ELECTROCHEMISTRY

Electrochemistry mostly deals with two things. How can be used to bring about a chemical reaction. This is one and the second is how can we get electricity by means of a chemical reaction. The processes are reverse of each other. In the first, electricity is used and in the second electricity is produced. These two processes are carried out in a special kind of vessel called cells. The first process, i.e use of electricity to carry out a chemical change is done in electrolytic cell and the second process, i.e use of chemical reaction to generate electricity is is done in electrochemical cell or Galvanic cell or Voltaic cell. The chemical process which is carried out in electrolytic cell is called and electrochemical cells are commonly called batteries. A non-spontaneous reaction, which is otherwise not possible, is forcefully carried out in electrochemical cell. For example, when is passed through a molten NaCl solution in an electrolytic cell, metallic is obtained at and Cl2 gas is obtained at . Without the use of electricity, this reaction could not have been possible.

+ - electrolysis 2 Na Cl 2 Na + Cl2 Similarly when electric current is passed through an aqueous solution of any electrolyte, say

Na2SO4, we get hydrgen gas at cathode and gas at anode by the decomposition of . electrical energy 2 H2O 2 H2 + O2 H2O cannot decompose of its own. It has become possible by supplying electrical energy. Hence during electrolysis, the electrical energy is converted to chemical energy. On the other hand, when we allow a lump of sodium metal to react with Cl2 gas, spontaneous reaction occurs to form NaCl. Na atom loses one electron to become Na+ while Cl atom accepts that electron to become Cl–. This the two ions are held together by ionic bond. This is a spontaneous reaction. Na → Na+ + e Cl + e → Cl– The transference of electron from Na atom to Cl atom takes place in the same vessel without giving any extra benefit to us. Only it produces some heat due to friction. In stead of carrying out this reaction in one container, if we keep the two reactants separate from each other and connect them by a conducting wire, the electron will spontaneously flow from Na to Cl2 through the conducting wire and thus electric current is now generated. We can harness this electricity in whichever way we like by putting any load in its path, like lighting a bulb, blowing a fan or putting any electric gadget. This is the principle of galvanic cell or a battery. A spontaneous reaction is used to generate electricity. To conclude, an electrochemical cell i.e a battery is used to bring about electrolysis in an electrolytic cell. We shall take up first the electrolysis and electrolytic cells.

1 ELECTROLYTIC CELLS AND THE PROCESS OF ELECTROLYSIS:

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An electrolytic cell is used to carry out a chemical process by the help of electric current. A chemical compound is decomposed to neutral elements in this process which is called electrolysis. If you take a small battery in your house and connect two electric wires to the two terminals of the battery and put other ends of these wires in pure water (say distilled water), you will not find anything occurring. But when you add a pinch of common salt(NaCl) in that water, you will find that gas bubbles are coming out from the ends of the two electric wires dipped in the solution. On analysis you will find that one gas is H2 and the other gas is O2. In this process H2O is decomposed to its elements H2 and O2 by passing electric current through it. NaCl acts as an electrolyte which increases the conductivity of water. Water in its pure form is a bad conductor of electricity but when some electrolyte like NaCl, H2SO4 etc. is added, water not only conducts electricity but also decomposed to give H2 and O2 at the two ends of the wire dipped in solution, called electrodes. An electrolytic cell consists of the following components. (i) Electrolysis bath or tank: This is vessel in which electrolysis is carried out. Two electrodes called anode and cathode are dipped or suspended inside the bath. Electrodes are the good conductors of electricity usually made of metals. Sometimes a nonmetallic conductor such as graphite is used as electrode. (ii) Cathode: The electrode which is connected to the negative (-ve) end of the battery is called the cathode. This carries negative potential and is called the negative plate. (iii) Anode: The electrode which is connected to the positive (+ve) end of the battery is called the anode. This carries positive potential and is called the positive plate. (iv) Electrolyte: An electrolyte is taken inside the electrolytic bath either in the form of aqueous (water) solution or molten state (liquid state). For example, you can take NaCl solution in water or you can use molten NaCl. In the above diagram, molten NaCl has been shown as electrolyte. You know that electrolyte contains free positive and negative ions which conduct

2 electricity. Na+ migrates to cathode(the negative electrode) and Cl– migrates to anode(the positve electrolde) and get themselves discharged their as show before to form neutral sodium metal at cathode and gas at anode. The reactions have been given before. (v) Battery and electrical conducting wires: A d.c battery is used to supply electric current through conducting wires to the electrolytic cell and carry out the chemical decomposition of the substance present in the bath.

ELECTROLYTES: There are two types of electrolytes which are used in electrolysis experiments. (i)Strong electrolytes: These are substances which are dissociated to a large extent to produce a large number of ions(+ve and -ve ions). Ionic salts like NaCl, KBr, CaCl2 etc. and strong bases like NaOH, KOH etc, strong acids like HCl, H2SO4, HNO3 etc. are included in this category. Na+Cl -(s) + aq. ------> Na+(aq) + Cl -(aq)

The term aq. stands for aqueous (H2O). The ionic solid dissolves in water giving freely moving hydrated ions. Please refer the chapter, " chemical bond" to revise the process of hydration of ions. These free ions are responsible to conduct electricity. The ionic compounds can also be used in the molten (liquid) form in stead of aqueous solution. Free ions are present in the liquid state in case of ionic substances which conduct electricity.

Strong acids like H2SO4, HNO3, HCl, HBr etc. although are covalent in their pure form, undergo dissociation to produce freely moving hydrated ions as follows. These free ions conduct electricity. + 2- H2SO4(l) + aq. ------> 2H (aq) + SO4 (aq) HCl(g) + aq. ------> H+(aq) + Cl -(g) When a strong electrolyte is used in the form of aqueous solution or in molten state, large amount of ions are produced and large current passes through the circuit and if a bulb is connected in the path of the circuit, we shall find that the bulb produces bright light. (ii)Weak Electrolyte: The substances which undergo weak dissociation and produce a small number of +ve and -ve ions are called weak electrolytes. All weak acids and weak bases fall into this category. Refer the chapter, 'acids and bases' for getting more details. HCN(g) + aq ------> H+(aq) + CN-(aq). - + CH3COOH(l) + aq ------> CH3COO (aq) + H (aq) These substances remain predominantly in the non-ionised form(LHS) and produce small number of ions(RHS). Therefore such weak electrolytes conduct electricity to a small extent. The electric bulb connected in the path of the circuit glows with a dim light when a weak electrolyte is used as electrolyte. MECHANISM OF ELECTROLYSIS: Before the current is allowed to pass through the electrolyte, the ions move in a random manner in all possible directions. But when current is passed through the electrolytic cell, the positive ions present in the electrolyte are attracted by the negative electrode called cathode and the negative ions are attracted by the positive electrode called anode. They get themselves discharged and become neutral at the respective electrodes and are liberated as free elements. Note that the positive ions are called cations and negative ions are called anions because they migrate and are discharged at the cathode and anode respectively. Let us see how they are discharged at the respective electrodes.

3 CATHODE REACTION: A cation contains less number of electrons than protons and thus carries positive charge, while the cathode (negative electrode) carries excess of electrons. While the cation reaches the cathode, it takes electron from the cathode and gets itself neutralised to free element. There is no more attraction with the electrode as soon it becomes neutral and is liberated at the cathode as neutral element. Na+ + e ------> Na Na+ was having one electron less (deficit of one electron) and so it picks up one electron from the cathode and becomes neutral (Na). Neutral sodium is deposited at the cathode. Can you say, what kind of reaction is this? Oxidation or reduction? Gain of electron is called reduction. So it is a reduction process. Remember that reduction occurs at cathode.

ANODE REACTION: The anion contains excess number of electrons than protons and thus carries -ve charge, while anode(positive electrode) is deficient in electron and carries +ve charge. When anion reaches the anode, it gives out the excess electrons present in it to the anode which is deficient in electrons. Thus the negative charge vanishes from the anion which is converted to free and neutral element. There is no more attraction with the electrode as soon it get itself discharged and is liberated at the anode as free element. – 2Cl ------> Cl2 +2e Cl– ion was having one excess electron and it gives this extra electron to anode and become neutral Cl atom. Two such Cl atom combine to form the diatomic Cl2 molecule. So Cl2 gas is liberated at anode. Since it is a gas, easily escapes out of the electrolyte bath and can be collected in a gas vessel if desired. What reaction occurs at the anode? Loss of electron is called oxidation. Hence oxidation occurs at anode. (To remember which process takes place at anode and cathode, remember this trick. The term oxidation begins with a vowel(o) and also the term anode begins with a vowel (a) while reduction starts with a consonant (r) and so also the term cathode, starts with a consonant(c). So vowel matches with vowel for oxidation, taking place at anode while consonant matches with consonant for reduction which takes place at cathode.)

ELECTROLYSIS OF WATER(HOFMANN ) : Hofmann designed for the first time an electrolytic cell for the . This is called Hofmann voltameter. It consists of three cylindrical tubes out of which the middle one is open for introducing water and electrolyte. The left tube is connected to +ve terminal of a 6 V d.c battery(anode) and the right tube is connected to the +ve terminal(cathode) at the bottom positions. Water is introduced into the voltameter and a little sulphuric acid is also mixed with it. When electricity is appled to the voltameter, O2 and H2 gases will start collecting at anode and cathode compartments respectively in the volume ration 1 : 2.

4 electrolysis 2 H2O2 H2 + O2 (cathode)(andode) If a few drops of methyl orage dye is mixed in both the anode and cathode compartments. We shall find pink colour in the anode compartment indicating the presece of H+(acidic solution) and yellow colour in the cathode compartment indicating the presence of OH–(basic solution). In stead of using methyl orange, we can add blue litmus paper in anode compartment which turn red inidicating acidic nature of anode compartment. When we add a red litmus in the cathode compartment, it turns blue indicating the basic nature of cathode compartment. MECHANISM OF CONDUCTON : Pure water is a poor conductor of electricity. If we take pure water in the voltameter, we wont get any gas in any electrode. Or if we connect a small bulb in the circuit, it wont glow. Because the in pure water the H+ and OH– concentraions are very small i.e 10–7 mole/L which is not capable for conducting electricity. You know that inside the aqueous solution, ions conduct electricity while free electrons conduct electricity in the external metallic conducting wire. But if we add a very small quanitity of a strong electrolyte lke NaCl, Na2SO4, KNO3 , H2SO4 etc. to the voltameter containing pure water, immediately we shall find the evolution of gases. The electrolyte added in fact does not undergo any chemical change, but it increases the conductivity of water and for that reason the decomposition of water takes place at the electrodes. Note that if you will use tap water in the voltameter, the elctrolysis will take place slowly as tap water contains dissolved salts as electrolytes. So electrolytes presence in water just act as a catalyst to enhance the conductivity and decompostion of water. Suppose let us add a small quantity of Na2SO4 to the voltameter containing water. The free Na+ 2- and SO4 start migrating to cathode and anode respectively. Due to this flow of ions, the electrical circuit is closed and water in contact with electrodes start decomposing into ions. Cathodic Reaction:

+ - 2 H2O H + OH (1) + 2 H + e H (2) 2H H2 (3) - 2H2O + 2e H2 + 2 OH (net equation)

+ – + H2O dissociates into H and OH ions. H ion accepts one electron from cathode and becomes + H atom. Two such H atoms combine to form H2 molecule. After the loss of H ions from water, the cathode compartment will be having excess of OH– for which it will be basic and so produce yellow colour with methyl orange, or turns red litmus blue. Anodic Reaction :

+ - 4 H2O H + OH (1) - 4 OH O2 + 2H2O + 4e (2) + 2H2O4H + O2 + 4e (net equation)

5 – + The OH produced from the dissociation of H2O gets discharged at anode to form O2 and H ions are left behind the anode compartment. Thus the anode compartment becomes acidic for which pink colour is obtained by methyl orange, or blue litmus will turn red. Note that the H+ ions formed in the anode compartment and OH– formed in the cathode compartment attract each other slowly migrate towards each other and finally react with each other to form H2O. Due to this continuous movement if ions, the electrolysis continues as long as there is water inside the voltmeter and the battery is not exhausted. + – → H (anode) + OH (cathode) H2O If we add up the cathode and anode reactions, we get the net reaction occuring in the voltameter, which is merly decompositionof water to H2 and O2 gases.

- 2 2H2O + 2e H2 + 2 OH (1) + 2H2O4H + O2 + 4e (2) + - 6H2O2H2 + O2 + 4H + 4OH

6H2O2H2 + O2 + 4H2O

2H2O2H2 + O2 (net equation) (cathode) (anode)

So we learnt that the added electrolyte, though absolutely required for initiating the conduction, but once electrolysis starts, it is water which plays the key role in sustaining conduction. + When H2SO4 is taken as electrolyte in voltameter, it is not H ions from acid that is discharges + at cathode, it is H ions produced from H2O which are discharged. The amount of H2SO4 used in the voltameter remains the same till end although the concentration of H2SO4 in aq. solution increases due to disappearnce of water. SAQ : What products are formed in the anode and cathode compartments in water voltameter? – + Answer: Cathode : H2 + OH Anode : O2 + H SAQ : During electrolysis experiment, wha the direction of electron(external wire) and ions(electrolytic cell). Also suggest the conventional direction of current ? Answer : Inside the electrolytic cell, +ve ion move in the direction from anode to cathode and – ions move in the direction of cathode to anode. In the external circuit electrons from anode to cathode. But the conventional direction of current is opposite to the direction of electron flow. In external circule current flow from cathode to anode and in the electrolyte from anode to cathode(movement of +ve ions)

PREFERENTIAL DISCHARGE OF IONS : In the aqueous solution of an electrolyte there is always a competion of more than one cation and more than one anion to be discharged at cathode and anode respectively.For example, in an aq.

6 + + solution of Na2SO4, two cations namely Na completes with H (from water) for cathodic reaction 2– – while SO4 and OH (from water) competes for anodic reaction. Out of the two cations, the one which has a greater tendency to undergo reduction (at cathode) is preferentially discharged at cathode. Similarly out of the two anions the one having greater tendency to undergo oxidation will be preferentially discharged at anode. These are called reduction potential and oxidation potentials of the ions respectively which are commonly called discharge potential. The cation having greater reduction potential will be prefentially discharged at cathode while the anion having greater oxidation potential will be preferentially discharged at anode. Note that out of more than one ions competing for discharge one ion gets preferentially discharged and the other ion remains as such as long as the preferred ion is not fully discharged.

REDUCTION POTENTIAL SERIES OF SOME COMMON CATIONS:

Now we shall see the reactions taking place in the electrolysis of different kinds of electrolytes.Before that first let us know about another aspect called metal activity series. Although this series originates from electrochemical process which will be discussed in the next section, we simply know the importance of the series now.

Li+ < K+ < Ca2+ < Na+ < Mg2+ < Al3+ < Zn2+ < Fe2+ < Sn2+ < Pb2+ < H+ < Cu2+ < Ag+ < Hg2+ < Au3+

Do you not find this series is similar to metal activity series ? Yes, although the sequence is same in both the seres, that was decreasing metal activity series, while this is increase reduction potential series. In metal activity series, they were metals, but here these are metal ion. Note that those metal ions lying before H+ have lower reduction potential than H+ while those metal ions which follow H+ have a greater reduction potential than H+.Take the following two examples. + + (a) Na2SO4 solution : In this solution Na competes with H . Tell which one will be discharged ? Yes, H+ which has greater reduction potential will be discharge while Na+ will remain in solution. → 2H+ + 2e H2 2+ + (a) CuSO4 solution : In this solution, Cu competes with H . Look to the above series, Cu2+ has a greater reduction potential. Hence Cu2+ will be discharged, not H+ from water. Cu2+ + 2e → Cu

SAQ: Suppose, an aq. solution conatains a mixture of Cu(NO3)2, AgNO3 and AuCl3. If the solution is subjected to electrolysis, which will be discharged first and which second and which third ? Answer : Look to the reduction potential series given before. . Au3+ > Ag+ > Cu2+ > H+ So first Au3+ will be discharged, then Ag+, then Cu2+ and finally H+. Au3+ + 3e → Au, then Ag+ + e → Agthen Cu2+ + 2e → Cu and finally + → 2H + 2e H2 Note that after the discharge of metal ions, the electrolysis will not stop, H+ from water will be discharge as long as there is water in the solution. Do not forget that in the discharge of anions at anode, eqivalent quantity of +ve ions will be left out to maintain electrical neutrality always.

7 OXIDATION POTENTIAL SERIES OF SOME COMMON ANIONS: Various anions have different tendency for undergoing oxidation. The increasing order of oxidation potentials of some common anions are given below. 2– – – – – – SO4 < NO3 < Cl < OH < Br < I – 2- – – Note that the oxidation potential of OH is greater than those of SO4 , NO3 and Cl . Hence – during the electrolysis of KNO3, K2SO4 and NaCl, preferential discharge of OH from water takes place at anode to form O2 gas. – → 4OH O2 + 2H2O + 4e

ELECTROLYSIS OF DIFFERENT ELECTROLYTES USING INERT ELECTRODES:

(A) USING INERT OR NON-ATTACKABLE ELECTRODES: When electrode (cathode and anode) materials do not take part in the electrode reaction, we consider the electrodes to be inert or non-attackable. Electrodes made up of , graphite etc. are taken to be inert and do not generally take part in the reaction occurring at cathode and anode. The electrolysis of some common electrolyte solutions using these inert electrodes is discussed below. 1. Electrolysis of NaCl solution: (a) Dilute NaCl solution: In aqueous solution, besides the metal ion of the electrolyte (i.e Na+ in this case) there is also present the H+ ions from water. Note that water is partially dissociated to H+ ion and OH- ion and there is always a small percentage of H+ ion and OH- in aqueous solution of any electrolyte. At cathode: Between Na and H, H comes later in the activity series. So H+ ion is preferentially discharged at cathode. Na+ ions remain as such in the solution. As H+ ions get discharged at cathode, more and more water molecules are dissociated to form more and more H+ and OH- and therefore the liberation H2 gas continues at cathode until there is no water in the electrolyte solution. + 2H + 2e ------> H2 At anode: There are two anions to compete, namely OH- produced by water and Cl- produced by the electrolyte (NaCl). At dilute solution when the number of Cl- ions is small, OH- ions produced from H2O have a greater tendency to be discharged and liberate O2 gas. You know that oxidation occurs at anode. The tendency to undergo oxidation is more with OH- ions than Cl- particularly when the NaCl solution is dilute. - 4OH ------> O2 + 2H2O + 4e

Cell reaction: Let us add the reactions taking place at cathode and anode to get a balanced cell reaction. + Cathode reaction: [2H + 2e ------> H2} X 2 - Anode reaction: 4OH ------> O2 + 2H2O + 4e [cancel 4e] ______+ - 4 H + 4 OH ------> 2 H2 + O2 + 2H2O ⇒ 4 H2O ------> 2H2 + O2 + 2H2O ⇒ 2H2O ------> 2H2 + O2

8 You find that the net effect of electrolysis of NaCl solution is the electrolysis of H2O producing

H2 at cathode and O2 at anode. Then the question arises if electric current is passed through pure H2O, will we get H2 and O2 gases? The answer is NO. Pure H2O is a bad conductor of electricity because it produces very very few H+ and OH- ions. But in presence of a strong electrolyte like NaCl, it conducts electricity and once the electric current flows through the solution, the competition between two cations and two anions start and ultimately H+ and OH- + - ions produced from H2O win the while Na and Cl produced from NaCl lose the game. The dissociation of H2O proceeds continuously till all water present in the solution is discharged into

H2 and O2.

(b) Concentrated NaCl solution: At cathode: In concentrated solution also H+ ion is preferentially discharged at cathode as before. + 2H +2e -----> H2 At anode: Anode reaction is different as compared to dilute NaCl solution. In concentrated solution - large number of Cl ions are present and hence they preferentially get discharged to produce Cl2 – – gas. O2 gas is not evolved here. Note that the oxidation potential of OH is close to Cl . At higher concentration of Cl–, the latter possesses greater oxidation potential, so it is discharged. Note that concentration of the ion also has an effect on potentials. You shall know about it later. - 2Cl -----> Cl2 + 2e (ii) Electrolysis of molten NaCl: In molten NaCl, there is no water. It is only liquid sodium chloride which is obtained by melting NaCl at about 8000C. In such a case, there is no other cation to compete with Na+ and no other anion to compete with Cl-. Sodium metal is obtained at cathode and chlorine gas is evolved at anode. At cathode: Na+ + e ------> Na - At anode: 2Cl ------> Cl2

(iii) Electrolysis of KNO3 solution:

In aqueous solution of KNO3, the two cations which will compete for discharge at cathode are K+ from the electrolyte and H+ from water. You know that H lies after K in the activity series and so H+ ion will be preferentially discharged. Again for anode there will be competition - - between NO3 produced by the electrolyte and OH produced by water. The tendency to undergo - - oxidation is more in case of OH , and NO3 being a bulky ion has little tendency to undergo oxidation. + At cathode: 2H + 2e ------> H2 - At anode: 4OH ------> O2 + 2H2O + 4e Can you not write the balanced cell reaction by adding the two half reactions taking place at anode and cathode after cancelling the electrons? Do it yourself. Note that some authors write the cathode reaction in such cases like this. K+ + e ------> K

2K + 2H2O ------>2KOH + H2 This means that first the metal K will be discharged at cathode which subsequently shall

9 react with H2O to produce H2 gas. This way of writing is incorrect. Hence you are advised to remember that metal ions like Na+, K+, Zn2+, Mg2+, Ca2+ which lie before H+ cannot + discharge to their neutral elements as long as H2O is there. H2O provides H ion to discharge preferentially.

(iv) Electrolysis of CuSO4 solution: Since Cu lies after H in the series, Cu2+ is discharged first not H+ produced by water. But 2- - between the bulky SO4 provided by the electrolyte and OH provided by water, it is the latter which has greater tendency to undergo oxidation at anode. At cathode: Cu2+ + 2e ------> Cu - At anode: 4OH ------> O2 + 2H2O + 4e

So Cu will be deposited at cathode and O2 gas will be evolved at anode. SAQ 1: Write the products obtained at cathode and anode by the electrolysis of the following:

(i)molten KBr (ii)MgCl2 solution (concentrated) (iii)AgNO3 solution

(iv)NaOH solution (v)molten CaF2 (vi) molten Al2O3

(vii) Al2(SO4)3 solution.

(B) ELECTRODES USING ACTIVE OR ATTACKABLE ELECTRODES: When electrode material takes part in the reactions taking place at cathode and anode, it is called an active or attackable electrode. Look to the following cases.

(i) Electrolysis of CuSO4 solution using Cu electrodes: When both cathode and anode are made up of (not platinum or graphite), then reaction taking place at anode will be different from that taking place when inert electrode is used. The reaction at cathode however will remain same. At anode, Cu metal from the anode will dissolve 2+ 2- - to form Cu and pass into the solution. Neither SO4 ion form the electrolyte nor OH produced by H2O will be discharged (oxidised). Cu metal has a relatively greater tendency(greater oxidation 2+ – 2+ potential) to undergo oxidation to form Cu than OH to form O2. At cathode Cu will be discharged as usual because Cu2+ has a greater reduction potential than H+.

At anode: Cu ------> Cu2+ + 2e At cathode: Cu2+ + 2e ------> Cu This principle is used in the purification (refining) of metals like Cu. Impure copper is made as anode and thin strip of pure copper is made as cathode. CuSO4 solution is used as electrolyte. Copper from the impure anode starts dissolving by forming Cu2+ at anode and these Cu2+ ions go to the other side of the electrolytic cell and are discharged at cathode as pure Cu and is deposited over the pure copper rod which is used as cathode. Thus the anode becomes thinner and thinner as more and more copper dissolves from it while cathode becomes thicker and thicker as more and more pure copper deposits on it. The impurities present in the impure anode does not dissolve and so settle down near the anode and is called anode mud. The anode mud contains some noble metals like silver(Ag), gold(Au)etc. which are present as impurities with Cu in small quantities. Note that Ag and Au have lower oxidation potential than Cu and hence they do not dissolve out as Ag+ or Au+.

(ii) Electrolysis of AgNO3 solution using Ag electrodes: + - - The reactions at anode will be similar as in case (i). Ag will dissolve to form Ag . NO3 or OH ions will not be discharged as Ag+ ion has a greater oxidation potential. At cathode, however Ag+ ion will be discharged as Ag.

10 At anode: Ag ------> Ag+ + e At cathode: Ag+ + e ------> Ag

(iii) NaCl solution using Hg cathode: + If we electrolyse NaCl solution using inert electrodes like graphite or platinum, H ion from H2O + is reduced at cathode to produce H2 gas. Na ion is never reduced as H appears after Na in the activity series. In presence of mercury cathode, however, H+ will not be reduced, rather. Na+ will be reduced. It won't give free Na metal but it will react with Hg(cathode) to produce sodium amalgam since Hg is used here as active electrode. Amalgam is an alloy(alloy is a homogenous mixture of two metals) of a metal with mercury. Subsequently the sodium amalgam is hydrolysed by water to produce NaOH and H2 and thereby Hg is set free.

At cathode: Na+ + Hg + e ------> Na(Hg) (sodium amalgam)

2Na(Hg) + 2H2O ------> 2NaOH + H2 + 2Hg

The anode reaction is same as in case of inert electrode i.e at higher concentration Cl2 gas will be evolved and at lower concentration O2 gas will be evolved.

SAQ 2: What substances are liberated at anode and cathode if the following are electrolysed.

(i)acidulated water (ii)molten CaCl2 (iii)molten Al2O3

(iv) Hg2(NO3)2 solution using Hg cathode

FARADAY'S LAWS OF ELECTROLYSIS: Faraday found an important relationship between the mass of the substance liberated at any electrode during electrolysis with the electrical charge that has passed through the electrolyte solution. Say for example, when we electrolyse dilute NaCl solution, we get H2 at cathode and

O2 at anode. When we electrolyse AgNO3 solution using inert electrode, we get, Ag at cathode and O2 gas at anode. What amount of these substances are liberated when a given quantity of charge(Q) is passed through the electrolyte solution? Faraday made an interesting observation. He found that for the passage of 96500 coulombs of charge, one gm. equivalent mass any substance is liberated at each electrode. Let us recapitulate what is equivalent mass of an element(revise the chapter equivalent mass). Equivalent mass is equal to atomic mass of the element divided by the valency(more correctly the change in oxidation number during the discharge of an ion at the electrode). The equivalent mass of H is 1.008, that of O is 8, and of Ag is 108gms and so on. Before you proceed further, revise equivalent mass.

Faraday found that by the passage of 96500 coulombs of charge through different electrolytes 1 gm. equivalent mass of an element was obtained.

For instance if it is H2, we get 1.008gms, if O2 we get 8 gms, and if Ag 108gms and so on. This was indeed an epoch making discovery since the same charge(96500 couls) liberates so small a quantity of H2 and so large a quantity of Ag. This quantity of charge is called 1 Faraday(1F) which is equal to 96500 coulombs. SAQ 3:

(i) How much of Cu will be deposited by the electrolysis of CuSO4 solution when 96500 couls of charge is passed through it?(Atomic Mass of Cu = 63.5)

11 (ii) How much of Al be deposited at the cathode by the passage of 1 Faraday of Charge by the electrolysis of Al2O3 taken in a suitable solvent like cryolite?(Atomic mass of Al=27)

(iii) How much of Cl2 gas be evolved at anode when ½ Faraday of charge is passed through molten NaCl?

(iv) What volume of O2 gas will be evolved at NTP when 19300 couls of charge is passed through acidulated water? (v) How many Faraday's of charge is required to liberate 127gms of Cu by the electrolysis of CuSO4 solution (atomic mass of Cu = 63.5).

FARADAY'S FIRST LAW OF ELECTROLYSIS: If Q amount of charge is passed through an electrolyte what mass of the substance is liberated? Let us calculate it. Let the equivalent mass of an element liberated is E and the mass of a substance deposited at any electrode is m gms. 1F or 96500 couls of charge liberates E gms of the substance 1 coul of charge liberates E/F gms of the substance So Q coul of charge liberates (E/F) X Q of the substance According to our assumption, m = (E/F) X Q (1)

⇒ m ∝ Q

This is Faraday's 1st law of electrolysis. Mass of any substance deposited (m) at any electrode is directly proportional to the amount of charge passed through the electrolyte(Q).

m ∝ Q ⇒ m = Z Q (2) (where z is the proportionality constant)

Comparing Equations (1) and (2), we know that Z = E/F z is called the Electrochemical Equivalent which is equal to the mass of the substance deposited for the passage of 1 coulomb of charge. Z = m/Q = E/F. The unit of Z is Kg/coul or g/coul. Usually g/coul unit is frequently used. We know that charge is equal to the product of current and time. Q = I X t ( where I=current and t=time)

Once again taking Faraday's first law: m = Z X Q = Z X I X t

The unit of current(I) is ampere and time is second. The charge that results by the passage of 1 ampere of current for 1 sec is 1 coulomb. Let us take an example to make use of Faraday's 1st law.

Example: Determine the mass of copper deposited from the solution of CuSO4 by the passage of a 10A current for 965 sec?(atomic mass of Cu = 63.5)

12 Solution: 1st method: Let us apply the Faraday's 1st law equation m = (E/F) X I X t , Equivalent mass(E) of Cu = 63.5/2= 31.75, (since the Cu2+ is reduced to Cu0, the change in oxidation number is 2, so we divided the atomic mass by 2 to get the equivalent mass) I = 10 amp., t = 965 secs, F= 96500 coul. Let us put all these in the above equation. m = (31.75/96500) X 10 X 965 = 3.175gms. (note that the units of I= amp., time= sec and E=gm, so mass = gm) Alternative Method: Let us find the charge(Q) first. Q = IX t = 10 A X 965 sec = 9650 couls and E = 31.75(found before) We know that 96500coul(1F) of charge deposits 1 gm equivalent mass = 31.75gms of Cu 9650 couls of charge must deposit ( 31.75/96500) X 9650 = 3.175gms. Now you decide which method you liked more, the formula method(first method) or by method of first principle (alternative)? You can choose any method from the two for such calculations. SAQ 4:

(i) A current of 5.0 Amp. flowing through a solution of molten ZnCl2 for 30 minutes deposits 3.048 gm of at the cathode. Calculate the equivalent mass of zinc.

(ii) A certain value of current is passed through NiCl2 solution for 3 hours and 20 minutes and deposited 100gm of nickel at the cathode. Find the value of the current.(Atomic mass of Ni=58.7) (iii) What mass of Bi metal is deposited electrolytically from Bi3+ solution in 30 minutes by a current of 40A?( Atomic mass of Bi = 209)

(iv) Electric current was passed in a CuSO4 solution for 30 mintues. Graphite electrodes were used. 6.35 g of copper was deposited at cathode. Find the current passing through the electrolyte.

(v) If 19300 Coul of charge passes through dilute NaCl solution what masses of H2 and O2 be liberated at the electrodes. (vi) Find the quantity of charge which when passes through dilute NaCl soltion to produce

11.2 L of O2 gas at anode at STP. (vii) If 8A current passes through a sulphuric acid solution for 20 minutes, what is the volume of O2 evolved at anode ? (viii) How many faradays of charge will bring about deposition of 270 g of Al from the electrolysis of molten alumina ? (ix) 9.5 g of a metal was deposited by the passage of 4 A current for 2 hours. What is the equivalent mass of the metal ?

FARADAY'S 2ND LAW:

Let us take two electrolyte solutions CuSO4 and AgNO3 in two different cells and connect them in series. What is the meaning of series connection? Series connection means same quantity of current and hence charge is passed through both the electrolytes. Let the amount of charge passed through these two electrolytes for a certain time period is Q coulombs. You know that in

13 CuSO4 electrolyte, Cu is deposited at cathode and O2 gas is liberated at anode and in AgNO3 electrolyte, Ag is deposited at cathode and O2 is liberated at anode. Let us apply first law to each electrolyte and compare between the masses of Cu in the 1st electrolytic cell and that of Ag in the second cell.

Let m1 is the mass of Cu deposited in the 1st cell and m2 is the mass of Ag deposited in the 2nd cell.

According to 1st law of Faraday, m1 = Z1 X Q = (E1/F) X Q (1)

m2 = Z2 X Q = (E2/F) X Q (2)

(where Z1 and Z2 are the electrochemical equivalents of Cu and Ag respectively and E1 and E2 are their equivalent masses respectively, while Q is the amount of fixed charge passed through them). Dividing equation (1) by equation (2), we get m E 1 = 1 m2 E2 This is Faraday's second law, which states that masses of substances deposited at electrodes in different electrolytes are directly proportional to their equivalent masses if same quantity of charge is passed through them. Out of the four parameters (m1, m2, E1 and E2), if any three are known then the fourth one is calculated out. See this example.

Example: Same quantity of current is passed through a CuSO4 solution and a AgNO3 solution for a certain time period, it was found that 1.08gm of Ag was deposited at the cathode of the second electrolyte. Calculate the mass of Cu deposited at the cathode of first electrolyte. (Atomic masses of Cu=63.5 and Ag=108)

Solution: Mass of Cu = m1 = ?, Mass of Ag = m2 = 1.08gm, 2+ 0 Equivalent mass of Cu =E1 = 63.5/2=31.75(since copper has gone from Cu to Cu ), + Equivalent mass of Ag=E2 = 108/1=108(since silver has gone from Ag to Ag). Now let us apply the equation for the Faraday's 2nd law ⇒ ⇒ m1/m2 = E1/E2 m1/1.08 = 31.75/108 m1 = 0.3175gm So the mass of Cu deposited is 0.3175gm. SAQ 5: (i) Two electrolytes namely acidulated water and concentrated NaCl solution were connected in series. After certain time of electrolysis, the volume of H2 gas liberated at the cathode of the 1st electrolyte was found to be 1.12 litres at NTP. What volume of Cl2 be liberated at the anode of the second electrolyte at NTP?

(ii) During electrolysis of dil H2SO4, 5.6 L of H2 were evolved at cathode after certain time, what volume of O2 be evolved at the anode at the same time. (iii) 2.5 A of current was passed through an acidulated water for 30 minutes. What is volume of H2 gas evolved at cathode at STP ? Also what is the volume of O2 liberated at anode during the same time ?

14 (iv) What amount of charge in Coul and Faraday be passed to get following ? (a) 1 mole of Al from 1 mole of Al3+ (b) 1 mole of Mg from 1 mole of Mg2+ 2+ (c) 0.5 mole of O2 from 1 mole of H2O (d) 1 mole of Cu from 1 mole of Cu 2+ – (e) 1 mole of Mn from 1 mole of MnO4 3+ 2– (f) 2 moles of Cr from 1 mole of Cr2O7 . (v) 0.2 A current was used to electrolyse molten AlCl3. 3 gm of Al was deposited at cathode. Answer the following questions. (a) How many g. equivalents of Al formed. (b)How many g. atoms(mole of atoms) of Al formed (c) How many atoms of Al formed. (d) How many electrons were used in this reduction. (e) How long(time) electrolysis took place. (f) How many Faradays charge was used.

(g) What volume of Cl2 produced at anode during that time.

SOME APPLICATIONS OF ELECTROLYSIS: (1) Extraction of active metals by electrolytic method: Alkali metals (Li, Na, K, Rb, Cs), alkaline earth metals(Be, Mg, Ca, Sr, Ba) and Al from group 13 are very active. The extraction of these metals from their minerals(oxides) cannot be done by any chemical method. Because these metals themselves are strong reducing agents. So no other reducing agents can be able to reduce the metal ion to metal. Hence they are extracted by the electrolysis of their molten compounds. For example, Na metal is obtained by the electrolysis of molten NaCl at the cathode. (2) Electrolytic Refining of impure metals: When metals like Zn, Ag, Ni, Cu, Pb, Al are extracted from their ores contain some amount of impurity. The impure metals are purified to 100% pure state by making the impure metal as anode and a strip of pure metal as cathode. The compound of the metal is used as electrolyte. We have discussed this under active or attackable electrodes for the purification of impure copper. the metal dissolves from the anode and gets deposited at cathode. (3) : The depositon of a thin coating of one metal on the surface of another metal by electrolysis process is called electroplating. Usually a superior metal is coated on the surface of an inferior metal so as to increase inferior metal's self life and look more attractive. To prevent corrosion in steel materials, coating of Ni or Cr is applied by this method. Spoons are electroplated by Ag or Au look more attractive and increase its durability. The following properties are enhanced due to electroplating. (a) Increase in corrosion resistance : Metals like Fe get corroded when exposed to moist air. The corrosion of Fe is called rusting(Fe2O3.xH2O). To prevent rusting of iron, it is electroplated with superior metal like Ni, cr, Ag or Au etc. (b) Inrease in abrasive strength and self life : The self life and abrasive strength of the base material increase due to electroplating. Abrasive strength is resistance to wear and tear due to friction. (c) Attractive Appearace: The electroplated article gets an improved appearance.

15 Working Process of Electroplating : During electroplating the metal or the article to be electroplated( like spoon, key ring, plate etc.) is made as cathode and the superior metal(like Ni, Cr, Ag, Au etc) whose coating to be applied on the above articles is made as anode. The compound of the superior metal is used as electrolyte. In this case the anode becomes an active or attackable electrode. When current passes through the electrolyte, the superior metal from anode dissolves by losing electron and the correponding metal ion gets deposited uniformly on the article acting as cathode. Silver plating on Steel spoon: A silver rod is made as anode and steel spoon as cathode. argentocyanide{ K[Ag(CN)2]}is used as electrolyte.

AgNO3 is not used as electrolyte because, in that case the silver coating will be non-uniform as the electrolysis takes place rapidly. But potassium argentocyanide produces very small number Ag+ and the electrolysis takes place slowly to produce a uniform coating on the article. → + + - K[Ag(CN)2 K + Ag + 2CN Anode: Ag → Ag+ + e Cathode: Ag+ + e → Ag Factors responsible for smooth electroplating with with uniform thickness: For a smooth and uniform electroplating depend on the time of electrolysis, temperature, concentration of electrolyte and current density. (1) Moderate Current Density : For good electroplating neither high nor low current density is used. Current density is the current passed per unit cross sectional area of the article to be electroplated. The current density should be moderate. (2) The concentration of the electrolyte : The electrolyte should have good solubility in water but the metal ion concentration should be low. This is possible by preparing a complex compound of the electrolyte with KCN. For example AgNO3 is mixed with KCN to produce K[Ag(CN)2] which is taken as electrolyte. This produces a low concentration of Ag+. → AgNO3 + KCN AgCN + KNO3 → AgCN + KCN K[Ag(CN)2] (complex compound)

The table below gives the anode and the electrolyte used for electroplating of a few metals. Metal Coating Anode Electrolyte Nickeling Ni Nickel ammonium sulphate

Copper plating Cu acidulated CuSO4

Silver plating Ag K[Ag(CN)2] Potassium argentocyanide

Gold plating Au Potassium auricyanide K[Au(CN)4]

16 (4) ELECTROTYPING: This is a old method of making metal plates used for printing purpose. With the advent of modern printing technology, this process has been obselete. The engraving of any art work or text on a blook of wood or carving on a metal can be converted to a metal plate. A block of wax is placed over the wooden carving and by applying pressure, a mold of the art work or text is made on the wax. If the carving has hollow grooves in the art work/tex, the wax mold will have mounts(bulging out) in the same place. Graphite powder is mixed with graphite to make it a good conductor. This mold is made the cathode in a cell carrying copper as anode and CuSO4 solution as electrolyte. On passing current, a thin layer of copper is electro-deposited on the wax mold(bulged out portions) on the artwork/text and is called the reverse plate. This sheet of copper is separated from wax by treating with boiling water and mounted on a block of lead for giving strength. This plate when used for printing gives the impression of the original arit work or text on the paper. Because the reverse of reverse is original work. Other metals like Ni, Cr and alloys like steel can be electrodeposited and their metal plate can be prepared. A large number of duplicate plates can also be done easily by this method by repeated electrolysis on different wax molds.

ELECTROCHEMICAL CELL/ GALVANIC CELL/VOLTAIC CELL

Electrolytic cell which we discussed before makes use of electric current to bring about a chemical change, which is otherwise a non-spontaneous reaction. It performs a redox reaction, oxidation at anode and reduction at cathode, thus decomposes the positive and negative ions to their neutral substances by the use of electrical energy. In contrast to this, electrochemical cell which we shall study now, make use of a spontaneous chemical reaction to produce electric current. In principle, it is the opposite to electrolytic cell. Here a redox reaction takes place, but not by using current, rather the reaction is spontaneous i.e it proceeds of its own. Just like water stored at a higher level spontaneously flows to a lower level and during its flowing down we can make use of the falling water in many ways, in the similar manner the electron flows spontaneously from one electrode remaining at a higher potential to another electrode remaining at lower potential, when the two electrodes are connected by an external conducting wire. This flow of electrons through a metallic conductor is nothing but the generation of current. We can call it chemical electricity like hydroelectricity, thermal electricity etc. Spontaneous reactions take place at the electrodes the sooner there is external connection between them. These are also called galvanic cell or a voltaic cells commonly called a battery. We shall study the principles involved and the chemical reactions taking place in the functioning of the batteries (electrochemical shells) now. Just see how it works. Suppose, you dip a zinc rod in a copper sulphate solution. What will happen ? Zn will be oxidised to Zn2+ while Cu+2 present in the soluion will be reduced to Cu. The blue colour of the solution will fade and there will be a red (Cu) deposit on the zinc rod. Zn will lose two electrons which Cu2+ will take. This give and take becomes so rapid in the solution that we cannot get electricity. In stead, we get heat energy out of this reaction due to resistance

17 of the electrolyte offered to the electrons. In stead of doing this, if we keep the reactants separately connect them by metellic conducting wire electrons will spontaneously flow from Zn rod to CuSO4 solution and thus produces electric current, which we harness in so many different ways. The side diagram gives the simplest galvanic cell, which consists of two electrodes also called the half cells. The left hand side(LHS) electrode is called the oxidation electrode or oxidation half cell and the right hand side(RHS) electrode is called the reduction electrode or reduction half cell. In the above example, the oxidation half cell consists of a Zn rod dipped in ZnSO4 solution and the reduction half cell consists of a Cu rod dipped in CuSO4 solution. The former is called as Zn electrode while the latter as Cu electrode. Some authors refer the oxidation electrode as anode and reduction electrode as cathode. But we shall not use the terms anode and cathode here in electrochemical cells as these terms are already used in electrolytic cells where electrolysis is carried out. The use of these terms often create confusion in the minds of young learners as the signs are opposite for the two cells. Anode is + electrode in electrolytic cell while it is –ve electrode in galvanic cell. Cathode is –ve in electrolytic and +ve in galvanic cell. So we avoid using them here. We shall say oxidation half cell for anode and reduction half cell for cathode in electrochemical cells. The Zn electrode here is at higher potential and Cu electrode is at a lower potential. The moment we make a connection between the two externally by means of a metallic wire, spontaneous reaction takes place at these electrodes and electrons flow from Zn electrode to Cu electrode.Thus we can make use of this current flowing through the external conductor for meeting our needs such lighting a bulb, running a fan, TV or putting any other load. Let us see what reaction takes place at the electrodes.

Oxidation Half Cell( OHC)/ Anode/ – Electrode Zn electrode here is the oxidation half cell. Conventionally the oxidation half cell is kept and written in the left hand side. Oxidation occurs at this half cell and therefore it is called oxidation half cell. You know that oxidation is the process in which electron is lost. In this case Zn atom dissolves from the Zn rod forming Zn2+ ions and thereby loses two electrons as per the following equation. These two electrons flow through the external wire spontaneously to the reduction electrode kept in the RHS. Zn ------> Zn2+ + 2e Reduction half Cell(RHC)/Cathode / +ve Electrode: Cu electrode here is the reduction half cell. Conventionally, the reduction half cell is kept and written in the right hand side. Reduction occurs at this half cell and therefore is called reduction half cell. You also know that reduction is the process in which the electron is gained. In this case Cu2+ present in solution takes electrons which have come to it from the oxidation half cell through the external metallic conductor and get itself reduced to Cu atom. This Cu atom is deposited on

18 the copper rod present in the Cu electrode. Cu2+ + 2e ------> Cu Note that there is an internal connection between two electrodes through a bent tube as you see in the diagram. This is called the salt bridge. The discussion on the function of the salt bridge is beyond the scope of this book. Cell reaction: Let us write the electrode(half cell) reactions once again and add the two half cell reactions to get the overall cell reaction.

2+ Oxidation half cell: Zn Zn + 2e 2+ Reduction half cell: Cu + 2e Cu 2+ 2+ Cell reaction: Zn + Cu Zn + Cu While adding the half cell equations, we cancelled the electrons from both the sides and got the net equation called the cell reaction. Now a question arises in our mind. Why the electrons do not flow from Cu electrode to Zn electrode i.e in the reverse direction to what we have seen before? In other words why the reverse reaction does not take place? For that to happen, oxidation has to occur at Cu electrode and reduction has to take place in the Zn electrode. This is not possible, because Zn electrode is at a greater potential than Cu electrode. In other words, Zn has a greater tendency to be oxidised than Cu. Therefore when a connection is made between these two electrodes, oxidation occurs at Zn electrode and reduction occurs at Cu electrode. Zn dissolved forming Zn2+ ions by losing two electrons and these two electrons flow through the external conductor and are accepted by Cu2+ ions present in the Cu electrode and is deposited as neutral Cu on the copper electrode. We shall know more about this electrode potential a little later. First, let us see the notation used for a galvanic cell and the cell reactions of some more cells.

FUNCTON OF SALT BRIDGE : You know that current does not flow if the electrical circuit is open (i.e not closed). In a galvanic cell, the two electrodes are externally connected by a conducting wire. The two electrolytes are connected by the salt bridge, so that the circuit is closed and there is continuous follow of current in the external wire. When Zn metal dissolves in the OHC the concentration of Zn2+ increases in 2- the ZnSO4 electrolyte and there are no equivalent quantity of the anions(SO4 ) to neutralise it. Similarly Cu2+ is discharged in the RHC for which concentration of Cu2+ decreases, i.e the 2- concentration of anion(SO4 ) increases. Salt bridge contains a strong electrolyte like NaNO3, KCl etc mixed in a viscouse hydrocarbon jelly.. The moment cation concentration(Zn2+) increases – – in OHC compartment, the negative ions in the salt bridge(NO3 or Cl etc.) migrate from there to the OHC compartment to neutralise it. Similarly the positve ions in the salt bridge (K+ or Na+) migrate to the RHC compartment to neutralise the excess negative charge. Due to the continuous movement of ions from salt bridge in opposite directions, the circuit is closed to maintain the flow of electric current in the external wire. That is why, salt bridge is replaced from to time as all electrolyte is exhausted from it. So if the salt bridge is not used, the flow of current stops after some time because circuit is not closed. In the first version of Daniel cell, salt bridge was not used. In stead a porous pot was used to

19 separate the two electrolytes. A zinc rod is dippsed in a dilute sulphuric acid taken in a porous pot.The porous pot is immersed in a bigger size copper pot 2+ containing CuSO4 solution. The excess Zn ions 2– formed the porous pot(OHC) and the excess SO4 formed in the copper pot migrate in opposite direction through the porous partition to close the circuit. Although such cell worked well, but had many drawbacks(not to be discussed here) and hence such version was rejected and was replaced by salt bridge version which has been explained before.

CELL NOTATION: A Cell has two half cells or electrodes. Each electrode has two components, i.e two species remaining together, one remaining in the oxidised state and the other in the reduced state. Say for example, Zn electrode that we discussed before consists of metallic Zn rod dipped in ZnSO4 solution which contains Zn2+ ions. Zn is in the reduced state and Zn2+ is in its oxidised state. We represent the electrode as Zn/ZnSO4. Similarly Cu electrodes consists of Cu (which is in its 2+ reduced state) dipped in CuSO4 solution(Cu ion is in its oxidised state). We represent this 2+ 2+ electrode as Cu/CuSO4. We can also represent these electrodes as Zn/Zn and Cu/Cu respectively picking up only the ions from the electrolyte solution. Therefore while writing the cell, we have to keep the two electrolyte solutions face to face like this. The external connection is not shown. 2+ 2+ Zn/ZnSO4 CuSO4/Cu OR Zn/Zn Cu /Cu Where the parallel lines in the middle stand for the salt bridge which connects the two cells internally. Any electrode written in the LHS means, there will be oxidation in that electrode and obviously there will be reduction in the electrode written in the RHS.This is because the electrode which has a higher oxidation potential is kept in the LHS and the electrode having lower oxidation potential is kept in the RHS. Let us take some more cells. SAQ 6: Indicate which species undergoes oxidation and which reduction in each of the following pairs. Give the balanced ion-electron equation of oxidation and reduction reactions. 2+ - 2+ 3+ + (i)Cu/Cu (ii)Cl2/Cl (iii)Fe /Fe (iv)H2/H

EXAMPLES OF MORE GALVANIC CELLS:

2+ + (i)Sn/SnCl2 HCl/H2(Pt) or Sn/Sn H /H2(Pt)

Note that for gas electrode like the H2 as shown above, the electrical contact into the electrolyte solution is achieved by a metallic conductor dipped inside the solution. Often platinum(Pt) wire is used for the purpose and therefore while writing the electrode, we also write (Pt) close to the electrode. Note that Pt is not the electrode and does not react. It only carries current from and to the solution. In metallic electrodes, the metal not only acts as an electrode but also carries the current from and to the solution. Therefore we do not take the help of Pt in these cases for electrical contact. In the above example, since Sn/Sn2+ electrode is written in the left hand side,

20 + oxidation occurs in this electrode and reduction occurs in H2/H electrode as it is written in the right hand side. Can you guess which species of the Sn electrode undergoes oxidation and which species of the H2 electrode undergoes reduction? Note that the species in reduced state (Sn) in + the Sn electrode undergoes oxidation while the species in the oxidised state(H ) of the H2 electrode undergoes reduction. Cell Reaction:

2+ Oxidation half cell: Sn Sn + 2e + Reduction half cell: 2 H + 2e H2 + 2+ Cell reaction: Sn + 2 H Sn + H2

+ + (ii) H2(Pt)/HCl AgNO3/Ag or H2(Pt)/H Ag /Ag + In this cell, oxidation occurs at the H2/H (LHS) electrode while reduction occurs at the + Ag/Ag (RHS) electrode. H2 is in the reduced state(ON of H=0) and will be oxidised in the LHS electrode, while Ag+ is in the oxidised state(ON of Ag=+1) and will be reduced. The half cell reactions and net cell reaction are as follows.

+ Oxidation half cell: H2 2 H + 2e Reduction half cell: Ag+ + e Ag X 2 + + Cell reaction: 2 Ag + H2 2 H + 2 Ag

2+ - (iii) Cu/CuSO4 NaCl/Cl2(Pt) or Cu/Cu Cl /Cl2(Pt) In this cell, the oxidation half cell(LHS) is the Cu electrode where oxidation occurs while the reduction half cell(RHS) is the Cl2 electrode where reduction occurs. Cu is in the reduced state and will be oxidised in the LHS electrode and Cl2 is in its oxidised state and will be reduced. Note that Pt is used for an electrical contact in the RHS since Cl2 is a gas electrode.

2+ Oxidation half cell: Cu Cu + 2e - Reduction half cell: Cl2 + 2e 2 Cl 2+ - Cell reaction: Cu + Cl2 Cu + 2 Cl

SAQ 7: (i) Write the half cell reactions if each of the following electrodes undergoes reduction: - 2+ 2+ 3+ 3+ (a)Cl /Cl2(Pt) (b)Cu/Cu (c)Fe /Fe (d)Al/Al (ii) Write the half cell reactions if each of the electrodes given in (i) undergoes oxidation. SAQ 8: Write the cell notation in ionic form and give the half cell reactions and also the cell reactions of the following electrochemical cells.

2+ (i)Mg/MgCl2 HCl/H2(Pt) (ii) H2(Pt)/HCl CuCl/Cu (iii)I2(Pt)/KI Fe / Fe3+(Pt)

21 CONSTRUCTING CELL FROM THE CELL REACTION: Supposing you are given a cell reaction. Say for example, Zn + Cu2+ ------> Zn2+ + Cu Looking to the reaction, can you not construct the half cells and full cell which give rise to the above cell reaction? Zn is oxidised to Zn2+, so it makes the oxidation half cell(LHS) and its notation is Zn/Zn2+. Cu2+ is reduced to Cu and so it makes the reduction half cell(RHS) and its notation is Cu2+/Cu. Now we shall join the two half cells by means of a salt bridge and write the cell notation. Zn/Zn2+ Cu2+/Cu Look to another example. Example: Write the half cell reactions and construct the cell from the following cell reaction. + 3+ 2Al + 6H ------> 2Al + 3H2 3+ + Solution: Here Al is oxidised to Al and H is reduced to H2. The half cell reactions are Oxidation half cell: Al ------> Al3+ + Reduction half cell: 2H + 2e ------> H2 Adding the above two reactions after cancelling electrons, we get the cell reaction given in the 3+ + question. The half cells are Al/Al and H /H2(Pt). Pt is used for electrical contact since H2 is a gas electrode.

3+ + Al/Al H /H2(Pt) SAQ 9 : Construct the cell which represent the following cell reaction. Also write the half cell reactions. 2+ - - - (i) Cu + Cl2 ------Cu + 2Cl (ii) 2I + Br2 ------> 2Br + I2 2+ 2+ + 2+ (iii) Mg + Sn ------> Mg + Sn (iv) Ca + 2H ------> Ca + H2

STANDARD ELECTRODE POTENTIAL: We know that water spontaneously flows from a higher level where its potential energy is high to a lower level where its potential energy is low. In the same way, electron spontaneously flows from the electrode having higher potential to the electrode having lower potential when the two electrodes are connected externally by a metallic conductor. Loss of electron (oxidation) occurs at the electrode(oxidation electrode) from which electron flows to the external conductor and gain of electron(reduction) occurs at the electrode(reduction electrode) to which electron enters from the external conductor. Let us discuss something more about this potential which is called standard electrode potential. The word standard means that the electrode potential is measured in the standard states which are taken to be 250C temperature, 1 M(molar) concentration of the 2+ 2+ electrolyte(say Zn in ZnSO4 solution or Cu in CuSO4 solution etc) and for gas electrode, the pressure is 1 atmosphere pressure of the gas. There are two types of electrode potential, namely (i)standard oxidation potential and (ii)standard reduction potential. We shall discuss first the standard oxidation potential.

STANDARD OXIDATION POTENTIAL(SOP): The standard oxidation potential of an electrodes give the relative ability of an electrode to undergo oxidation in the standard states (mentioned earlier) and standard reduction potential gives the relative ability of the electrodes to undergo reduction. While we compare between two electrodes, we have to consider any one of the potential. First let us consider the oxidation

22 potential. Let us take an analogy. Supposing there are 100 students in a class and each one is a boxer. A merit list is prepared on the basis of their ability to perform boxing. The best boxer is given rank 1 and the worst of them is given the last rank(100) and all the remaining are ranked in the decreasing order of boxing capability. When there is a boxing competition between rank 1 with rank 2 of the list, rank 1 will give blows and rank 2 will have to receive it. Giving of a blow by a boxer is analogous to loss of electron by one electrode(LHS) having higher potential and undergoing oxidation while receiving blows by a boxer is analogous to gain of electron by the other electrode(RHS) having lower potential and undergoing reduction. Blow giving capacity of No.1 boxer is more hence it gives the blow, and blow giving capacity of No.2 boxer is less than No. 1, hence it receives the blow from No.1. Note that No.2 can give the blow only when it fights with boxers having rank lower than 2 but not with No.1. The blow giving(analogous to tendency to lose electron) capacity is thought to be analogous to the potential of the electrode called standard oxidation potential. The electrodes are arranged in the order of their ability to undergo oxidation which is measured by the parameter standard oxidation potential. If the standard oxidation potential of one electrode is greater than the other, oxidation will occur at the first electrode and consequently reduction will occur at the second electrode. It is not that oxidation will occur at both the electrodes or reduction will occur at both the electrodes. When boxer No. 2 fights with No. 3 or any boxer of lower rank than 2, who will give blow and who will receive blow? No. 2 will give blow and the no. 3 or boxer of any lower rank will receive the blow. Note that the same No.2 which was receiving blow from No.1 is now giving blows to others. So the electrode in which reduction occurs is at a lower oxidation potential than the other where oxidation occurs. But oxidation will occur in the same electrode if it is coupled with an electrode having lower oxidation potential. Compare the half cell reactions given

2+ + + + in the example Sn/Sn H /H2(Pt)and example H2(Pt)/H Ag /Ag given before. In the first 2+ + cell, the Sn/Sn electrode has a higher standard oxidation potential than H2/H electrode, hence oxidation was taking place in the Sn electrode(LHS) and reduction was taking place at the H2 + electrode(RHS). But the in the second cell, H2/H has a higher standard oxidation potential and + Ag/Ag has a lower potential. Therefore oxidation was occurring at H2 electrode and reduction was occurring at the Ag electrode. From all these analogies and examples we can conclude that between the two electrodes that make an electrochemical cell, the one having higher standard oxidation potential will undergo oxidation and will be kept and written in the LHS while the other having lower standard oxidation potential will undergo reduction and will be kept and written in the RHS.

STANDARD REDUCTION POTENTIAL(SRP) Supposing another list is prepared in respect of the 100 students according to the blow receiving capacity i.e the one who receives the blow from everybody is assigned rank No. 1. In other words, the No. 100 in the previous blow giving list becomes No. 1 in the blow receiving list and No. 1 in the blow giving list will rank 100th in the blow receiving list. In fact No. 100 in this list will not receive blow from anybody. Can you tell who will give blow and who will receive when no.4 and no. 9 from this merit list are allowed to fight? No. 4 will receive the blow and no. 9 will give the blow because no. 4 has a greater blow receiving capacity than no. 9. In the same manner, if the electrodes are ranked according to their ability to undergo reduction, we express it by the term standard reduction potential. The electrode having greater reduction potential

23 will undergo reduction and is written in the RHS and the electrode having lower reduction potential will undergo oxidation and is written in the LHS. The standard reduction potential of electrodes are same in magnitude as their standard oxidation potential but only their signs are opposite. In our subsequent discussion, we shall make use of standard oxidation potential of electrodes for comparison most of the time The two concepts are same and we should use any one of them for writing the cell notation and find the EMF.

ELECTROCHEMICAL SERIES: When the electrodes are arranged in the decreasing order of their standard oxidation potential or increasing order of standard reduction potential we get a series called + electrochemical series. The unit of electrode potential is volt. electrode[H2(Pt)/H ] has been arbitrarily given as 0 volt and is located somewhere in the middle and many electrodes have higher oxidation potential than H2 elctrode and are kept above it and they have +ve oxidation potential. Many other electrodes have lower oxidation potential than H2 electrode and are kept below it and they have -ve oxidation potential. The metal activity series that we knew before is similar to this series. In metal activity series, only metals are written while in electrochemical series the whole electrodes(e.g Na/Na+) are written. The order is same. Note that when any electrode, located anywhere in the series having higher oxidation potential is coupled with any other electrode placed anywhere in the series having lower oxidation potential, oxidation will occur at the former electrode and reduction will occur at the latter electrode. Let us first look at the electrochemical series and then shall analyse it further.

Electrochemical Series (Decreasing SOP) Electrode Standard Oxidation Electrode Standard Oxidation Potential (in volt) Potential (in volt)

+ + Li/Li 3.03 H2/H 0 K/K+ 2.93 Sn2+/Sn4+ -0.13 Cs/Cs+ 2.92 Cu+/Cu2+ -0.15 Ba/Ba2+ 2.90 Cu/Cu2+ -0.34 2+ 4- Ca/Ca 2.87 [Fe(CN)6] / + 3- Na/Na 2.71 [Fe(CN)6] -0.35 Mg/Mg2+ 2.37 Cu/Cu+ -0.52 3+ - Al/Al 1.66 I /I2 -0.53 Mn/Mn2+ 1.18 Fe2+/Fe3+ -0.77 2+ 2+ Zn/Zn 0.76 Hg/Hg2 -0.79 Cr/Cr3+ 0.74 Ag/Ag+ -0.8 2+ 2+ 2+ Fe/Fe 0.44 Hg2 /Hg -0.85 2+ - Co/Co 0.28 Br /Br2 -1.09 2+ 3+ 2- + Ni/Ni 0.25 Cr /Cr2O7 /H -1.33 2+ - Sn/Sn 0.14 Cl /Cl2 -1.36 Pb/Pb2+ 0.126 Au/Au3+ -1.5 2+ - + Mn /MnO4 /H -1.7 Co2+/Co3+ -1.82 - F /F2 -2.87

24 The electrochemical series in terms of increasing SRP will be the as this but the signs are opposite. In the above electrochemical series, the standard oxidation potential of some of the important electrodes are given in volt in the decreasing order of their values. Standard Potential from the Half Cell sequence : From sequence of the pair of species used for a particular electrode, it can be known whether the standard potential value given is oxidation potential or reduction potential. 0 0 = + 0.76 V = - 0.76 V E 2+ E 2+ Zn Zn Zn Zn The first one is oxidation potential as the sequence is Zn → Zn2+ which is the sequence of oxidation. But the 2nd one is reduction potential as the sequence Zn2+ → Zn is the sequence of reduction.

0 0 E = - 1.36 V E = +1.36 V Cl Cl2 Cl2 Cl

– → → – The first one is SOP as Cl Cl2 is oxidation while the 2nd one is SRP as Cl2 Cl is reduction. N.B. Note that active metals which lie above the hydrogen electrode have a +ve SOP and –ve SRP while the metals lying below hydrogen and other nonmetallic electrodes have –ve SOP and +ve SRP. For example Cu/Cu2+ has –0.34V SOP and +0.34V SRP while Zn/Zn2+ has +0.76V SOP and –0.76V SRP. APPLICATION OF THE ELECTROCHEMICAL SERIES:

1. Determination of oxidation and reduction electrodes (half cells): Construction of Cells: By looking to the relative locations of the electrodes in the electrochemical series, we can know which electrode will undergo oxidation and which reduction. This will enable us to write and place the electrodes in proper positions, i.e oxidation half cell in the LHS and reduction half cell in RHS. Take for instance, if we want to construct a cell by coupling Ag/Ag+ electrode with Ca/Ca2+ electrode, which electrode will involve oxidation and which reduction, that we have to make sure.Try to find out these electrodes in the above series. Ca electrode lies far ahead of the Ag electrode and the standard oxidation potential of the former is greater than the latter. So oxidation will occur at Ca electrode and reduction will occur at Ag electrode. You know already that oxidation occurs at the electrode which has greater oxidation potential and reduction occurs at the electrode which has lower oxidation potential. So Ca/Ca2+ elctrode is placed in the left side and Ag/Ag+ electrode is kept in the right side while constructing the cell. Cell Notation: Ca/Ca2+ Ag+/Ag Let me remind you that while writing the cell notation, the electrolytes are kept facing each other on either side of the parallel line(salt bridge).

2+ Oxidation half cell: Ca Ca + 2e + Reduction half cell: Ag + e Ag X 2 Cell reaction: Ca + 2 Ag+ Ca2+ + 2Ag

25 SAQ 1O: Determine in each of the following pairs, which electrode will undergo oxidation and which reduction.Give the reason. Show it by giving proper cell notation. Also write the half cell reactions and cell reactions. Refer the values of standard oxidation potential from the series. - 3+ + + - - (i) Cl /Cl2(Pt) and Al/Al (ii)K/K and Na/Na (iii)Cl /Cl2 and Br /Br2 SAQ 11: Check if the following cell notations are correct or wrong. If wrong, give the correct cell notations.

2+ 2+ 3+ 3+ - - (i) Cu/Cu Fe /Fe (ii) Al/Al Cr /Cr (iii)F2(Pt)/F Br /Br2(Pt)

2. Determination of Cell EMF or Voltage: The voltage of an electrochemical cell or battery is the potential difference between the two electrodes. This was formerly called electromotive force (EMF). Now also the term EMF is widely used to denote the potential difference or voltage of a cell. Standard Cell EMF(voltage) = Standard Oxidation Potential of OHC(LHS) - Standard Oxidation Potential of RHC (RHS)

0 0 0 E cell = E LHS - E RHS (SOP) OR Standard Cell EMF(voltage) = Standard Reduction Potential of RHC(RHS) - Standard Reduction Potential of OHC (LHS) 0 0 0 E cell = E RHS - E LHS (SRP)

(where 0 refers to the standard state and LHS and RHS denote the left hand side or the oxidation half cell and right hand side or the reduction half cell respectively) Note that cell EMF is always a +ve quantiy. If by an chance, we find the cell EMF is negative, then the cell notation is wrong. Just reverse the electrodes, the cell EMF will be positve. Example: Find the standard EMF of the cell: Zn/Zn2+ Cu2+/Cu. Refer the SOP series. Solution: Refer the electrochemical series and find the standard oxidation potential of the 0 0 above two electrodes designated as E Zn and E Cu respectively. 0 0 E Zn = 0.76V E Cu = - 0.34V 0 0 0 E cell = E Zn - E Cu = 0.76 - (-0.34) = 0.76 + 0.34 = 1.1 V. the standard EMF of the cell is 1.1V. Example : The standard potentials of two half cells are given. Construct the cell and write the cell notation. E0 = - 0.44V (SRP) E0 = + 0.34V (SRP) (Fe2+/Fe) (Cu2+/Cu) Solution: From the sequence of the species, we can know that these potential values are SRP. Since SRP of Cu electrode is greater, reduction occurs there and is written in RHS while SRP of Fe electrode is lower, oxidation occurs at this electrode and is written in the LHS.

26 Cell Notation : Fe/Fe2+//Cu2+/Cu E0(Cell) = E0(RHS) - E0(LHS) = +0.34 - (- 0.44) = 0.78 V

Example: Find the standard EMF of the cell: Al/Al3+ Cr3+/Cr

0 0 Solution:EAl = 1.66V, E Cu = 0.74V 0 0 0 E cell = E Al - E Cr = 1.66 - 0.74 = 0.92 V. + - Example: Find the standard EMF of the cell: Ag/Ag Cl /Cl2(Pt) 0 0 Solution:EAg = -0.8V E Cl = - 1.36V 0 0 0 E cell = E Ag - E Cl = -0.8 - (-1.36) = -0.8 + 1.36= 0.56V. Example : F

3. Relative strengths of Oxidising and Reducing Agents: Comparing the values of standard oxidation potential it is possible to say which is a better oxidising agent and which is poor oxidising agent, similarly which is a better reducing agent and which a poor reducing agent. REDUCING AGENTS: We know that if the oxidation potential of an electrode is high, it has a higher tendency to be oxidised and hence it is a strong reducing agent. Similarly if the oxidation potential of an electrode is low, it has a lower tendency to be oxidised and hence high tendency to be reduced. So it is a strong oxidising agent. The electrodes which have +ve values of standard oxidation potential like Na/Na+, K/K+, Mg/Mg2+, Al/Al3+, /Zn/Zn2+ etc.(refer the series) have a higher tendency to undergo oxidation. The greater the value of the potential the greater is the tendency of the metal to undergo oxidation. Therefore the metals like Na, K, Mg, Al, Zn etc. are very good reducing agents. But if you are asked to arrange these metals in the order of decreasing strengths as reducing agent, you will compare the standard oxidation potential and the one having greater value of the potential has greater strength as reducing agent. Hence the order among them is K>Na>Mg>Al>Zn Note that all are good reducing agents as they have high tendency to be oxidised, but K is the best among them and Zn is poorest among them.

OXIDISING AGENTS: The electrodes which have low or -ve values of oxidation potential have low tendency to undergo oxidation, rather they have high tendency to undergo reduction. In other words, if the SRP of an electrode is higher, it has greater tendency to undergo reduction. Therefore they are good oxidising - - - - 3+ 2- agents. Electrodes like Cl /Cl2, Br /Br2,, I /I2, F /F2, Cr /Cr2O7 have negative values of oxidation potential(+ve value of SRP) and hence the corresponding species existing in the oxidised state 2- such as Cl2, Br2, I2, F2, Cr2O7 etc. are good oxidising agents. Between them, the one having lower oxidation potential(greater SRP) is a better oxidising agent. So if we arrange the above in the order of decreasing strength as oxidising agent, it will be 2- F2 > Cr2O7 >Cl2>Br2>I2 Example: Predict whether the following reaction is possible or not.

Br2 + KCl ------> Cl2 + KBr

27 Solution:

Look to the order of halogens as oxidising agents given before. Br2 is a less strong oxidising - agent than Cl2, so it cannot oxidise Cl to Cl2. Rather the reverse reaction,

KBr + Cl2 ------> Br2 + KCl is possible. - Cl2 being a better oxidising agent can oxidise Br to Br2. That is why we have learnt in the first chapter of Chemical Arithmetic the halogen activity series, F2>Cl2>Br2>I2. The halogen placed earlier in the series can displace the other halogen placed latter from the salt of the latter. Now we know the logic of it. SAQ 12: Between the following pairs which is a better reducing agent and why? (i) Na, K (ii) Fe, Al (iii) Mg, Ca (iv) Sn, Pb (v)Fe2+, Sn2+ SAQ 13: Between the following pairs which is the better oxidising agent and why?

(i)Cl2, Br2 (ii) F2, I2 (iii) Cl2, I2

(iv)Br2, F2 - 2- + 3+ (v)MnO4 , Cr2O7 (vi)Ag , Au SAQ 14: Predict whether the following reactions are possible or not. Give reason and your comments. (i)Mg2+ + Sn ------> Mg + Sn2+ (ii)Ag+ + Fe ------> Fe2+ + Ag - 2+ - - (iii)2Br + Zn ------> Br2 + Zn (iv)2F + Cl2 ------> F2 + 2Cl (v)K + Cu2+ ------> Cu + K+ (vi)Fe2+ + Mn2+ -----> Fe3+ + Mn

TYPES OF ELECTROCHEMICAL CELLS: Broadly there are two types of electrochemical cells which produce electricity out of a chemical reaction. (i) Dry Cells (ii) Wet Cells

DRY CELLS: In dry cells, the electrolyte used is almost dry in theform of a paste or jelly and contain limited liquid contents for which there is no leakage. Hence these are portable and easy to handle and used extensively in electronic goods. The bigger (Eveready, Nippon etc.) batteries and smaller batteries(pencil battery as is commonly called) that you purchase from the market for household purposes belong to this category. Button cells used in writst watch, calculators and other electronic equipments belong to this category. The oldest among them is the Carbon-Zinc cell or Lechlanche cell. WET CELLS: In wet cells, the electrolyte used contains liquid(often water). Such cells are not portable and is used at one place. Lead-acid battery belongs to this category.

Electrochemical cells are also categorized as Primary or Irriversible or Non-rechargeable Cell (b) Secondary or Reversible or Chargeable Cells Primary cells once used cannot be reused. After the chemical reaction is complete and the cell attains an equilibrium state, the current stops flowing through the external conductor. This is because the two electrodes attain the same potential. Like water cannot flow if the heights of the levels are same, in the same way electrons stop flowing when the potentials of the electrodes

28 become same after long use. We thus throw the battery to dust bin. These types of cells cannot be recharged. Button cells, Zinc-carbon battery and many others belong to this category. Secondary cells can be recharged after the discharge of the cell and hence can be reused. The cell reaction undergoing in the discharge mode of the cell(during its use) can be reversed by applying it a current in opposite direction with a DC source having greater EMF that the cell. This is charging process in which the reverse reaction occurs and the reactants reappear during the process and the cell is ready for reuse. Lead-acid battery belongs to this category. We shall discuss more about it later. Below you find the details of a few cells.

LECHLANCHE CELL: This is also called carbon-zinc cell. The outer casing is made of Zn which is surrounded by a paste of NH4Cl and

ZnCl2.. Zn casing acts as –ve electrode or OHC. At the middle there is a graphite rod which shows its head portion outside. This is surrounded at the inner side by a mixture MnO 2 and carbon powder. The + MnO2/NH4 couple acts as the +ve electrode or RHC. Graphite rod serves as the carrier of current from outside to the half cell. The cell notation and half cell reactions are as follows.

2+ Cell Notation: Zn/Zn NH4Cl/MnO2(graphite)

OHC: Zn → Zn2+ + 2e + → RHC: 2NH4 + 2MnO2 + 2e Mn2O3 + H2O + 2NH3 ______+ → 2+ Zn + 2NH4 + 2MnO2 Zn + Mn2O3 + H2O + 2NH3 The EMF of each such cell is 1.5V. This is a non-rechargeable cell. There is a competing side reaction at the graphite electrode. + → 2NH4 (aq.) + 2e H2(g) + 2NH3(g)

H2 and NH3 produced react with MnO2 and ZnCl2 so that the pressue of the gases built up is reduced. → ZnCl2(aq.) + 2NH3(g) Zn(NH3)2Cl2(s) → 2MnO2(s) + H2(g) Mn2O3(s) + H2O(l)

Carbon-Zinc Alkaline Battery : This is similary to Lechlanche cell. But the difference is - KOH(an alkali) is used in stead of NH4Cl. The life of this cell is greater than Lechlanche cell. Moreover, gas pressure does not build up here and hence are more extensive used as primary cell.

29 – → OHC : Zn + OH Zn(OH)2 + 2e → – RHC : 2 MnO2 + H2O + 2e Mn2O3 + 2OH

Silver Battery: This is a dry and non-rechargeable cell. – -ve electrode : Zn/OH +ve electrode : Ag/Ag2O – → OHC : Zn + 2OH Zn(OH)2 + 2e → – Ag2O + H2O + 2e 2Ag + 2OH Mercury Battery: -ve electrode : Zn/KOH +ve electrode : Hg/HgO Due to mercury pollution, the use of this cell is banned. – → OHC : Zn + 2OH Zn(OH)2 + 2e → – RHC: HgO + H2O + 2e Hg + 2OH LEAD -ACID BATTERY : These can be recharged and can be reused for many times. Such cell becomes temporarily defunct like a primary cell when the potential difference between the two electrodes becomes equal to zero. But again it can be brought t o life and can be reused after a process called charging. Have you seen a black and big car battery or a small, good looking battery used in motor bikes? These are called reversible batteries. Do you know that when such battery becomes defunct, is subjected to the process of charging? In charging process, stronger current from the town supply(made DC) is passed in the opposite direction. The oxidation half cell receives the electron and the reduction half cell loses electron. So during charging, reverse process takes place as compared to the process when the battery is used. When the battery is in use, it is said to remain in the discharging mode and when it is defunct and subjected to charging, it is said to be in the charging mode. In charging mode, reduction occurs at the oxidation half cell of the battery by receiving electron from the town supply electric current and oxidation occurs at the reduction half cell. Thus the entire reaction that takes place in the discharging mode of the cell is reversed. Supposing the cell is Zn/Zn2+ Cu2+/Cu . In the discharging mode Zn vanishes in the LHS electrode as it forms Zn2+ while Cu2+ vanishes in the RHS electrode as it forms Cu. But in the charging mode, Zn is produced by the reduction of Zn2+ and Cu2+ is produced by the oxidation of Cu. Thus during charging the lost Zn and Cu2+ are reformed and we can then stop charging and find that the battery is now alive. It has developed a potential difference and can now be used in the discharging mode in our motor bike or car. Note that during the charging mode, the cell act as an electrolytic cell in which a non- spontaneous reaction is brought about by using an external DC source.

30 This is the oldest kind of reversible cell used in cars and trucks. This was black in colour. But nowadays white plastic casings have replaced the old black casings. The cell notation of such battery is as follows.

Oxidation half cell: Pb/H2SO4 and Reduction Half Cell: PbO2/H2SO4

Since H2SO4 is common to both the electrodes, we write the cell as follows.

Pb/H2SO4/PbO2 During the use of such battery i.e in the discharging mode, the half cell reactions are as follows. Discharging Mode:

2- Oxidation half cell: Pb + SO 4 PbSO4 + 2e 2- + 2 H O Reduction half cell: PbO2 + SO4+4 H + 2e PbSO4 + 2 + 2- Cell reaction: Pb + PbO 2 + 4 H + 2 SO4 2 PbSO4 + 2 H2 O

Pb + PbO2 + 2 H2SO4 2 PbSO4 +2H2O

Charging Mode: During its charging, opposite reaction takes place. At the oxidation half cell(Pb electrode), reduction 2- occurs i.e the reaction PbSO4 + 2e -----> Pb + SO4 takes place. At the reduction half cell, 2- + (PbO2 electrode) oxidation occurs, i.e the reaction, PbSO4 + 2H2O ------> PbO2 + SO4 + 4H +2e takes place. Thus the overall reaction during the charging mode becomes:

2PbSO4 + 2H2O ------> Pb + PbO2 + 2 H2SO4 which is opposite to cell reaction that takes place during discharging mode. Note that H2SO4 is consumed in the discharging mode and is regenerated in the charging mode.

Construction of Cell:

Usually this battery consists of several pairs(usuall six pair) of Pb-PbO2 plates close to each other. In fact PbO2 plate is also a lead plated on which PbO2 coating has been applied. The six

Pb plates are connected to make the –ve electrode and the six PbO2 plates are connected to make the +ve electrode. This assmbly of plates is immersed in a plastic container containing 37% sulphuric acid which serves as the electrolyte. Each pair of plates produce a voltage of 2V, so a battery of 6 pairs will produce 12V EMF. Overcharging with higher charging voltage makes some water to decompose to hydrogen and oxygen(electrolysis) and so the amount of water in the electrolyte decreases. So periodically the distilled water is to be added to it to maintain the strength of sulphuric acid. After a few repeated discharging-charging cycles, the cell becomes useless and cannot be further used. Out of the several reasons for it, the principal reason is 'sulphation'. After a prolonged use, PbSO4 sticking on the lead plates drop down into the electrolyte in the form of course crystal grains and form a suspension. That is why the electrical conductivity stops inside the electrolyte. Morever over in winter season, the PbO2 coating slips out of the lead plate. NICKEL-CADMIUM(NICAD) Rechargeable cells: You must have seen that batteries which are used in electronic gadgets like telephone, mobile phone etc. are recharged. These are although dry cells but are rechargeable. This is the discovery of the recent times.The cell notation for the nickel-cadmium(NICAD) cell is as follows. The voltage of this battery is 1.2V.

31 - Cd/OH NiO2 The half cell reactions are as follows: - Oxidation half cell: Cd + 2 OH ------> Cd(OH)2 + 2e - Reduction half cell: NiO2 + 2H2O + 2e -----> Ni(OH)2 + 2 OH ______

Cell Reaction: Cd + NiO2 + 2H2O ------> Cd(OH)2 + Ni(OH)2

Nickel Metal Hydride/NiMH Battery: This is similar to NICAD battery. The difference is - in place of Cd for –ve electrode, hydride of an alloy which consists of one from La, Ce or Nd and another from Mn or Co. This cell is 2-3 times more efficient than NICAD and is rechargeable. Nowadays NiMH batteries are used most extensively. Ion Battery: This is nowadays very popular as it is used in most of the electronic deivices. This is also rechargeableand can be prepared in any shape.. The movement of Li+ ions from –ve to +ve electrode is responsible for the cell voltage. The cell reaction is very complicated and is not discussed here. Lithium Battery : These batteries are different from Lithium ion battery. Lithium batteries are non-rechargeable while Lithium ion batteries are chargeable. Several types of lithium batteries are available, the most widely used among them is Li-MnO2 battery. Besides these, Li-SOCl2, Li-SO2, Li-I2, Li-

Ag|CrO4, Li-CuO, Li-FeS2, Li-V2O5, Li-CoO2 etc are available. These batteries are available in the shape of bottons for which these are also called BOTTON CELLS. Lithium is the –ve electrode and the the other component is the +ve electrode. These are used in many electronic deivices, pacemakers in heart and in many medical instruments, wrist watches, calculators, digital thermometer. All the botton cells contain some amount of mercury to absorb the gas produce and eliminate the buiding up of gas pressure inside the cell. Since mercury causes pollution, the use of mercury in these cells has been banned. Zinc-air button cell is one example where mercury is no used. Zinc-air button cell : This cell consist of several pores through which air enters inside the cell and oxygen present in air takes part in cell reaction. This cell used in hearing aids and its voltage is 1.65V. – – –ve electrode : Zn/OH +ve electrode : O2/OH .

OHC : Zn + OH– → Zn(OH)2 + 2e → – RHC: O2 + 2H2O + 4e 4OH ______→ 2 Zn + O2 + 2H2O 2Zn(OH)2

32 RESPONSE TO SAQs SAQ 1: + - (i)At cathode: K + e ------> K; At anode: 2Br ------> Br2 + 2e + - (ii)At cathode: 2H + 2e ------> Mg; At anode: 2Cl ------> Cl2 + 2e + - (iii)At cathode: Ag + e ------> Ag; At anode: 4OH ------> O2 + 2H2O + 4e + - (iv)At cathode: 2H + 2e -----> H2 At anode: 4OH ------> O2 + 2H2O + 4e

2+ - (v)At cathode: Ca +2e ------> Ca At anode: 2F ------> F2 + 2e SAQ 2: + - (i) At cathode: 2H +2e ------> H2; At anode: 4OH ------> O2 + 2H2O + 4e 2+ - (ii) At cathode: Ca + 2e ------> Ca; At anode: 2Cl ------> Cl2 + 2e 3+ 2- (iii) At cathode: Al +3e ------> Al; At anode: 2O ------> O2 + 4e 2+ 2+ (iv) At cathode: Hg2 + 2e ------> 2Hg; At anode: 2Hg ------> Hg2 + 2e (since Hg lies below H) (since Hg is an active electrode) SAQ 3: (i) Eq. Mass of Cu = 63.5/2= 31.75. So 1 Faraday(96500 couls) of charge will deposit 31.75gms of Cu. (ii) Eq. Mass of Al = 27/3=9. So 9gms of Al will be deposited by the passage of 1 Faraday of charge. (iii) Eq. Mass of Cl = 35.5/1 =35.5. 1 Faraday of charge will produce 35.5gms. of Chlorine gas; So ½ Faraday of charge will produce 35.5/2=17.75gm. (iv) Eq. Mass of O =8; 96500 couls of charge will evolve 8 gms of oxygen gas So 19300 couls of charge will evolve 1.6gms.

32 gms of O2 occupy 22.4 litres at NTP;

So 1.6gms of O2 will occupy 1.12 litres at NTP. (v) Eq. Mass of Cu = 63.5/2=31.75 31.75gms of Cu is deposited by 1 Faraday of charge, So 127gms of Cu must be deposited by 4 Faraday of charge. SAQ 4: (i) Q= I X t = 5 A X (30X60)sec = 9000 couls. 9000 couls of charge deposits 3.048 gms of Zn; So 96500 couls of charge must deposit 32.68gms of Zn So the equivalent mass of Zn =32.68 as 1F(96500 couls) of charge deposits 1 gm. equivalent mass of any substance. Alternatively: m = (E/F) X I X t (1st law). All the data are given and we can find E. (ii) Eq. Mass of Ni = 58.7/2= 29.35 m = (E/F) X I X t, ⇒ 100 = (29.35/96500) X I X ( 200 X 60)(since 3 hrs. 20mins=200min) ⇒ I = 27.4 amp. Solve this problem also by 1st principle. (iii) Eq. Mass of Bi = 209/3 = 69.66

33 m = (E/F) X I X t, ⇒ m = (69.66/96500) X 40 X (30X60) = 51.97gms. Try this problem also by first principle method. SAQ 5:

The volume of H2 gas at NTP = 1.12 litres; so its mass will (2/22.4)X 1.12 = 0.1 gm

mH = 0.1 gm.; mCl = ?;EH = 1.008; ECl = 35.5 ⇒ Apply the Faraday's 2nd law: mH/mCl = EH/ECl mCl = 3.521gm

71gms of Cl2 occupies 22.4 litres at NTP; so 3.521gm will occupy 1.1108 litres at NTP SAQ 6: In each pair the species remaining in the reduced state will be oxidised to the species remaining in the oxidised state and not to any other state. Similarly the species remaining in the oxidised state will be reduced to the species remaining in the reduced state and not to any other species.For example, Cu will b oxidised to Cu2+ and Cu2+ will be reduced to Cu. Cu2+ will not be reduced to Cu+1. (i) Cu is in the reduced state(oxidation number of Cu=0) and will undergo oxidation by the loss of electrons. Cu ------> Cu2+ + 2e Cu2+ is in the oxidised state(oxidation number of Cu=+2) and will undergo reduction by gain of electron. The reduction is the reverse to the above reaction. Cu2+ + 2e ------> Cu (ii) Cl - is in the reduced state(ON of Cl = -1) and will undergo oxidation. - 2Cl ------> Cl2 + 2e

Cl2 is in the oxidised state(ON of Cl=0 which is greater than -1) and will undergo reduction. - Cl2 + 2e ------> 2Cl (iii) Fe2+ is in the reduced state(ON of Fe=+2) and will be oxidised. Fe2+ ------> Fe3+ + e Fe3+ is in the oxidised state (ON of Fe=+3) and will be reduced. Fe3+ +e ------> Fe2+

(iv) H2 is in the reduced state(ON of H=0) and will be oxidised. + H2 ------> 2H + 2e H+ is in the oxidised state(ON of H =+1) and will be reduced. + 2H + 2e ------> H2 SAQ 7: - 2+ (i) (a) Cl2 + 2e ------> 2Cl (b) Cu + 2e ------> Cu (c) Fe3+ + e ------> Fe2+ (d) Al3+ + 3e ------> Al - 2+ (ii) (a) 2Cl ------> Cl2 + 2e (b) Cu ------> Cu + 2e (c) Fe2+ ------> Fe3+ + e (d) Al ------> Al3+ + 3e Note that the oxidation process is just the reverse of the reduction process for each electrode. SAQ 8:

2+ + (i) Mg/Mg H /H2(Pt) Oxidation Half Cell(LHS): Mg ------> Mg2+ + 2e + Reduction Half Cell(RHS): 2H + 2e ------> H2 ______+ 2+ Cell Reaction: Mg + 2H ------> Mg + H2

34 + + (ii) H2(Pt)/H Cu /Cu + Oxidation Half Cell(LHS): H2 ------> 2H + 2e Reduction Half Cell(RHS): 2X [Cu+ + e ------> Cu] ______+ + Cell Reaction: H2 + 2Cu ------> 2H + 2Cu - 2+ 3+ (iii) I2(Pt)/I Fe /Fe (Pt) (Pt is used in LHS electrode as I2 is a bad conductor. Pt is also used in RHS electrode because both Fe2+ and Fe3+ remain in aqueous solution and cannot conduct electricity to the external metallic conductor). - Oxidation Half Cell(LHS): 2I ------> I2 + 2e Reduction Half Cell(RHS): 2X [Fe3+ + e ------> Fe2+] ______- 3+ 2+ Cell Reaction: 2I + 2Fe ------> I2 + 2Fe SAQ 9 : 2+ - (i) LHS: Cu ------> Cu + 2e; RHS: Cl2 + 2e ------> 2Cl 2+ - Cell: Cu/Cu Cl /Cl2(Pt) - - (ii) LHS: 2I ------> I2 + 2e RHS: Br2 + 2e ------> 2Br - - Cell: I2(Pt)/I Br /Br2(Pt) (iii) LHS: Mg ------> Mg2+ + 2e RHS: Sn2+ + 2e ------> Sn Cell: Mg/Mg2+ Sn2+/Sn

2+ + (iv) LHS: Ca ------> Ca + 2e LHS: 2H + 2e ------> H2 2+ + Cell: Ca/Ca H /H2(Pt) SAQ 1O: 3+ - (i) Oxidation will occur at Al/Al electrode and reduction at Cl /Cl2 electrode as the standard oxidation potential of the former is 1.66V and of the latter is -1.36V.

3+ - Cell Notation: Al/Al Cl /Cl2(Pt)

3+ Oxidation half cell: Al Al + 3e X 2 - Reduction half cell: Cl 2 + 2e 2 Cl X 3 3+ - Cell reaction: 2Al + 3Cl2 2Al + 6Cl (ii)K/K+ has a greater standard oxidation potential(2.93V) than Na/Na+(2.71V). So oxidation will occur at the former electrode and reduction at the latter electrode. Cell Notation: K/K+ Na+/Na

+ Oxidation half cell: K K + e + Reduction half cell: Na + e Na + + Cell reaction: K + Na K + Na

35 - - (iii)Br /Br2 has a greater standard oxidation potential(-1.09V) than Cl /Cl2(-1.36V). So oxidation will occur at the former electrode and reduction at the latter electrode.

- - Cell Notation: Br2(Pt)/Br Cl /Cl2(Pt) (Note that Br2 is a liquid and Cl2 is a gas and therefore Pt wire is used to make electrical contact between the external conductor and the electrolyte solution) - Oxidation half cell: 2 Br Br + 2e 2- Reduction half cell: Cl 2 + 2e 2 Cl - - Cell reaction: 2 Br + Cl2 Br2 + 2 Cl SAQ 11: (i) This cell notation is wrong. Fe/Fe2+ electrode has a greater oxidation potential(0.44V) than the Cu/Cu2+(-0.34V). So oxidation will occur in the former and reduction in the latter. Hence the electrodes are to be exchanged. The correct cell notation is as follows. Fe/Fe2+ Cu2+/Cu (ii) This cell notation is correct. Al/Al3+ electrode has a greater oxidation potential(1.66V) than Cr/Cr3+ electrode(0.74V).Hence oxidation will occur at the former electrode and reduction at the latter. - - (iii) This cell notation is wrong. Br /Br2 has a greater oxidation potential (-1.09V) than F

/F2 electrode(-2.87V). So oxidation will occur at the former and reduction at the latter. The electrodes are to be exchanged to get the correct cell notation. The correct cell notation is

- - Br2(Pt)/Br F /F2(Pt) SAQ 12: (i)K is better, since K/K+ has greater oxidation potential than Na/Na+ (ii)Al is better, since Al/Al3+ has greater oxidation potential than Fe/Fe2+ (iii)Ca is better, since Ca/Ca2+ has greater oxidation potential than Mg/Mg2+ (iv)Sn is better, since Sn/Sn2+ has greater oxidation potential than Pb/Pb2+ (v)Sn2+ is better since Sn2+/Sn4+ has greater oxidation potential than Fe2+/Fe3+ SAQ 13: - - (i)Cl2 is better, since Cl /Cl2 has lesser oxidation potential than Br /Br2 - - (ii)F2 is better, since F /F2 has lesser oxidation potential than I /I2 - - (iii)Cl2 is better, since Cl /Cl2 has lesser oxidation potential than I /I2 - - (iv)F2 is better, since F /F2 has lesser oxidation potential than Br /Br2 - 2+ - 3+ 2- (v)MnO4 is better, since Mn /MnO4 has lesser oxidation potential than Cr /Cr2O7 (vi)Au3+ is better since Au/Au3+ has lesser oxidation potential than Ag/Ag+ SAQ 14: (i) Not possible, since Mg/Mg2+ has a greater oxidation potential than Sn/Sn2+, Mg should undergo oxidation and Sn2+ should undergo reduction. So the reverse reaction, Mg + Sn2+ -----> Mg2+ + Sn is possible. (ii)Possible, since Fe/Fe2+ has a greater oxidation potential than Ag/Ag+, henc Fe will be oxidised and Ag+ is to be reduced. 2+ - (iii)Not possible, since Zn/Zn has a greater oxidation potential than Br /Br2, hence Zn

36 should be oxidised and Br2 should be reduced. The reverse reaction is thus possible. - - - (iv)Not possible, since Cl /Cl2 has a greater oxidation potential than F /F2, hence Cl

should be oxidised and F2 should be reduced. The reverse reaction is thus possible. (v)Possible, since K/K+ has a greater oxidation potential than Cu/Cu2+, hence K is oxidised and Cu2+ is reduced. (vi)Not possible, since Mn/Mn2+ has a greater oxidation potential than Fe2+/Fe3+, hence Mn should be oxidised and and Fe3+ should be reduced. Therefore the reverse reaction is possible.

37