Chapter 3 Electrochemistry free study material by TEACHING CARE online tuition and coaching classes

 

Electrochemistry is the branch of physical chemistry which deals with the relationship between electrical energy and chemical changes taking place in redox reactions i.e., how chemical energy produced in a redox reaction can be converted into electrical energy or how electrical energy can be used to bring about a redox reaction which is otherwise non-spontaneous. Chemical changes involving production or consumption of electricity are called electrochemical changes.

• Definition : “The substances whose aqueous solution undergo decomposition into ions when electric current is passed through them are known as electrolytes and the whole process is known as electrolysis or electrolytic decomposition.”

Solutions of acids, bases, salts in water and fused salts etc. are the examples of electrolytes. Electrolytes may be weak or strong. Solutions of cane sugar, glycerine, alcohol etc., are examples of non-electrolytes.

• Electrolytic cell or Voltameter : The device in which the process of electrolysis or electrolytic decomposition is carried out is known as electrolytic cell or voltameter. Following are the important characteristics of voltameter,
• Voltameter consist of a vessel, two electrodes and electrolytic
• Voltameter convert electrical energy into chemical energy e., electrical energy is supplied to the electrolytic solution to bring about the redox reaction (i.e., electrolysis) which is non- spontaneous and takes place only when electrical energy is supplied.
• In voltameter, both the electrodes are suspended in only one of the electrolytic solution or melt of the electrolyte in the same
• The electrodes taken in voltameter may be of the same or different
• The electrode on which oxidation takes place is called anode (or +ve pole) and the electrode on which reduction takes place is called cathode (or –ve pole)
• During electrolysis in voltameter cations are discharged on cathode and anions on anode.
• In voltameter, outside the electrolyte electrons flow from anode to cathode and current flow from

cathode to anode.

 

• For voltameter, E

cell

= -ve

and ΔG = +ve.

 

(3)  Mechanism of electrolysis

• The net chemical change that takes place in the cell is

called the cell reaction, which is non-spontaneous in case of electrolytic cell.

 

 

 

• The mechanism of electrolysis can be easily explained on the basis of ionisation theory according to which, the electrolytes are present in the form of ions in solution and the function of electricity is only to direct these ions to their respective
• During electrolysis cations move towards the cathode (–vely charged) while anions move towards the anode (+vely charged).
• The anions on reaching the anode give up their electrons and converted into the neutral

At anode : A – ¾¾¾® A + e – (Oxidation)

On the other hand cations on reaching the cathode take up electrons supplied by battery and converted to the neutral atoms.

At cathode : B+ + e – ¾¾¾® B (Reduction)

This overall change is known as primary change and products formed is known as primary products.

• The primary products may be collected as such or they undergo further change to form molecules or These are called secondary products and the change is known as secondary change.
• The products of electrolysis depend upon,

• Nature of electrolyte,
• Concentration of electrolyte,
• Charge density flown during electrolysis,
• Nature of electrodes used,
• Electrodes which do not take part in chemical change involved in a cell are known as Inert electrode. Graphite, Platinum.
• Electrodes which take part in chemical change involved in a cell knows as active electrode. Hg, Au, Cu, Fe, Zn and Ni electrode etc.
• Over voltage : For metal ions to be deposited on the cathode during electrolysis, the voltage required is almost the same as the standard electrode However for liberation of gases, some extra voltage is required

 

than the theoretical value of the standard electrode potential

voltage or bubble voltage.

(E ⁰ ).

This extra voltage required is called over

 

• The deposition of different ions at the electrodes takes place only for the time of electricity is passed and stops as soon as electricity is switched

Note : ® The electrolyte as a whole remains neutral during the process of electrolysis as equal number of charges are neutralised at the electrodes.

• If the cathode is pulled out from the electrolytic solution of the cell then there will be no passage of current and ions will show simply diffusion and state moving
• If electrode is active at cathode, metal goes on depositing on cathode and at anode metal is

(4)  Preferential discharge theory

• According to this theory “If more than one type of ion is attracted towards a particular electrode, then the ion is discharged one which requires least energy or ions with lower discharge potential or which occur low in the electrochemical series”.
• The potential at which the ion is discharged or deposited on the appropriate electrode is termed the

discharge or deposition potential, (D.P.). The values of discharge potential are different for different ions.

• The decreasing order of discharge potential or the increasing order of deposition of some of the ions is given below,

For cations : Li ⁺ , K ⁺ , Na ⁺ , Ca ²⁺ , Mg ²⁺ , Al ³⁺ , Zn²⁺ , Fe ²⁺ , Ni ²⁺ , H ⁺ , Cu²⁺ , Hg ²⁺ , Ag ⁺ , Au³⁺ .

 

 

 

For anions : SO^2– , NO^– , OH ^– , Cl ^– , Br ^– , I ^– .

4            3

• Electrolysis and electrode processes : The chemical reactions which take place at the surface of electrode are called electrode reactions or electrode processes. Various types of electrode reactions are described below,

 

• Electrolysis of molten sodium chloride : Molten sodium chloride contains

(l)

(l)

NaCl(l) ⇌ Na⁺ + Cl ^–

Na ⁺

and Cl ^– ions.

 

On passing electricity,

Na⁺ ions move towards cathode while

Cl ^– ions move towards anode. On reaching

 

cathode and anode following reactions occur.

 

At cathode :

+

Na

(l)

• e^– ® Na(l)

¯ (Reduction, primary change)

 

 

(l)

At anode : Cl^–

– e^– ®

Cl (Oxidation, primary change)

 

 

• ·

Cl+ Cl ® Cl2(g ) ­ (Secondary change)

 

Overall reaction :

+

2Na

(l)

• 2Cl ^–

¾¾Elec¾troly¾sis ® 2Na(l)

• Cl

2(g)

 

(l)

The sodium obtained at the cathode is in the molten state.

• Electrolysis of molten lead bromide : Molten lead bromide contains

Pb ²⁺ and

Br ^– ions.

 

(l)

PbBr2

⇌ Pb 2+   + 2Br –

 

(l)

(l )

(l )

 

On passing electricity,

Pb 2+

ions move towards cathode while

Br ^– ions move towards anode. On reaching

 

cathode and anode following reactions occur,

 

At cathode :

Pb²⁺ + 2e^– ® Pb

¯ (Reduction, primary change)

 

(l)                            (l)

 

 

At anode :

(l )

Br – – e – ® Br

(Oxidation, primary change)

 

 

• ·

Br + Br ® Br2(g ) ­ (Secondary change)

 

Overall reaction :

Pb2+  + 2Br –  ¾¾Elec¾troly¾sis ® Pb

• Br

 

(l)

(l)

(l)

2(g)

 

The lead obtained at the cathode is in the molten state.

• Electrolysis of water : Water is only weakly ionized so it is bad

 

conductor of electricity but the presence of an acid

(H₂SO₄ ) increases the

 

degree of ionization of water. Thus, in solution, we have three ions, i.e.,

4

H ⁺ , OH ^– and SO^2– produced as follows,

 

H2O(l) ⇌ H +(aq)  + OH -(aq) ;

H2SO4

 

(aq)

® 2H +(aq) + SO2-

4

(aq)

 

When electric current is passed through acidified water,

H ⁺ ions move

 

4

towards cathode,

OH ^– and SO^2–

ions move towards anode. Now since the

 

4

discharge potential of

OH ^– ions is much lower than that of

SO2-

ions,

 

 

 

 

therefore,

OH ^–

ions are discharged at anode while

SO2-

ions remain in solution. Similarly,

H ⁺ ions are

 

4

discharged at the cathode. The reactions, occurring at the two electrodes may be written as follows,

At cathode : H ⁺(ag) + e ^– ® H.(Reduction, primary change)

• ·

 

H+ H ® H 2(g )

(Secondary change)

 

At anode : OH ^–(aq) + e ^– ® OH (Oxidation, primary change) 4OH ® 2H 2 O(l) + O2 (Secondary change)

 

Overall reaction :

4 H +(aq) + 4OH -(aq) ® 2H 2        + 2H 2 O(l)  + O2       .

 

(aq)                                 (g)

4

• Electrolysis of aqueous copper sulphate solution using inert electrodes : Copper sulphate and water ionize as under,

 

CuSO4

(aq)

® Cu2+(aq)  + SO2-

(aq)

(almost completely ionized)

 

H 2 O(l)  ⇌ H +(aq)  + OH -(aq)

(only slightly ionized)

 

4

On passing electricity, Cu²⁺ (aq) and H⁺(aq) move towards the cathode while SO ^2– ions and OH ^– ions move

 

towards the anode. Now since the discharge potential of Cu ²⁺

ions is lower than that of H ⁺

ions, therefore,

 

Cu²⁺ ions are discharged at cathode. Similarly, OH ^– ions are discharged at the anode. the reactions, occuring at the two electrodes may be written as follows,

At cathode : Cu²⁺ + 2e ^– ® Cu (Reduction, primary change)

 

At anode : 4OH ^– – 4e ^– ® 4OH

(Oxidation, primary change)

 

4OH ® 2H 2 O(l ) + O2 (g ) (Secondary change)

4

• Electrolysis of an aqueous solution of copper sulphate using copper electrode : Copper sulphate and water ionize as under:

 

CuSO4

(aq)

® Cu2+(aq)  + SO2-

(aq)

(almost completely ionized)

 

H ₂O _(l) ⇌ H ⁺ _(aq)+ OH ^–(aq) (weakly ionized)

 

On passing electricity,

Cu²⁺ and H ⁺

move towards cathode while

OH ^– and SO^2–

ions move towards anode.

 

4

Now since the discharge potential of

Cu2+

ions is lower than that of

H ⁺ ions, therefore, Cu²⁺

ions are discharged

 

in preference to

H ⁺ ions at cathode. Unlike electrolysis of

CuSO₄ using platinum electrodes, no ions are

 

4

liberated here, instead, anode itself undergoes oxidation (i.e., loses electrons) to form

Cu 2+

ions which go into the

 

solution. This is due to the reason that Cu is more easily oxidised than both occuring at the two electrodes may be written as follows,

At cathode : Cu²⁺(aq) + 2e ^– ® Cu_(s) (Reduction, primary change).

At anode : Cu_(s) ® Cu²⁺(aq) + 2e ^– (Oxidation, primary change).

OH ^– and SO^2–

ions. The reactions,

 

Products of electrolysis of some electrolytes

Electrolyte	Electrode	Product	at	cathode	Product at anode
Aqueous NaOH Fused NaOH Aqueous NaCl Fused NaCl Aqueous CuCl2 Aqueous CuCl2 Aqueous CuSO4	Pt or Graphite Pt or Graphite Pt or Graphite Pt or Graphite Pt or Graphite Cu electrode Pt or Graphite	H2  Na  H2  Na  Cu  Cu  Cu			O2  O2  Cl2  Cl2  Cl2  Cu oxidised to Cu2+ ions  O2

 

 

 

 

Note : ® Alkali metals, alkaline earth metals and other metals having

E ⁰ lower than hydrogen can not be

 

obtained during electrolysis of their aqueous salt solutions because of their strong electropositive nature.

• The electrolysis of aqueous solution of electrolytes is some what more complex because of the ability of water to be oxidised as well as reduced.
• Application of electrolysis: Electrolysis has wide applications in Some of the important applications are, as follows,
• Production of hydrogen by electrolysis of
• Manufacture of heavy water (D₂O) .
• Electrometallurgy : The metals like Na, K, Mg, Al, , are obtained by electrolysis of fused electrolytes.

• Manufacture of non-metals : Non-metals like hydrogen, fluorine, chlorine are obtained by
• Electro-refining of metals : It involves the deposition of pure metal at cathode from a solution containing the metal Ag, Cu etc. are refined by this method.
• Electrosynthesis : This method is used to producing substances through non-spontaneous reactions

 

carried out by electrolysis. Compounds like NaOH, KOH,

by this method.

Na2CO3 , KClO3 , white lead,

KMnO₄ etc. are synthesised

 

• Electroplating : The process of coating an inferior metal with a superior metal by electrolysis is known as electroplating. The aim of electroplating is, to prevent the inferior metal from corrosion and to make it more attractive in The object to be plated is made the cathode of an electrolytic cell that contains a solution of ions of the metal to be deposited.

 

For electroplating	Anode	Cathode	Electrolyte
With copper	Cu	Object	CuSO4 + dilute H 2SO4
With silver	Ag	Object	K[Ag(CN)2]
With nickel With gold	Ni Au	Object Object	Nickel ammonium sulphate  K[Au(CN)2]

 

 

 

Thickness of coated layer : Let the dimensions of metal sheet to be coated be (a cm ´ b cm). Thickness of coated layer = c cm

Volume of coated layer = (a ´ b ´ c) cm³

 

Mass of the deposited substance = Volume ´ density

 

\

= (a ´ b ´ c) ´ dg

 

 

Using above relation we may calculate the thickness of coated layer.

Note : ®          Sometimes radius of deposited metal atom is given instead of density.

For example, radius of silver atom = 10^-8 cm ; Atomic mass of Ag = 108

Mass of single silver atom =             108       g

6.023 ´ 10²³

 

Volume of single atom = 4 ´ p R³

3

= 4 ´ 3.14 ´ (10^-8 )³ cm³

3

 

Density of Ag =  Mass of single atom  = 108 / 6.023 ´ 10²³

= 42.82 g / cm³

 

Volume of single atom

4 ´ 3.14 ´ (10^-8 )³

3

 

 

The laws which govern the deposition of substances (In the form of ions) on electrodes during the process of electrolysis is called Faraday’s laws of electrolysis. These laws given by Michael Faraday in 1833.

• Faraday’s first law : It states that,

“The mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed.”

i.e., W µ Q

Where,  W= Mass of ions liberated in gm,

Q = Quantity of electricity passed in Coulombs = Current in Amperes (I) × Time in second (t)

W µ I ´ t or W = Z ´ I ´ t

\

In case current efficiency (h) is given, then

where, Z = constant, known as electrochemical equivalent (ECE) of the ion deposited.

When a current of 1 Ampere is passed for 1 second (i.e., Q = 1 ), then, W = Z

Thus, electrochemical equivalent (ECE) may be defined as “the mass of the ion deposited by passing a current of one Ampere for one second (i.e., by passing Coulomb of electricity)”. It’s unit is gram per Coulomb. The ECE values of some common elements are,

 

 

 

Element			Hydrogen	Oxygen	Copper	Silver	Iodine	Mercury
	Z, g C–1	1.045 ´ 10–5		8.29 ´ 10–5	3.294 ´ 10–4	1.18 ´ 10–3	1.315 ´ 10–3	1.039 ´ 10–3

Note : ®          Coulomb is the smallest unit of electricity.

 

• 96500 Coulombs

= 6.023 ´ 10²³

electrons.

 

• 1 Coulomb =

6.023 ´ 10²³

96500

= 6.28 ´ 10¹⁸

electrons, or 1 electronic charge = 1.6 ´ 10

-19

Coulomb.

 

• Faraday’s second law : It states that,

“When the same quantity of electricity is passed through different electrolytes, the masses of different ions liberated at the electrodes are directly proportional to their chemical equivalents (Equivalent weigths).” i.e.,

 

W1 = E1 W2      E2

or Z1 It

Z2 It

= E1 or

E2

Z1 = E1 Z2          E2

(Q W = ZIt)

 

Thus the electrochemical equivalent (Z) of an element is directly proportional to its equivalent weight (E), i.e.,

where, F = Faraday constant = 96500 Cmol ^-1

So, 1 Faraday = 1F =Electrical charge carried out by one mole of electrons.

1F = Charge on an electron × Avogadro’s number. 1F = e – ´ N = (1.602 ´ 10 -19 c) ´ (6.023 ´ 10 23 mol -1 ).

Number of Faraday = Number of electrons passed

6.023 ´ 10 23

• Faraday’s law for gaseous electrolytic product : For the gases, we use

 

where,

V = Volume of gas evolved at S.T.P. at an electrode

Ve = Equivalent volume = Volume of gas evolved at an electrode at S.T.P. by 1 Faraday charge

 

Examples : (a) O₂ : M = 32, E = 8 ; 32 g O₂ º 22.4 L at S.T.P. [M = Molecular mass, E = Equivalent mass]

8 g O2 º 5.6L at S.T.P.; Thus Ve of O2 = 5.6L.

 

(b) H2 :

M = 2, E = 1;

2 g H₂ º 22.4 L

at S.T.P.

 

1 g H₂ º 11.2L

at S.T.P.

 

Thus

Ve of H2 = 11.2L.

 

(c) Cl2 : M = 71, E = 35.5 ; 71 g Cl2 º 22.4 L at S.T.P.

 

35.5 g Cl₂ º 11.2L

at S.T.P.

 

Thus Ve of Cl₂ = 11.2L.

• Quantitative aspects of electrolysis : We know that, one Faraday (1F) of electricity is equal to the charge carried by one mole (6.023 ´ 10²³ ) of electrons. So, in any reaction, if one mole of electrons are involved, then that reaction would consume or produce 1F of electricity. Since 1F is equal to 96,500 Coulombs, hence 96,500 Coulombs of electricity would cause a reaction involving one mole of

 

Q = nF = n ´ 96,500 C

If in any reaction, n moles of electrons are involved, then the total electricity given by,

Thus, the amount of electricity involved in any reaction is related to,

• The number of moles of electrons involved in the reaction,
• The amount of any substance involved in the

(Q)

involved in the reaction is

 

 

 

Therefore, 1 Faraday or 96,500 C or 1 mole of electrons will reduce,

 

• 1 mole of monovalent cation, (b)

1 mole of divalent cation,

2

 

(c)

1 mole of trivalent cation,                  (d)

3

1 mole of n valent cations.

n

 

Example : 1       By passing 0.1 Faraday of electricity through fused sodium chloride, the amount of chlorine liberated is

[Tamil Nadu CET 2002]

(a) 35.45 g                              (b) 70.9 g                                    (c) 3.55 g                            (d) 17.77 g

 

Solution: (c)

Cl ^– ® 1 Cl2 + e ^– i.e., 1Faraday liberate Cl2 = 1 mol = 1 ´ 71 = 35.5

(Q wt.of Cl = 35.5)

 

2                                                        2           2

\ 0.1Faraday will liberate Cl₂ = 35.5 ´ 0.1 = 3.55 g.

Example : 2 Silver is monovalent and has atomic mass of 108. Copper is divalent and has an atomic mass of 63.6. The same electric current is passed for the same length of time through a silver coulometer and a copper coulometer. If 27.0 g of silver is deposited, then the corresponding amount of copper deposited is

[Kerala CEE 2002]

(a) 63.60 g                              (b)  31.80 g                                  (c) 15.90 g                          (d) 7.95 g

 

Solution: (d)

Wt.of Cu deposited

= Eq. Wt. of Cu ;

Wt. of Cu deposited

 

= 63.6 / 2

 

Wt.of Ag deposited

Eq. Wt. of Ag

27.0 g

108

 

Wt. of Cu deposited = 63.6 / 2 ´ 27.0 g = 7.95 g.

108

Example : 3       How much charge must be supplied to a cell for the electrolytic production of 245 g

Because of a side reaction, the anode efficiency for the desired reaction is 60%

 

NaClO4

 

from

 

NaClO₃ ?

 

(a)

6.43 ´ 10⁵ C

(b)

6.67F

(c)

6.43 ´ 10⁶ C

(d)

66.67F

 

Solution: (a,b) ClO^– + 2H ⁺ + 2e ^– ® ClO^– + H O

4                                                               3               2

 

Eq. wt. of

NaClO4

= 23 + 35 + 64 = 61.25

2

 

No. of equivalents of

NaClO4

= 245 = 4 º 4.0F 61.25

 

The anode efficiency is 60%

No. of Faradays = 4.0 ´ 100 = 6.67F

60

6.67F = 6.67 ´ 9.65 ´ 10⁴ C = 6.43 ´ 10⁵ C.

Example : 4 When 9.65 Coulombs of electricity is passed through a solution of silver nitrate (At. wt. of Ag =107.87 taking as 108) the amount of silver deposited is                                                                               [EAMCET 1992] (a) 10.8 mg                                                    (b) 5.4 mg                                   (c) 16.2 mg                         (d) 21.2 mg

 

Solution: (a)

W =    EIt 96500

= 108 ´ 9.65 = 0.0108 g = 10.8 mg.

96500

 

Example : 5       Three faradays of electricity are passed through molten

Al2O3 , aqueous solution of

CuSO4

and molten

 

NaCl taken in different electrolytic cells. The amount of Al Cu and Na deposited at the cathodes will be in the ratio of                                                                                                                                             [BHU 1990]

 

(a (c)

1mole : 2mole : 3mole 1mole : 1.5mole : 3mole

(b)

(d)

3mole : 2mole : 1mole 1.5mole : 2mole : 3mole

 

Solution: (c)       Eq. of Al = Eq. of Cu = Eq. of Na, or 1 mole Al = 1 mole Cu =1 mole Na

 

3                 2

or 2 : 3 : 6 or 1 : 1.5 : 3 mole ratio.

 

 

 

 

Example : 6       When electricity is passed through the solution of

 

AlCl₃, 13.5 gm of Al are deposited. The number of Faraday

 

must be                                                                                        [NCERT 1974; MP PET 1992; PM PMT 1994]

(a) 0.50                           (b) 1.00                              (c) 1.50                        (d) 2.00

 

Solution (c)        Eq. of Al =

13.5

 

27 / 3

= 1.5 ; Thus 1.5 Faraday is needed.

 

Example : 7       A metal wire carries a current of 1 A. How many electrons move past a point in the wire in one second

 

(a)

6.02 ´ 10 ²³

(b)

3.12 ´ 10¹⁸

(c)

3.02 ´ 10 ²³

(d)

6.24 ´ 10¹⁸

 

Solution (d)       Total charge passed in one second ‘ Q‘ = I ´ t = 1 ´ 1 = 1c

Q 96500 current carried by 6.02 ´ 10 ²³ electrons

 

\ 1C current carried by

6.02 ´ 10 ²³

96500

= 6.24 ´ 10¹⁸

 

Example : 8       How long will it take for a current of 3 Amperes to decompose 36 g of water? (Eq. wt. of hydrogen is 1 and that of oxygen 8)                                                                                                                                          [BHU 1992]

• 36 hours (b) 18 hours                                (c) 9 hours                          (d) 5 hours

 

Solution (a)       Eq. wt. of

H₂O = 1 + 8 = 9. Hence 9 g is decomposed by 96500 C. So 36 g of water will be decomposed by

 

Q = 96500 ´ 4C

Hence t = Q = 96500 ´ 4 S = 128666.66 sec . = 35.7 hrs ⋍ 36 hrs.

I               3

Example : 9       Three Ampere current was passed through an aqueous solution of an unknown salt of metal M for 1hour.

 

2.977 g of M ⁿ⁺

was deposited at cathode. The value of n is (At wt. of M =106.4).

 

(a) 1                                (b) 2                                   (c) 3                             (d) 4

 

Solution (d)

M ⁿ⁺ + ne ^– ® M ;

W =  I ´ t   Þ

2.977

= 3 ´ 1 ´ 60 ´ 60

(1 hour = 60×60);

n = 4.

 

E       96500

106.4 / n

96500

 

Example : 10 Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume of the two gases (dry and S.T.P.) produced will be approximately (in litres)                        [AIIMS1993] (a) 22.4          (b) 44.8                                 (c) 67.2                              (d) 89.4

 

Solution (c)        At cathode: 4 H ⁺ + 4e ^– ® 2H2

or 2 ×22.4 litres at S.T.P.

 

(4 F )

2moles

 

At anode: 2O ^2– ®

O2  + 4e ^–

or 22.4 litres at S.T.P.

 

1mole

(4 F )

 

\ Total volume of the two gases produced at S.T.P. = 2 ´ 22.4 + 22.4 = 67.2 litres

Example : 11 A current of 2 Amperes is passed through a litre of 0.100 molar solution of copper sulphate for 4825 seconds. The molarity of copper sulphate at the end of experiment is                                                             [ISM, Dhanbad 1995] (a) 0.075                                       (b) 0.050                            (c) 0.025                      (d) 0

 

Solution (b)

Cu²⁺ + 2e ^– ® Cu

Q = 2 ´ 4825 = 9650 Coulombs

\ 96500 Coulombs of electricity deposits 0.5 mole copper

\ 9650 C = 0.5 ´ 9650 = 0.05 mole of copper

96500

\ Molarity at the end of the experiment = 0.1 – 0.05 = 0.050

 

Example : 12 2.5 Faradays of electricity is passed through a solution of CuSO₄ . The number of gram equivalents of copper deposited on the cathode would be                                                           [DPMT 1982; CPMT 1973; MP PMT 2001] (a) 1     (b) 2     (c) 2.5                                                                             (d) 1.25

Solution (c)  Number of Faraday passed = Number of equivalents deposited.

 

 

 

Example : 13 How many c.c. of oxygen will be liberated by 2 Ampere current flowing for 193 seconds through acidified water                                                                                                                                           [AFMC 1981]

(a) 11.2 c.c.                             (b) 33.6 c.c.                                 (c)  44.8 c.c.                         (d) 22.4 c.c.

 

Solution (d)

Q = I ´ t = 2 ´ 193 = 386 C

H2O ® H2 + 1 O2  i.e.,  O^2– ® 1 O2 + 2e ^–

 

i.e., 2F = 2 ´ 96500 gives

 

2                        2

2

O = 1 mole =11200 c.c.

2

 

\ 386 c will give O₂

=     11200

2 ´ 96500

´ 386 = 22.4 c.c.

 

4

Example : 14 How many Coulombs of electricity are required for reduction of 1 mole of MnO^– to Mn²⁺

 

(a) 96500 C                              (b)

1.93 ´ 10⁵ C

(c)

4.83 ´ 10⁵ C

(d)

9.65 ´ 10⁶ C

 

Solution (c)

(MnO4 )^– ® Mn²⁺

i.e., Mn⁷⁺ + 5e ^– ® Mn²⁺

 

Thus 1 mole of

MnO^– requires 5 F = 5 ´ 96500 C = 482500 ≃ 4.83 ´ 10⁵ C.

 

4

Example : 15 The quantity of electricity required to reduce 12.3 g of nitrobenzene to aniline assuming 50% current efficiency is

• 115800 Coulombs (b) 57900 Coulombs            (c) 231600 Coulombs    (d) 28950 Coulombs

 

Solution (a)

C6 H5 NO2 + 6H ⁺ + 6e ^– ® C6 H5 NH2

i.e., 1 mole (123 g) require

6 ´ 96500 C. Hence 12.3 g will require

 

= 6 ´ 9650C. As current efficiency is 50%, quantity of electricity actually required = 2 ´ 6 ´ 9650 = 115800C.

 

Example : 16 A current of 3 ampere was passed for 2 hours through a solution of at cathode. The current efficiency is (At. wt. of Cu=63.5)

CuSO4 .3g of Cu ²⁺ ions were discharged

 

(a) 20%                           (b) 42.2%                           (c) 60%                        (d) 80%

 

Solution (b)

WCu2+

= E ´ I ´ t 96500

 

3 = 63.5 ´ I ´ 2 ´ 60 ´ 60

2 ´ 96500

I = 1.266 Ampere

Current efficiency =

 

 

Current passed actually

 

 

´ 100

 

= 1.266 ´ 100 = 42.2%

 

Total current passed experimentally                     3

• Conductors and Non – conductors : All substances do not conduct electrical current. The substances which allow the passage of electric current are called conductors. The best metal conductors are such as copper, silver, tin, On the other hand, the substances which do not allow the passage of electric current through them are called non-conductors or insulators. Some common examples of insulators are rubber, wood, wax, etc.
• Types of conductors : The conductors are broadly classified into two types,
  • Metallic conductors or electronic conductors
• In metallic conductors, flow of electricity takes place without the decomposition of the
• Flow of electricity is due to the flow of electrons only e., there is no flow of matter.
• In addition to metals, graphite and certain minerals also conduct electricity due to presence of free electrons in them, hence they are collectively called as electronic conductors.
• Metallic conduction decreases with increase of This is because kernels start vibrating which produce hinderance in the flow of electrons.

 

 

 

• The resistance offered by metals is also due to vibrating
• Metallic conductors obey Ohm’s
  • Electrolytic conductors or Ionic conductors
• In electrolytic conductors flow of electricity takes place by the decomposition of the substance (Electrolyte).
• Flow of electricity is due to the movement of ions and hence there is flow of
• Solutions of acids, bases and salts are the examples of electrolytic
• The electrolytic conduction will not occur unless the ions of the electrolyte are free to move. Therefore, these substances do not conduct electricity in the solid state but conduct electricity in the molten state or in their aqueous
• The electrical conduction increases with increase of temperature. This is generally due to increase in dissociation or decrease in the interionic
• The resistance shown by an electrolytic solution is due to factors like interionic attractions, viscosity of solvent etc.
• Electrolytic conductors also obey Ohm’s
• All electrolytes do not ionise to the same extent in solution. On this basis, electrolytes are broadly divided into two types: strong electrolytes and weak

Strong electrolytes : The electrolytes which are almost completely dissociated into ions in solution are

 

called strong electrolytes. For example,

NaCl, KCl, HCl, NaOH, NH 4 NO3 , etc.

 

Weak electrolytes : The electrolytes which do not ionise completely in solution are called weak electrolytes.

 

For example,

CH3 COOH, H 2 CO3 , H3 BO3 , HCN, HgCl2 , ZnCl2 , NH 4 OH, etc. Thus in case of weak electrolytes, an

 

equilibrium is established between the unionised electrolyte and the ions formed in solution. The extent of ionisation of a weak electrolyte is expressed in terms of degree of ionisation or degree of dissociation. It is defined as the fraction of total number of molecules of the electrolyte which ionise in the solution. It is generally denoted by alpha (a).For strong electrolytes, a is almost equal to 1 and for weak electrolytes, it is always less than 1.

The electrical conductivity of the solutions of electrolytes depends upon the following factors,

• Interionic attractions : These depend upon the interactions between the ions of the solute molecules, e., solute-solute interactions. If the solute-solute interactions are large, the extent of dissociation will be less. These interactions are also responsible for the classification of electrolytes as strong electrolytes and weak electrolytes.
• Solvation of ions : These depend upon the interactions between the ions of the solute and the molecules of the solvent and are called solute-solvent interactions. If the solute-solvent interactions are strong, the ions of the solute will be highly solvated and their electrical conductivity will be

• Viscosity of the solvent: The viscosity of the solvent depends upon the solvent-solvent interactions. Larger the solvent-solvent interactions, larger will be the viscosity of the solvent and lower will be the electrical conductivity of the

When a voltage is applied to the electrodes dipped into an electrolytic solution, ions of the electrolyte move and, therefore, electric current flows through the electrolytic solution. The power of the electrolytes to conduct electric current is termed conductance or conductivity.

I µ V

• Ohm’s law : This law states that the current flowing through a conductor is directly proportional to the potential difference across it, e.,

where I is the current strength (In Amperes) and V is the potential difference applied across the conductor (In Volts)

 

 

 

 

or I = V

R

or V = IR

 

where R is the constant of proportionality and is known as resistance of the conductor. It is expressed in Ohm’s and is represented as W. The above equation is known as Ohm’s law. Ohm’s law may also be stated as,

“the strength of current flowing through a conductor is directly proportional to the potential difference applied across the conductor and inversely proportional to the resistance of the conductor.”

• Resistance : It measures the obstruction to the flow of current. The resistance of any conductor is directly proportional to the length (l) and inversely proportional to the area of cross-section (a) so that

where r (rho) is the constant of proportionality and is called specific resistance or resistivity. The resistance depends upon the nature of the material.

Units : The unit of resistance is ohm (W). In terms of SI, base unit is equal to (kgm 2 ) / (s 3 A 2 ).

• Resistivity or specific resistance : We know that resistance R is

R = r l

a

Now, if l = 1 cm, a = 1 cm² then R = r

Thus, resistivity is defined as the resistance of a conductor of 1 cm length and having area of cross-section equal to 1 cm².

 

Units : The units of resistivity are

r = R. a = Ohm cm2   = Ohm. cm

 

l                   cm

Its SI units are Ohm metre (W m). But quite often Ohm centimetre (W cm) is also used.

• Conductance : It is a measure of the ease with which current flows through a It is an additive property. It is expressed as G. It is reciprocal of the resistance, i.e.,

Units : The units of conductance are reciprocal Ohm (ohm^-1) or mho. Ohm is also abbreviated as W so that

Ohm-1 may be written as W^–1.

According to SI system, the units of electrical conductance are Siemens, S (i.e., 1S = 1 W^–1).

• Conductivity : The inverse of resistivity is called conductivity (or specific conductance). It is represented by the symbol, k (Greek kappa). The IUPAC has recommended the use of term

conductivity over specific conductance. It may be defined as, the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. In other words, conductivity is the conductance of one centimetre cube of a solution of an electrolyte. Thus,

Units : The units of conductivity are k =       1      = Ohm-1 cm^–1 or W^–1 cm^-1

Ohm. cm

In SI units, l is expressed in m area of cross-section in m² so that the units of conductivity are S m^-1.

 

 

 

• Molar conductivity or molar conductance : Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in

It is denoted by L (lambda). Molar conductance is related to specific conductance (k ) as,

 

L =	k
	M

where, M is the molar concentration. If M is in the units of molarity i.e., moles per litre (mol L^-1), the L may be expressed as,

For the solution containing 1 gm mole of electrolyte placed between two parallel electrodes of 1 sq. cm area of cross-section and one cm apart,

But if solution contains 1 gm mole of the electrolyte therefore, the measured conductance will be the molar conductivity. Thus,

(L) = k ´ V

In other words,

where V is the volume of the solution in cm³ containing one gram mole of the electrolyte. If M is the concentration of the solution in mole per litre, then

M mole of electrolyte is present in 1000 cm³

1 mole of electrolyte is present in = 1000 cm³ of solution

M

Thus, L = k ´ Volume in cm³ containing 1 mole of electrolyte. or

Units of Molar Conductance : The units of molar conductance can be derived from the formula ,

L = k ´ 1000

M

The units of k are Scm^-1 and units of L are, Λ = S cm^-1 ´ cm3 = Scm² mol^-1 = S cm²mol^-1

mol

 

According to SI system, molar conductance is expressed as

mol m^-3. This is because

mol m^-3 = 1000æ L ö´ molarityæ mol ö

Sm²mol ^-1, if concentration is expressed as

 

ç m3 ÷               ç L   ÷

 

è      ø

Now, L = k   =

è        ø

k (Sm^-1 )

= Sm² mol ^-1

 

M        (1000 Lm^-3 ) ´ (Molarity mol L^-1 )

 

Thus, the units of molar conductivity are literature and are related to each other as

Sm² mol ^-1 (SI) and

S cm²mol ^-1.

Both types of units are used in

 

1Sm²mol ^-1 = 10⁴ Scm²mol ^-1 or 1S cm²mol ^-1 = 10^-4 Sm²mol ^-1

 

 

 

• Equivalent conductivity : It is defined as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte in

 

It is expressed as L e

and is related to specific conductance as

(M is Molarity of the solution)

 

where C is the concentration in gram equivalent per litre (or Normality). This term has earlier been quite frequently used. Now it is replaced by molar

conductance. The units of equivalent conductance are Ohm-1 cm2 (gmequiv)-1 .

(8)         Experimental measurement of conductance

• The conductance of a solution is reciprocal of the resistance, therefore, the experimental determination of the conductance of a solution involves the measurement of its

We know that unknown resistance can be measured on a Wheatstone bridge. However, for

measuring the resistance of an ionic solution, we face following two main difficulties,

• Direct current (DC) can not be passed because it may change the composition of the solution by electrolysis and
• A solution of unknown resistance can not be connected to the bridge like a metallic wire or other solid

First difficulty can be solved by passing an alternating current (AC) from ac source of power. The second difficulty can be solved by using a specially designed vessels called conductivity cell.

• Calculation of conductivity : We have seen that conductivity (k) is reciprocal of resistivity (r) , e.,

 

k = 1

r

and

r = R a     \

l

 

where G is the conductance of the cell, l is the distance of separation of two electrodes having cross section

 

area

a cm².

The quantity

æ l ö

ç a ÷

is called cell constant and is expressed in

cm^-1.

Knowing the value of cell constant

 

è ø

and conductance of the solution, the specific conductance can be calculated as,

Conductivity = Conductance ´ Cell constant

k = G ´ Cell constant i.e.,

• Determination of cell constant : The cell constant is generally not calculated from the values of l and a because these are difficult to measure for a given However, it is usually determined accurately by measuring the conductance of a standard solution whose conductivity is known. For this purpose, a standard solution of KCl is used whose conductivity is known at different concentrations and temperatures. The conductivities of different KCl solutions at 298 K are given in table.

Conductivity and molar conductivity of KCl solutions at 298.15 K

Molarity            Concentration                         Conductivity                               Molar conductivity
(mol L–1)	(mol m–3)	S cm–1	S m–1	S cm2 mol–1	S m2 mol–1
1.000	1000	0.1113	11.13	111.3	111.3×10–4
0.100	100.0	0.0129	1.29	129.0	129.0×10–4
0.010	10.00	0.00141	0.141	141.0	141.0×10–4

 

 

 

Example : 17 If 0.01 M      solution of an electrolyte has a resistance of 40 Ohm’s in a cell having a cell constant of

0.4 cm^–1, then its molar conductance in Ohm^–1cm²mol ^–1 is                                                                 [CBSE 1996]

 

(a) 10                              (b)

10²

(c)

10³

10⁴

 

Solution: (c)

k = 1 ´ l

(where k = specific conductance, R = Resistance, l

= cell constant)

 

R     a                                                                                                  a

\ k = 1 ´ 0.4 = 0.01ohm^–1cm^–1 40

Strength of the solution = 0.01M

\ Volume of solution containing one mole of the electrolyte = 1,00,000 cc.

\ Molar conductance = k ´ V in cc. = 0.01 ´ 1,00,000 = 10³ Ohm^–1cm²mol ^–1

Example : 18 The resistance of 1 N solution of acetic acid is 250 Ohm, when measured in a cell of cell constant 1.15cm^–1 .

The equivalent conductance (in Ohm^–1cm²equiv ^–1 ) of 1 N acetic acid is                                          [CBSE PMT 1994]

(a) 4.6                             (b) 9.2                                (c) 18.4                        (d) 0.023

 

 

Solution: (a)

k = G ´ cell constant

= Cell constant = 1.15

 

R

L      = k ´ 1000 = 1.15 ´ 1000

250

 

= 4.6W^–1cm²eq ^–1 .

 

^eq       Normality      250       1

Example : 19 The cell constant of a given cell is 0.47cm^–1. The resistance of a solution placed in this cell is measured to be

31.6 Ohm. The conductivity of the solution (in Scm^–1 where S has usual meaning) is                             [Pb CET 1995]

(a) 0.15                           (b) 1.5                                (c) 0.015                      (d) 150

 

 

Solution: (c)

k = 1 ´ cell constant =  1  ´ 0.47 = 0.015 Scm^–1 .

 

R                                  31.6

 

Example : 20 A conductivity cell has two platinum electrodes of 1.2 cm²

area, separated by a distance of 0.8 cm. The cell

 

constant is,                                                                                                                          [Pb CET 1989]

 

(a)

0.66cm^–1

(b)

1.5cm^–1

(c)

0.96cm^–1

(d)

0.66cm

 

Solution: (a)       Cell constant = l

a

= 0.8 = 0.66cm^–1 . 1.2

 

Example : 21 The resistance of 0.01 N solution of an electrolyte was found to be 210 Ohm at 298 K. Its conductance is,

 

(a)

4.76 ´ 10 ^–3 mho

(b)

4.76 mho

(c)

210 mho

(d) None of these

 

Solution: (a)       Conductance = 1 = 1 = 4.76 ´ 10^–3mho.

R      210

Example : 22 The resistance of 0.01 N solution of an electrolyte was found to be 210 Ohm at 298 K, using a conductivity cell of cell constant 0.66cm^–1 . The specific conductance of solution is,

 

(a)

(c)

3.14 ´ 10 ^–3 mho cm^–1

3.14 mho cm^–1

(b)

(d)

3.14 ´ 10 ^–3 mho ^–1 cm

3.14 mho ^–1 cm^–1

 

Solution: (a)

k = 1 ´ l =  1 ´ 0.66 = 3.14 ´ 10 ^–3 mho cm^–1 .

 

 

R     a      210

Example : 23 Molar conductivity of a solution is  1.26 ´ 10² W^–1cm² mol ^–1 . Its molarity is 0.01. Its specific conductivity will be                                                                                                                                  [Manipal PMT 2002]

 

 

 

 

(a)

1.26 ´ 10 ^–5

(b)

1.26 ´ 10 ^–3

(c)

1.26 ´ 10 ^–4

(d)

0.0063

 

 

Solution: (b)

L     = k ´ 1000

^m       Molarity

or k =  L m ´ Molarity

1000

= 1.26 ´ 10 ^–2 ´ 0.01 = 1.26 ´ 10 -3.

1000

 

 

 

 

In general, conductance of an electrolyte depends upon the following factors,

(1) Nature of electrolyte , (2) Concentration of the solution, (3) Temperature

• Nature of electrolyte : The conductance of an electrolyte depends upon the number of ions present in the Therefore, the greater the number of ions in the solution the greater is the conductance. The number of ions produced by an electrolyte depends upon its nature. The strong electrolytes dissociate almost completely into ions in solutions and, therefore, their solutions have high conductance. On the other hand, weak electrolytes, dissociate to only small extents and give lesser number of ions. Therefore, the solutions of weak electrolytes have low conductance.
• Concentration of the solution : The molar conductance of electrolytic solution varies with the concentration of the electrolyte. In general, the molar conductance of an electrolyte increases with decrease in concentration or increase in The molar conductance of a few electrolytes in water at different concentrations are given in table

C	HCl	KCl	KNO3	CH3COOH	NH4OH
0.1	391.3	129.0	120.4	5.2	3.6
0.05	399.1	133.4	126.3	–	–
0.01	412.0	141.3	132.8	16.3	11.3
0.005	415.8	143.5	131.5	–	–
0.001	421.4	146.9	141.8	49.2	34.0
0.0005	422.7	147.8	142.8	67.7	46.9
0(Infinite	426.2	149.9	146.0	390.7	271.0
dilution)

Inspection of table reveals that the molar conductance of strong electrolyte

 

( HCl, KCl, KNO₃ )

as well as weak electrolytes ( CH

₃COOH, NH4

1. OH) increase

 

with decrease in concentration or increase in dilution. The variation is however different for strong and weak electrolytes.

• Variation of conductivity with concentration for strong electrolytes : In case of strong electrolytes, there is a tendency for molar conductivity to approach a certain limiting value when the concentration approaches zero e., when the dilution is infinite. The molar conductivity when the

concentration approaches zero (Infinite dilution) is called molar conductivity at infinite dilution. It is denoted by L0 .

 

Thus,

L = L0 when C ® 0 (At infinite dilution)

It has been observed that the variation of molar conductivity with concentration may be given by the

 

expression

L = L0 – Ac 1 / 2

where, A is a constant and L0 is called molar conductivity at infinite dilution.

 

 

 

 

The variation of molar conductivity with concentration can be studied by plotting the values of L m

square root of concentration ( C) . The plots of variation of molar conductivity

 

against

 

with         for KCl and HCl     are given in fig. It has been noticed that the

 

variation of L m

with concentration,

is small (Between 4 to 10% only) so

 

that the plots can be extrapolated to zero concentration. This gives the limiting value of molar conductance when the concentration approaches zero, called molar conductivity at infinite dilution.

• Variation of molar conductivity with concentration for weak electrolytes : The weak electrolytes dissociate to a much lesser extent as

compared to strong electrolytes. Therefore, the molar conductivity is low as compared to that of strong electrolytes.

 

However, the variation of L m

with

is very large and so much so that we cannot obtain molar

 

conductance at infinite dilution

(L0 )

by extrapolation of the L m

versus

plots. The behavior of weak

 

electrolytes such as CH3 COOH

is shown in figure.

 

Note : ®           The L0

Kohlrausch law.

value for weak electrolytes can be obtained by an indirect method  based upon

 

Explanation for the variation : The variation of molar conductance with concentration can be explained on the basis of conducting ability of ions for weak and strong electrolytes.

For weak electrolytes the variation of L with dilution can be explained on the bases of number of ions in solution. The number of ions furnished by an electrolyte in solution depends upon the degree of dissociation with dilution. With the increase in dilution, the degree of dissociation increases and as a result molar conductance

 

increases. The limiting value of molar conductance whole of the electrolyte dissociates.

(L0 )

corresponds to degree of dissociation equal to 1 i.e., the

 

Thus, the degree of dissociation can be calculated at any concentration as,

a = Lc

L0

 

where a is the degree of dissociation, Lc

conductance at infinite dilution.

is the molar conductance at concentration C and L0

is the molar

 

For strong electrolytes, there is no increase in the number of ions with dilution because strong electrolytes are completely ionised in solution at all concentrations (By definition). However, in concentrated solutions of strong electrolytes there are strong forces of attraction between the ions of opposite charges called inter-ionic forces. Due to these inter-ionic forces the conducting ability of the ions is less in concentrated solutions. With dilution, the ions become far apart from one another and inter-ionic forces decrease. As a result, molar conductivity increases with dilution. When the concentration of the solution becomes very-very low, the inter-ionic attractions become negligible and the molar conductance approaches the limiting value called molar conductance at infinite dilution. This value is characteristic of each electrolyte.

• Temperature : The conductivity of an electrolyte depends upon the temperature. With increase in temperature, the conductivity of an electrolyte

Electricity is carried out through the solution of an electrolyte by migration of ions. Therefore,

• Ions move toward oppositely charged electrodes at different
• During electrolysis, ions are discharged or liberated in equivalent amounts at the two electrodes, no matter what their relative speed is.
• Concentration of the electrolyte changes around the electrode due to difference in the speed of the

 

 

 

• Loss of concentration around any electrode is proportional to the speed of the ion that moves away from the electrode, so

The relation is valid only when the discharged ions do not react with atoms of the electrodes. But when the ions combine with the material of the electrode, the concentration around the electrode shows an increase. For

 

example, during electrolysis of

AgNO₃ solution using Ag electrodes, the concentration of

AgNO3

around the anode

 

increases, because every nitrate ion that reaches at the anode dissolve from it one

Ag + ion to form

AgNO3 .

 

 

• Definition : “The fraction of the total current carried by an ion is known as transport number, transference number or Hittorf number may be denoted by sets symbols like t₊ and t_– or t_c and t_a or n_c and n_a.”

From this definition,

 

t   =            Current carried by an anion           ;

^a       Total current passed through the solution

t c  =

Current carried by a cation

 

Total current passed through the solution

 

evidently,

ta + tc = 1.

 

• Determination of transport number : Transport number can be determined by Hittorf’s method, moving boundary method, emf method and from ionic mobility.

(3)  Factors affecting transport number

• Temperature : A rise in temperature tends to bring the transport number of cation and anion more closer to 0.5. It means that the transport number of an ion, if less than 0.5 at the room temperature increases and if greater than 5 at room temperature decreases with rise in temperature.
• Concentration of the electrolyte : Transport number generally varies with the concentration of the

 

electrolyte. In case of partially dissociated

CdI₂ , the value decreases from 0.49 at low concentration to almost zero

 

at higher concentration and becomes negative at still higher concentration. The transport numbers of

Cd 2+

ions in

 

0.01 N, 0.05 N, 0.02 N and 0.50 N

CdI 2

at 25°C are 0.449, 0.402, 0.131 and 0.005 respectively. This abnormal

 

behavior may be explained by assuming,

 

• That in very dilute solution :

number.

CdI ₂ ionises to

Cd 2+

ions and

I ^– ions. Thus

Cd 2+

shows the usual transport

 

• That with increase in concentration :

CdI₂ takes on

I ^– ions and form complex, CdI2 + 2I ^– ⇌ [CdI4 ]2-.

This

 

explains the negative value for transport number of Cd ²⁺

ions at higher concentration.

 

• Nature of the other ions present in solution : The transport number of anion depends upon the

 

speed of the anion and the cation and vice versa. For example, the transport number of Cl ^–

ion in NaCl is 0.0004

 

but in HCl it is 0.16. This is because

H ⁺ moves faster than

Na ⁺ .

 

• Hydration of ion : In general, a decrease in the degree of hydration of anion will increase its transport

 

number. For example, the transport number of

Li⁺ , Na⁺ , K ⁺

ions in

LiCl, NaCl, KCl

solutions are 0.328, 0.396,

 

0.496 respectively. Thus the ionic mobility of the cations is in the order

Li +   £ Na +   £ K +

which is in the reverse

 

 

 

 

order than that expected from the size of the ions. This anomaly can be explained by saying that

Li ⁺ is hydrated to

 

greater extent than

Na ⁺

which in turn is more than

K ⁺ . Thus the effective size of

Li ⁺

is more than that of

Na ⁺

 

which in turn is more than that of

K ⁺ .

 

• Transport number and Ionic mobility : Ionic mobility or Ionic conductance is the conductivity of a solution containing 1 g ion, at infinite dilution, when two sufficiently large electrodes are placed 1 cm

Unit of ionic mobility is Ohm^–1 cm² or V^–1S^-1cm² : Ionic mobility and transport number are related as,

Absolute ionic mobility is the mobility with which the ion moves under unit potential gradient. It’s unit is

cm sec^-1 .

 

• Kohlrausch law states that, “At time infinite dilution, the molar conductivity of an electrolyte can be

expressed as the sum of the contributions from its individual ions” i.e., L^¥ = n l^¥ + n l^¥ , where, n             and n                                                                                                                                                                are

l

m              +     +              –    –                                    +                        –

 

+

the number of cations and anions per formula unit of electrolyte  respectively and, l^¥

and

^¥ are the molar

 

–

conductivities of the cation and anion at infinite dilution respectively. The use of above equation in expressing the molar conductivity of an electrolyte is illustrated below,

• Molar conductivity of HCl : The molar conductivity of HCl at infinite dilution can be expressed as,

 

L

¥

HCl

= n H

+ l¥ +

+ n Cl

– l^¥ – ; For HCl, n +

= 1 and n

Cl ^–

= 1. So,

 

H

Cl

H

L^¥    = (1 ´ l^¥ + ) + (1 ´ l^¥ – ); Hence, L^¥         = l^¥ + + l^¥ –

HCl                           H                              Cl                                             HCl                H                 Cl

 

 

as,

• Molar conductivity of MgCl₂ : For

MgCl₂ , the molar conductivity at infinite dilution can be expressed

 

L

¥

MgCl2

= n Mg

2+ l¥

2+  + n

– l^¥ – ; For

MgCl₂ , n

Mg ²⁺

= 1 and n

Cl –    = 2

 

Mg

Cl

Cl

So, L^¥       = 1 ´ l^¥ 2+  + 2 ´ l^¥ – ; Hence, L^¥           = l^¥ 2+ + 2l^¥ –

MgCl2                          Mg                              Cl                                         MgCl2                Mg                      Cl

 

• Molar conductivity of CH₃COOH :

CH₃ COOH

in solution ionises as,

 

CH3 COOH ⇌ CH3 COO^– + H ⁺ ; So, the molar conductivity of CH3 COOH (acetic acid) at infinite dilution

 

can be expressed as,

¥

L

CH ₃COOH

= n H

+ l^¥ +  + n

CH 3 COO

¥

l–

CH 3 COO

• ; For, CH₃COOH, n H +

= 1 and n

CH3COO^–   = 1

 

H

So, L^¥

= 1 ´ l^¥ +  + 1 ´ l^¥

–    or L^¥

= l^¥ + + l^¥          –

 

CH3COOH                            H

CH3COO

CH3COOH                   H

CH3COO

 

The molar conductivities of some ions at infinite dilution and 298 K

Cation	λ¥ S cm2mol – +	1         Cation	λ	¥ S cm 2mol +	–1	Anion	λ¥S cm 2mol –1 –	Anion	λ¥ S cm2mol –	–1
H +  Tl +	349.8 74.7	Ba 2+  Ca 2+	127.2 119.0		OH –  Br –		198.0 78.4	ClO – 4 SO2- 4	68.0 159.2
K =	73.5	Sr 2+	119.0		I –		76.8	CH3 COO –	40.9

 

 

 

• Applications of Kohlrausch’s law : Some typical applications of the Kohlrausch’s law are described below,

L

L

m

• Determination of ¥  for weak electrolytes : The molar conductivity of a weak electrolyte at infinite

 

dilution

(L^¥ )

cannot be determined by extrapolation method. However,

¥     values for weak electrolytes can be

 

m

m

determined by using the Kohlrausch’s equation. For acetic acid this is illustrated below. According to the Kohlrausch’s law, the molar conductivity of acetic acid (CH ₃COOH) is given by,

 

L

¥

CH₃COOH

= n H

+ l^¥ +  + n

CH3COO

¥

l–

CH3COO

–  = 1 ´ l^¥ +

+   ´ ^¥

1   l

CH3COO

• ; Taking the values of l^¥ +

and

¥

l          –

CH3COO

 

H

H

H

CH COOH

(from table), we can write, l^¥

3

= (349.8 + 40.9) Scm²mol ^-1 = 390.7 Scm²mol ^-1 .

 

Sometimes, the molar conductivity values for the ions may not be available. In such cases, the following procedure may be followed,

• Select a series of strong electrolytes, so that their sum or difference gives the weak electrolyte. Generally, three strong electrolytes are

 

• Measure L _m values of these salts at various concentrations (Cm ) and plot Lmagainst

for each salt

 

m

separately. Determine L^¥  for each salt (Strong electrolyte) by extrapolation method.

m

• Add or subtract the equations to get the L^¥ of the weak

Suppose we have to determine the molar conductivity of a weak electrolyte MA at infinite dilution. For this

m

purpose, we take three salts, viz., MCl, NaA and NaCl, and determine their L^¥ values by extrapolation method.

 

Then, according to the Kohlrausch’s law, L^¥

= l^¥ + + l^¥ – ; L^¥

= l^¥  +   + l^¥–  ;   L^¥

= l^¥ + + l^¥ –

 

 

From these equations, we can write,

MCl               M               Cl

NaA              Na                A

NaCl               Na               Cl

 

L^¥    + L^¥

– L^¥

= (l^¥ +   + l^¥ – ) + (l^¥  +   + l^¥– ) – (l^¥  +   + l^¥ – ) = l^¥ +   + l^¥–    = L^¥

 

MCl

 

MA

So, L^¥

NaA

= L

¥

MCl

NaCl

+ L

¥

NaA

M               Cl

– L

¥

NaCl

Na               A

Na               Cl

M                A                  MA

 

L

m

Thus, we can obtain the molar conductivity of a weak electrolyte at infinite dilution from the three suitable strong electrolytes.

^¥ values of

 

• Determination of the degree of ionisation of a weak electrolyte : The Kohlrausch’s law can be used

l

l

m

for determining the degree of ionisation of a weak electrolyte at any concentration. If          ^c is the molar conductivity

 

L

L

m

of a weak electrolyte at any concentration C and, Then, the degree of ionisation is given by,

^¥ is the molar conductivity of a electrolyte at infinite dilution.

 

Thus, knowing the value of

^c , and

^¥ (From the Kohlrausch’s equation), the degree of ionisation at any

 

m

m

concentration (a _c )  can be determined.

• Determination of the ionisation constant of a weak electrolyte : Weak electrolytes in aqueous solutions ionise to a very small extent. The extent of ionisation is described in terms of the degree of ionisation (a).In solution, the ions are in dynamic equilibrium with the unionised molecules. Such an equilibrium can be described by a constant called ionisation constant. For example, for a weak electrolyte AB, the ionisation

 

 

 

 

equilibrium is, AB ⇌

A⁺ + B^– ; If C is the initial concentration of the electrolyte AB in solution, then the equilibrium

 

concentrations of various species in the solution are, [AB] = C(1 – a),

[A⁺ ] = Ca

and [B^– ] = Ca

 

 

Then, the ionisation constant of AB is given by,

K = [A⁺ ][B] = Ca.Ca  =

Ca ²

 

 

[AB]       C(1 – a)

(1 – a)

 

m

m

We know, that at any concentration C, the degree of ionisation (a ) is given by, a = L^c / L^¥

 

C(L^c

/ L^¥ )²

C(L^c )²                                    _{¥             c}

 

Then,

K =            ^{m       m}      =                  ^m           ; Thus, knowing

Lm   and Lm

at any concentration, the ionisation

 

[1 – (L^c / L^¥ )]      L^¥ (L^¥ – L^c )

m            m                    m         m              m

constant (K) of the electrolyte can be determined.

• Determination of the solubility of a sparingly soluble salt : The solubility of a sparingly soluble salt in a solvent is quite Even a saturated solution of such a salt is so dilute that it can be assumed to be at infinite

m

L

l

dilution. Then, the molar conductivity of a sparingly soluble salt at infinite dilution (L^¥ ) can be obtained from the

 

relationship,

^¥ = n

₊ l^¥ + n _– ^¥

……..(i)

 

+

m

–

The conductivity of the saturated solution of the sparingly soluble salt is measured. Form this, the conductivity

 

of the salt (k _salt ) can be obtained by using the relationship,

k salt = k sol – k water , where, k water

is the conductivity of

 

the water used in the preparation of the saturated solution of the salt.

 

 

L

¥

salt

= 1000k _salt

Cm

……..(ii)

 

From equation (i) and (ii) ;

 

,    Cm is the molar concentration of the sparingly soluble salt in its saturated solution.

 

Thus, Cm is equal to the solubility of the sparingly soluble salt in the mole per litre units. The solubility of the salt in

gram per litre units can be obtained by multiplying Cm with the molar mass of the salt.

 

 

Example : 24 The equivalent conductivity of 0.1 M weak acid is 100 times less than that at infinite dilution. The degree of dissociation is                                                                                                       [Tamil Nadu CET 2001,02]

(a) 100                            (b) 10                                 (c) 0.01                        (d) 0.001

 

 

Solution: (c)

Lc

a =    ^m =

L0

x / 100

 

x

=   1

100

= 0.01.

 

Example : 25 Equivalent conductances  of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and

 

90 Ohm^–1cm²

respectively. The equivalent conductance of CH₃COOH

at infinite dilution would be

[CBSE PMT 1997]

 

(a)

101.38 Ohm^–1cm²

(b)

253.62 Ohm^–1cm²

(c)

390.71ohm^–1cm²

678.90 ohm^–1cm²

 

Solution: (c)

L⁰ (CH 3 COOH) = L⁰ (CH 3 COONa) + L⁰ (HCl) – L⁰ (NaCl)

= 91 + 426.16 – 126.45

= 390.71 Ohm^–1cm² .

 

Example : 26 Molar ionic conductivities of a bivalent electrolyte are 57 and 73. The molar conductivity of the solution will be

[DPMT 2002]

 

(a)

130 S cm²mol ^–1

(b)

65 S cm²mol ^–1

(c)

260 S cm²mol ^–1

(d)

187 S cm²mol ^–1

 

L

m

Solution: (a)          ⁰ = 57 + 73 = 130 Scm²mol ^–1 .

 

 

 

Example : 27 The ionic mobilities of the cation and the anion of a salt are 140 and 80 mho respectively. The equivalent conductivity of the salt at infinite dilution will be                                                                                     [Haryana CET 1991]

• 160 mho (b) 280 mho                                (c) 60 mho                          (d) 220 mho

 

Solution: (d)

L⁰   = l⁰ + l⁰ = 140 + 80 = 220 mho.

 

eq             c             a

Example : 28 The ionization constant of a weak electrolyte is 25 ´ 10 ^–6 while the equivalent conductance of its 0.01 M

 

solution is 19.6 S cm²eq ^–1 .

will be

The equivalent conductance of the electrolyte at infinite dilution (in

Scm²eq ^–1 )

 

(a) 250                            (b) 196                               (c) 392                         (d) 384

 

 

Solution: (c)

HA  ⇌ H ⁺ + A^– ; K = Ca .Ca = Ca 2  ≃ Ca ²

 

C                     0        0

C(1 – a)

1 – a

 

C(1-a )         Ca         Ca

25 ´ 10^–6 = 10^–2 ´a ² or a = 5 ´ 10^–2

 

a = Lc

L0

or L⁰ = Lc

a

=     19.6

5 ´ 10^–2

= 392W^–1cm².

 

 

“Electrochemical cell or Galvanic cell is a device in which a spontaneous redox reaction is used to convert chemical energy into electrical energy i.e. electricity can be obtained with the help of oxidation and reduction reaction”.

• Characteristics of electrochemical cell : Following are the important characteristics of electrochemical cell,
• Electrochemical cell consists of two vessels, two electrodes, two electrolytic solutions and a salt
• The two electrodes taken are made of different materials and usually set up in two separate vessels.
• The electrolytes are taken in the two different vessels called as half –
• The two vessels are connected by a salt bridge/porous
• The electrode on which oxidation takes place is called the anode (or – ve pole) and the electrode on which reduction takes place is called the cathode (or + ve pole).
• In electrochemical cell, ions are discharged only on the
• Like electrolytic cell, in electrochemical cell, from outside the electrolytes electrons flow from anode to cathode and current flow from cathode to
• For electrochemical cell,

 

Ecell = +ve,

DG = –ve.

 

• In a electrochemical cell, cell reaction is

 

(2) Salt bridge and its significance

• Salt bridge is U – shaped glass tube filled with a gelly like substance, agar – agar (plant gel) mixed with an electrolyte like KCl, KNO₃, NH₄NO₃
• The electrolytes of the two half-cells should be inert and should not react chemically with each

 

 

 

• The cation as well as anion of the electrolyte should have same ionic mobility and almost same transport

 

number, viz.

KCl, KNO₃ , NH ₄ NO₃ etc.

 

• The following are the functions of the salt bridge,

• It connects the solutions of two half – cells and completes the cell
• It prevent transference or diffusion of the solutions from one half cell to the
• It keeps the solution of two half – cells electrically
• It prevents liquid – liquid junction potential e. the potential difference which arises between two solutions when they contact with each other.

Note : ® Salt bridge can be replaced by a porous partition which allows the migration of ions without intermixing of solution.

• KCl (aq) cannot be used as a salt bridge for the cell, Cu(s) CuSO₄ (aq) AgNO₃ (aq) Ag(s) .

 

Because AgCl is precipitated as follows,

(3) Representation of an electrochemical cell

AgNO3 + KCl ¾¾® AgCl ¯ +KNO3

(ppt.)

 

• The interfaces across which a potential difference exists are shown by a semicolon (;) or a single vertical line (|). For example, the two half- cells of the following electrochemical cell can be represented as follows,

 

Zn(s) + Cu²⁺ (aq) ¾¾® Zn²⁺ (aq) + Cu(s);

Zn ; Zn ²⁺

or Zn| Zn²⁺

and

Cu²⁺ ; Cu  or Cu²⁺ | Cu

 

These indicate that potential difference exists at the Zn and

Zn 2+

ions interface, and similarly at the

Cu 2+

 

and Cu interface. Sometimes coma or plus signs is observed in the formulation of half-cells. For example,

Ag, AgCl | Cl ^– or Ag + AgCl | Cl ^– . These indicate that Ag and AgCl together constitute the electrode.

• The contact between two solutions by means of a salt bridge is indicated by double vertical line

(||) between them e.g.; Cu ²⁺ || Zn²⁺

• The anode half-cell (or the oxidation half cell) is always written on the left hand side and the cathode half cell (or the reduction half-cell) on the right hand side, with the respective metal electrode on the outside extremes,

 

e.g.,

Zn ; Zn²⁺ || Cu²⁺ ; Cu     or

Zn| Zn²⁺ || Cu²⁺ | Cu . Sometimes negative and positive sings are put on the

 

electrodes to show that they are negative (anode) and positive (cathode) electrodes.

e.g.  (–) Zn ; Zn²⁺ || Cu²⁺ ; Cu (+)

• An arrow when drawn below the cell formulation gives the direction of the current inside the cell while an arrow drawn above the formulation gives the direction of the electrons flown in the outer

 

e.g.,

Zn ; Zn²⁺

¾¾¾¾®

||

¾¾¾¾®

Cu²⁺  ; Cu

 

• The potential difference between the electrodes (e., the cell potential or EMF or emf ) is stated in volts

 

along with the temperature at which it is applicable. For example,

E25o C   = 1.130 volt

or E298 K

= 1.130 volt

 

With the usual above conventions, the emf of the cell will have a positive value. However, when the cell is formulated in the reverse order, the emf will have a negative value. In other words, the negative value of emf indicates that the oxidation process expected on the left hand electrode will not occur spontaneously and in case oxidation must be made to occur on the left hand electrode, an emf of a value of somewhat larger then that of the specified cell potential will be required from an external source.

• The concentration of solutions, pressure of gases and physical state of solids and liquids involved, are

 

indicated in the cell formation. For example,

Pt, H ₂ (0.9 atm) ;

H ⁺ (a = 0.1) ||Cu²⁺(a = 0.1); Cu

 

 

 

Note : ® Sometimes we get confused in the nomenclature of electrodes. As a memory aid keep in mind the alphabetical order of the first e.g. A (anode) comes before C (cathode). The cell may be written by arranging each of the pair left – right, anode – cathode, oxidation – reduction, negative and positive in the alphabetical order as,

 

• Reversible and irreversible cells : A cell is said to be reversible if the following two conditions are fulfilled
  • The chemical reaction of the cell stops when an exactly equal external emf is
  • The chemical reaction of the cell is reversed and the current flows in opposite direction when the external

emf is slightly higher than that of the cell. Any other cell, which does not obey the above two conditions, is termed

 

as irreversible. Daniell cell is reversible but

Zn| H ₂ SO₄ |Ag

cell is irreversible in nature

 

 

are,

• Types of electrochemical cells : Two main types of electrochemical cells have been reported, these

 

• Chemical cells : The cells in which electrical energy is produced from the energy change accompanying

 

a chemical reaction or a physical process are known as chemical cells. Chemical cells are of two types,

• Chemical cells without transference : In this type of chemical cells, the liquid junction potential is neglected or the transference number is not taken into consideration. In these cells, one electrode is reversible to cations while the other is reversible to the anions of the
• Chemical cells with transference : In this type of chemical cells, the liquid-liquid junction potential or diffusion potential is developed across the boundary between the two This potential develops due to the

 

difference in mobilities of + ve

and

• ve

ions of the electrolytes.

 

• Concentration cells : “A cell in which electrical energy is produced by the transference of a substance from a system of high concentration to one at low concentration is known as concentration cells”. Concentration cells are of two
  • Electrode concentration cells : In these cells, the potential difference is developed between two electrodes at different concentrations dipped in the same solution of the electrolyte. For example, two hydrogen electrodes at different gaseous pressures in the same solution of hydrogen ions constitute a cell of this

 

Pt, H ₂ (pressure p₁ ) Anode

| H ⁺ |

H ₂ (pressure p₂ ) Pt ;

Cathode

at 25^o C If

p1 > p2

, oxidation

 

occurs at L. H. S. electrode and reduction occurs at R. H. S. electrode.

In the amalgam cells, two amalgams of the same metal at two different concentrations are immersed in the

 

same electrolytic solution.

M(Hg C1 ) |Mⁿ⁺ | Zn(Hg C2 )

 

 

The emf of the cell is given by the expression,                                         at 25 ^o C

 

 

 

• Electrolyte concentration cells : In these cells, electrodes are identical but these are immersed in solutions of the same electrolyte of different concentrations. The source of electrical energy in the cell is the tendency of the electrolyte to diffuse from a solution of higher concentration to that of lower concentration. With the expiry of time, the two concentrations tend to become Thus, at the start the emf of the cell is maximum and it gradually falls to zero. Such a cell is represented in the following manner ( C₂ is greater then C₁ ).

 

 

+n                            n+

or   Zn| Zn²⁺ (C1 )    Zn²⁺ (C2 )| Zn

 

M | M

(C₁ )|| M

(C2 )|M

Anode    ||   Cathode

 

The emf of the cell is given by the following expression,                                                  at 25^o C

 

The concentration cells are used to determine the solubility of sparingly soluble salts, valency of the cation of the electrolyte and transition point of the two allotropic forms of metal used as electrodes, etc.

Note : ® In concentration cell net redox change is zero and the decrease in free energy during transfer of substance from one concentration to other is responsible for production of electrical energy.

• Heat of reaction in an electrochemical cell : Let n Faraday charge flows out of a cell of emf E, then

– DG = nFE                                                                       …….(i)

Gibbs – Helmholtz equation from thermodynamics may be given as

 

DG = DH + Tæ ¶DG ö

 

…….(ii)

 

è

¶T

ç        ÷

ø P

 

From equation (i) and (ii) we get, – nFE = DH + T é¶(–nFE)ù

= DH – nFTæ ¶E ö

; DH = –nFE + nFTæ ¶E ö   ,

 

ê    ¶T      ú

ç ¶T ÷

ç ¶T ÷

 

ç      ÷

where æ ¶E ö

¶T

ë             û P

= Temperature coefficient of cell

è      ø P

è      ø P

 

è      ø _P

 

ç      ÷

Case I: When æ ¶E ö

¶T

= 0, then DH = –nFE

 

è      ø _P

 

ç      ÷

Case II: When æ ¶E ö > 0, then nFE > DH,

¶T

i.e. process inside the cell is endothermic.

 

è      ø

ç      ÷

Case III: When æ ¶E ö < 0, then nFE < DH, i.e., process inside the cell is exothermic.

¶T

è      ø

One of the main use of galvanic cells is the generation of portable electrical energy. These cells are also popularly known as batteries. The term battery is generally used for two or more Galvanic cells connected in series. Thus, a battery is an arrangement of electrochemical cells used as an energy source. The basis of an electrochemical cell is an oxidation – reduction reaction. However, for practical purposes there are some limitations to the use of redox reactions. A useful battery should also fulfil the following requirements;

• It should be light and compact so that it can be easily
• It should have reasonably long life both when it is being used and when it is not
• The voltage of the battery should not vary appreciably during its

Types of commercial cells : There are mainly two types of commercial cells,

 

 

 

• Primary cells : In these cells, the electrode reactions cannot be reversed by an external electric energy In these cells, reactions occur only once and after use they become dead. Therefore, they are not chargeable. Some common example are, dry cell, mercury cell, Daniell cell and alkaline dry cell.

 

 

 

 

 

• Voltaic cell

Cu           –  +

Zn rod

 

 

 

Dil.

Cathode : Cu rod Anode : Zn rod Electrolyte : dil. H 2 SO4

Emf : 1.08 V

 

 

Polarisatio n

Cu                        Local At cathode : Cu ²⁺ + 2e ^– ® Cu At Anode : Zn ® Zn²⁺ + 2e^– Over all

Zn                                     reaction : Zn + Cu ²⁺ ® Zn²⁺ + Cu

 

 

 

 

 

 

 

• Daniall cell

e–

–

 

Anode (Zn)

 

 

 

 

 

1M ZnSO4 (aq)

 

Electrons flow

e–

Key

Ammeter         Current

 

Salt bridge

 

 

 

Cotton Plugs

 

 

 

 

 

 

 

 

1M CuSO4 (aq)

(Depolariser)

 

 

+

 

Cathode (Cu)

 

Cathode : Cu rod Anode : Zn rod Electrolyte : dil. H 2 SO4 Emf : 1.1 V

At cathode : Cu ²⁺ + 2e ^– ® Cu

At Anode : Zn ® Zn²⁺ + 2e^–

Over all reaction : Zn + Cu ²⁺ ® Zn²⁺ + Cu

 

 

 

• Lechlanche cell (Dry cell)

+               Seal

Graphite (cathode) MnO₂ +C

(Depolariser)

Cathode : Graphite rod

Anode : Zn pot

Electrolyte : Paste of NH4 Cl + ZnCl2 in starch

Emf : 1.2 V to 1.5 V

 

Paste of

At cathode : NH ⁺ + MnO₂

+ 2e ^– ® MnO(OH)^– + NH ₃

 

4

NH₄Cl+ZnCl₂

 

 

Zinc anode

–

At Anode : Zn ® Zn²⁺ + 2e ^–

Over all reaction :

Zn + NH ⁺ + MnO  ® Zn²⁺ + MnO(OH)^– + NH

 

4                     2                                                                          3

 

• Mercury cell

Zn rod

 

 

Glass pot

 

C           –        A     +                  Mercury oxide Porous pot

Cathode : Mercury (II) oxide

Anode : Zn rod

Electrolyte :  Paste of KOH + ZnO

Emf : 1.35 V

At cathode : HgO     + H O    + 2e ^–  ® Hg

 

 

 

• 2OH ^–

 

(s)

2  (l)

(l)

(aq)

 

Paste of

KOH+ ZnO

(Electrolyte)

At Anode :

Zn(s)

(amalgam)

–

+ 20H

(aq)

® ZnO(s)

• H 2

O(l)

+ 2e ^–

 

Over all reaction : Zn(s) + HgO(s) ® ZnO(s) + Hg(l)

 

 

 

Note : ®             In a dry cell

ZnCl₂ combines with

NH ₃ produced to form the complex

[Zn(NH 3 )2 Cl2 ] ,

 

otherwise the pressure developed due to

NH ₃ would crack the seal of the cell.

 

• Mercury cell give a constant voltage throughout its life because the electrolyte KOH is not consumed in the
• Secondary cells: In the secondary cells, the reactions can be reversed by an external electrical energy Therefore, these cells can be recharged by passing electric current and used again and again. These are also celled storage cells. Examples of secondary cells are, lead storage battery and nickel – cadmium storage cell.

 

In charged                                     Lead storage cell                                                          Alkali cell

 

 

+                                –                                      +

Glass vessel

 

PbO₂

 

Pb

dil. H₂SO₄

Ni(OH)₂

–

 

Fe(OH)₂

Perforated steel grid

 

KOH 20%

+ Li(OH), 1%

 

 

 

Positive electrode        Perforated lead plates coated with PbO₂                          Perforated steel plate coated with Ni(OH)₄ Negative electrode       Perforated lead plates coated with pure lead                    Perforated steel plate coated with Fe Electrolyte                 dil. H₂SO₄                                                             20% solution of KOH + 1% LiOH

 

During charging          Chemical reaction

At cathode : PbSO₄ + 2H⁺ + 2e^– ® Pb + H₂SO₄ At anode : PbSO₄ + SO ^{– –} + 2H O – 2e^– ® PbO

Chemical reaction

At cathode :    Ni (OH)₂ + 2OH⁺ – 2e^– ®

Ni(OH)₄

 

4                       2                                            2

+

2H₂SO₄

At anode : Fe(OH)₂ + 2K⁺ + 2e^– ® Fe + 2KOH

 

Specific gravity of H₂SO₄ increases and when specific gravity becomes 1.25 the cell is fully charged.

Emf of cell: When cell is fully charged then E =

2.2 volt

During discharging Chemical reaction

Emf of cell : When cell is fully charged then E = 1.36 volt

 

 

 

Chemical reaction

 

4

At cathode : Pb + SO ^{– –}

– 2e^–

® PbSO₄

At cathode : Fe + 2OH^–

– 2e^–

® Fe(OH)₂

 

At anode : PbO₂ + 2H⁺ – 2e^– + H₂SO₄ ® PbSO₄

+

2H₂O

Specific gravity of H₂SO₄ decreases and when specific gravity falls below 1.18 the cell requires recharging.

Emf of cell : When emf of cell falls below 1.9 volt the cell requires recharging.

At anode : Ni(OH)₄ + 2K⁺ + 2e^– ® Ni(OH)₂

+

2KOH

Emf of cell : When emf of cell falls below 1.1 V it requires charging.

 

Efficiency                  80%                                                                      60%

 

 

 

These are Voltaic cells in which the reactants are continuously supplied to the electrodes. These are designed

 

to convert the energy from the combustion of fuels such as

H ₂ , CO, CH ₄ , etc. directly into electrical energy. The

 

common example is hydrogen-oxygen fuel cell as described below,

In this cell, hydrogen and oxygen are bubbled through a porous carbon electrode into concentrated aqueous sodium hydroxide or potassium hydroxide. Hydrogen (the fuel) is fed into the anode compartment where it is oxidised. The oxygen is fed into cathode compartment where it is reduced. The diffusion rates of the gases into the cell are carefully regulated to get maximum efficiency. The net reaction is the same as burning of hydrogen and oxygen to form water. The reactions are

 

At anode :

2[H

(g) + 2OH – ](aq) ¾¾® 2H

O(l) + 2e –

 

2

2

At cathode :

O2 (g) + 2H 2 O(l) + 4e ^– ¾¾® 4OH ^– (aq)

 

Overall reaction :

2H2(g) + O2(g) ¾¾® 2H2O(l)

 

Each electrode is made of porous compressed carbon containing a

small amount of catalyst (Pt, Ag or CoO). This cell runs continuously as

long as the reactants are fed. These fuel cells are more efficient than conventionally used methods of generating electricity on a large scale by burning hydrogen, carbon, fuels because these fuel cells convert the energy of the fuel directly into electricity. This cell has been used for electric power in the Apollo space programme. Fuel cells offer great promises for

energy conversion in future. The important advantages of fuel cells over ordinary batteries are

• High efficiency : The fuel cells convert the energy of a fuel directly into electricity and therefore, they are more efficient than the conventional methods of generating electricity on a large scale by burning hydrogen, carbon

fuels. Though we expect 100 % efficiency in fuel cells, so far 60 – 70% efficiency has been attained. The conventional methods of production of electrical energy involve combustion of a fuel to liberate heat which is then used to produce electricity. The efficiency of these methods is only about 40%.

• Continuous source of energy : There is no electrode material to be replaced as in ordinary battery. The fuel can be fed continuously to produce power. For this reason, H ₂ – O₂ fuel cells have been used in space
• Pollution free working : There are no objectionable byproducts and, therefore, they do not cause pollution problems. Since fuel cells are efficient and free from pollution, attempts are being made to get better commercially practical fuel

Thermodynamic characteristics of some Fuel cells

 

Fuel cells	Cell reaction	ΔGo (KJ mol– )		ΔHo (KJ mol– )	E (volt)
H2 – O2	2H2 + O2 ® 2H2O	– 237.2		– 258.9	1.23
C – O2	C + O2 ® CO2	– 137.3		– 110.5	0.71
CH4 – O2	CH4 + 2O2 ® CO2  + H2O	– 818.0		– 890.4	1.060

 

 

 

• When a metal (M) is placed in a solution of its ions (M⁺⁺), either of the following three possibilities can occurs, according to the electrode potential solution pressure theory of
• A metal ion M ⁿ⁺ collides with the electrode, and undergoes no
• A metal ion M ⁿ⁺ collides with the electrode, gains n electrons and gets converted into a metal atom M, (e. the metal ion is reduced).

Mⁿ⁺ (aq) + ne^– ¾¾® M(s)

• A metal atom on the electrode M may lose a electrons to the

 

electrode, and enter to the solution as

Mⁿ⁺ , (i.e. the metal atom is oxidised).

 

M(s) ¾¾® Mⁿ⁺ (aq) + ne ^– .

Thus, “the electrode potential is the tendency of an electrode to lose or gain electrons when it is in contact with solution of its own ions.”

• The magnitude of electrode potential depends on the following factors,
  • Nature of the electrode, (ii) Concentration of the ions in solution, (iii)
• Types of electrode potential : Depending on the nature of the metal electrode to lose or gain electrons, the electrode potential may be of two types,
  • Oxidation potential : When electrode is negatively charged with respect to solution, e., it acts as anode.

Oxidation occurs. M ¾¾® Mⁿ⁺ + ne

• Reduction potential : When electrode is positively charged with respect to solution, e. it acts as

 

cathode. Reduction occurs.

Mⁿ⁺ + ne ^– ¾¾® M

 

• Standard electrode potential : “If in the half cell, the metal rod (M) is suspended in a solution of one molar concentration, and the temperature is kept at 298 K, the electrode potential is called standard electrode

potential, represented usually by E^o ”. ‘or’

The standard electrode potential of a metal may be defined as “the potential difference in volts developed in a cell consisting of two electrodes, the pure metal in contact with a molar solution of one of its ions and the normal hydrogen electrode (NHE)”.

(5) Reference electrode or reference half – cells

It is not possible to measure the absolute value of the single electrode potential directly. Only the difference in potential between two electrodes can be measured experimentally. It is, therefore, necessary to couple the electrode with another electrode whose potential is known. This electrode is termed as reference electrode or reference half – cells. Various types of half – cells have been used to make complete cell with spontaneous reaction in forward direction. These half – cells have been summarised in following table,

Various Types of Half – cells

Type	Example	Half – cell reaction	Q =	Reversible to	Electrode Potential (oxidn), E =
Gas ion half – cell	Pt(H2)| H + (aq)  Pt(Cl2)| Cl – (aq)	1 H (g) ® H +(aq) + e– 2 2 Cl–(aq) ® 1 Cl (g) + e– 2   2	[H + ] 1 [Cl –]	H + Cl –	E0 – 0.0591log[H + ]  E0 + 0.0591log[Cl –]

 

 

 

• “The difference in potentials of the two half – cells of a cell known as electromotive force (emf) of the cell or cell potential.”

The difference in potentials of the two half – cells of a cell arises due to the flow of electrons from anode to cathode and flow of current from cathode to anode.

• The emf of the cell or cell potential can be calculated from the values of electrode potentials of two the half

– cells constituting the cell. The following three methods are in use :

• When oxidation potential of anode and reduction potential of cathode are taken into account

 

E

0

cell

= Oxidation potential of anode + Reduction potential of cathode = E ⁰

(anode) + E ⁰

(cathode)

 

ox

red

• When reduction potentials of both electrodes are taken into account

 

E

0

cell

= Reduction potential of cathode – Reduction potential of anode = E ⁰

0

–  E

Anode

0

= E

right

0

–   E

left

 

Cathode

• When oxidation potentials of both electrodes are taken into account

E^o   = Oxidation potential of anode – Oxidation potential of cathode = E ⁰ (anode) – E ⁰ (cathode)

cell                                                                                                                                                ox                         ox

• Difference between emf and potential difference : The potential difference is the difference between the electrode potentials of the two electrodes of the cell under any condition while emf is the potential generated by a cell when there is zero electron flow, e., it draws no current. The points of difference are given below

 

 

 

Emf	Potential difference
It is the potential difference between two electrodes when	It is the difference of the electrode potentials of the two
no current is flowing in the circuit.	electrodes when the cell is under operation.
It is the maximum voltage that the cell can deliver.	It is always less then the maximum value of voltage which
	the cell can deliver.
It is responsible for the steady flow of current in the cell.	It is not responsible for the steady flow of current in the
	cell.

 

 

 

 

Ecell

(4)    Cell EMF and the spontaneity of the reaction : We know,

DG = –nFEcell

• For a spontaneous process, DG is negative. Then, according to the equation for a spontaneous process, should be Thus, the cell reaction will be spontaneous when the cell emf is positive.
• For a non – spontaneous process, DG is Then, according to equation for a non – spontaneous

 

process,

Ecell

should be negative. Thus, the cell reaction will be non – spontaneous when the cell emf is negative.

 

• For the process to be at equilibrium,

DG = 0 . Then, according to the equation Ecell

should be zero. Thus,

 

the cell reaction will be at equilibrium when the cell emf is zero. These results are summarized below

 

Nature of reaction	ΔG(or ΔGo )	Ecell(or	Eo cell	)
Spontaneous	–	+
Equilibrium	0	0
Non – spontaneous	+	–

 

 

Example : 29 Calculate the emf of the following cell at 25 ^o C .

Ag(s)| AgNO3 (0.01mol kg ^–1 )|| AgNO3 (0.05 mol kg ^–1 )| Ag(s)

 

(a) – 0.414 V                          (b) – 0.828 V                              (c) – 0.414 V                      (d) – 0.0414 V

Solution: (d)       It is a concentration cell. The emf of a concentration cell is

 

Ecell = + 0.0592 log C1

at 25^o C and n = 1

 

n              C2

 

= +0.0592 log 0.01 = 0.0591 log 0.2

0.05

= 0.0591 ´ (-0.699) = -0.0414 V

 

Example : 30 The standard electrode potentials of the two half calls are given below

 

Ni ²⁺ + 2e ^– ⇌ Ni ; E ⁰ = -0.25

volt ;

Zn²⁺ + 2e ^– ⇌ Zn ; E ⁰ = -0.77 volt

 

The voltage of cell formed by combining the two half – cells would be                                                   [BHU 1987]

(a) – 1.02 V                              (b) + 0.52 V                                (c)  + 1.02 V                        (d) – 0.52 V

 

Solution: (b)

Ni ²⁺ + 2e ^– ⇌

Ni ; Eo

= -0.25

volt

 

Zn²⁺ + 2e ^– ⇌

Zn ; E ⁰ = -0.77 volt

 

Ecell

= Reduction potential of cathode – Reduction potential of anode = -0.25 – (-0.77)

= – 0.25 + 0.77 = 0.52 V

 

Example : 31 The EMF of the cell,

Ni | Ni ²⁺ || Cu ²⁺ | Cu

is 0.59 volt. The standard electrode potential (reduction

 

potential) of copper electrode is 0.34 volt. The standard electrode potential of nickel electrode will be (a) 0.25 volt                                             (b) – 0.25 volt                            (c) 0.93 volt                       (d) – 0.93 volt

 

Solution: (b)

E ⁰ = E ⁰

(RHS) – E ⁰

(LHS)

= E 0 2+            – E 0 2+

 

cell

Re d

Re d

Cu      / Cu

Ni      / Ni

 

 

 

 

Ni

i.e. 0.59 = 0.34 – E ⁰ 2+

/ Ni

or E^o 2+

/ Ni

= 0.34 – 0.59 = – 0.25 Volt

 

 

 

(1)

Ni

Nernst’s equation for electrode potential

The potential of the electrode at which the reaction,

Mⁿ⁺ (aq) + ne^– ® M(s)

 

 

takes place is described by the equation,

 

or

EMn+  / M

0

= E

Mn+ / M

• RTln

nF

[M(s)]

 

[Mn+ (aq.)]

 

 

above eq. is called the Nernst equation.

 

Where,

EMn+  / M  = the potential of the electrode at a given concentration,

E ⁰ n+           = the standard electrode potential

 

M     / M

R = the universal gas constant, 8.31 J K ^-1 mol ^-1 ,

 

T= the temperature on the absolute scale,

n = the number of electrons involved in the electrode reaction,

F = the Faraday constant : (96500 C),

[M(s)]= the concentration of the deposited metal,

[Mⁿ⁺ (aq)] = the molar concentration of the metal ion in the solution,

The concentration of pure metal M(s) is taken as unity. So, the Nernst equation for the written as,

 

 

 

 

Mⁿ⁺ / M

 

 

 

 

 

electrode is

 

At 298 K, the Nernst equation for the

Mⁿ⁺ / M

electrode can be written as,

 

 

For an electrode (half – cell) corresponding to the electrode reaction, Oxidised form + ne ^– ® Reduced form

The Nernst equation for the electrode is written as,

At 298 K, the Nernst equation can be written as,

(2) Nernst’s equation for cell EMF

For a cell in which the net cell reaction involving n electrons is,

aA + bB ® cC + dD

The Nernst equation is written as,

 

 

 

E      = E 0     – RT ln [C]c [D]d

E

 

 

cell

cell          nF

[A]a [B]b

 

Where,

0

E

cell

 

or

0

= E

cathode

0

–  E

anode

. The

o cell

is called the standard cell potential.

 

 

At 298 K, above eq. can be written as, or

It may be noted here, that the concentrations of A, B, C and D referred in the eqs. are the concentrations at the time the cell emf is measured.

• Nernst’s equation for Daniells cell : Daniell’s cell consists of zinc and copper electrodes. The electrode reactions in Daniell’s cell are,

 

At anode :

At cathode :

Net cell reaction :

Zn(s) ® Zn²⁺ (aq) + 2e ^–

Cu²⁺ (aq) + 2e ^– ® Cu(s)

 

Zn(s) + Cu²⁺ (aq) ® Cu(s) + Zn²⁺ (aq)

 

 

Therefore, the Nernst equation for the Daniell’s cell is,

Since, the activities of pure copper and zinc metals are taken as unity, hence the Nernst equation for the Daniell’s cell is,

The above eq. at 298 K is,

For Daniells cell, E ⁰   = 1.1 V

 

 

(4)

cell

Nernst’s equation and equilibrium constant

For a cell, in which the net cell reaction involing n electrons is,

aA + bB ® cC + dD

The Nernst equation is

 

 

 

 

 

…..(i)

 

 

 

At equailibrium, the cell cannot perform any useful work. So at equilibrium, the ratio

ECell is zero. Also at equilibrium,

 

 

 

 

[C]^c [D]^d

[A]^a [B]^b

é[C]^c [D]^d ù

=

ê[A]^a [B]^b ú

= Kc

 

ë              û equil

 

Where, Kc

is the equilibrium constant for the cell reaction. Then, at equilibrium eq……. (i) becomes,

 

0 = E ⁰   – RT ln K

…..(ii)

 

cell

nF          ^c

nFE ⁰

 

or        In

 

or

Kc   =           cell  

RT

 

…..(iii)

 

 

 

At 298 K, eq. (ii) can be written as,

0 = E ⁰ – 0.0592 log K

 

 

This gives, log Kc

cell

 

= (nE

0

cell

n                  c

 

/ 0.0592)

 

cell

K    = 10^(nE0

/ 0.0592)

….. (iv)

 

c

 

1. eq. (iii) and (iv) may be used for obtaining the value of the equilibrium constant from the

 

0

E

cell

value.

 

 

 

 

Example : 32 The standard emf for the cell reaction

Zn + Cu²⁺ ® Cu + Zn²⁺ is 1.1 volt at 25 ^o C .

The emf for the cell reaction, when 0.1 M Cu²⁺ and 0.1 M Zn ²⁺ solution are used, at 25^o C is

[MNR 1994, Manipal PMT 2001]

(a) 1.10 V                                (b) 0.10 V                                    (c) –1.10V                           (d) –0.110 V

 

 

Solution : (a)

Ecell

0

cell

0.0591

= E       –

n

[Zn²⁺ ]

log [Cu ²⁺ ]

 

Ecell = 1.10 – 0.0591 log é 0.1ù

 

 

Ecell

= 1.10V

2          êë 0.1úû

(Q log 1 = 0)

 

Example : 33 The hydrogen electrode is dipped in a solution of

pH 3 at

25^o C . The potential of the cell would be

 

 

 

 

Solution : (b)

(the value of 2.303RT/F is 0.059V)                                                                                               [KCET 1993]

(a) 0.177 V                              (b) –0.177 V                               (c) 0.087 V                          (d) 0.059 V

H ⁺ + e ^– ® 1 H 2

2

 

0.0591

P 1 / 2

ìQ Because pressure is not given ü

 

E = E ⁰ –

log   ^H2       í      [   ]                           ý

 

n             [H + ] îSo H ₂ = 1 then                          þ

 

 

 

 

0                               1          0

 

æ                          1   ö

 

= E   – 0.0591 log          = E

[H ⁺ ]

– 0.0591pH çQ pH = log [H + ] ÷

= 0 – 0.0591 ´ 3 = -0.177V

 

Example : 34

E ⁰ for the cell

è

Zn / Zn²⁺(aq) // Cu²⁺(aq) / Cu is 1.10V at

ø

25^o C . The equilibrium constant for the reaction

 

Zn + Cu²⁺ ⇌ Cu + Zn²⁺

is of the order of                                                                          [CBSE PMT 1997]

 

 

(a)

(aq)

 

10 -28

(aq)

(b)

10 -37

10 +18

1037

 

Solution : (d)

Zn + Cu²⁺ ⇌

Zn²⁺ + Cu; E^o = 1.10V

 

E = Eo  – 0.0591 log [Zn²⁺ ]

2          [Cu²⁺ ]

 

 

At equlibrium E= 0 and

[Zn²⁺] eqm  = K

[Cu²⁺] eqm

 

\ 0 = 1.10 – 0.0591 log K

2

Example : 35 For the half cell reaction,

or 1.10 = 0.0591 log K

2

H ⁺ (aq) + e ® 1 H ₂ (g)

or K = 1.94 ´ 10³⁷

 

 

(a)

(c)

E(H ⁺ / H ₂ ) = -59.2mV ´ pH

E ⁰ (H ⁺ / H 2 ) = pH log [H ⁺ ]

2 1atm

(b)

(d)

E ⁰ (H ⁺ / H 2 ) = -59.2mV ´ pH

E(H ⁺ / H ₂ ) = – pH log [H ⁺ ]

 

 

Solution : (a)

EH ⁺ / H2

0

= E   +

H   / H2

– 0.0592

1

P 1 / 2

log   ^H2 

[H ⁺ ]

 

Because SHE =0; then, = 0+0.0592 log [H ⁺ ] = -0.0592pH = – 59.2 mV × pH (mV = milli Volt)

Example : 36 A Galvanic cell is constructed as follows. A half- cell consists of a platinum wire immersed in a solution

 

containing 1.0 M of solution of Tl ⁺

Sn²⁺ and 1.0M of

Sn⁴⁺ and another half- cell has a thallium rod immersed in a 1.0 M

 

Given Sn⁴⁺ (aq) + 2e ® Sn²⁺(aq);        E⁰ = +0.13V

and Tl ⁺ (aq) + e ® Tl(s); E⁰ = -0.34V,

What is the cell voltage if the Tl ⁺ concentration is increased tenfold

(a) 0.411V                               (b) 4.101 V                                  (c) 0.492 V                          (d) 0.222 V

Solution : (a) The cell is represented as

-0.34 V                                                                                                 + 0.13 V

 

Tl(s)| Tl ⁺ (1.0M)|| Sn⁴⁺ (1.0M), Sn⁺² (1.0 M)| Pt

The cell reaction is [ Tl(s) ® Tl ⁺ + e ´ 2 ]

Sn⁴⁺ + 2e ^– ® Sn²⁺

Overall reaction : 2Tl(s) + Sn⁴⁺ ® 2Tl ⁺ + Sn²⁺

(Sn⁴⁺) = 1M ]

E = (E 0 – E 0 ) – 0.0592 log [Tl ⁺ ]²[Sn²⁺ ]

 

 

 

 

[Concentration (Tl⁺) = (10M), (Sn²⁺ ) = 1M ,

 

R              L                   2

[Sn⁴⁺ ]

 

= 0.47V – 0.0296 log (10)²            [Tl ⁺ concentration increases tenfold]

 

 

 

= 0.47 – 0.0592 = 0.411V

 

The electrical work (electrical energy) is equal to the product of the EMF of the cell and electrical charge that flows through the external circuit i.e.,

 

Wmax = nFEcell

……(i)

According to thermodynamics the free energy change (DG)

is equal to the maximum work. In the cell work is

 

done on the surroundings by which electrical energy flows through the external circuit, So

 

–Wmax, = DG

DG = –nFEcell

from eq. (i) and (ii)

In standard conditions

……(ii)

 

Where DG⁰ = standard free energy change

2.303

 

E       =

But

0

cell

nF     RT log Kc

2.303

 

\ DG⁰ = –nF ´

nF

RT log Kc

 

 

DG⁰ = -2.303 RT log Kc

 

DG⁰ = –RT ln Kc

(2.303 log X = ln X)

 

 

 

Example : 37 Calculate the standard free energy change for the reaction, 2Ag + 2H ⁺ ® H ₂ + 2Ag ⁺

E⁰ for

 

Ag ⁺ + e ^– ® Ag

is 0.80 V                                                                                                                         [ISM Dhanbad 1994]

 

(a)

+154.4kJ

(b)

+308.8kJ

(c)

-154.4kJ

(d) – 308.8kJ

 

cell

Solution : (a) For the given reaction E ⁰       = -0.80V

 

cell

DG = –nFE ⁰

=–2×96500×(–0.80)J=154400J =154.4kJ

 

Example : 38 The standard EMF of Daniell cell is 1.10 volt. The maximum electrical work obtained from the Daniell cell is

[MP PET 2002]

cell

(a) 212.3kJ                               (b) 175.4 kJ                                  (c)  106.15 kJ                       (d) 53.07kJ

 

Solution : (a) Electrical work obtained = –Wmax

= DG = –nE ⁰

F = 2 ´ 96500 ´ (1.1)J = 212300J = 212.3kJ

 

Example: 39      The    logarithm    of    the   equilibrium    constant,

log Keq, of    the   net   cell   reaction   of    the   cell,

 

cell

X(s)| X ²⁺ ||Y ⁺ |Y(s)(given E⁰

= 1.20V), is

 

(a) 47.2                           (b) 40.5                              (c) 21.4                        (d) 12.5

 

Solution : (b)

DG⁰ = –nE⁰F = –RT ln Keq

Þ ln Keq

= nE ⁰ F =

RT

nE ⁰

 

RT / F

 

 

 

 

 

log Keq

=       nE ⁰

2.303 RT

F

= nE ⁰

0.0592

(at 25⁰ C) ;

log K eq

= 2 ´ 1.20 = 40.5

0.0592

 

Example : 40 Give that at

25 ^o C , for Cr ³⁺ (aq) + e ^–  ® Cr ²⁺ (aq), E ⁰ = -0.424V

 

Cr ²⁺ (aq) + 2e ^–  ® Cr(s), E ⁰ = -0.900V

find E ⁰ at 25⁰ C for Cr ³⁺ (aq) + 3e ® Cr(s)

(a) – 0.741V                            (b) +0.741 V                               (c) –1.324 V                       (d) –0.476 V

Solution : (a) Adding the first two reactions, we get the third equation and using the free energy concept, we have

 

DG⁰ + DG⁰ = DG⁰ ; – n E⁰F – n  E⁰F = –n  E⁰F

(n = number of electron involved)

 

1           2           3          1   1          2   2             3 3

 

₀     n E⁰ + n E⁰

1´(-0.424) + 2 ´(-0.900)

 

E3 =   1   1        2   2 =

n3

= -0.741V

3

 

Example : 41 Given              the          half          –          cell          reactions,

Cu⁺(aq) + e^– ® Cu(s),

E⁰ = +0.52V ;

 

Cu ²⁺ (aq) + e ^– ®

+

1

Cu(aq),

E ⁰ = +0.16V

 

2

the equilibrium constant for the disproportionation reaction

2Cu⁺ (aq) ® Cu(s) + Cu ²⁺ (aq) at 298 K is

 

(a)

6 ´ 10 ⁴

(b)

6 ´ 10⁶

(c)

1.2 ´ 10⁶

(d)

1.2 ´ 10 ^–6

 

Solution : (c)

E ⁰ = E ⁰ – E ⁰ = 0.52 – 0.16 = 0.36V

 

1         2

 

 

log Keq

=    nE ⁰

0.0592

= 1 ´ 0.36 = 6.081 ,

0.0592

\ K eq

= 1.20 ´ 10⁶

 

Example : 42 Given standard electrode potentials

E 0

Fe ²⁺ + 2e ^– ® Fe – 0.440V ;      Fe ³⁺ + 3e ^– ® Fe – 0.036V

The standard electrode potential (E ⁰ ) for         Fe ³⁺ + e ® Fe ²⁺ is                                                   [AIIMS 1982]

 

 

Solution : (d)

(a) – 0.476 V                           (b) –0.404 V                               (c)  + 0.404V                       (d)+0.772V

DG⁰ = –nFE⁰

 

Fe²⁺ + 2e^– ® Fe

…..(i)

 

DG⁰ = -2 ´ F ´ (-0.440V) = 0.880F ;

 

Fe³⁺ + 3e^– ® Fe

…..(ii)

 

DG⁰ = -3 ´ F ´(-0.036) = 0.108 F , On subtracting equation (i) from equation (ii) we get

 

DG⁰ = 0.108F – 0.880F

=– 0.772F

 

E ⁰ for the reaction = DG0 = – (–0.772F) = + 0.772V

nF                1´ F

• The standard reduction potentials of a large number of electrodes have been measured using standard hydrogen electrode as the reference electrode. These various electrodes can be arranged in increasing or decreasing order of their reduction potentials. The arrangement of elements in order of increasing reduction potential values is called electrochemical series.

 

 

 

The electrochemical series, also called activity series, of some typical electrodes is being given in Table.

Standard reduction electrode potentials at 298K

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(2) Characteristics of Electrochemical series

• The negative sign of standard reduction potential indicates that an electrode when joined with SHE acts as anode and oxidation occurs on this electrode. For example, standard reduction potential of zinc is –0.76 volt, When zinc electrode is joined with SHE, it acts as anode (–ve electrode) e., oxidation occurs on this electrode. Similarly, the +ve sign of standard reduction potential indicates that the electrode when joined with SHE acts as cathode and reduction occurs on this electrode.
• The substances, which are stronger reducing agents than hydrogen are placed above hydrogen in the series and have negative values of standard reduction potentials. All those substances which have positive values of reduction potentials and placed below hydrogen in the series are weaker reducing agents than
• The substances, which are stronger oxidising agents than H ⁺ ion are placed below hydrogen in the
• The metals on the top (having high negative value of standard reduction potentials) have the tendency to lose electrons These are active metals. The activity of metals decreases form top to bottom. The non-metals on the bottom (having high positive values of standard reduction potentials) have the tendency to accept electrons readily. These are active non-metals. The activity of non-metals increases from top to bottom.

(3)    Application of Electrochemical series

• Reactivity of metals: The activity of the metal depends on its tendency to lose electron or electrons, e.,

tendency to form cation (M ⁿ⁺ ) . This tendency depends on the magnitude of standard reduction potential. The

metal which has high negative value (or smaller positive value) of standard reduction potential readily loses the electron or electrons and is converted into cation. Such a metal is said to be chemically active. The chemical

 

 

 

reactivity of metals decreases from top to bottom in the series. The metal higher in the series is more active than the metal lower in the series. For example,

• Alkali metals and alkaline earth metals having high negative values of standard reduction potentials are chemically active. These react with cold water and evolve hydrogen. These readily dissolve in acids forming corresponding salts and combine with those substances which accept
• Metals like Fe, Pb, Sn, Ni, Co, etc., which lie a little down in the series do not react with cold water but react with steam to evolve hydrogen.
• Metals like Cu, Ag and Au which lie below hydrogen are less reactive and do not evolve hydrogen from
  • Electropositive character of metals : The electropositive character also depends on the tendency to lose electron or electrons. Like reactivity, the electropositive character of metals decreases from top to bottom in the electrochemical On the basis of standard reduction potential values, metals are divided into three groups
• Strongly electropositive metals : Metals having standard reduction potential near about – 2.0 volt or more negative like alkali metals, alkaline earth metals are strongly electropositive in
• Moderately electropositive metals : Metals having values of reduction potentials between 0 and about –

2.0 volt are moderately electropositive Al, Zn, Fe, Ni, Co, etc., belong to this group.

• Weakly electropositive : The metals which are below hydrogen and possess positive values of reduction potentials are weakly electropositive Cu, Hg, Ag, etc., belong to this group.
  • Displacement reactions
• To predict whether a given metal will displace another, from its salt solution: A metal higher in the series will displace the metal from its solution which is lower in the series, e., The metal having low standard reduction potential will displace the metal from its salt’s solution which has higher value of standard reduction potential. A metal higher in the series has greater tendency to provide electrons to the cations of the metal to be precipitated.
• Displacement of one nonmetal from its salt solution by another nonmetal: A non-metal higher in the series (towards bottom side), e., having high value of reduction potential will displace another non-metal with lower reduction potential, i.e., occupying position above in the series. The non-metal’s which possess high positive reduction potentials have the tendency to accept electrons readily. These electrons are provided by the ions of the

 

nonmetal having low value of reduction potential,. Thus, iodides.

Cl 2

can displace bromine and iodine from bromides and

 

Cl₂ + 2KI ® 2KCl + I₂

2I ^– ® I 2 + 2e ^–

Cl2 + 2e ^– ® 2Cl ^–

…..(Oxidation)

…..(Reduction)

 

[The activity or electronegative character or oxidising nature of the nonmetal increases as the value of reduction potential increases.]

 

• Displacement of hydrogen from dilute acids by metals : The metal which can provide electrons to H ⁺

present in dilute acids for reduction, evolve hydrogen from dilute acids.

ions

 

Mn ® Mnⁿ⁺ + ne ^–

2H ⁺ + 2e ^– ® H 2

…..(Oxidation)

…..(Reduction)

 

The metal having negative values of reduction potential possess the property of losing electron or electrons.

Thus, the metals occupying top positions in the electrochemical series readily liberate hydrogen from dilute acids and on descending in the series tendency to liberate hydrogen gas from dilute acids decreases.

 

 

 

The metals which are below hydrogen in electrochemical series like Cu, Hg, Au, Pt, etc., do not evolve hydrogen from dilute acids.

• Displacement of hydrogen from water : Iron and the metals above iron are capable of liberting hydrogen from water. The tendency decreases from top to bottom in electrochemical series. Alkali and alkaline earth metals liberate hydrogen from cold water but Mg, Zn and Fe liberate hydrogen from hot water or
  • Reducing power of metals: Reducing nature depends on the tendency of losing electron or More the negative reduction potential, more is the tendency to lose electron or electrons. Thus reducing nature decreases from top to bottom in the electrochemical series. The power of the reducing agent increases, as the standard reduction potential becomes more and more negative. Sodium is a stronger reducing agent than zinc and zinc is a stronger reducing agent than iron. (decreasing order of reducing netur)

 

Element :

Reduction potential :

Na   >

– 2.71

Zn

– 0.76

• Fe

– 0.44

 

Alkali and alkaline earth metals are strong reducing agents.

• Oxidising nature of non-metals : Oxidising nature depends on the tendency to accept electron or More the value of reduction potential, higher is the tendency to accept electron or electrons. Thus, oxidising nature increases form top to bottom in the electrochemical series. The strength of an oxidising agent increases as the value of reduction potential becomes more and more positive.

 

F2 (Fluorine) is a stronger oxidant than Cl 2 , Br2

and

I 2 . Cl 2

(Chlorine) is a stronger oxidant than Br₂ and I ₂

 

Element :

Reduction potential :

I 2              Br2                 Cl2                         F2

+ 0.53    + 1.06     + 1.36     + 2.85

 

¾¾¾¾¾¾¾¾¾¾¾¾¾¾¾®

Oxidising nature increases

Thus, in electrochemical series

 

 

 

 

 

 

 

 

 

 

 

 

 

• Thermal stability of metallic oxides : The thermal stability of the metal oxide depends on its electropositive nature. As the electropositivity decreases form top to bottom, the thermal stability of the oxide also decreases from top to bottom. The oxides of metals having high positive reduction potentials are not stable towards The metals which come below copper form unstable oxides, i.e., these are decomposed on heating.

BaO ù

Ag  O ¾¾D  ® 2Ag + 1 O  ;  2HgO ¾¾D  ® 2Hg + O   ;  Na  O ú ¾¾D  ® No decomposition

2                                                  2    2                                                                            2                  2       ú

Al2 O3 úû

• Extraction of metals : A more electropositive metal can displace a less electropositive metal from its salt’s This principle is applied for the extraction of Ag and Au by cyanide process. silver from the solution

 

 

 

 

containing sodium argento cyanide,

 

NaAg(CN)₂ , can be obtained by the addition of zinc as it is more electro-

 

positive than

Ag.

2NaAg(CN)2 + Zn ® Na2 Zn(CN)4   + 2Ag

 

 

• When metals are exposed to atmospheric conditions, they react with air or water in the environment to form undesirable compounds (usually oxides). This process is called corrosion. Almost all metals except the least active metals such as gold, platinum and palladium are attacked by environment e., undergo corrosion. For example, silver tarnishes, copper develops a green coating, lead or stainless steel lose their lustre due to corrosion. Corrosion causes enormous damage to building, bridges, ships and many other articles made of iron.

Thus corrosion is a process of deterioration of a metal as a result of its reaction with air or water (environment) surrounding it.

 

In case of iron, corrosion is called rusting. Chemically, rust is hydrated form of ferric oxide,

Fe2 O3 .

xH 2 O .

 

Rusting of iron is generally caused by moisture, carbon dioxide and oxygen present in air. It has been observed that rusting takes place only when iron is in contact with moist air. Iron does not rust in dry air and in vacuum.

• Factors which affect corrosion : The main factors which affect corrosion are
  • Position of metals in emf series : The reactivity of metal depends upon its position in the electrochemical More the reactivity of metal, the more will be the possibility of the metal getting corroded.
  • Presence of impurities in metals : The impurities help in setting up voltaic cells, which increase the speed of corrosion
  • Presence of electrolytes : Presence of electrolytes in water also increases the rate of corrosion For example, corrosion of iron in sea water takes place to larger extent than in distilled

 

• Presence of

CO2 in water : Presence of CO2 in natural water increase rusting of iron. Water containing

 

CO2

acts as an electrolyte and increases the flow of electrons from one place to another

• Presence of protective coatings : When the iron surface is coated with layers of metals more active

 

than iron, then the rate of corrosion is retarded. For example, coating of zinc on iron prevents rusting.

• Temperature : A rise in temperature (with in a reasonable limit) increases the rate of

• Classification of corrosion process : Depending upon the nature of corrosion, and the factors affecting it, the corrosion may be classified as
• Chemical corrosion : Such corrosion, generally takes place when
• Reactive gases come in contact with metals at high temperatures g., corrosion in chemical industry.
• Slow dissolution of metal takes place when kept in contact with non conducting media containing organic

acids.

• Bio-chemical corrosion or Bio-corrosion: This is caused by the action of Soils of

definite composition, stagnant water and certain organic products greatly favour the bio-corrosion.

• Electrochemical corrosion : It occurs in a gaseous atmosphere in the presence of moisture, in soils and in For example, the following corrosions are electrochemical in nature
• Corrosion of insoluble anodes
• Corrosion of pipelines with current carrying liquids flowing through
• Corrosion of underground metal structure
• Mechanism of rusting of iron : Electrochemical theory of rusting. The phenomenon of corrosion can be explained with the help of electrochemical theory which involves oxidation and reduction reactions. According to electrochemical theory of corrosion, it is believed that non-uniform surface of metal or impurities

 

 

 

present in iron behave like small electric cells (called corrosion couples) in the presence of water containing

 

dissolved oxygen or carbon dioxide. A film of moisture with dissolved

CO₂ constitutes electrolytic solution covering

 

the metal surface at various places. The schematic representation of mechanism of rusting of iron is shown in Fig.

In the small electrolytic cells, pure iron acts as anode while cathodes are impure portions. The overall rusting involves the following steps,

• Oxidation occurs at the anodes of each electrochemical Therefore, at each anode neutral iron atoms are oxidised to ferrous ions.

At anode :        Fe(s) ¾¾® Fe ²⁺ (aq) + 2e ^– .

Thus, the metal atoms in the lattice pass into the solution as ions, leaving electrons on the metal itself. These electrons move towards the cathode region through the metal.

• At the cathodes of each cell, the electrons are taken up

 

by hydrogen ions (reduction takes place). The H ⁺

ions are

 

obtained either from water or from acidic substances (e.g. in water

CO2 )

 

3

H 2 O ¾¾® H ⁺ + OH ^– or CO2 + H 2 O ¾¾® H ⁺ + HCO^–

At cathode :      H ⁺ + e ^– ¾¾® H

The hydrogen atoms on the iron surface reduce dissolved

 

oxygen.

4 H + O₂ ¾¾® 2H ₂O

 

Therefore, the overall reaction at cathode of different electrochemical cells may be written as,

4 H ⁺ + O2 + 4e ^– ¾¾® 2H 2 O

• The overall redox reaction may be written by multiplying reaction at anode by 2 and adding reaction at cathode to equalise number of electrons lost and gained e.

 

Oxidation half reaction        :

Fe(s) ¾¾® Fe ²⁺ (aq) + 2e ^– ] ´ 2

(E –    = -0.44V)

 

 

 

Reduction half reaction        :

 

Overall cell reaction             :

4 H +   + O   + 4e – ¾¾® 2H O(E –   = 1.23V)

2                                                     2

2                                                                       2               cell

2Fe(s) + 4 H +   + O   ¾¾® 2Fe 2+ (aq) + 2H O(E     – = 1.67V)

 

The ferrous ions are oxidised further by atmospheric oxygen to form rust.

 

4 Fe ²⁺ (aq) + O2 (g) + 4 H 2 O ¾¾® 2Fe2 O3 + 8H ⁺

and

Fe2 O3  + xH 2 O ¾¾® Fe2 O3 . xH 2 O

Rust

 

It may be noted that salt water accelerates corrosion. This is mainly due to the fact that salt water increases the electrical conduction of electrolyte solution formed on the metal surface. Therefore, rusting becomes more serious problem where salt water is present.

• Corrosion protection : Corrosion of metals can be prevented in many Some commonly used methods are described below.
  • By surface coating : Corrosion of metals can be prevented by coating their surfaces with any of the following
• By applying, oil, grease, paint or varnish on the

 

 

 

• By coating/depositing a thin layer of any other metal which does not For example, iron surface can be protected from corrosion by depositing a thin layer of zinc, nickel or chromium on it. Copper/brass can be protected by coating it with a thin layer of tin. Tinning of brass utensils is a very common practice in our country.
• By Galvanization : Prevention of corrosion of iron by Zn
  • By connecting metal to a more electropositive metal : A metal can be protected from corrosion by connecting it to a more electropositive metal. As long as the more electropositive metal is there, the given metal does not get corroded. For example, iron can be protected from corrosion by connecting it to a block/plate of zinc or This method of corrosion protection is called cathodic protection.
  • By forming insoluble phosphate or chromate coating : Metal surfaces are treated with phosphoric acid to form an insoluble phosphate By connecting metal to a more coating on the surface. This phosphate coating protects the metal from Formation of a thin chromate layer also prevents the corrosion of metals.
  • Using anti – rust solutions : To retard the corrosion of iron certain anti – rust solutions are used. For example, solutions of alkaline phosphates and alkaline chromates are generally used as anti – rust Due to

 

the alkaline nature of these solutions, the H ⁺

ions are removed from the solutions, and rusting is prevented. For

 

example, iron articles are dipped in boiling alkaline sodium phosphate solutions, when a protective insoluble sticking film of iron phosphate is formed.

 

***