Basic concepts of electrochemistry, Formulas and forms for Computer science. De Nicola Piove di Sacco
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Basic concepts of electrochemistry, Formulas and forms for Computer science. De Nicola Piove di Sacco

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1. ELECTROCHEMISTRY

Electrochemistry is the study of production of electricity

from the energy released during a spontaneous chemical

reaction and the use of electrical energy to bring about

non-spontaneous chemical transformations.

2. ELECTROCHEMICAL CELLS

A spontaneous chemical process is the one which can

take place on its own and in such a process the Gibb’s

energy of the system decreases. It is this energy that

gets converted to electrical energy.The reverse process

is also possible in which we can make non-spontaneous

processes occur by supplying external energy in the form

of electrical energy. These inter conversions are carried

out in equipments called Electrochemical Cells.

3. TYPES

Electrochemical Cells are of two types:

3.1 Galvanic Cells

Converts chemical energy into electrical energy

3.2 Electrolytic Cells

Converts electrical energy into chemical energy.

4. GALVANIC CELL

Cell energy is extracted from a spontaneous chemical

process or reaction and it is converted to electric current.

For example, Daniell Cell is a Galvanic Cell in which Zinc

and Copper are used for the redox reaction to take place.

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

Oxidation Half : Zn (s) Zn2+ (aq) + 2e–

Reduction Half : Cu2+(aq) + 2e– Cu(s)

Zn is the reducing agent and Cu2+ is the oxidising

agent.The half cells are also known as Electrodes. The

oxidation half is known as Anode and the reduction half is

called Cathode. Electrons flow from anode to cathode in

the external circuit. Anode is assigned negative polarity

and cathode is assigned positive polarity. In Daniell Cell,

Zn acts as the anode and Cu acts as the cathode.

5. ELECTROLYTIC CELL

These electrodes are dipped in and electrolytic solution

containing cations and anions. On supplying current the

ions move towards electrodes of opposite polarity and

simultaneous reduction and oxidation takes place.

5.1 Preferential Discharge of ions

Where there are more than one cation or anion the process

of discharge becomes competitive in nature. Discharge

of any ion requires energy and in case of several ions

being present the discharge of that ion will take place first

which requires the energy.

6. ELECTRODE POTENTIAL

It may be defined as the tendency of an element, when it is

placed in contact with its own ions to either lose or gain

electrons and in turn become positively or negatively charged.

The electrode potential will be named as oxidation or

reduction potential depending upon whether oxidation or

reduction has taken place.

oxidation n

reduction M s M aq ne

 

reductionn

oxidation M aq ne M s  

6.1 Characteristics

(a) Both oxidation and reduction potentials are equal in

magnitude but opposite in sign.

(b) It is not a thermodynamic property, so values of E are

not additive.

7. STANDARD ELECTRODE POTENTIAL (EO)

It may be defined as the electrode potential of an electrode

determined relative to standard hydrogen electrode under

standard conditions. The standard conditions taken are :

7. ELECTROCHEMISTRY

Mahesh Tutorials Science

(i) 1M concentration of each ion in the solution.

(ii) A temperature of 298 K.

(iii) 1 bar pressure for each gas.

8. ELECTROCHEMICAL SERIES

The half cell potential values are standard values and are

represented as the standard reduction potential values

as shown in the table at the end which is also called

Electrochemical Series.

9. CELL POTENTIAL OR EMF OF A CELL

The difference between the electrode potentials of two

half cells is called cell potential. It is known as electromotive

force (EMF) of the cell if no current is drawn from the cell.

E cell = E

cathode + E

anode

For this equation we take oxidation potential of anode and

reduction potential of cathode.

Since anode is put on left and cathode on right, it follows

therefore,

= E R + E

L

For a Daniel cell, therefore

2 2

o o o

cell Cu / Cu Zn / Zn E E E 0.34 0.76 1.10V     

10. CELL DIAGRAM OR REPRESENTATION

OF A CELL

The following conventions or notations are applied for writing

the cell diagram in accordance with IUPAC recommendations.

The Daniel cell is represented as follows :

Zn(s) | Zn2+ (C 1 ) || Cu2+ (C

2 ) | Cu (s)

(a) Anode half cell is written on the left hand side while

cathode half cell on right hand side.

(b) A single vertical line separates the metal from aqueous

solution of its own ions.

2 2

Cathodic chamberAnodic chamber

Zn s | Zn aq ; Cu aq | Cu s  

(c) A double vertical line represents salt bridge

(d) The molar concentration (C) is placed in brackets after

the formula of the corresponding ion.

(e) The value of e.m.f. of the cell is written on the extreme

right of the cell. For example,

Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu EMF = +1.1V

(f) If an inert electrode like platinum is involved in the

construction of the cell, it may be written along with the

working electrode in bracket say for example, when a zinc

anode is connected to a hydrogen electrode.

2

1 2 2 Zn s | Zn C || H C | H | Pt s 

11. SALT BRIDGE

Salt bridge is used to maintain the charge balance and to

complete the circuit by facilitating the flow of ions through it.

It contains a gel in which an inert electrolyte like Na 2 SO

4 or

KNO 3 etc are mixed. Negative ions flow to the anode and

positive ions flow to the cathode through the salt bridge and

charge balance is maintained and cell keeps on functioning.

12. SPONTANEITY OF A REACTION

G = – nFE CELL

For a spontaneous cell reaction ΔG should be negative

and cell potential should be positive.

If we take standard value of cell potential in the above

equation we will obtain standard value of ΔG as well.

ΔGo = – nFE0 CELL

13. TYPES OF ELECTRODES

13.1 Metal-Metal Ion electrodes

A metal rod/plate is dipped in an electrolyte solution

containing metal ions. There is a potential difference

between these two phases and this electrode can act as a

cathode or anode both.

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ELECTROCHEMISTRY 213

214 ELECTROCHEMISTRY

Mahesh Tutorials Science

Anode: M Mn+ + ne–

Cathode: Mn+ + ne– M

13.2 Gas Electrodes

Electrode gases like H 2 , Cl

2 etc are used with their respective

ions. For example, H 2 gas is used with a dilute solution of HCl

(H+ ions). The metal should be inert so that it does not react

with the acid.

Anode: H 2

2H+ + 2e–

Cathode: 2H+ + 2e– H 2

The hydrogen electrode is also used as the standard to

measure other electrode potentials. Its own potential is

set to 0 V as a reference. When it is used as a reference

the concentration of dil HCl is taken as 1 M and the

electrode is called “Standard Hydrogen Electrode (SHE)”.

13.3 Metal-Insoluble salt electrode

We use salts of some metals which are sparingly soluble

with the metal itself as electrodes. For example, if we use

AgCl with Ag there is a potential gap between these two

phases which can be identified in the following reaction:

AgCl(s) + e– Ag(s) + Cl–

This electrode is made by dipping a silver rod in a solution

containing AgCl(s) and Cl– ions.

13.4 Calomel Electrode

Mercury is used with two other phases, one is a calomel

paste (Hg 2 Cl

2 ) and electrolyte containing Cl– ions.

Cathode :

Hg 2 Cl

2 (s) + 2e– 2Hg(l) + 2Cl–(aq)

Anode :

2Hg(l) + 2Cl–(aq) Hg 2 Cl

2 (s) + 2e–

This electrode is also used as another standard to measure

other potentials. Its standard form is also called Standard

Calomel Electrode (SCE).

13.5 Redox Electrode

In these electrodes two different oxidation states of the

same metal are used in the same half cell. For example,

Fe2+ and Fe3+ are dissolved in the same container and an

inert electrode of platinum is used for the electron transfer.

Following reactions can take place:

Anode: Fe2+ Fe3+ + e–

Cathode: Fe3+ + e– Fe2+

14. NERNST EQUATION

It relates electrode potential with the concentration of ions.

Thus, the reduction potential increases with the increase

in the concentration of ions. For a general electrochemical

reaction of the type.

neaA bB cC dD 

  

Nernst equation can be given as

c d

o

cell cell a b

C DRT E E n

nF A B l 

ELECTROCHEMISTRY 215

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c d

o

cell cell a b

C D2303 E E RT log

nF A B  

Substituting the values of R and F we get

c d

o

cell cell a b

C D0.0591 E E log ,at 298 K

n A B  

15. APPLICATIONS OF NERNST EQUATION

15.1 Equilibrium Constant from Nernst Equation

For a Daniel cell, at equilibrium

2

o

cell cell 2

Zn2.303RT E 0 E log

2F Cu





      

or

2

o

cell 2

Zn2.303RT E log

2F Cu





    

But at equilibrium,

2

c2

Zn K

Cu





    

o

cell c

2.303RT E log K

2F 

o

cell c

2.303 8.314 298 E log K

2 96500



c

0.0591 logK

2 

In general, o

cell c

0.0591 E log K

n 

or, log K C =

o

celln E

0.0591

16. CONCENTRATION CELLS

If two electrodes of the same metal are dipped separately

into two solutions of the same electrolyte having different

concentrations and the solutions are connected through

salt bridge, such cells are known as concentration cells.

For example

H 2 | H+(c

1 ) || H+ (c

2 ) | H

2

Cu | Cu+2 (c 1 ) || Cu2+(c

2 ) | Cu

These are of two types :

16.1 Electrode concentration cells

H 2 (P

1 ) | H+ (C) || H+ (C) | H

2 (P

2 )

E cell = 0 – 2

1

P0.059 log

n P

where 2 1p p for spontaneous reaction

16.2 Electrolyte concentration cell

The EMF of concentration cell at 298 K is given by

Zn | Zn2+ (c 1 ) || Zn2+ (c

2 ) | Zn

2

cell

1 1

c0.0591 E log ,

n c 

where c 2 > c

1 for spontaneous reaction

17. CASES OF ELECTROLYSIS

17.1 Electrolysis of molten sodium chloride

2NaCl (l )  2Na + (l ) + 2Cl– (l )

The reactions occurring at the two electrodes may be

shown as follows :

At cathode :

2Na+ + 2e–  2Na E0 = – 2.71 V

At anode :

2Cl–  Cl 2 + 2e– E0 = – 1.36V

Overall reaction :

2Na+ (l ) + 2 Cl– (l ) 2Na (l ) + Cl 2 (g)

or 2NaCl (l ) 2Na (l ) + Cl 2 (g)

At cathode At anode

216 ELECTROCHEMISTRY

Mahesh Tutorials Science

17.2 Electrolysis of an aqueous solution of sodium chloride

NaCl (aq) Na+(aq) + Cl– (aq)

(almost completely ionized)

H 2 O (l )  H

+(aq) + OH–(aq)

(only slightly ionized)

At cathode :

2Na+ + 2e– 2Na E0 = – 2.71V

2H 2 O + 2e– H

2 + 2OH– E0 = – 0.83V

Thus H 2 gas is evolved at cathode value Na+ ions remain

in solution.

At anode :

2H 2 O O

2 + 4H+ + 4e– E0 = – 1.23V

2Cl– Cl 2 + 2e– E0 = – 1.36V

Thus, Cl 2 gas is evolved at the anode by over voltage

concept while OH– ions remain in the solution.

18. BATTERIES

When Galvanic cells are connected in series to obtain a

higher voltage the arrangement is called Battery.

18.1 Primary Batteries

Primary cells are those which can be used so long the

active materials are present. Once they get consumed the

cell will stop functioning and cannot be re-used. Example

Dry Cell or Leclanche cell and Mercury cell.

18.2 Dry cell

Anode : Zn container

Cathode : Carbon (graphite) rod surrounded by powdered

MnO 2 and carbon.

Electrolyte : NH 4 Cl and ZnCl

2

Reaction :

Anode : Zn Zn2+ + 2e–

Cathode : MnO 2 + 

4NH  e – MnO (OH) + NH

3

The standard potential of this cell is 1.5 V and it falls as

the cell gets discharged continuously and once used it

cannot be recharged.

18.3 Mercury cells

These are used in small equipments like watches, hearing aids.

Anode : Zn – Hg Amalgam

Cathode : Paste of HgO and carbon

Electrolyte : Paste of KOH and ZnO

Anode : Zn (Hg) + 2OH– ZnO (s) + H 2 O + 2e–

Cathode : HgO (s) + H 2 O + 2e– Hg (l) + 2OH–

Overall Reaction : Zn (Hg) + HgO (s) ZnO (s) + Hg (l)

The cell potential is approximately 1.35V and remains

constant during its life.

18.4 Secondary Batteries

Secondary cells are those which can be recharged again

and again for multiple uses. e.g. lead storage battery and

Ni – Cd battery.

18.5 Lead Storage Battery

Anode : Lead (Pb)

Cathode : Grid of lead packed with lead oxide (PbO 2 )

Electrolyte : 38% solution of H 2 SO

4

Discharging Reactions

Anode: Pb(s) + SO 4

2–(aq) PbSO 4 (s) + 2e–

Cathode: PbO 2 (s) + 4H+(aq) + SO

4

2–(aq) + 2e–

PbSO 4 (s) + 2H

2 O(l)

Overall Reaction : Pb(s) + PbO 2 (s) + 2H

2 SO

4 (aq)

2PbSO 4 (s) + 2H

2 O(l)

To recharge the cell, it is connected with a cell of higher

potential and this cell behaves as an electrolytic cell

and the reactions are reversed. Pb(s) and PbO 2 (s) are

regenerated at the respective electrodes.These cells

deliver an almost consistent voltage.

Recharging Reaction : 2PbSO 4 (s) + 2H

2 O(l) Pb(s) +

PbO 2 (s) + 2H

2 SO

4 (aq)

19. FUEL CELLS

A fuel cell differs from an ordinary battery in the sense

that the reactants are not contained inside the cell but are

ELECTROCHEMISTRY 217

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externally supplied from an external reservoir. Fuel cell is

used in space vehicles and in this cell the two gases are

supplied from external storages. In this cell carbon rods

are used as electrodes with KOH as the electrolyte.

Cathode : O 2 (g) + 2H

2 O (l) + 4e– 4OH– (aq)

Anode : 2H 2 (g) + 4OH– (aq) 4H

2 O (l) + 4e–

overall Reaction : 2H 2 (g) + O

2 (g) 2H

2 O (l)

20. CORROSION

It involves a redox reaction and formation of an

electrochemical cell on the surface of iron or any other metal.

At one location oxidation of iron takes place (anode) and

at another location reduction of oxygen to form water takes

place (cathode). First Fe gets oxidised to Fe2+ and then in

the presence of oxygen it forms Fe3+ which then reacts

with water to form rust which is represented by Fe 2 O

3 .xH

2 O.

Anode : 2Fe (s)  2 Fe2+ + 4e– Eº = + 0.44 V

Cathode : O 2 (g) + 4H+ + 4e–  2H

2 O (l) Eº = 1.23 V

Overall R × N :

2Fe (s) + O 2 (q) + 4H+  2Fe2+ + + 2H

2 O Eº

cell = 1.67 M

Rusting of iron can be avoided by painting it or by coating

it with some other metals like Zinc. The latter process is

known as Galvanisation. As the tendency of Zn to get

oxidised is more than iron it gets oxidised in preference

and iron is protected. This method of protecting one metal

by the other is also called Cathodic Protection.

21. CONDUCTANCE (G)

It is the reciprocal of resistance and may be defined as the

ease with which the electric current flows through a

conductor.

1 G

R 

SI unit is Siemen (S).

1 S = 1 ohm–1 (mho)

22. CONDUCTIVITY ()

It is the reciprocal of resistivity (). 1 1

G R A A

    

 

Now if  = 1 cm and A = 1 cm2, then  = G.

Hence, conductivity of an electrolytic solution may be

defined as the conductance of a solution of 1 cm length

with area of cross-section equal to 1 cm2.

23. FACTORS AFFECTING ELECTROLYTIC

CONDUCTANCE

23.1 Electrolyte

An electrolyte is a substance that dissociates in solution to

produce ions and hence conducts electricity in dissolved

or molten state.

Examples : HCl, NaOH, KCl (Strong electrolytes).

CH 3 –COOH, NH

4 OH (Weak electrolytes).

The conductance of electricity by ions present in the

solutions is called electrolytic or ionic conductance. The

following factors govern the flow of electricity through a

solution of electrolyte.

218 ELECTROCHEMISTRY

Mahesh Tutorials Science

(i) Nature of electrolyte or interionic attractions : Lesser

the solute-solute interactions, greater will be the freedom

of movement of ions and higher will be the conductance.

(ii) Solvation of Ions : Larger the magnitude of solute-solvent

interactions, greater is the extent of solvation and lower

will be the electrical conductance.

(iii) The nature of solvent and its viscosity : Larger the solvent-

solvent interactions, larger will be viscosity and more will

be the resistance offered by the solvent to flow of ions

and hence lesser will be the electrical conductance.

(iv) Temperature : As the temperature of electrolytic solution

rises solute-solute, solute-solvent and solvent-solvent

interactions decreases, this results in the increase of

electrolytic conductance.

24. MEASUREMENT OF CONDUCTANCE

As we know, 1

R A  

 The value of  could be known,

if we measure l, A and R. The value of the resistance of

the solution R between two parallel electrodes is

determined by using ‘Wheatstones’ bridge method

(Fig.)

It consists of two fixed resistance R 3 and R

4 , a variable

resistance R 1 and the conductivity cell having the unknown

resistance R 2 . The bridge is balanced when no current

passes through the detector. Under these conditions,

31 1 4

2

2 4 3

RR R R or R

R R R  

25. MOLAR CONDUCTIVITY ( m )

It may be defined as the conducting power of all the ions

produced by dissolving one mole of an electrolyte placed

between two large electrodes at one centimeter apart.

Mathematically,

m m

1000 V,

C

     

where, V is the volume of solution in cm3 containing 1

mole of electrolyte and C is the molar concentration.

Units :

1

m 3

1000 S cm

C mol cm





   

= ohm–1 cm2 mol–1 or S cm2 mol–1

26. EQUIVALENT CONDUCTIVITY ( eq )

It is conducting power of one equivalent of electrolyte

placed between two large electrodes at one centimeter apart.

Mathematically :

eq v   

eq

1000

N

  

Where, v is the volume of solution in cm3 containing

1 equivalent of electrolyte and N is normality.

Units :

eq

1000

N

  

1 1 2 1

3 2 1

S cm Ohm cm equivalent

equivalent cm S cm equivalentor

  

   

27. VARIATION OF CONDUCTIVITY AND MOLAR

CONDUCTIVITY WITH DILUTION

Conductivity decreases with decrease in concentration,

this is because the number of ions per unit volume that

carry the current in the solution decreases on dilution.

Molar conductivity m V   increases with decrease

in concentration. This is because the total volume V of

solution containing one mole of electrolyte also increases.

ELECTROCHEMISTRY 219

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It has been found that the decrease in  on dilution of a solution is more than compensated by increases in its

volume.

Graphic representation of the variation of m vs c

28. LIMITING MOLAR CONDUCTIVITY ( m )

The value of molar conductivity when the concentration

approaches zero is known as limiting molar conductivity or

molar conductivity at infinite dilution. It is possible to

determine the molar conductivity at infinite dilution o

m  in

case of strong electrolyte by extrapolation of curve of

m vs c. On contrary, the value of molar conductivity of weak electrolyte at infinite dilution cannot be determined

by extapolation of the curve as the curve becomes almost

parallel to y-axis when concentration approaches to zero.

The mathematical relationship between m

 and o m for

strong electrolyte was developed by Debye, Huckel and

Onsagar. In simplified form the equation can be given as

1/ 2

m m b c    

where m

 is the molar conductivity at infinite dilution

and b is a constant which depends on the nature of the

solvent and temperature.

29. KOHLRAUSCH’S LAW

It states that 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. In general, if an

electrolyte on dissociation gives v + cations and v

– anions

then its limiting molar conductivity is given by

o o

m v v 

       

Here, o and o

 are the limiting molar conductivities of

cations and anions respectively.

30. APPLICATIONS OF KOHLRAUSCH’S LAW

30.1 (i) Calculation of molar conductivities of weak

electrolyte at infinite dilution

For example, molar conductivity of acetic acid at infinite

dilution can be obtained from the knowledge of molar

conductivities at infinite dilution of strong electrolyte like

HCl, CH 3 COONa and NaCl as illustrated below.

3

o o o o o

CH COO Na H Cl Na Cl                      

i.e. 3

o

m CH COOH 3 o o o

m HCl m NaClm CH COONa    

30.2 (ii) Determination of Degree of Dissociation

of Weak Electrolytes

Degree of dissociation

c

m

o

m

  



30.3 (iii) Determination of Dissociation Constant

(K) of Weak Electrolytes:

2c K

1

 



also

c

m

m



  



2 2 c c

m m m

c c m m m m m

c / C K

1 /



  

     

     

31. USE OF G IN RELATING EMF VALUES OF HALF CELL REACTIONS

When we have two half cell reactions such that on adding

them we obtain another half cell reaction then their emfs

cannot be added directly. But in any case thermodynamic

functions like G can be added and emf values can be related through them. Consider the following three half cell reactions:

Fe2+ + 2e–  Fe E 1

Fe3+ + 3e–  Fe E 2

Fe3+ + e–  Fe2+ E 3

We can easily observe that the third reaction can be

obtained by subtracting the first reaction from the second.

But the same relation does not apply on the emf values.

That is, E 3  E

2 – E

1 . But the G values can be related

according to the reactions. That is,

G 3 = G

2 – G

1

– n 3 FE

3 = – n

2 FE

2 + n

1 FE

1

– E 3 = – 3E

2 + 2E

1

 E 3 = 3E

2 – 2E

1

NOTE

We should always remember that emf values are additive

only when two half cell reactions are added to give a

complete balanced cell reaction. In any other case we

will be using G values to obtain relations between emf

values.

32. FORMULAE

1. R A

        

 Cell constant

where, R = Resistance

A = Area of cross-section of the electrodes.

 = Resistivity

2. 1

R   cell constant

where,  = Conductivity or specific conductance

3. m 1000

M

  

where, m

 = Molar conductivity

M = Molarity of the solution.

4. m

 (A x B

y ) = x

m

 (Ay+) + y m  (Bx–)

where, m

 = Molar conductivity at infinite dilution x and y

are the number of cations and anions produced by one

formula unit of the electrolyte on complete dissociation.

5.

c

m

m



  



where,  = Degree of dissociation

c

m = Molar conductivity at a given

concentration

6. For a weak binary electrolyte AB

2 c2 m

c

m m m

cc K

1  

  

    

where, K = Dissociation constant

o o o

cell cathode anode E E E 

= Eo Right + Eo left

7. Nernst equation for a general electrochemical reaction

aA + bB ne 

 cC + dD

c d

o

cell cell a b

C DRT E E In

nF A B  

c d

o

cell cell a b

C D2.303RT E E log

nF A B  

a b

o

cell cell c d

A B0.059 E E log

n C D   at 298 K

8. log K c =

o

cell

n E

0.0591

where, K c = Equilibrium constant.

9. o o r cellG nFE   (Creterion of spontaneity)

o

r c G 2.303RT log K  

where,  r Go = Standard Gibbs energy of the reaction.

10. Q = I × t

where Q = Quantity of charge in coulombs

I = Current in amperes

t = Time in seconds

11. m = Z × I × t

where m = mass of the substance liberated at the

electrodes

Z = Electrochemical equivalent.

where E = Equivalent weight = E/96500

STANDARD REDUCTION POTENTIALS AT 298 K.

IN ELECTROCHEMICAL ORDER

H 4 XeO

6 + 2H+ + 2e–  XeO

3 + 3H

2 O + 3.0

F 2 + 2e–  2F– + 2.87

O 3 + 2H+ + 2e–  O

2 + H

2 O + 2.07

  24 2 82 SO2e2OS + 2.05

Ag2+ + e–  Ag+ + 1.98

Co3+ + e–  Co2+ + 1.81

H 2 O

2 + 2H+ + 2e–  2H

2 O + 1.78

Au+ + e–  Au + 1.69

Pb4+ + 2e–  Pb2+ + 1.67

2HClO + 2H+ + 2e– Cl 2 + 2H

2 O + 1.63

Ce4+ + e–  Ce3+ + 1.61

2HBrO + 2H+ + 2e– Br 2 + 2H

2 O + 1.60

OH4Mne5H8MnO 2 2

4   + 1.51

Mn3+ + e–  Mn2+ + 1.51

Au3+ + 3e–  Au + 1.40

Cl 2 + 2e– 2Cl– + 1.36

OH7Cr2e6H14OCr 2 32

72   + 1.33

O 3 + H

2 O + 2e–  O

2 + 2OH– + 1.24

O 2 + 4H+ 4e– 2H

2 O + 1.23

OHClOe2H2ClO 234   + 1.23

MNO 2 + 4H+ + 2e–  Mn2+ + 2H

2 O + 1.23

Br 2 + 2e– 2Br– + 1.09

Pu4+ + e–  Pu3+ + 0.97

OH2NOe3H4NO 23   + 0.96

  22 2

Hge2Hg2 + 0.92

ClO– + H 2 O + 2e– Cl– + 2OH– + 0.89

Hg2+ + 2e–  Hg + 0.86

OHNOeH2NO 223   + 0.80

Ag+ + e–  Ag + 0.80

Hg2e2Hg22   + 0.79

Fe3+ + e–  Fe2+ + 0.77

BrO– + H 2 O + 2e–  Br– + 2OH– + 0.76

  2442 SOHg2e2SOHg + 0.62

  OH4MnOe2OH2MnO 22 2 4 + 0.60

  244 MnOeMnO + 0.56

I 2 + 2e– 2I– + 0.54

Cu+ + e–  Cu + 0.52

  I3e2I3 + 0.53

NiOOH + H 2 O + e– Ni(OH)

2 + OH– + 0.49

  2442 CrOAg2e2CrOAg + 0.45

O 2 + 2H

2 O + 4e–  4OH– + 0.40

  OH2ClOe2OHClO 324 + 0.36

[Fe(CN) 6 ]3– + e–  [Fe(CN)

6 ]4– + 0.36

Cu2+ + 2e–  Cu + 0.34

Hg 2 Cl

2 + 2e– 2Hg + 2Cl– + 0.27

AgCl + e–  Ag + Cl– + 0.22

Bi + 3e– Bi + 0.20

Cu2+ + e–  Cu+ + 0.16

Sn4+ + 2e–  Sn2+ + 0.15

AgBr + e–  Ag + Br– + 0.07

Ti4+ + e–  Ti3+ 0.00

2H+ + 2e–  H 2 0, by definition

Fe3+ + 3e–  Fe – 0.04

O 2 H

2 O + 2e–   OHHO2 – 0.08

Pb2+ + 2e–  Pb – 0.13

In+ + e–  In – 0.14

Sn2+ + 2e–  Sn – 0.14

AgI + e–  Ag + I– – 0.15

Ni2+ + 2e– Ni – 0.23

Co2+ + 2e–  Co – 0.28

In3+ + 3e–  In – 0.34

Tl+ e–  Tl – 0.34

PbSO 4 + 2e–  Pb + 2

4SO – 0.36

221 ELECTROCHEMISTRY

Ti3+ + e–  Ti2+ – 0.37

Cd2+ + 2e–  Cd – 0.40

In2+ + e–  In+ – 0.40

Cr3+ + e– Cr2+ – 0.41

Fe2+ + 2e–  Fe – 0.44

In3+ + 2e–  In+ – 0.44

S + 2e–  S2– – 0.48

In3+ + e–  In2+ – 0.49

U4+ + e–  U3+ – 0.61

Cr3+ + 3e– Cr – 0.74

Zn2+ + 2e–  Zn – 0.76

Cd(OH) 2 + 2e– Cd + 2OH– – 0.81

2H 2 O + 2e– H

2 + 2OH– – 0.83

Cr2+ + 2e– Cr – 0.91

Mn2+ + 2e–  Mn – 1.18

V2+ + 2e– V – 1.19

Ti2+ + 2e–  Ti – 1.63

Al3+ + 3e–  Al – 1.66

U3+ + 3e– U – 1.79

Sc3+ + 3e–  Sc – 2.09

Mg2+ + 2e–  Mg – 2.36

Ce3+ + 3e– Ce – 2.48

La3+ + 3e– La – 2.52

Na+ + e–  Na – 2.71

Ca2+ + 2e– Ca – 2.87

Sr2+ + 2e–  Sr – 2.89

Ba2+ + 2e– Ba – 2.91

Ra2+ + 2e– Ra – 2.92

Cs+ + e–  Cs – 2.92

Rb+ + e–  Rb – 2.93

K+ + e– K – 2.93

Li+ + e– Li – 3.05

REDUCTION

POTENTIALS IN ALPHABETICAL ORDER

Ag+ + e–  Ag + 0.80

Ag2+ + e–  Ag+ + 1.98

AgBr + e–  Ag + Br– + 0.0713

AgCl + e–  Ag + Cl– + 0.22

Ag 2 CrO

4 + 2e–  2Ag + 2

4CrO + 0.45

AgF + e–  Ag + F– + 0.78

AgI + e–  Ag + I– – 0.15

Al3+ + 3e–Al – 1.66

Au+ + e–  Au + 1.69

Au3+ + 3e– Au + 1.40

Ba2+ + 2e– Ba + 2.91

Be2+ + 2e– Be – 1.85

Bi3+ + 3e– Bi + 0.20

Br 2 + 2e– 2Br– + 1.09

BrO– + H 2 O + 2e–  Br– + 2OH– + 0.76

Ca2+ + 2e– Ca – 2.87

Cd(OH) 2 + 2e–  Cd + 2OH– – 0.81

Cd2+ + 2e–  Cd – 0.40

Ce3+ + 3e– Ce – 2.48

Ce4+ + e–  Ce3+ + 1.61

Cl 2 + 2e–  2Cl– + 1.36

ClO– + H 2 O + 2e–  Cl– + 2OH– + 0.89

OHClOe2H2ClO 234   + 1.23

  OH2ClOe2OHClO 324 + 0.36

Co2+ + 2e–  Co – 0.28

Co3+ + e–  Co2+ + 1.81

Cr2+ + 2e– Cr – 0.91

OH7Cr2e6H14OCr 2 32

72   + 1.33

Cr3+ + 3e– Cr – 0.74

Cr3+ + e– Cr2+ – 0.41

ELECTROCHEMISTRY 222

Cs+ e–  Cs – 2.92

Cu+ + e–  Cu + 0.52

Cu2+ + 2e–  Cu + 0.34

Cu2+ + e–  Cu+ + 0.16

F 2 + 2e–  2F– + 2.87

Fe2+ + 2e–  Fe – 0.44

Fe3+ + 3e–  Fe – 0.04

Fe3+ + e–  Fe2+ + 0.77

[Fe(CN) 6 ]3– + e–  [Fe(CN)

6 ]4– + 0.36

2H+ + 2e–  H 2 0, by definition

2H 2 O + 2e– H

2 + 2OH– – 0.83

2HBrO + 2H+ + 2e– Br 2 + 2H

2 O + 1.60

2HClO + 2H+ + 2e– Cl 2 + 2H

2 O + 1.63

H 2 O

2 + 2H+ + 2e–  2H

2 O + 1.78

H 4 XeO

6 + 2H+ + 2e– XeO

3 + 3H

2 O + 3.0

Hg2e2Hg22   + 0.79

Hg 2 Cl

2 + 2e–  2Hg + 2Cl– + 0.27

Hg2+ + 2e–  Hg + 0.86

2Hg2+ + 2e– 2 2Hg + 0.92

Hg 2 SO

4 + 2e–  2Hg + 2

4SO + 0.62

I 2 + 2e– 2I– + 0.54

  I3e2I3 + 0.53

In+ + e–  In – 0.14

In2+ + e–  In+ – 0.40

In3+ + 2e–  In+ – 0.44

In3+ + 3e–  In – 0.34

In3+ + e–  In2+ – 0.49

K+ + e–  K – 2.93

La3+ + 3e– La – 2.52

Li+ e– Li – 3.05

Mg2+ + 2e–  Mg – 2.36

Mn2+ + 2e–  M n

– 1.18

Mn3+ + e–  Mn2+ + 1.51

MnO 2 + 4H+ + 2e–  Mn2+ + 2H

2 O + 1.23

OH4Mne5H8MnO 2 2

4   + 1.51

  244 MnOeMnO + 0.56

  OH4MnOe2OH2MnO 224 + 0.60

Na+ + e–  Na – 2.71

Ni2+ + 2e– Ni – 0.23

NiOOH + H 2 O + e–  Ni(OH)

2 + OH– + 0.49

OHNOeH2NO 223   – 0.80

OH2NOe3H4NO 23   + 0.96

  OH2NOe2OHNO 223 + 0.10

O 2 + 2H

2 O + 4e– 4OH– + 0.40

O 2 + 4H+ + 4e– 2H

2 O + 1.23

  22 OeO – 0.56

O 2 + H

2 O + 2e–  

2HO + OH – – 0.08

O 3 + 2H+ + 2e–  O

2 + H

2 O + 2.07

O 3 + H

2 O + 2e–  O

2 + 2OH– + 1.24

Pb2+ + 2e–  Pb – 0.13

Pb4+ + 2e–  Pb2+ + 1.67

PbSO 4 + 2e–  Pb + 24SO – 0.36

Pt2+ + 2e–  Pt + 1.20

Pu4+ + e–  Pu3+ + 0.97

Ra2+ + 2e– Ra – 2.92

Rb+ + e–  Rb – 2.93

S + 2e–  S2– – 0.48

  24 2 82 SO2e2OS + 2.05

SC3+ + 3e– Sc – 2.09

Sn2+ + 2e–  Sn – 0.14

Sn4+ + 2e–  Sn2+ + 0.15

Sr2+ + 2e–  Sr – 2.89

Ti2+ + 2e–  Ti – 1.63

Ti3+ + e– Ti2+ – 0.37

Ti4+ + e– Ti3+ 0.00

Tl+ + e–  Tl – 0.34

U3+ + 3e– U – 1.79

U4+ + e–  U3+ – 0.61

V2+ + 2e– V – 1.19

V3+ + e– V2+ – 0.26

Zn2+ + 2e–  Zn – 0.76

223 ELECTROCHEMISTRY

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SOLVED EXAMPLES

Example – 1

Give the relationship between equivalent and molar

conductance ?

Sol. m 1000

Molarity    and eq

1000

Normality   

m

eq

Normality

Molarity

  



Example – 2

Can nickel spatula be used to stir a copper sulphate solution ?

Support your answer with a reason

2 2

o o

Ni / Ni Cu /Cu E 0.25 V, E 0.34 V.    

Sol. o o o cell cathode anode

E E E 

2 2

o o o

cell Cu / Cu Ni / Ni E E E 0.34 V 0.25 0.59 V        

As o cell

E is +ve, G = – ve, because G = – nEo F, i.e,

reaction will take place. Therefore, we cannot stir a

copper sulphate solution with nickel spatula.

Example – 3

State two advantages of H 2 —O

2 fuel cell over ordinary cell.

Sol. The two advantages of H 2 —O

2 fuel cell over ordinary

cell are :

(i) They do not cause any pollution.

(ii) They have high efficiency of 60-70%.

Example – 4

What is galvanisation ?

Sol. The process of coating zinc over iron is called

galvanisation.

Example – 5

Which type of a metal can be used in cathodic protection

of iron against rusting ?

Sol. A metal which is more electropositive than iron such as

Al, Zn, Mg can be used in cathode protection of iron

against rusting.

Example – 6

Write the chemical equations for all the steps involved in the

rusting of iron, Give any one method to prevent rusting of iron.

Sol. Anode: 2Fe s Fe aq 2e ,   2

o

Fe / Fe E 0.44V  

Cathode : 2 2O g 4H aq 4e 2H O,    

2 2

o

H / O / H O E 1.23V 

Overall reaction

Fe(s)  Fe2+ (aq) + 2e– ] × 2

O 2 (g) + 4H+ + 4e–  2H

2 O

2Fe + O 2 + 4H+(aq)  2Fe2+ + 2 H

2 O o

cellE 1.67V

Further 4Fe2+(aq) + O 2 (g) + 4H

2 O(l) 

2Fe 2 O

3 + 8H+(aq)

Fe 2 O

3 + xH

2 O 

Fe 2 O

3 . xH

2 O

Hydrated ferric oxide (Rust)

Galvanisation is used to prevent rusting of iron.

Example – 7

The following chemical reaction is occurring in an

electrochemical cell.

Mg(s) + 2 Ag+ (0.0001 M)  Mg2+ (0.10M) + 2Ag(s)

The Eo electrode values are

Mg2+ / Mg = – 2.36 V

Ag+ / Ag = 0.81 V

For this cell calculate/write

(a) (i) Eo value for the electrode 2Ag+/2Ag.

(ii) Standard cell potential o cellE .

(b) Cell potential (E) cell

(c) (i) Symbolic representation of the above cell.

(ii) Will the above cell reaction be spontaneous ?

Sol. (a) (i) o Ag / Ag

E 0.81V 

ELECTROCHEMISTRY 224

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(ii) o o o cell cathode anodeE E E 

2

o o

Ag / Ag Mg / Mg E E 0.81 2.36    

o

cellE 3.17V

(b) At anode : Mg Mg2+ + 2e–

A cathode : Ag+ + e–  Ag ] × 2

Mg + 2Ag+ Mg2+ + 2Ag

E cell

=

2

o

cell 2

Mg0.059 E log

n Ag





    

2 4

0.059 0.1 3.17 log

2 10  

= 3.17 – 0.0295 log 107

= 3.17 – 0.0295 × 7 = 3.17 – 0.21

E cell

= 2.96 V

(c) (i) Mg | Mg2+ (0.1 M) || Ag+ (0.0001M) | Ag

(ii) Yes, as the cell potential is positive.

Example – 8

(a) Current of 1.50 A was passed through an electrolytic

cell containing AgNO 3 solution with inert electrodes.

The weight of Ag deposited was 1.50g. How long did

the current flow ?

(b) Write the reactions taking place at the anode and

cathode in the above cell if inert electrodes are used.

(c) Give reactions taking place at the two electrodes if

these are made up of Ag.

Sol. (a) According to Faraday’s first law, charge required to

deposit 1.50 g.

Ag 96500

1.50 1331.70 coulombs 108

 

Time taken 1331.70

893.5s 1.50

 

(b) Inert electrodes

Anode: 2 22H O O g 4H aq 4e    

Cathode : Ag+ (aq) + e– Ag s

(c) Ag electrodes

Anode : Ag(s)  Ag+ (aq) + e–

Cathode : Ag+ (aq) + e–  Ag(s)

Example – 9

Explain Kohlrausch’s law of independent migration of ions.

Mention one application of Kohlrausch’s law.

Sol. Kohlrausch’s law of independent migration of ions: The

molar conductivity of an electrolyte at infinite dilution

is the sum of the individual contributions of the anion

and cation of the electrolyte.

0 0 0v v       

where, 0 and 0

 are the limiting molar

conductivities of the cation and anion respectively

and v + and v

– are the number of cations and anions

formed from a formula unit of the electrolyte. For

example, one formula unit of Al 2 (SO

4 ) 3 gives two Al3+

ions and three sulphate ions. Therefore,

3 2 2 4 3 4

o o o

m Al (SO Al SO 2 3     

Application : It can be used to determine molar

conductivity of weak electrolytes at infinite dilution :

Consider acetic acid as the example of a weak electrolyte.

3 a 3

o o o

m CH COON CH COO Na      

o o o

m HCl H Cl      

o o o

m NaCl Na Cl      

From (i) + (ii) – (iii) we get

3

o o o o o o

CH COO Na H Cl Na Cl                

33

o o o

CH COOHCH COO H       

Example – 10

The electrical resistance of a column of 0.05 mol L–1 NaOH

solution of diameter 1 cm and length 50 cm is 5.55 × 103

ohm. Calculate its resistivity, conductivity and molar

conductivity.

Sol. A = r2 = 3.14 × 0.52 cm2 = 0.785 cm2 = 0.785 × 10–4 m2,

l = 50 cm = 0.5 m

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Mahesh Tutorials Science

R A

  

or

3RA 5.55 10 0.785

50 cm

    



= 87.135  cm

Conductivity = 1 11 1 S cm 0.01148 S cm

87.135

        

Molar conductivity, m 1000

c

  

1 3 1

1

0.01148 S cm 1000 cm L

0.05molL

 



 = 229.6 S cm2 mol–1

Example – 11

The measured resistance of a conductance cell containing

7.5 × 10–3 M solution of KCl at 25oC was 1005 ohms. Calculate

(a) specific conductance (b) molar conductance of the

solution. Cell constant = 1.25 cm–1.

Sol. Specific conductance 1

cell constant R

 

1 1 11 1.25 cm 0.001244 cm 1005

     

Molar conductance m 1000

Molarity

  

1 1 3 1

3 1

0.001244 cm 1000cm L

7.5 10 mol L

  

 

 

1 2 1165.87 cm mol .  

Example – 12

m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm 2

mol–1 respectively. Calculate 0 m

 for HAc.

Sol. 0 o o o o o o o oAcm HAc H H Cl Ac Na Cl Na                 

o o o

m HCl m NaAc m NaCl     

= (425.9 + 91.0 – 126.4) S cm2 mol–1

= 390.5 S cm2 mol–1.

Example – 13

The conductivity of 0.0011028 mol L–1 acetic acid is

4.95 × 10–5 S cm–1. Calculate its dissociation constant if o m

for acetic acid is 390.5 S cm2 mol–1.

Sol.

5 1 3

m 1

4.95 10 Scm 1000cm

c L0.001028molL

 



   

= 44.88 S cm2 mol–1

2 1

m

o 2 1

m

44.88 Scm mol 0.115

390.5 Scm mol





    



212 0.001028 mol L 0.115c K

1 0.115

   



= 1.65 × 10–5 mol L–1

Example – 14

A cell is prepared by dipping copper rod in 1 M copper

sulphate solution and zinc rod in 1 M zinc sulphate solution.

The standard reduction potentials of copper and zinc are

0.34 V and – 0.76 V respectively.

(i) What will be the cell reaction?

(ii) What will be the standard electromotive force (EMF)

of the cell ?

(iii) Which electrode will be positive ?

(iv) How will the cell be represented ?

Sol. (i) The cell reaction can be

2 2 Zn Cu Zn Cu

   

or 2 2Cu Zn Cu Zn    

The EMF comes out to be positive for the 1st reaction.

Hence, the cell reaction is 2 2

Zn Cu Zn Cu    

(ii) 2 2 o o o o o

cell cathode anode Cu /Cu Zn / Zn E E E E E    

= 0.34 + 0.76 = 1.10 V

(iii) reduction takes place on copper electrode. Hence

it is positive

(iv) Zn | Zn2+ (1 M) || Cu2+ (1 M) | Cu

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Example – 15

Represent the cell in which the following reaction takes place

Mg (s) + 2Ag+ (0.0001 M)  Mg2+ (0.130 M) + 2 Ag 9 (s)

Calculate its E (cell) if

o

cell E = 3.17 V.

Sol. The cell can be written as Mg | Mg2+ (0.130 M) || Ag+

(0.0001 M) | Ag

2 2

o o

cell cell cell2

Mg MgRT 2.303RT E E In E log

nF 2F AgAg

 



          

2

0.059V 0.130 3.17V log

2 0.0001  

= 3.17 V – 0.21 V = 2.96 V

Example – 16

A zinc rod is dipped in 0.1 M solution of ZnSO 4 . The salt is

95% dissociated at this dilution at 298 K. Calculate the

electrode potential 2 o

Zn / Zn E 0.76V .  

Sol. The electrode reaction written as reduction reaction is

Zn2+ + 2e–  Zn (n = 2)

Applying Nernst equation, we get

2 2

o

Zn / Zn Zn / Zn 2

0.0591 1 E E log

2 Zn   

   

As 0.1 M ZnSO 4 solution is 95% dissociated, this means

that in the solution,

2 95Zn 0.1M 0.095M 100

   

2 Zn / Zn

0.0591 1 E 0.76 log

2 0.095    

= – 0.76 – 0.02955 (log 1000 – log 95)

= – 0.76 – 0.02955 (3 – 1.9777)

= – 0.76 – 0.03021

= – 0.79021 V

Example – 17

Calculate the potential (emf) of the cell

Cd | Cd2+ (0.10 M) || H+ (0.20 M) | Pt, H 2 (0.5 atm)

(Given Eo for Cd2+ / Cd = – 0.403 V, R = 8.14 JK–1 mol–1, F =

96,500 C mol–1).

Sol. The cell reaction is

Cd + 2H+ (0.20 M)  Cd2+ (0.10 M) + H 2 (0.5 atm)

2 2

o o o

cell H /1/ 2H Cd / Cd E E E 0    –(– 0.403) = 0.403 V

Applying Nernst equation to the cell reaction,

2

2

Ho

cell cell 2

Cd P2.303 RT E E log

nF H





     

2

2.303 8.314 298 0.1 0.5 0.403 log

2 96,500 0.2

 

= 0.403 – 0.003 = 0.400 V

Example – 18

Calculate the equilibrium constant of the reaction

Cu (s) + 2Ag+ (aq)  Cu2+ (aq) + 2Ag(s) o cellE 0.46V

Sol. o

cell c

0.059V E log K 0.46V

2  

or c

0.46V 2 log K 15.6

0.059V

   K

c = Antilog 15.6

K c = 3.92 × 1015

Example – 19

Calculate the standard free energy change and maximum

work obtainable for the reaction occurring in the cell :

(Daniell cell).

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

[Given 2 2 o o

Zn / Zn Cu / Cu E 0.76V,E  

10.34V, F 96,500 Cmol ]  

Also calculate the equilibrium constant for the reaction.

Sol. (i) 2 2 o o o

cell Cu / Cu Zn / Zn E E E 0.34 0.76    

= 1.10 V

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Mahesh Tutorials Science

The reaction taking place in the Daniell cell is

Zn(s) + Cu2+ (aq)  Cu(s) + Zn 2+ (aq)

For this reaction, n = 2

Go = – o cell

nFE

= – 2 × 96500 C mol–1 × 1.10 V

= – 212300 CV mol–1

= – 212300 J mol–1 (1 CV = 1 J)

= – 212.300 kJ mol–1

Thus, the maximum work that can be obtained from

the Daniel cell = 212.3 kJ.

(ii) Go = – RT In K c = – 2.303 RT log K

c

 –212300 = – 2.303 × 8.14 × 298 × log K c

or log c

212300 K 37.2704

2.303 8.314 298  

 K c = Antilog 37.2074 = 1.6 ×1037

Example – 20

Calculate the equilibrium constant, K c for the reaction.

3Sn4+ + 2Cr  3Sn2+ + 2Cr3+

Given Eo = 0.885 V.

Sol. o

cell c

0.059 E log K , n 6

n  

c

0.059 0.885 log K

6 

log c

6 0.885 K

0.059



K c = Antilog 90 = 1 × 1090

Example – 21

Chromium metal can be plated out from an acidic solution

containing CrO 3 according to the following equation:

CrO 3 (aq) + 6H+ (aq) + 6e– Cr (s) + 3H

2 O

Calculate (i) how many grams of chromium will be plated

out by 24,000 coulombs and (ii) how long will it take to

plate out 1.5 g of chromium by using 12.5 amp current ? (At.

mass of Cr = 52).

Sol. (i) 6 × 96, 500 coulomb deposit Cr = 1 mole = 52 g

 24,000 coulomb deposit 52 24000

Cr g 2.1554g 6 965000

 

(ii) 52 g of Cr is deposited by electricity = 6 × 96500 C

 1.5 g require electricity 6 96500

1.5C 16071C 52

 

 Time for which the current is required to be passed

16071.9 1336s.

12.5 A  

Example – 22

(a) Calculate the equilibrium constant for the reaction

Cd2+ (aq) + Zn(s)  Zn2+ (aq) + Cd(s)

If 2 o

Cd / Cd E 0.403V  

2

o

Zn / Zn E 0.763V  

(b) When a current of 0.75A is passed through a CuSO 4

solution for 25 min, 0.369 g of copper is deposited at

the cathode. Calculate the atomic mass of copper.

(c) Tarnished silver contains Ag 2 S. Can this tarnish be

removed by placing tarnished silver ware in an

aluminium pan containing an inert electrolytic solution

such as NaCl. The standard electrode potential for

half reaction:

Ag 2 S(s) + 2e–  2Ag (s) + S2– is –0.71V

and forAl3+ + 3e–  2Al (s) is – 1.66 V

Sol. (a) o o o cell c aE E E 0.403 0.763 0.360V     

As log Kc = o

cellnE 2 0.360

0.059 0.059

         

0.720 12.20

0.059

     

K c = antilog (12.20) = 1.585 × 1012

(b) M = Z I t

x 0.369 0.75 25 60

2 96500 

(x = molar mass of copper)

x = 63.3 g/mol

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(c) o cellE for reaction of tarnished silver ware with

aluminium pan is (– 0.71 V) + 1.66 V i.e., + 0.95 V

Tarnished silver ware, therefore, can be cleaned by

placing it in an aluminium pan as o cellE is positive.

Example – 23

(a) Calculate the standard free energy change for the

following reaction at 25oC.

Au (s) + Ca2+ (aq, 1M) Au3+ (aq, 1M) + Ca(s)

3 2

o o

Au / Au Ca /Ca E 1.50V, E 2.87 V    

Predict whether the reaction will be spontaneous or

not at 25oC. Which of the above two half cells will act

as an oxidizing agent and which one will be a reducing

agent?

(b) The conductivity of 0.001 M acetic acid is 4 × 10–5 S/cm.

Calculate the dissociation constant of acetic acid, if

o

m for acetic acid is 390. 5 S cm 2/mol.

Sol. (a) Eo cell = (– 2.87 V) – (1.50 V) = – 4.37 V

Go cell = – 6 × 96500 × – 4.37 V = + 2530.230 kJ/mol

Since  r Go is positive, reaction is non-spontaneous.

Au3+/Au half cell will be a reducing agent, Ca2+/Ca half

cell will be an oxidising agent.

(b) c

m

1000 K

molarity  

K = Specific conductance

5 2 14 10 S/ cm 1000 40S cm mol

0.001

   

m

o

m

40 0.103

390.5

    



22 5

c

0.001 0.103C K 1.19 10

1 1 0.103

     

Example – 24

(a) Depict the galvanic cell in which the following reaction

takes place :

Zn (s) + 2Ag+ (aq)  Zn2+ (aq) + 2Ag (s)

Also indicate that in this cell

(i) which electrode is negatively charged.

(ii) what are the carrier of the current in the cell.

(iii) what is the individual reaction at each electrode.

(b) Write the Nernst equation and determine the e.m.f. of

the following cell at 298 K:

Mg (s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu (s)

(Given : 2 2 o o

Mg / Mg Cu / Cu E 2.375 V, E 0.34V     )

Sol. (a) Zn | Zn2+ (conc.) || Ag+ (conc) | Ag

(i) Zn electrode is negatively charged.

(ii) Current carriers of cell are

- electrons in external wire

- Zn2+ ions in anodic half cell.

- Ag+ ions in cathodic half cell.

- Ions of salt bridge, i.e., K+ and Cl–.

(iii) At anode 2Zn Zn 2e   

At cathode 2Ag e 2Ag  

(b) 2Mg Mg 2e  

2 Cu 2e Cu

  

2 2Mg Cu Cu Mg   

Nernst equation

2

o

cell cell 2

Mg0.059 E E log

n Cu





     

2 2

o o

cell Cu /Cu Mg /Mg 2

Mg0.059 E E E log

2 Cu  





      

230 ELECTROCHEMISTRY

Mahesh Tutorials Science

3

4

0.059 10 0.34 2.375 log

2 10



   

= 0.34 + 2.375 – 0.0295 log 10

E cell = 2.6855 V

E cell = 2.685 V

Example – 25

(a) Define molar conductivity of a substance and describe

how weak and strong electrolytes’ molar conductivity

changes with concentration of solute. How is such

change explained ?

(b) A voltaic cell is set up at 25 oC with the following half

cells:

Ag+ (0.001 M) | Ag and Cu2+ (0.10 M) | Cu

What would be the voltage of this cell ?

(Eo cell = 0.46 V)

Sol. Molar Conductivity m : It may be defined as the

conductance of a solution containing 1 mole of

electrolyte such that the entire solution is placed in

between two electrodes one centimetre apart.

m k v 

or

m

k 1000

M

 

Molar conductivity increases with decrease in

concentration or increase in dilution as number of ions

as well as mobility of ions increased with dilution.

For strong electrolytes, the number of ions do not

increase appreciably on dilution and only mobility or

ions increases due to decrease in interionic attractions.

Therefore, m increases a little as shown in graph by a

straight line.

For weak electrolytes, the number of ions as well as

mobility of ions increases on dilution which results in a

very large increase in molar conducvity especially near

infinite dilutuion as shown by curve in the figure.

At anode : Cu(s) Cu2+ (aq) + 2e

At cathode : 2Ag+ (aq) + 2e  2Ag(s)

2 Cu s 2Ag aq Cu aq 2Ag s

   

Here,

2

o

cell cell 2

Cu0.0591 E E log

n Ag





     

Here, o cell

E 0.46 V, n 2 

[Ag+] = 0.001M = 1 × 10–3 M, [Cu2+] = 0.1 M

cell 2 3

0.0591 0.1 E 0.46 log

2 10  

5

cell

0.0591 0.0591 E 0.46 log 10 0.46

2 2     × 5 log 10

E cell = 0.46 – 0.0591 × 2.5 × 1 = 0.46 – 0.14775 = 0.31225V

E cell = 0.312 V

Example – 26

(a) State the relationship amongst cell constant of cell, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solute related to conductivity of its solution ?

(b) A voltaic cell is set up at 25oC with the following half- cells:

Al | Ag3+ (0.001 M) and Ni | Ni2+ (0.50 M)

Calculate the cell voltage

2 3

o o

Ni |Ni Al |Al E 0.25V, E 1.66V     

Sol. (a) 1

R A

l      

ELECTROCHEMISTRY 231

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where,   Conductivity

1

A = Cell constant

R = Resistance

m

1000

M

  

where, m

 = Molar conductivity

 = Conductivity

M = Molarity of Solution

(b) At anode : 3Al s Al aq 3e] 2 

At cathode : Ni2+ (aq) + 2e Ni s ] 3  

2Al (s) + 3Ni2+ (aq)  2Al3+ (aq) + Ni(s)

2 3

o

cell cell 3 2

Al0.0591 E E log

n Ni





     

Here, n = 6, [Al3+] = 0.001 M = 1 × 10–3 M,

[Ni2+] = 0.5M

o

cell E 1.41V

2 3 6

o

cell 3

100.0591 0.0591 10 E 1.41V log 1.41 log

6 6 0.1250.5

 

   

6 6 30.0591 0.05911.41 log 10 8 1.41 log10 log 2 6 6

     

0.0591 0.0591 1.41 6 log10 3log 2 1.41 6 3 0.3010

6 6        

0.0591 0.3012 1.41 5.097 1.41

6 6     

= 1.41 + 0.0502 = 1.4602V

E cell = 1.46 V

Example – 27

(a) What type of a cell is the lead storage battery ? Write

the anode and the cathode reactions and the overall

reaction occurring in a lead storage battery while

operating.

(b) A voltaic cell is set up at 25oC with the half-cells Al |

Al3+ (0.001 M) and Ni | Ni2+ (0.50 M). Write the equation

for the reaction that occurs when the cell genrates an

electric current and determine the cell potential.

2 3

o o

Ni |Ni Al | Al (Given :E 0.25V, E 1.66V).    

Sol. (a) The lead storage battery is a secondary cell.

The cell reactions when the battery is in use are

given below

At anode: 24 4Pb s SO aq PbSO s 2e    

At cathode: 22 4PbO s SO aq 4H aq 2e     

 4 2PbSO s 2H O 

Overall cell reaction: Pb(s) + 2H 2 SO

4 (aq) 

PbSO 4 (s) + 2H

2 O(  )

(b) 2Al (s) + 3Ni+2 (aq)  3Ni (s) + 2Al+3 (aq)

o

cellE = 1.41 V

3 2

cell 3

0.0591 (10 ) E 1.41 log

6 (0.5)

    !

= 1.46 V

Example – 28

(a) Express the relationship amongst cell constant,

resistance of the solution in the cell and conductivity

of the solution. How is molar conductivity of a solute

related to conductivity of its solution.

(b) Calculate the equilibrium constant for the reaction.

2 2 Fe s Cd aq Fe aq Cd s

  

2 2

o o

Cd |Cd Fe |Fe (Given :E 0.40V, E 0.44V).   

Sol. (a) Conductivity 1

Resistance R   Cell constant (G)

232 ELECTROCHEMISTRY

Mahesh Tutorials Science

m

1000 ,

M

   where, m = Molar conductivity

(b) Fe(s) + Cd2+ (aq)  Fe2+ (aq) + Cd(s)

o

cell

c

E log k n

0.059 

Here, n = 2

o o o

cell cathode anodeE E E 

2 2

o o

Cd / Cd Fe / Fe E E  

= – 0.4 + 0.44

o

cell E 0.04V

c

2 0.04 0.08 log k

0.059 0.059

 

c log k 1.3536

k c = Antilog 1.3536

k c = 22.57

Example – 29

(a) Define the term molar conductivity. How is it related

to conductivity of the related solution?

(b) One half-cell in a voltaic cell is constructed from a

silver wire dipped in silver nitrate solution of unknown

concentration. Its other half-cell consists of zinc

electrode dipping in 1.0 M solution of Zn(NO 3 ) 2 . A

voltage of 1.48 V is measured for this cell. Use this

information to calculate the concentration of silver

nitrate solution used.

2 2

o o

Zn / Zn Ag / Ag (E 0.76V, E 0.80V)    

Sol.

(a) Molar conductivity m : It may be defined as the

conductivity of one molar electrolytic solution placed

between two electrodes one centimetre apart and have

enough area of cross section to hold entire volume.

m c

  

where,  = Conductivity

c = Concentration of solution in mol L–1

(b) At anode : 2Zn s Zn aq 2e  

At cathode : Ag+(aq) + e– Ag(s)] × 2

Zn(s) + 2Ag+ (aq)  Zn2+ + 2Ag(s)

2 o

cell cell 2

0.0591 [Zn ] E E log

n Ag



  

 

Here, n = 2, [Zn2+] = 1 M

Eo cell = 2

o o

Ag /Ag Zn / Zn E E 0.80V 0.76V   

Eo cell = 1.56 V

2

0.0591 1 1.48 1.56 log

2 Ag  

 

2

0.0591 1 0.08 log

2 Ag   

 

2

1 0.16 log 2.7072 2.7072

0.0591Ag   

 

2

log1 log Ag 2.7072  

0 2 log Ag 2.7072  

log Ag 1.3536 2.6464     

2Ag Anti log 2.6464 4.43 10 M    

Ag 0.044M   

Example – 30

(a) Corrosion is essentially an electrochemical

phenomenon. Explain the reactions occurring during

corrosion of iron kept in an open atmosphere.

(b) Calculate the equilibrium constant for the equilibrium

reaction.

2 2 Fe s Cd aq Fe aq Cd s

  

2 2

o o

Cd / Cd Fe / Fe (Given : E 0.40V, E 0.44V)    

Sol. (a) At anode : Oxidation of Fe atoms takes place

Fe  Fe2+ + 2e– 2oFe / FeE 0.44V  

At cathode : Reduction of oxygenin the presence of H+

ions. The H+ ions are produced by either H 2 O or H

2 CO

3

(formed by dissolution of CO 2 in moisture)

ELECTROCHEMISTRY 233

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2H aq 2e 2H   

2 2

1 2H O g H O

2  

Net reaction at cathodic area

2 2

1 2H aq O 2e H O

2

   

2 2

o

H / O / H O E 1.23V 

The overall reaction

2

2 2

1 Fe s 2H aq O g Fe aq H O

2

     

o

cellE 1.67V

The ferrous ions are further oxidised by atmospheric

oxygen to ferric ions which come out as rust in the form

of hydrated ferric oxide (Fe 2 O

3 . xH

2 O).

(b) Fe(s) + Cd2+ (aq)  Fe 2+ (aq) + Cd(s)

log k c =

oE cell n 0.059

Here, n = 2

o o o

cell cathode anode E E —E

2 2

o o

Cd / Cd Fe / Fe E E 40 0.44      

Eo cell = 0.04V

log c

2 0.04 0.08 k

0.059 0.059

 

log k c = 1.3536

k c = Antilog 1.3536

k c = 22.57

Example – 31

Two half cell reactions of an electrochemical cell are given

below :

2

4 MnO aq 8H aq 5e Mn     

o

2 aq 4H O ,E 1.51V 

Sn2+ (aq)  Sn4+ (aq) + 2e–, Eo = + 0.15 V

Construct the redox equation from the two half cell reactions

and predict if this reaction favours formation of reactants

or product shown in the quation

Sol. At cathode : 2

4MnO aq 8H 5e Mn      

o

2aq 4H O ] 2 E 1.15V  

At anode: Sn2+ 4Sn aq 2e ] 5   Eo = + 0.15 V

Overall reaction :

2 2

42MnO aq 5Sn aq 16H aq 2Mn      

(aq) + 5Sn4 (aq) + 8H 2 O 

4 2 2 4

o o

Sn / Sn Sn / Sn E E 0.15V      

2 4 2 4

o o o o o

cell cathode anode MnO /Mn Sn / Sn E E E E E      

= 1.51 – (– 0.15)

o

cellE 1.66V

As o cellE is +ve therefore the reaction will take place in

forward direction, i.e., favours the formation of products.

Example – 32

(a) Account for the following

(i) Alkaline medium inhibits the rusting of iron

(ii) Iron does not rust even if the zinc coating is broken

in a galvanized iron pipe.

(b) Cu2+ + 2e– Cu ; Eo = + 0.34 V

Ag+ + e–  Ag; Eo = + 0.80 V

(i) Construct a galvanic cell using the above data.

(ii) For what concentration of Ag+ ions will the emf of

the cell be zero at 25oC, if the concentration of Cu2+

is 0.01 M? [log 3.919 = 0.593]

Sol. (a) (i) The alkalinity of the solution prevents the

availability of H+ ions.

(ii) Zinc is more electropositive than iron.

Therefore, zinc coating acts anode and the

exposed iron portions act as cathode. If zinc

coating is broken, zinc undergoes corrosion,

protecting iron from rusting. No attack

occurs on iron till all the zinc is corroded.

(b) At anode Cu  Cu2+ + 2e–

At cathode [Ag+ + e–  Ag] × 2

Cu + 2 Ag+  Cu2+ + 2Ag

Cell representation

Cu | Cu2+ (conc.) || Ag+ (conc.) | Ag

2

o

cell cell 2

Cu0.059 E E log

n Ag





     

O = (0.80 – 0.34) 2 0.059 0.01

log 2 x

  !

2

0.01 15.59 log

x

    

 

–9x = 1.597 ×10 M

[Ag+] = 1.597 × 10–9M

Example – 33

(a) State advantages of H 2 -O

2 fuel cell over ordinary

cell.

(b) Silver is electrodeposited on a metallic vessel of total

surface area 500 cm2 by passing a current of 0.5 amp

for two hours. Calculate the thickness of silver

deposited.

[Given: Density of silver = 10.5 g cm–3, Atomic mass of

silver = 108 amu, F = 9,500 C mol–1]

Sol. (a) Advantages Fuels Cells:

1. It is a pollution-free device since no harmful

products are formed.

2. This is very efficient cell. Its efficiency is

about 75% which is considerably higher than

conventional cells.

3. These cells are light in weight as compared

to electrical generators to produce

corresponding quantity of power.

4. It is a continuous source of energy if the

supply of gases is maintained.

(b) Mass of silver deposited

m = z I t.

108 0.5 2 3600

96500 

m = 4.029 g

m m d v

v d   

34.029 V 0.3837 cm

10.5  

Let the thickness of silver deposited be x cm.

 V = A × x

 V

x A



0.3837 x

500 

 x = 7.67 × 10–4 cm.

Example – 34

(a) Give reasons for the following:

(i) Rusting of iron is quicker in saline water than in

ordinary water.

(ii) Resistance of a conductivity cell filled with 0.1 M

KCl solution is 100 ohm.If the resistance of the same

cell when filled with 0.02 M KCl solution is 520

ohms, calculate the conductivity and molar

conductivity of 0.02 M KCl solution.

(Conductivity of 0.1 M KCl solution is 1.29 Sm–1).

Sol. (a) (i) It is because in saline water, there is more H+ ions.

Greater the number of H+ ions, quicker the rusting.

(ii) Due to higher reduction potential of hydrogen

we get hydrogen at cathode.

(b) 1

R   cell constant

 cell constant = R

= 1.29 Sm–1 × 100 ohm

= 129 m–1 = 1.29 cm–1

For second solution

1

R   cell constant

3 11 1.29 2.48 10 S cm 520

   

m

1000

M

  

3 2.48 10 1000 248

0.02 2

  

2 1

m 124 S cm mol 

Example – 35

(a) Explain why electrolysis of aqueous solution of NaCl

gives H 2 at cathode and Cl

2 at anode. Write overall

reaction.

( 2 2 2

o o o

H O/ HNa / Na Cl / 2Cl E 2.71 V; E 0.83 V, E     

2 2

o

H O / H O 1.36 V;E 1.23 V)  

(b) Calculate the emf of the cell of Zn / Zn2+ (0.1 M) || Cd2+

(0.01 M) / Cd at 298 K,

[ 2 2 o o

Zn / Zn Cd / Cd Given E 0.76V and E    – 0.40 V]

Sol. (a) Because of higher reduction potential of water, water

is reduced in preference to sodium at therefore instea of

deposition of sodium metal, hydrogen is discharged at

cathode.

2 2H O 2e H g 2OH    

22Cl Cl g  

2 2 2 H O 2Cl H g Cl g 2OH    

At anode Cl 2 gas is liberated because of overpotential

of oxygen.

(b)

2

2

Zn Zn 2e Half cell reactions

Cd 2e Cd

 

 

"  # $

  #%

Zn + Cd2+  Zn2+ + Cd} cell reaction

o o o

cell cathode anodeE = E + E

= 0.76 – 0.40 = 0.36 V

o

cell

0.0591 E = E log Q

n 

2

2

0.0591 Zn 0.36 log

2 Cd





    !

0.0591 0.1 0.36 log

2 0.01

   ! = 0.33V

Example – 36

Three iron sheets have been coated separately with three

metals A, B and C whose standards reduction potentials

are given below.

metal A B C iron

o

ValueE – 0.46 V –0.66V –0.20V –0.44V

Identify in which case rusting will take place faster when

coating is damaged.

Sol. As iron (–0.44V) has lower standard reduction potential

than C (– 0.20V) only therefore when coating is broken,

rusting will take place faster.

235 ELECTROCHEMISTRY

236 ELECTROCHEMISTRY

Mahesh Tutorials Science

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

Basics of electrochemical Cell

1. Which of the following has been universally accepted as a

reference electrode at all temperatures and has been

assigned a value of zero volt ?

(a) platinum electrode (b) copper electrode

(c) graphite electrode (d) standard hydrogen electrode

2. The reaction 1/2H 2 (g) + AgCl(s) = H

+ (aq) + Cl

– (aq) + Ag(s)

occurs in the galvanic cell :

(a) Ag | AgCl(s) | KCl(sol.) || AgNO 3 ) (sol.) | Ag

(b) Pt | H 2 (g) | HCl(sol. || AgNO

3 (sol) | Ag

(c) Pt | H 2 (g) | HCl (sol.) || AgCl(s) | Ag

(d) Pt | H 2 (g) | KCl (sol.) || AgCl(s) | Ag

3. The equation representing the process by which standard

reduction potential of zinc can be defined is :

(a) Zn 2+ (s) + 2e

–  Zn

(b) Zn (g)  Zn2+(g) + 2e–

(c) Zn 2+ (g) + 2e

–  Zn

(d) Zn 2+ (aq.) + 2e

–  Zn (s)

4. Which of the following statement is wrong about galvanic

cell ?

(a) cathode is positive charged

(b) anode is negatively charged

(c) reduction takes place at the anode

(d) reduction takes place at the cathode

5. Which are used as secondary reference electrodes ?

(a) Calomel electrode (b) Ag/AgCl electrode

(c) Hg/Hg 2 Cl

2 – KCl electrode

(d) All of the above

Applications of Electrochemical Series

6. The standard electrode potentials (reduction) of

Pt/Fe 3+ , Fe

2+ and Pt/Sn

4+ , Sn

2+ are + 0.77 V and 0.15 V

respectively at 25°C. The standard EMF of the reaction

Sn 4+ + 2Fe

2+  Sn2+ + 2Fe3+ is

(a) – 0.62 V (b) – 0.92 V

(c) + 0.31 V (d) + 0.85 V

7. Adding powdered Pb and Fe to a solution containing 1.0 M

is each of Pb 2+ and Fe

2+ ions would result into the formation

of (Eº Pb +2 /Pb = – 0.13V, Eº Fe

+2 /Fe = –0.44V)

(a) More of Pb and Fe 2+ ions

(b) More of Fe and Pb 2+ ions

(c) More of Fe and Pb (d) More of Fe 2+ and Pb

2+ ions

8. Strongest reducing agent is :

(a) K (b) Mg

(c) Al (d) I

9. Zn can not displace following ions from their aqueous

solution :

(a) Ag +

(b) Cu 2+

(c) Fe 2+

(d) Na +

10. Which of the following displacement does not occur :

(a) Zn + 2H +  Zn2+ + H

2&

(b) Fe + 2Ag +  Fe2+ + Ag'

(c) Cu + Fe 2+  Cu2+ + Fe'

(d) Zn + Pb 2+  Zn2+ + Pb '

11. The oxidation potential of Zn, Cu, Ag, H 2 and Ni are 0.76,

–0.34, – 0.80, 0, 0.55 volt respectively. Which of the following

reaction will provide maximum voltage ?

(a) Zn + Cu 2+  Cu + Zn2+

(b) Zn + 2Ag +  2Ag + Zn2+

(c) H 2 + Cu

2+  2H+ + Cu

(d) H 2 + Ni

2+  2H+ + Ni

12. The position of some metals in the electrochemical series in

decreasing electropositive character is given as

Mg > Al > Zn > Cu > Ag. What will happen if a copper

spoon is used to stir a solution of aluminium nitrate ?

(a) The spoon will get coated with aluminium

(b) An alloy of copper and aluminium is formed

(c) The solution becomes blue

(d) There is no reaction

ELECTROCHEMISTRY 236

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