+2 UNIT 5 PAGE- 3

ELECTROLYSIS
The passage of electricity through electrolytes in their molten or dissolved state cause chemical changes under suitable conditions. For example, the passage of electricity through acidified water results in the formation of hydrogen and oxygen gases. The process of chemical decomposition of the electrolyte by the passage of electricity through its molten or dissolved state is called electrolysis.
Electrolytic cell
The device in which the process of electrolysis is carried out is called electrolytic cell. It consists of :
(i) Electrolytic tank, which is made of some non-conducting material like glass, wood or bakelite.

Fig.11 Electrolytic Cell
ii) Electrolyte in its dissolved state or in the molten state.
iii) Source of electricity : an electrochemical cell or battery.
iv) Two metallic strips or rods, suspended in the electrolyte and connected to the battery through conducting wires. These rods are called electrodes.
The electrode connected to the negative terminal of the battery is called cathode while the other one which is connected to the positive terminal is called anode. The apparatus used to constitute electrolytic cell has been shown in Fig 11.
Criteria of product formation in electrolysis
The process of electrolysis can be explained on the basis of the theory of ionisation. When an electrolyte is dissolved in water, it splits up into charged particles called ions. The positively charged ions are called cations while the negatively charged ions are called anions. The ions are free to move about in aqueous solution. When electric current is passed through the solution, the ions respond to the applied potential difference and their movement is directed towards the oppositely charged electrodes. The cations move towards the negatively charged electrode, while anions move towards the positively charged electrode. The formation of products at the respective electrodes is due to the oxidation (loss of electrons) at the anode and reduction (gain of electrons) at the cathode.
For example, when electricity is passed through the molten sodium chloride, sodium is deposited at the cathode while chlorine is evolved at the anode. The process can be represented as :
NaCl(ℓ)  Na+ + Cl
Anode : Cl  ½ Cl2 + e  ( Oxidation)
Cathode : Na+ + e  Na ( Reduction)
Similarly , electrolysis of molten lead bromide [PbBr2(ℓ)] produces lead at the cathode and Br2 at the anode.
In case there is a possibility of formation of more than one product at the electrodes, or there is a competition between the liberation of ions at the electrodes, then the product formed depends upon their respective electrode potentials, as described below:
(a) At the cathode
Reduction process occurs at the cathode. Therefore, for different competing reduction processes, the one with higher reduction potential , will preferably take place. For example, during the electrolysis of aqueous solution of sodium chloride there is possibility of following reactions at the cathode:
Reduction of Na+ ions
Na+(aq) + e Na(s) : E0Red = 2.71 V
or Reduction of H2O molecules:
H2O(ℓ)+ 2 e   ½ H2 + OH(aq) : E0Red = 0.41 V
Thus, we find that during electrolysis of aqueous NaCl, the reduction of water will preferentially take place at the cathode because E0Red of water is higher. Hence, the product of electrolysis at the cathode will be H2 gas instead Na(s).
Now let us compare the E0Red values of copper and H2O.
Cu2+(aq) + 2 e  Cu(s) ; E0Red = 0.34 V
H2O( l ) + 2 e ½ H2 + OH(aq) : E0Red = 0.41 V
Thus during electrolysis of aqueous solution of copper sulphate reduction of Cu2+ ions will take place at the cathode in preference to the reduction of H2O molecules.
(b) At anode
Oxidation processes occurs at the anode. Therefore, for different competing oxidation processes, the one with higher oxidation potential (or lower reduction potential) will preferably occur. For example, if we carry out electrolysis of aqueous solution of copper sulphate, the competing oxidation processes at the anode are :
Oxidation of SO42 ion :
2 SO42(aq)  S2O82+ 2 e : E0oxi = 2.01 V
or Oxidation of water molecules :
H2O( ) ½ O2(g) + 2 H+(aq) + 2 e : E0oxi = 1.23 V
As the oxidation potential of water is higher, the product formed at the anode will be O2 gas instead of S2O82 ion.
Similarly, if electrolysis of copper sulphate solution is carried out using copper electrodes, then the process occurring at the anode will be oxidation of copper atoms to copper ions instead of oxidation of water because oxidation potential of copper is higher.
Cu(s)  Cu2++ 2 e : E0oxi = 0.34 V
H2O (ℓ) ½ O2(g) + 2 H+(aq)+2 e : E0oxi = 1.23 V
Thus, in such a case copper from anode will go on dissolving into solution as Cu2+ ions while Cu2+ ions from solution will go on depositing at the cathode as copper atoms.
The above discussion leads us to a conclusion that for different competing reactions at the electrodes :
Cathode reaction will be the one with higher E0Red value.
Anode reaction will be the one with higher E0oxi value.
It may be noted that in some cases the unexpected results are obtained due to overvoltage. The transference of electrons from water is kinetically slow process as it requires relatively large activation energies. The slowness of electrode reaction creates some electrical resistance at the electrode surface. Hence extra voltage is required to provide activation energy and to overcome resistance in case of oxidation of water molecules. The extra voltage required for the oxidation of water is called overvoltage. Due to overvoltage, the oxidation of Cl ions occurs in preference to H2O.) For example, let us compare the oxidation potentials of Cl ion and water.
H2O(ℓ)  ½O2 (g)+ 2 H+(aq) + 2 e ; E0oxi = 1.23 V
Cl (aq)  ½ Cl2 (g)+ e  ; E0oxi = 1.36 V
Although the oxidation potential of water is more than that of Cl ions , yet during the electrolysis of concentrated solution of sodium chloride, the chloride ions oxidise in preference to H2O molecules at the anode giving Cl2 gas as the product.
In light of the above discussion, let us discuss the product formed during electrolysis of some of the electrolytes.
1. Electrolysis of aqueous solution of NaCl
NaCl in aqueous solutions ionise as :
NaCl(aq)  Na+(aq) + Cl(aq)
Reaction at the anode : Cl (aq)  ½ Cl2 (g)+ e
Reaction at the Cathode : H2O(ℓ )+ 2 e  ½ H2 + OH(aq)
Thus Cl2 gas is evolved at the anode whereas H2 gas is liberated at the cathode.
2. Electrolysis of copper sulphate using platinum electrodes
Copper sulphate ionises in aqueous solution :
CuSO4(aq)  Cu2+(aq) + SO42 (aq)
Reaction at the anode : H2O(ℓ) ½ O2 (g)+ 2 H+(aq) +2 e
Reaction at the cathode : Cu2+(aq) + 2 e  Cu(s)
Thus , copper is deposited at the cathode and O2 is liberated at the anode.
3. Electrolysis of copper sulphate solution using copper electrodes
Copper sulphate ionises as :
CuSO4(aq)  Cu2+(aq) + SO42  (aq)
Reaction at anode : Cu(s)  Cu2++ 2 e
Reaction at cathode : Cu2+(aq) + 2 e  Cu(s)
Thus copper dissolves at the anode and is deposited at the cathode.
QUANTITATIVE ASPECTS OF ELECTROLYSIS
Faraday performed a large number of experiments on electrolysis and summarised the results of his experiment in the form of two laws known as Faraday's laws of electrolysis. These are :
1. First Law of electrolysis
The amount of any substance liberated at the electrode is directly proportional to the quantity of electricity passed through the electrolyte solution.
The quantity of electricity (Q )is equal to the product of the current strength and time for which it is passed. In other words :
Q = Ampere x second
The practical unit of electricity is coulomb, which is the current in ampere(I) multiplied by the time in seconds(t), i.e.,
Coulomb = Ampere x second
If w is the weight of a substance liberated or deposited at the electrode during electrolysis, then from the first law, we get :
w  Q
 I t
= z I t Since Q = I t
where I = Current in ampere ; Q = quantity of electricity ;
t = time in seconds. z = is a constant of proportionality, called electrochemical equivalent of the substance.
When Q = 1 Coulomb, then w = z. Consequently, electrochemical equivalent of an element may be defined as the weight of the substance deposited/evolved by the passage of 1 coulomb of electricity.
2. Second Law of electrolysis
When the same quantity of electricity is passed through different electrolytic solutions connected in series, the weights of different substances produced at the electrodes are proportional to their equivalent weights.
Mathematically:
w  E
where w = weight of the substance liberated or deposited.
E= Equivalent of the substance liberated or deposited.
Thus if w1 and w2 are the weights of two different substances liberated by the passage of same quantity of electricity and E1 and E2 are their respective equivalent weights, then :

Let us consider, the passage of electricity through solutions of copper sulphate and silver nitrate, connected in series, so that the same amount of electricity is passed through them, then from the second law, we get :

Illustrations
For example , during the passage of electric current through molten NaCl, sodium gets deposited at the cathode and chlorine is liberated at anode.
NaCl  Na + ½ Cl2
(At cathode) (At anode)
During electrolysis, sodium ions move towards the cathode , accept electrons and get deposited as :
Na++ e Na(s)
Similarly, at anode, chloride ions give up electrons and produce Cl atoms as :
Cl  e  ½ Cl2(g)
or 2 Cl  2 e Cl2(g)
It is clear that 2 moles of electrons produce 1 mole of Cl2 or 1 mole of electrons produce ½ mole of Cl2.
Charge on an electron = 1.602 x 1019 C
1 mole of electrons = 6.023 x 1023 electrons.
Charge on 1 mole of electrons=
= 6.023 x 1023 mol-1 x 1.602 x 10-19 C
= 96485 C mol1.
The charge on one mole of electrons is called 1 Faraday, F. Thus,
1 F = 96485 C or approximately 96,500 C. Thus, charge on n moles of electrons will be equal to:
Q = n F
Now, 1 mole of Cl2 is obtained by 2 moles of electrons or
2 x 96,500 C of charge during electrolysis of NaCl. Similarly, 1 mole of copper will be produced by 2 mole of electrons or
2 x 96,500 C of charge:
Cu2++ 2 e  Cu
From the knowledge of moles of electrons required during the electrode reaction, the amount of substance deposited or evolved can be calculated. For example, aluminium gets deposited as :
Al3+ + 3 e  Al
1 mole 3 mole 1 mole
Thus, 1 mole of Al will be deposited by 3 mole of electrons or 3 Faraday ( or 3 x 96500 C) of electricity.
Problems
40. How many coulombs are required for the following reactions ?
(i) 1 mol of Al3+ to Al. (ii) 1 mol of Cu2+ to Cu
(iii) 1 mol of MnO4 to Mn2+.
41. How many coulombs are liberated during the following oxidations ?
i) 1 mol of H2O to O2
ii) 1 mol of FeO to Fe2O3.
42. How many coulombs are required to produce :
(i) 20 g of calcium from molten CaCl2.
(ii) 50 g of Aluminium from molten Al2O3.
43. The charge on an electron is 1.602 x 1019 C. Calculate the value of Faraday constant.
44. For how long a current of 1.5 A has to be passed through the electrolyte in order to deposit 1 g of Al when electrode reaction is : Al3+ + 3 e  Al
45. Calculate the mass of hydrogen evolved by passing a current of 0.5 ampere for 40 minutes through acidified water.
46. A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ?
47. A solution of nickel nitrate, Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5.0 amps for 20 minutes. What mass of Ni will be produced at the cathode ?
48. Three electrolytic cells A, B and C containing ZnSO4 , AgNO3 and CuSO4 respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow ? What mass of copper and zinc were deposited ?
Difference between electrochemical cell and electrolytic Cell
Electrochemical cells convert chemical energy into electrical energy. On the other hand , electrolytic cell is a device which converts electrical energy into chemical energy. The two cells differ significantly with respect to the charges on the electrodes. For example, in electrochemical cell , anode is negative whereas in electrolytic cells the anode is positive. Similarly cathode is positive in electrochemical cell whereas it is negative in electrolytic cell. The main points of difference have been summarised in the following TABLE.
Differences between Galvanic and Electrolytic cells
Galvanic cell Electrolytic cell
1.In Galvanic cell electrical energy is produced.
2. In Galvanic cell, reaction taking place is spontaneous. 1.In electrolytic cell electrical energy is consumed.
2.In electrolytic cell, reaction taking place is non-spontaneous.
3. The two half-cells are set up in different containers and are connected through salt bridge or porous partition. 3. Both the electrodes are placed in the solution or molten electrolyte in the same container.
4. In a Galvanic cell, anode is negative and cathode is positive. 4. In electrolytic cell, the anode is positive and the cathode is negative.
5. The electron move from anode to cathode in external circuit. 5. The electrons are supplied by the external source. They enter through cathode and come out through anode.
CORROSION
It is a common observation that certain metals(except those which are least reactive like Au, Pt, Pd etc) are slowly eaten up on long exposure to atmosphere. For example, silver gets tarnished, copper develops green coating on its surface, iron rusts and lead looses its lustre. In fact, such metals react with gases or moisture present in the environment to form undesirable compounds. This process in general is referred to as corrosion. Corrosion may, thus be defined as the process of slow conversion of metals into their undesirable compounds(usually oxides) by reaction with moisture and other gases present in the atmosphere.
FACTORS WHICH AFFECT CORROSION
The factors which affect corrosion are :
1. Reactivity of the metal : The more active metals are prone to corrosion .
2. Presence of impurities : Presence of impurities helps in setting up a corrosion cell and makes the corrosion to occur rapidly. For example, pure iron does not rust.
3. Air and moisture : Air and moisture are quite helpful in corrosion. The presence of gases like CO2 and SO2 in air makes it still rapid. For example no rusting is caused if iron is kept under vaccum.
4. Strains in metal also help in corrosion. For example, in iron articles, rusting is more pronounced on the areas having bends, scratches, nicks and cuts.
5. Presence of electrolytes : The presence of electrolytes also make the corrosion process faster. For example, iron rusts more rapidly in saline water in comparison to pure water.
Let us study the mechanism of corrosion by studying the familiar example of rusting of iron. Chemically, rust is hydrated iron(III) oxide, F2O3.x H2O. It is generally caused by moisture, CO2 and O2 of air. Rust is non-sticking brown coloured material which can be easily removed by scratching. There are a number of theories about the mechanism of rusting. The most widely accepted theory is electrochemical theory.
ELECTROCHEMICAL THEORY OF RUSTING
According to this theory, the impure surface behaves like a small electrochemical cell in presence of water containing dissolved oxygen or carbon dioxide. Moisture having dissolved oxygen or carbon dioxide in it constitutes electrolytic solution.
At anode, oxidation of Fe atoms takes place. Thus, Fe atoms pass into solutions as Fe2+ ions leaving behind electrons in the metal which are pushed into the cathode area.
Fe  Fe2+ + 2 e : E0ox = + 0.44 V
At cathode the electrons are picked up by H+ ions which are produced either from H2O or from H2CO3 ( formed by dissolution of CO2 in moisture).
CO2 + H2O  H+ + HCO3
2 H+(aq) + 2 e  2 H
The H atoms, thus formed reduce the dissolved oxygen as :
2 H + ½ O2  H2O
The net reduction process at the cathodic area is :
2 H+ + ½ O2 + 2 e  H2O: E0Red = 1.23 V
The net reaction of of corrosion cell can be put as :
Fe + 2 H+ + ½ O2  Fe2+ + H2O : E0cell =1.67 V
The ferrous ions so formed move through water and come at the surface of iron object where these are further oxidised to ferric state by atmospheric oxygen and constitute rust which is hydrated iron(III) oxide.
2 Fe2++ ½ O2 + 2 H2O  Fe2O3 + 4 H+
Fe2O3 + x H2O  Fe2O3 . x H2O
(rust)
2 Fe2+ + ½ O2 +(2 + x ) H2O  Fe2O3 . x H2O + 4 H+

Fig 12. Rusting Of Iron.
PREVENTION OF CORROSION
There are several methods for protecting metals from corrosion(iron from rusting ).Some of these methods are discussed below :
1. Barrier Protection
In this method, a barrier film is introduced between iron and atmospheric oxygen and moisture. Barrier protection can be achieved by any of the following methods.
i) Painting the surface.
ii) By coating the surface with a thin film of oil or grease.
iii) By electroplating iron by some non-corrosive metal such as nickel, chromium, copper etc.
In this type of protection, if scratches or cracks appear in the protective layer, then the surface of iron may get exposed. In this region, moisture and oxygen may come in contact with iron and rusting starts. This rusting extends beneath the protective layer and eventually peels off the protective layer.
2. Sacrificial Protection
Sacrificial protection means covering of the surface of iron with a layer of metal which is more active than iron and thus prevents the loss of electrons from iron. The active metal loses electrons in preference of iron and goes into the ionic state. With the passage of time, the covering metal gets consumed. But as long as it is present on the surface of iron, it will protect the iron from rusting and does not allow even nearly exposed surface of iron to corrode. Zinc is commonly used to coat iron surfaces. The process of coating a thin film of zinc on iron is known as galvanisation. Since zinc is more electropositive than iron, it loses electrons preferentially and goes into solution as Zn2+ ion.



Fig 13. Protection of Rusting of iron by coating of Zn
As long as zinc is present on the surface, iron will not corrode. If some scratches occur on the protective film of zinc, both iron and zinc are exposed to oxygen. But due to the lower reduction potential of Zn compared to that of iron, zinc undergoes oxidation in preference to Fe. Thus, iron does not corrode even if the film of zinc is broken at some point.
3. Electrical Protection
This method is used for protecting iron articles which are in contact with water such as underground water pipes. The article of iron is connected with more active metals like magnesium or zinc. The active metal has higher oxidation potential (e.g., oxidation potential of magnesium , Mg(s)  Mg2++ 2 e; E0= 2.67 V) than iron and will act as anode. Consequently, the active metal loses electrons in preference to iron and therefore , protects iron from being rusted.

Fig 14. Electrical Protection of Iron pipes.
4. Using anti-rust Solutions
To retard the corrosion of iron, certain anti-rust solutions are used. For example, solution of alkaline phosphates and alkaline chromates are generally used as anti-rust solutions. Due to the alkaline nature of these solutions, the H+ ions are removed from 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.
QUESTIONS
1. What are conductors and insulators ?
2. What is meant by the term :
i) metallic conduction.
ii) Electrolytic conduction.
3. Explain the following terms :
i) Weak and strong electrolytes.
ii) Specific conductivity.
iii) Equivalent conductivity.
iv) Molar conductivity.
4. Define Cell constant.
5. Describe an experiment for measuring conductance of electrolytic solution.
6. State the factors which determine the magnitude of conductance of an electrolyte.
7. How is molar conductivity related to concentration of an electrolyte ?
8. How will you explain a weak and strong electrolyte based on their conductivity values ?
9. Explain the conductance behaviour of strong and weak electrolytes.
10. How does specific conductance of an electrolytic solution vary with dilution ?
11. Define the term degree of ionisation of an electrolyte.
12. How can you determine the equivalent conductance of strong electrolyte by extrapolation method ?
13. Why can't we determine the equivalent conductance of a weak electrolyte by extrapolation method ?
14. State and explain Kohlrausch's law of independent migration of ions.
15. Discuss the applications of Kohlrausch's law.
16. Write down the expression for equivalent conductivity.
17. State and explain Faraday's laws of electrolysis.
18. What products will be obtained by the electrolysis of :
i) molten sodium chloride.
ii) Aqueous solution of sodium chloride.
Support your answer in terms of electrode potential.
19. What is electrode potential ?
20. Discuss standard electrode potential.
21. Differentiate between oxidation electrode potential and reduction electrode potential.
22. Why are single electrode potentials not measurable ?
23. Define E.M.F of a cell or cell potential.
24. Explain the difference between E.M.F and potential difference.
25. What does positive value of E0cell in the case of Daniel cell indicate ?
26. What is normal hydrogen electrode ?
27. Describe the construction of a normal hydrogen electrode.
28. What is the significance of normal hydrogen electrode ?
29. How is electrode potential measured ?
30. What is the significance of the statement that the standard reduction potential of zinc is 0.76 V ?
31. What is the significance of the statement that the standard reduction potential of copper is +0.34 V ?
32. What is meant by electrochemical series ?
33. List the important uses of electrochemical series.
34. What is the importance of electrochemical series ?
35. The standard electrode potential of a number of half-cells are provided in the increasing order. Mention three tendencies in properties of elements from such data.
36. How does electrochemical series help us in predicting whether a redox reaction is feasible in a given direction or not ?
37. Why colour of copper sulphate is discharged after some times , when iron rod is dipped into it ?
38. Can we store copper sulphate solution in silver vessel ? Give suitable explanation.
39. What happens when , if copper sulphate solution is stored in a zinc vessel and why ?
40. Can we store copper sulphate in a zinc vessel ? Give suitable explanation.
41. Explain the differences between the positive and the negative value of potential of an electrode.
42. When zinc dust is added to copper sulphate solution, copper is deposited. Why does the reverse not occur ?
43. Explain why zinc displaces lead from lead nitrate solution ; while gold does not.
44. Explain why iron will liberate hydrogen from dilute hydrochloric acid and sulphuric acids ; while copper will not.
45. Explain with reason, why zinc and iron decompose steam; while copper and mercury do not.
46. Describe the construction of Daniel cell and the reaction involved. How is it represented ?
47. Give a short account of 'salt bridge' . State its function.
48. Explain with diagram, the role of a salt bridge.
49. How are Galvanic cells different from electrolytic cells ?
50. Explain the difference between electrochemical cell and electrolytic cell.
51. Write down the modern sign conventions used in writing galvanic cell reactions.
52. The cell potential of a spontaneous reaction is always positive. What do you think of the occurrence of a reaction for which cell potential is negative ?
53. Can 1M solution of ferrous sulphate be stored in a nickel vessel ?
54. Predict the products of electrolysis of each of the following :
i) An aqueous solution of silver nitrate with silver electrodes.
ii) An aqueous solution of silver nitrate with inert electrodes.
iii) A dilute aqueous solution of sulphuric acid.
iv) An aqueous acidified solution of cupric chloride using platinum electrodes.
55. Arrange the following metals in the order in which they displace each other.
Al, Cu, Fe, Mg, Zn.
56. Depict the electrochemical cell in which the reaction is :
Zn(s) + 2 Ag+(aq)  Zn2+(aq) + 2 Ag(s)
Also show :
a) Cathode and anode reactions.
b) Movement of ions and electrons.
c) Electrode reactions.
57. What is Nernst Equation ?
58. How can hydrogen electrode may be used to measure the pH of a solution ?
59. When molten sodium chloride is electrolysed, sodium and chlorine are evolved at the cathode and anode respectively. However, when concentrated aqueous solution of sodium chloride is electrolysed, chlorine is produced at the anode, but hydrogen(and not sodium) is obtained at the cathode. Explain why ?
60. What is an electric cell ? How does it differ from a battery ?
61. What is a primary cell ?
62. Write a note on dry cell.
63. What is a secondary cell ?
64. Describe the construction and working of lead storage battery.
65. What is a lead-storage cell ? Give reactions involved during charging and discharging. Mention its uses.
66. Write a note on nickel-cadmium accumulator.
67. Write a note on fuel cells.
68. Write a note on photo-electrolysis of water.
69. Write a note on sensistiser for photochemical decomposition of water.
70. Explain the term corrosion.
71. Explain the electrochemical theory of corrosion.
72. What is rusting ? Indicate the chemical changes in rust formation . Give chemical equations for steps of rusting.
73. Explain briefly the mechanism of corrosion of iron.
74. What is rust ? State one similarity between rusting and burning.
75. Describe briefly the techniques used for preventing corrosion of metals.
76. How can you prevent corrosion.
77. Explain how corrosion of iron is prevented by galvanisation.
78. Iron is galvanised for protecting it from rusting. Explain why ?
79. We can use aluminium in place of zinc for cathodic protection of rusting. Comment.
80. What are the factors which affect corrosion ?
81. Carbon dioxide is always present in natural water. Explain its effect on rusting of iron.
82. Rusting of iron is quicker in saline water than in ordinary water. Explain.
83. Why is galvanisation of iron is preferred to tinning of iron ?
84. Iron may be protected from rusting by coating with zinc or tin. By referring to the data given below ,explain why zinc protects iron more effectively than tin, once the protective coating has been scratched.
Zn2+ + 2 e Zn : E0 = 0.76 V
Fe2+ + 2 e Fe : E0 = 0.44 V
Sn2+ + 2 e Sn : E0 = 0.14 V
85. Describe the fuel cell used in Appolo space programme.