pole.

Cell potential is 1.5V

Lead storage Battery :

It consists of lead anode and a grid of

lead packed with lead dioxide (PbO2) as

cathode.

Electrolyte is 38% solution of sulphuric

acid.

The reactions taking place when the

battery is in use are

2

( ) 4 ( ) 4( )

2

2 4 ( ) ( ) 4 ( ) 2 ( )

2

( ) 4 2 2

s aq s

aq aq s l

Anode Pb SO PbSO e

Cathode PbO s SO H e PbSO H O

The overall reaction is

( ) 2 ( ) 2 4 ( ) 4 ( ) 2 ( ) 2 2 2 s s aq s lPb PbO H SO PbSO H O

On charging the battery the entire

reaction is reversed.

Fuel cells

Galvanic cells that are designed to

convert the energy of combustion of

fuels like hydrogen, methane etc

directly into electrical energy are called

fuel cells.

Eg Hydrogen-Oxygen Fuel cell.

In this hydrogen and oxygen gases are

bubbled through porous carbon

electrodes into concentrated aqueous

sodium hydroxide solution. Catalyst like

finely divided platinum or palladium is

incorporated into the electrodes for

increasing the rate of electrode reaction

Reaction taking place are

Overall reaction is

This is the most common cell used in

Apollo space programme to get electricity

- -2 (g) 2 (l) (aq)

- -2(g) (aq) 2 (l)

Cathode O + 2H O + 4e 4OH

Anode 2H + 4OH 4H O + 4e

2 (g) 2 (g) 2 (l)2H + O 2H O

So that the bi-product water formed is

used as drinking water after distillation

for the astronauts.

Fuel cells are highly efficient (about 70%)

pollution free and is continuous source

of energy as long as fuel is supplied.

Corrosion:

When a metal is exposed to the

atmosphere it is slowly attacked by the

constituents of the environment as a

result of which the metal is slowly lost in

the form of its compound . This is called

corrosion.

e.g tarnishing of silver, development of a

bluish green coating on copper etc.

corrosion for iron is called rusting.

Rusting of iron is an electrochemical

phenomenon. Air and moisture are

essential for corrosion. Presence of

electrolytes help rusting. A tiny

electrochemical cell is set up on the surface

of iron where a small droplets of water sit on.

Reaction taking place are

H+ are produced from H2CO3 formed due to

dissolution of carbon dioxide from air into

water

2+ -(s)

-2 (g) 2 (l)

At Anode 2Fe 2Fe + 4e

At Cathode O + 4H +(aq)+ 4e 2H O

The Fe2+ ions are further oxidised by

atmospheric oxygen to ferric ion which

are ultimately converted to hydrated

ferric oxide called rust(Fe2O3.xH2O)

Rusting may be prevented by

barrier protection like painting, metal

plating etc. Underground iron pipes are

prevented from rusting by sacrificial or

cathodic protection by connecting the

iron pipe with pieces of more

electropositive metals like Mg. Mg gets

corroded first keeping iron pipe safe.

This is corroded magnesium pieces are

to replaced by new pieces timely.

Problems:

1. The resistance of a conductivity cell

containing 0.001 M KCl solution at

298K is 1500. What is the cell

constant if the conductivity of 0.001M

KCl solution at 298K is 0.146×10-3

Scm-1?

Cell constant G*= Rk

=resistance × conductivity

=0.146×10-3 Scm-1×1500S-1

= 0.219 cm-1

2. A conductivity cell when filled with

0.01M KCl has a resistance of 747.5

ohm at 25oC. When the same cell was

filled with an aqueous solution of

0.05M CaCl2 solution the resistance

was 876 ohm. Calculate

(i) Conductivity of the solution

(ii) Molar conductivity of the solution

(given conductivity of 0.01M KCl =

0.14114 sm-1)

Cell constant G* = Rk

= 747.5×0.14114

=0.105.5m-1

-1-1cell constant 105.5m

Conductivity k = = = 0.1204SmR 876 ohm

2 -1m

k 0.1204Molar conductivity λ = = = 0.00241sm mol

1000C 1000 × 0.05

3. The electrical resistance of a column

of 0.05M NaOH solution of diameter

1cm and length 50cm is 5.55×103 ohm.

Calculate its

(i) resistivity

(ii) conductivity

(iii) molar conductivity

Cell constant

l = 50 cm

Diameter = 1 cm radius = 0.5 cm

Area of cross section A = r2 = 3.14×(0.5)2

= 0.785 cm3

o lG =

a

* -150G = = 63.694 cm

0.785

-2

1 1Resistivity ρ = = = 87.135 Ω

k 1.148 ×10

m

-2

2 -1

1000kMolarconductivityλ =

C

1000 ×1.148 ×10 =

0.05

= 229.6 S cm mol