Redox Reactions PDF

REDOX RE ACTIONS 25 5 UNIT 8 REDOX REACTIONS...

REDOX RE ACTIONS 25 5 UNIT 8 REDOX REACTIONS d he Wher e ther e is oxidation, ther e is alw ays re duct ion – Chemist ry is essent ially a study of redox s ystems. is After studying this unit you will be able to Chemistry deals with varieties of matter and change of one bl · identify redox reactions as a class kind of matter into the other. Transformation of matter from of reactions in which oxidation one kind into another occurs through the various types of and reduction reactions occur reactions. One important category of such reactions is pu simultaneously; Redox Reactions. A number of phenomena, both physical · de fi ne the terms ox id atio n, as well as biological, are concerned with redox reactions. red uction , oxid ant ( oxidis ing These reactions find extensive use in pharmaceutical, be T agent) and reductant (reducing biological, industrial, metallurgical and agricultural areas. agent); The importance of these reactions is apparent from the fact re o R · exp lain mec hani sm o f re dox that burning of different types of fuels for obtaining energy reactions by electron transfer for domestic, transport and other commercial purposes, process; electrochemical processes for extraction of highly reactive tt E · use the concept of oxidation metals and non-metals, manufacturing of chemical number to identify oxidant and compounds like caustic soda, operation of dry and wet reductant in a reaction; batteries and corrosion of metals fall within the purview of C · cl as sif y red ox r eac ti on in to redox processes. Of late, environmental issues like c omb i nati on ( sy n the s i s) , deco mposition, di splace ment Hydrogen Economy (use of liquid hydrogen as fuel) and an d d i sp r o po r tio n ati o n development of ‘Ozone Hole’ have started figuring under no N reactions; redox phenomenon. · sug gest a c ompar ativ e or der among various reductants and 8.1 CLASSICAL IDEA OF REDO X REACTIONS – oxidants; OXIDATION AND REDUCTION REACTIONS © · balance c hemic al equatio ns Originally, the term oxidation was used to describe the usi ng ( i) o xi dati on n umbe r addition of oxygen to an element or a compound. Because (ii) half reaction method; of the presence of dioxygen in the atmosphere (~2 0%), · l earn the co nc ep t o f re do x many elements combine with it and this is the principal reactions in terms of electrode reason why they commonly occur on the earth in the processes. form of their oxides. The following reactions represent oxidation processes according to the limited definition of oxidation: 2 Mg (s) + O2 (g) ® 2 MgO (s) (8.1) S (s) + O2 (g) ® SO2 (g) (8.2)  25 6 CHE MIST RY In reactions (8.1) and (8.2), the elements been broadened these days to include removal magnesium and sulphur are oxidised on of oxygen/electronegative element from a acc ount of addition of oxygen to them. s ub stance or additio n of hy drogen/ Similarly, methane is oxidised owing to the electropositive element to a substance. addition of oxygen to it. According to the definition given above, the CH4 (g) + 2O2 (g) ® CO2 (g) + 2H2O (l) (8.3) following are the examp les of reduc tion processes: A careful examination of reaction (8.3) in which hydrogen has been replaced by oxygen 2 HgO (s) 2 Hg (l) + O2 (g) (8.8) d prompted chemists to reinterpret oxidation in (removal of oxygen from mercuric oxide ) terms of removal of hydrogen from it and, 2 FeCl3 (aq) + H2 (g) ® 2 FeCl2 (aq) + 2 HCl(aq) he therefore, the scope of term oxidation was (8.9) broadened to include the removal of hydrogen from a substance. The following illustration is (removal of electronegative element, chlorine another reaction where removal of hydrogen from ferric chloride) can also be cited as an oxidation reaction. CH2 = CH2 (g) + H2 (g) ® H3C – CH3 (g) (8.10) is 2 H2S(g) + O2 (g) ® 2 S (s) + 2 H2O (l) (8.4) (addition of hydrogen) As knowledge of chemists grew, it was 2HgCl2 (aq) + SnCl2 (aq) ® Hg2Cl2 (s)+SnCl4 (aq) bl natural to extend the term oxidation for (8.11) reactions similar to (8.1 to 8.4), which do not (addition of mercury to mercuric chloride) involve oxygen but other electronegative In reaction (8.11) simultaneous oxidation pu elements. The oxidation of magnesium with of stannous chloride to stannic chloride is also fluorine, chlorine and sulphur etc. occurs occ urr ing be c aus e of the ad dition of according to the following reactions : electronegative element chlorine to it. It was be T soon realised that oxidation and reduction Mg (s) + F2 (g) ® MgF2 (s) (8.5) always occ ur simultaneously (as will be re apparent by re-examining all the equations o R Mg (s) + Cl2 (g) ® MgCl2 (s) (8.6) given above), hence, the word “redox” was Mg (s) + S (s) ® MgS (s) (8.7) coined for this class of chemical reactions. tt E Incorporating the reactions (8.5 to 8.7) Problem 8.1 w ithin the f old of oxid ation r e ac tions In the reactions given below, identify the encouraged chemists to consider not only the s pe cies unde r going oxidation and C removal of hydrogen as oxidation, but also the reduction: re moval of e le ctr op os itive elements as oxidation. Thus the reaction : (i) H2S (g) + Cl2 (g) ® 2 HCl (g) + S (s) no N (ii) 3Fe3O4 (s) + 8 Al (s) ® 9 Fe (s) 2K4 [Fe(CN)6](aq) + H2O2 (aq) ®2K3[Fe(CN)6](aq) + 2 KOH (aq) + 4Al2O3 (s) is interpreted as oxidation due to the removal (iii) 2 Na (s) + H2 (g) ® 2 NaH (s) © of electropositive element potassium from Solution potassium ferrocyanide before it changes to potassium ferricyanide. To summarise, the (i) H2S is oxid ise d be cause a more electronegative element, chlorine is added term “oxidation” is defined as the addition of oxygen/electronegative element to a to hydrogen (or a more electropositive element, hydrogen has been removed s ub stance o r remo val o f hydro gen/ electropositive element from a substance. from S). Chlorine is reduced due to addition of hydrogen to it. I n the b e ginning, r e d uc tion w as (ii) Aluminium is oxidised because considered as remov al of oxygen from a compound. However, the term reduction has oxygen is added to it. Ferrous ferric oxide  REDOX RE ACTIONS 25 7 (Fe3O4) is reduced because oxygen has For convenience, each of the above been removed from it. processes can be considered as two separate (iii) With the careful application of the steps, one involving the loss of electrons and concept of electronegativity only we may the othe r the gain of ele ctr ons. As an infe r that s od ium is oxidise d and illustration, we may further elaborate one of these, say, the formation of sodium chloride. hydrogen is reduced. Reaction (iii) chosen here prompts us to 2 Na(s) ® 2 Na+(g) + 2e– think in terms of another way to define Cl2(g) + 2e– ® 2 Cl– (g) d redox reactions. Each of the above steps is called a half 8.2 REDOX REACTIONS IN TERMS OF reaction, which explicitly shows involvement he of electrons. Sum of the half reactions gives ELECTRON TRANSFER REACTIONS the overall reaction : We have already learnt that the reactions 2Na(s) + Cl2(g) ® 2NaCl (s) (8.12) 2 Na(s) + Cl2 (g) ® 2 Na+ Cl– (s) or 2 NaCl (s) 4Na(s) + O2(g) ® 2Na2O(s) Reactions 8. 12 to 8.14 suggest that half is (8.13) reactions that involve loss of electrons are 2Na(s) + S(s) ® Na2S(s) (8.14) called oxidation reactions. Similarly, the are redox reactions because in each of these half reactions that involve gain of electrons bl reactions sodium is oxid ised due to the are called reduction reactions. It may not ad d ition of e ithe r oxyge n or mor e be out of context to mention here that the new e le c tr one gativ e e le me nt to s od ium. way of defining oxidation and reduction has pu Simultaneously, chlorine, oxygen and sulphur b ee n ac hie ve d only by es tablishing a are reduced because to each of these, the correlation between the behaviour of species electropositive element sodium has been as per the classical idea and their interplay in be T added. From our knowledge of chemical electron-transfer change. In reactions (8.12 to re bonding we also know that sodium chloride, 8.14) sodium, which is oxidise d, acts as o R sodium oxide and sodium sulphide are ionic a reducing agent because it donates electron compounds and perhaps better written as to each of the elements interacting with it and Na+Cl – (s ), (Na+)2O 2– (s), and (Na+)2 S 2– (s ). thus helps in reducing them. Chlorine, oxygen tt E Deve lopment of c harges on the spec ies and sulphur are reduced and act as oxidising produced suggests us to rewrite the reactions agents because these accept electrons from (8.12 to 8.14) in the following manner : sodium. To summarise, we may mention that C Oxidation: Loss of electron(s) by any species. Reduction: Gain of electron(s) by any species. no N Oxidising agent : Acceptor of electron(s). Reducing agent : Donor of electron(s). Problem 8.2 Justify that the reaction : © 2 Na(s) + H2(g) ® 2 NaH (s) is a redox change. Solution Since in the above reaction the compound formed is an ionic compound, which may also be repre sented as Na+H– (s), this suggests that one half reaction in this process is : 2 Na (s) ® 2 Na+(g) + 2e–  25 8 CHE MIST RY and the other half reaction is: At this stage we may investigate the state H2 (g) + 2e– ® 2 H – (g) of equilibrium for the reaction represented by equation (8.15). For this purpose, let us place This splitting of the reaction under a strip of metallic copper in a zinc sulphate examination into two half re actions solution. No visible reaction is noticed and automatically reveals that here sodium is attempt to detect the presence of Cu2+ ions by oxidis ed and hydr oge n is re duce d, passing H2S gas through the solution to therefore, the complete reaction is a redox produce the black colour of cupric sulphide, change. CuS, does not succeed. Cupric sulphide has d such a low solubility that this is an extremely 8.2.1 Competitive Electron T rans fer sensitive test; yet the amount of Cu2+ formed Reactions he cannot be detected. We thus conclude that the Place a str ip of metallic zinc in an aqueous state of equilibrium for the reac tion (8.15) solution of copper nitrate as shown in Fig. 8.1, greatly favours the products over the reactants. for about one hour. You may notice that the strip becomes coated with reddish metallic Let us extend electron transfer reaction now is copper and the blue colour of the solution to copper metal and silver nitrate solution in disappears. Formation of Zn2+ ions among the water and arrange a set-up as shown in products can easily be judged when the blue Fig. 8.2. The solution develops blue colour due to the formation of Cu2+ ions on account of the bl colour of the s olution due to Cu2+ has disappeare d. If hydrogen sulphide gas is reaction: passed through the colourless solution pu containing Zn2+ ions, appearance of white zinc sulphide, ZnS can be seen on making the solution alkaline with ammonia. be T The reaction between metallic zinc and the (8.16) re aqueous solution of copper nitrate is : 2+ He re, Cu(s ) is oxidised to Cu (aq) and o R Zn(s) + Cu2+ (aq) ® Zn2+ (aq) + Cu(s) (8.15) Ag+(aq) is reduced to Ag(s). Equilibrium greatly In reaction (8.15), zinc has lost electrons favours the products Cu2+ (aq) and Ag(s). to form Zn2+ and, therefore, zinc is oxidised. By way of contrast, let us also compare the tt E Evidently, now if zinc is oxidised, releasing reaction of metallic cobalt place d in nickel ele ctrons, s omething must be r educ ed, sulphate solution. The reaction that occurs C accepting the electrons lost by zinc. Copper here is : ion is reduced by gaining electrons from the zinc. Reaction (8.15) may be rewritten as : no N (8.17) © Fig. 8.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker. C:\Chemistry XI\Unit-8\Unit-8(5)(reprint).pmd 27.7.6, 16.10.6 (reprint)  REDOX RE ACTIONS 25 9 Fig. 8.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker. d At equilibrium, chemical tests reveal that both However, as we shall see later, the charge Ni2+(aq) and Co2+(aq) are present at moderate transfer is only partial and is perhaps better he concentrations. In this case, neither the described as an electron shift rather than a reactants [Co(s) and Ni2+(aq)] nor the products complete loss of electron by H and gain by O. [Co2+(aq) and Ni (s)] are greatly favoured. What has b een said here with respect to This competition for release of electrons equation (8.18) may be true for a good number of othe r r e actions inv olv ing c ov alent is incidently reminds us of the competition for release of protons among acids. The similarity compounds. Two such examples of this class suggests that we might develop a table in of the reactions are: bl which metals and their ions are listed on the H2(s) + Cl2(g) ® 2HCl(g) (8.19) basis of their tendency to release electrons just and, as we do in the case of acids to indicate the CH 4(g) + 4Cl2(g) ® CCl4(l) + 4HCl(g) (8.20) pu strength of the acids. As a matter of fact we have already made certain comparisons. By In order to keep track of electron shifts in comparison we have come to know that zinc chemical re actions involving formation of be T releases e lectrons to copper and copper covalent compounds, a more practical method releases electrons to silver and, therefore, the of using o xidation numb er has b ee n re electron releasing tendency of the metals is in dev eloped. In this method , it is alw ays o R the order: Zn>Cu>Ag. We would love to make assumed that there is a complete transfer of our list more vast and design a metal activity electron from a less electronegative atom to a series o r electro chemical series. The more electonegative atom. For example, we tt E competition for electrons betwe en various rewrite equations (8.18 to 8.2 0) to show charge on each of the atoms forming part of metals help s us to design a clas s of cells, the reaction : C named as Galvanic cells in which the chemical 0 0 +1 –2 reactions b ecome the source of electrical energy. We would study more about these cells 2H2(g) + O2(g) ® 2H2O (l) (8.21) no N in Class XII. 0 0 +1 –1 H2 (s) + Cl2(g) ® 2HCl(g) (8.22) 8.3 OXIDATION NUMBER –4 + 1 0 +4 –1 +1 –1 A less obvious example of electron transfer is CH4(g) + 4Cl2(g) ® CCl4(l) +4HCl(g) (8.23) © realised when hydrogen combines with oxygen to form water by the reaction: It may be emphasised that the assumption 2H2(g) + O2 (g) ® 2H2O (l) (8.18) of electron transfer is made for book-keeping Though not simple in its approach, yet we purpose only and it will become obvious at a can visualise the H atom as going from a later stage in this unit that it leads to the simple neutral (zero) state in H2 to a positive state in description of redox reactions. H2O, the O atom goes from a zero state in O2 O x idatio n numb er deno tes the to a dinegative state in H2O. It is assumed that o xidatio n s tate of an element in a there is an electron transfer from H to O and compound ascertained according to a set consequently H2 is oxidised and O2 is reduced. of rules formulated on the basis that C:\Chemistry XI\Unit-8\Unit-8(5)(reprint).pmd 27.7.6, 16.10.6 (reprint)  26 0 CHE MIST RY electron pair in a covalent bo nd belongs of oxygen but this number would now be entirely to more electronegative element. a positive figure only. It is not always possible to remember or 4. The oxidation number of hydrogen is +1, make out e asily in a compound/ ion, which except when it is bonded to metals in binary element is more electronegative than the other. compounds (that is compounds containing Therefore, a set of rules has been formulated two elements). For example, in LiH, NaH, to determine the oxidation number of an and CaH2, its oxidation number is –1. element in a compound /ion. If two or more 5. In all its compounds, fluorine has an than two atoms of an element are present in oxidation number of –1. Other halogens (Cl, d 2– the molecule/ion such as Na2S2O3/Cr2O7 , the Br, and I) also have an oxidation number oxidation number of the atom of that element of –1, when they occ ur as halide ions in he will then b e the average of the oxidation their compounds. Chlorine, bromine and number of all the atoms of that element. We iodine whe n combined with oxygen, for may at this stage, state the rules for the example in oxoacids and oxoanions, have calculation of oxidation number. These rules are: positive oxidation numbers. 6. The algebraic sum of the oxidation number is 1. In elements, in the free or the uncombined of all the atoms in a compound must be state, each atom bears an oxidation zero. In polyatomic ion, the algebraic sum number of zero. Evidently each atom in H2, of all the oxidation numbers of atoms of bl O2, Cl2, O3, P4, S8, Na, Mg, Al has the the ion must equal the charge on the ion. oxidation number zero. Thus, the sum of oxidation number of three 2. For ions composed of only one atom, the oxygen atoms and one carbon atom in the pu oxidation number is equal to the charge carbonate ion, (CO3)2– must equal –2. on the ion. Thus Na+ ion has an oxidation By the application of above rules, we can number of +1, Mg2+ ion, +2, Fe3+ ion, +3, find out the oxidation number of the desired be T Cl– ion, –1, O2– ion, –2; and so on. In their element in a molecule or in an ion. It is clear c omp ound s all alk ali me tals hav e re that the metallic elements have positive o R oxidation number of +1, and all alkaline oxidation number and nonmetallic elements earth metals have an oxidation number of have positive or negative oxidation number. +2. Aluminium is regarded to have an The atoms of transition elements usually tt E oxid ation numb e r of +3 in all its display several positive oxidation states. The compounds. highest oxidation number of a representative 3. The oxidation number of oxygen in most element is the group number for the first two C compounds is –2. However, we come across groups and the group number minus 10 two kinds of exceptions here. One arises (following the long form of periodic table) for in the case of peroxides and superoxides, the other groups. Thus, it implies that the no N the compounds of oxygen in which oxygen highest value of oxidation number exhibited atoms are directly linked to e ach other. by an atom of an element generally increases While in peroxides (e.g., H2O2, Na2O2), each across the period in the periodic table. In the oxygen atom is assigned an oxidation third period, the highest value of oxidation © number of –1, in superoxides (e.g., KO2, number changes from 1 to 7 as indicated below RbO2) each oxygen atom is assigned an in the compounds of the elements. oxidation number of –(½). T he second A term that is often used interchangeably exception appears rarely, i.e. when oxygen with the oxidation number is the oxidation is bonded to fluorine. In such compounds state. Thus in CO2, the oxidation state of e.g., oxygen difluoride (OF2) and dioxygen carbon is +4, that is also its oxidation number difluor ide (O2F2), the oxygen is assigned and similar ly the oxidation state as well as an oxidation numb er of + 2 and + 1, oxidation number of oxygen is – 2. This implies respectiv ely. The number as signed to that the oxidation number de note s the oxygen will depend upon the bonding state oxidation state of an element in a compound.  REDOX RE ACTIONS 26 1 Group 1 2 13 14 15 16 17 Element Na Mg Al Si P S Cl Compound Na Cl MgSO4 AlF3 SiCl4 P4O10 SF6 HClO4 Highest oxidation +1 +2 +3 +4 +5 +6 +7 number state of the group element The oxidation number/state of a metal in a The idea of oxidation number has been d compound is sometimes presented according invariab ly app lied to d ef ine oxid ation, to the notation given by German chemist, reduction, oxidising agent (oxidant), reducing Alfred Stock. It is popularly known as Stock he agent (reductant) and the redox reaction. To notation. According to this, the oxidation summarise, we may say that: number is expressed by putting a Roman Oxidation: An increas e in the oxidation numeral representing the oxidation number number of the element in the given substance. in parenthesis after the symbol of the metal in Reduction: A dec reas e in the oxidation is the molecular formula. Thus aurous chloride and auric chloride are written as Au(I)Cl and number of the element in the given substance. Au(III)Cl3. Similarly, stannous chloride and Oxidising agent : A re age nt w hic h c an bl stannic chloride are written as Sn(II)Cl2 and increase the oxidation number of an element Sn(IV)Cl4. This change in oxidation number in a given substance. These reagents are called implies change in oxidation state, which in as oxidants also. pu turn helps to identify whether the species is Reducing agent: A reagent which lowers the present in oxidised form or red uced form. oxidation number of an element in a given Thus, Hg2(I)Cl2 is the reduced form of Hg(II) Cl2. substance. These reagents are also called as be T Problem 8.3 reductants. re Using Stock notation, represe nt the Redox reactions: Reactions which involve o R following compounds :HAuCl4, Tl2O, FeO, change in oxidation number of the interacting Fe2O3, CuI, CuO, MnO and MnO2. species. Solution tt E Problem 8.4 By applying various rules of calculating Justify that the reaction: the oxidation numb er of the desired C element in a compound, the oxidation 2Cu2O(s) + Cu2S(s) ® 6Cu(s) + SO2(g) number of each metallic element in its is a redox reaction. Identify the species compound is as follows: oxidised/ reduced, which acts as an no N HAuCl4 ® Au has 3 oxidant and which acts as a reductant. Tl2O ® Tl has 1 Solution FeO ® Fe has 2 Let us assign oxidation number to each © Fe2O3 ® Fe has 3 of the sp ecies in the reaction under CuI ® Cu has 1 examination. This results into: CuO ® Cu has 2 +1 –2 +1 –2 0 +4 –2 MnO ® Mn has 2 2Cu2O(s) + Cu2S(s) ® 6Cu(s) + SO2 MnO2 ® Mn has 4 We therefor e, conc lude that in this Therefore, these compounds may be reaction copper is reduced from +1 state represented as: to zero oxidation state and sulphur is HAu(III)Cl4, Tl2(I)O, Fe(II)O, Fe2(III)O3, oxidised from –2 state to +4 state. The Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O2. above reaction is thus a redox reaction.  26 2 CHE MIST RY Further, Cu2O helps sulphur in Cu2S to that all decomposition reactions are not redox increase its oxidation number, therefore, reactions. For example, decomp osition of Cu(I) is an oxidant; and sulphur of Cu2S calcium carbonate is not a redox reaction. helps copper both in Cu2S itself and Cu2O +2 + 4 –2 +2 –2 +4 –2 to d ec re as e its oxidation numbe r; CaCO3 (s) CaO(s) + CO2(g) therefore, sulphur of Cu2S is reductant. 3. Displacement reactions 8.3.1 Types of Redox Reactions In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an d 1. Combination reactions atom) of another element. It may be denoted A combination reaction may be denoted in the as: he manner: X + YZ ® XZ + Y A+ B ® C Displacement reactions fit into two categories: Either A and B or both A and B must be in the me tal d is p lac e me nt and non-me tal elemental form for such a reaction to be a redox displacement. is reaction. All combustion reactions, which make use of elemental dioxygen, as well as (a) Metal displacement : A metal in a other reactions involving elements other than compound can be displaced by another metal bl dioxygen, are redox reactions. Some important in the unc ombined state. We have already examples of this category are: discussed about this class of the reactions 0 0 +4 –2 under section 8.2.1. Metal displacement pu C(s) + O2 (g) CO2(g) (8.24) r e ac tions f ind many ap p lic ations in metallurgical processes in which pure metals 0 0 + 2 –3 are obtained from their compounds in ores. A be T 3Mg(s) + N2(g) Mg3N2(s) (8.25) few such examples are: re +2 +6 – 2 0 0 +2 +6 –2 –4 +1 0 +4 –2 +1 –2 CuSO4(aq) + Zn (s) ® Cu(s) + ZnSO4 (aq) o R CH4(g) + 2O2(g) CO2(g) + 2H2O (l) (8.29) 2. Decomposition reactions +5 –2 0 0 +2 –2 tt E Decomposition reactions are the opposite of V2O5 (s) + 5Ca (s) 2V (s) + 5CaO (s) c omb ination r e ac tions. P r e c is e ly, a (8.30) decomposition reaction leads to the breakdown C +4 –1 0 0 +2 –1 of a compound into two or more components at least one of which must be in the elemental TiCl4 (l) + 2Mg (s) Ti (s) + 2 MgCl2 (s) no N state. Examples of this class of reactions are: (8.31) +1 –2 0 0 +3 –2 0 +3 – 2 0 2H2O (l) 2H2 (g) + O2(g) (8.26) Cr2O3 (s) + 2 Al (s) Al2O3 (s) + 2Cr(s) (8.32) © +1 –1 0 0 2NaH (s) 2Na (s) + H2(g) (8.27) In each case, the reducing metal is a better reducing agent than the one that is being +1 +5 – 2 +1 –1 0 reduced which evidently shows more capability 2KClO3 (s) 2KCl (s) + 3O2(g) (8.28) to lose electrons as compared to the one that It may care fully be noted that there is no is reduced. change in the oxidation number of hydrogen (b) Non-metal displacement: The non-metal in methane under combination reactions and d is place me nt re d ox r e ac tions includ e that of potassium in potassium chlorate in hydrogen displacement and a rarely occurring reaction (8.28). This may also be noted here reaction involving oxygen displacement.  REDOX RE ACTIONS 26 3 All alkali metals and some alkaline earth order Zn> Cu>Ag. Like metals, activity series metals (Ca, Sr, and Ba) which are very good also exists for the halogens. The power of these reductants, will displace hydrogen from cold elements as oxidising agents decreases as we water. move down from fluorine to iodine in group 0 +1 –2 +1 –2 +1 0 17 of the p eriodic table. This implies that 2Na(s) + 2H2O(l) ® 2NaOH(aq) + H2(g) fluorine is so reactive that it can replace (8.33) chloride, bromide and iodide ions in solution. 0 +1 –2 +2 – 2 +1 0 In fact, fluorine is so reactive that it attacks Ca(s) + 2H2O(l) ® Ca(OH)2 (aq) + H2(g) water and displaces the oxygen of water : d (8.34) +1 –2 0 +1 –1 0 Less active metals such as magnesium and 2H2O (l) + 2F2 (g) ® 4HF(aq) + O2(g) (8.40) he iron react with steam to produce dihydrogen gas: It is for this reason that the displacement 0 +1 –2 +2 –2 +1 0 reactions of chlorine, bromine and iodine Mg(s) + 2H2O(l) Mg(OH)2(s) + H2(g) using fluorine are not generally carried out in (8.35) aqueous solution. On the other hand, chlorine is can displac e bromide and iodide ions in an 0 + 1 –2 +3 –2 0 aqueous solution as shown below: 2Fe(s) + 3H2O(l) Fe2O3(s) + 3H2(g) (8.36) 0 + 1 –1 +1 –1 0 bl Many metals, including those which do not Cl2 (g) + 2KBr (aq) ® 2 KCl (aq) + Br2 (l) react with cold water, are capable of displacing (8.41) hydrogen from acids. Dihydrogen from acids 0 +1–1 +1 –1 0 pu may even be produced by such metals which Cl2 (g) + 2KI (aq) ® 2 KCl (aq) + I2 (s) do not react with steam. Cadmium and tin are (8.42) the examples of such metals. A few examples As Br2 and I2 are coloured and dissolve in CCl4, be T for the displacement of hydrogen from acids can easily be identified from the colour of the solution. The above reactions can be written re are: o R 0 +1 –1 +2 –1 0 in ionic form as: Zn(s) + 2HCl(aq) ® ZnCl2(aq) + H2 (g) 0 –1 –1 0 (8.37) Cl2 (g) + 2Br– (aq) ® 2Cl– (aq) + Br2(l) (8.41a) tt E 0 +1 –1 +2 –1 0 0 –1 –1 0 Mg (s) + 2HCl (aq) ® MgCl2 (aq) + H2 (g) Cl2 (g) + 2I – (aq) ® 2Cl– (aq) + I2 (s) (8.42b) (8.38) C Reactions (8.41) and (8.42) form the basis 0 + 1 –1 +2 –1 0 of identifying Br– and I – in the laboratory Fe(s) + 2HCl(aq) ® FeCl2(aq) + H2(g) through the test popularly known as ‘Layer (8.39) no N Test’. It may not be out of place to mention Reactions (8.37 to 8.39) are used to here that bromine likewise can displace iodide prepare dihydrogen gas in the laboratory. ion in solution: Here, the reactivity of metals is reflected in the 0 –1 –1 0 © rate of hydrogen gas evolution, which is the Br2 (l) + 2I – (aq) ® 2Br– (aq) + I2 (s) (8.43) slowest for the least active metal Fe, and the fastest for the most reactive metal, Mg. Very The halogen displacement reactions have less active metals, which may occur in the a direct industrial application. The recovery native state such as silver (Ag), and gold (Au) of halogens from their halides requires an do not react even with hydrochloric acid. oxidation process, which is represented by: I n se ction (8.2.1 ) we hav e alre ad y 2X– ® X2 + 2e– (8.44) discussed that the metals – zinc (Zn), copper here X denotes a halogen element. Whereas (Cu) and silver (Ag) through tendency to lose chemical means are available to oxidise Cl –, electrons show their reducing activity in the Br– and I –, as fluorine is the strongest oxidising  26 4 CHE MIST RY – agent; there is no way to convert F ions to F2 fluorine shows deviation from this behaviour by chemical means. The only way to achieve when it reacts with alkali. The reaction that F2 from F– is to oxidise electrolytic ally, the takes place in the case of fluorine is as follows: details of which you will study at a later stage. 2 F2(g) + 2OH– (aq) ® 2 F – (aq) + OF2(g) + H2O(l) 4. Disproportionation reactions (8.49) Disproportionation reactions are a special type (It is to be noted with care that fluorine in of redox re actions. In a disprop ortionation reaction (8.49) will undoubtedly attack water reaction an element in one oxidation state is to produce some oxygen also). This departure d simultaneously oxidised and reduced. One of shown by fluorine is not surprising for us as the r e ac ting s ub s tanc e s in a we know the limitation of fluorine that, being disproportionation reaction always contains the most electronegative element, it cannot he an element that can exist in at least three exhibit any positive oxidation state. This oxidation states. The element in the form of means that among halogens, fluorine does not reacting substance is in the intermediate show a disproportionation tendency. oxidation s tate; and both highe r and lower Problem 8.5 is oxidation states of that element are formed in the reaction. The decomposition of hydrogen Which of the following species, do not peroxide is a familiar example of the reaction, show disp roportionation reaction and bl where oxygen experiences disproportionation. why ? – – – – +1 –1 +1 –2 0 ClO , ClO2 , ClO3 and ClO4 2H2O2 (aq) ® 2H2O(l) + O2(g) (8.45) Also write reaction for each of the species pu Here the oxygen of peroxide, which is present that disproportionates. in –1 state, is converted to zero oxidation state Solution be T in O2 and decreases to –2 oxidation state in Among the oxoanions of chlorine listed H2O. above, ClO4– does not dis proportionate re Phos phorous, s ulp hur and c hlorine because in this oxoanion chlorine is o R undergo dis proportionation in the alkaline present in its highest oxidation state that medium as shown below : is, +7. The disproportionation reactions for the other three oxoanions of chlorine tt E 0 –3 +1 P4(s) + 3OH– (aq)+ 3H2O(l) ® PH3(g) + 3H2PO2– are as follows: (aq) +1 –1 +5 C – – – (8.46) 3ClO ® 2Cl + ClO3 0 –2 +2 +3 +5 –1 S8(s) + 12 OH– (aq) ® 4S2– (aq) + 2S2O32– (aq) 6 ClO2 – 4ClO3 + 2Cl – – no N + 6H2O(l) +5 –1 +7 (8.47) – – – 4ClO3 ® Cl + 3 ClO4 0 +1 –1 Cl2 (g) + 2 OH– (aq) ® ClO– (aq) + Cl– (aq) + Problem 8.6 © H2O (l) Suggest a scheme of classification of the (8.48) following redox reactions The reaction (8.48) describes the formation (a) N2 (g) + O2 (g) ® 2 NO (g) of hous e hold b le ac hing age nts. T he hypochlorite ion (ClO– ) formed in the reaction (b) 2Pb(NO3)2(s) ® 2PbO(s) + 2 NO2 (g) + oxidises the colour-bearing stains of the ½ O2 (g) substances to colourless compounds. (c) NaH(s) + H2O(l) ® NaOH(aq) + H2 (g) – – It is of interest to mention here that whereas (d) 2NO2(g) + 2OH (aq) ® NO2(aq) + – bromine and iodine follow the same trend as NO3 (aq)+H2O(l) exhib ited by chlorine in reac tion (8. 48),  REDOX RE ACTIONS 26 5 Solution (c), hydrogen of water has been displaced In reaction (a), the compound nitric oxide by hydrid e ion into dihydr ogen gas. is formed by the c ombination of the T he r ef ore , this may b e calle d as ele mental subs tances , nitrogen and displacement redox reaction. The reaction oxygen; therefore, this is an example of (d) involv es disproportionation of NO2 c ombination r ed ox r e ac tions. The (+4 state) into NO2– (+3 state ) and NO3– reaction (b) involves the breaking down (+5 state). Therefore reaction (d) is an of lead nitrate into three c omponents; ther efore, this is categorised under examp le of disproportionation redox d decomposition redox reaction. In reaction reaction. he The Paradox of Fractional Oxidation Number Sometimes, we come across with certain compounds in which the oxidation nu mber of a particular element in th e compound is in fraction. Examples are: C3O2 [where oxidation number of carbon is (4/3)], is Br3O8 [where o xidation number of bromine is (16/3)] and Na2S 4O6 (where oxidation number of sulphur is 2.5). We know t hat the idea of fractional oxidat ion number is un convincing to us, because bl electron s are never shared/transferred in fraction. Act ually this fract ional oxidation state is the avera ge oxidation state of the element un der examination an d the structural parameters reveal t hat the element for whom fractio nal oxidation state is realised is present in different pu 2– oxidatio n states. Structure of the species C3O2, Br3O8 and S 4O6 reveal the fo llowing bonding situations: +2 0 +2 be T O = C = C*= C = O Structure of C3O2 re (carbon suboxide) o R tt E Structu re of Br3O8 (tribromo octaoxide) Structu re of S 4O62– (tetrathionate ion) The element marked with asterisk in each species is exhibiting the different oxidation C state (oxidation number) fro m rest of the atoms of the same element in each of th e species. This reveals that in C3O2, two carbon atoms are present in +2 oxidation state each , whereas the third one is present in zero oxidation state and th e average is 4/3. However, the realistic picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br3O8, each no N of the tw o terminal bromine atoms are present in +6 oxidation state and the middle bromine is present in +4 oxidation state. Once again t he average, that is different from reality, is 2– 16/3. In the same fash ion, in the species S 4O6 , each o f the two extreme sulphurs exhibits oxidation state of +5 and t he two middle sulphurs as zero. The average of four oxidation © numbers of sulphurs of the S 4O62– is 2.5, wh ereas the reality being + 5,0,0 and +5 oxidation number respectively for each sulphur. We may th us, in general, conclude that the idea of fractional oxidation state should be taken with care and the reality is revealed by the structures only. Fu rther, whenever we come across with fractional oxidation state of an y particular element in any species, we must understa nd that this is the average oxidation number o nly. In reality (revealed by st ructures only), the element in that particular species is present in more than one whole number oxidation states. Fe3O4, Mn 3O4, Pb3O4 are some of the other examples of the compounds, which are mixed oxides, where we come across with fractional oxidation states of the metal atom. However, the oxidation states may be in fraction as in O2+ and O2– where it is +½ and –½ respectively.  26 6 CHE MIST RY Problem 8.7 (a) Oxidation Number Method: In writing equations for oxidation-reduction reactions, Why do the following reactions proceed just as for other reactions, the compositions differently ? and f or mulas must b e know n for the Pb3O4 + 8HCl ® 3PbCl2 + Cl2 + 4H2O substances that react and for the products that and are formed. The oxidation number method is Pb3O4 + 4HNO3 ® 2Pb(NO3)2 + PbO2 + now best illustrated in the following steps: 2H2O Step 1: Write the correct formula for each Solution reactant and product. d Pb3 O4 is ac tually a s toic hiometric Step 2: Identify atoms which undergo change mixture of 2 mol of PbO and 1 mol of in oxid ation number in the reac tion by he PbO2. In PbO2, lead is pr esent in +4 assigning the oxidation number to all elements oxidation s tate , whe reas the stable in the reaction. oxidation state of lead in PbO is +2. PbO2 Step 3: Calculate the increase or decrease in thus can act as an oxidant (oxidising the oxidation number per atom and for the is agent) and, therefore, can oxidise Cl – ion entire molecule/ion in which it occurs. If these of HCl into chlorine. We may also keep in are not equal the n multiply by suitable mind that PbO is a basic oxide. Therefore, number so that these become equal. (If you bl the reaction realise that two substances are reduced and Pb3O4 + 8HCl ® 3PbCl2 + Cl2 + 4H2O nothing is oxidised or vice-versa, something can be splitted into two reactions namely: is wrong. Either the formulas of reactants or pu products are wrong or the oxidation numbers 2PbO + 4HCl ® 2PbCl2 + 2H2O have not been assigned properly). (acid-base reaction) Step 4: Ascertain the involvement of ions if be T +4 –1 +2 0 the reaction is taking place in water, add H+ or PbO2 + 4HCl ® PbCl2 + Cl2 +2H2O re OH – ions to the expression on the appropriate (redox reaction) o R side so that the total ionic charges of reactants Since HNO3 itself is an oxidising agent and products are equal. If the reaction is therefore, it is unlikely that the reaction carried out in acidic solution, use H+ ions in tt E may occur b etween PbO2 and HNO3. the equation; if in basic solution, use OH– ions. However, the acid-base reaction occurs Step 5 : Make the numbers of hydrogen atoms between PbO and HNO3 as: C in the expression on the two sides equal by 2PbO + 4HNO3 ® 2Pb(NO3)2 + 2H2O adding water (H2O) molecules to the reactants It is the passive nature of PbO2 against or products. Now, also check the number of no N HNO3 that makes the reaction different oxygen atoms. If there are the same number from the one that follows with HCl. of oxyge n atoms in the reactants and products, the equation then represents the 8.3.2 Balancing of Redox Reactions balanced redox reaction. © Two methods are used to balance chemical Let us now explain the steps involved in equations for redox processes. One of these the method with the help of a few problems methods is bas ed on the change in the given below: oxidation number of reducing agent and the oxidising agent and the other method is based Problem 8.8 on splitting the redox reaction into two half Write the net ionic equation f or the reactions — one involving oxidation and the reaction of potassium dichromate(VI), other involving reduction. Both these methods K2Cr2O7 with sodium sulphite, Na2SO3, are in use and the choice of their use rests with in an acid solution to give chromium(III) the individual using them. ion and the sulphate ion.  REDOX RE ACTIONS 26 7 Solution the oxidant and bromide ion is the Step 1: The skeletal ionic equation is: reductant. 2– 2– 3+ Cr2O7 (aq) + SO3 (aq) ® Cr (aq) Step 3: Calc ulate the inc re ase and 2– decrease of oxidation number, and make + SO4 (aq) the increase equal to the decrease. Step 2: Assign oxidation numbers for +7 –1 +4 +5 Cr and S – – – 2MnO4(aq)+Br (aq) ® 2MnO2(s)+BrO3(aq) +6 –2 +4 –2 +3 +6 –2 2– 2– Cr2O7 (aq) + SO3 (aq) ® Cr(aq)+SO4 (aq) 2– Step 4: As the reaction occurs in the basic d medium, and the ionic charges are not This indicates that the dichromate ion is equal on both sides, add 2 OH– ions on the oxidant and the sulphite ion is the he the right to make ionic charges equal. reductant. – – 2MnO4 (aq) + Br (aq) ® 2MnO2(s) + Step 3: Calc ulate the inc re ase and – – decrease of oxidation number, and make BrO3 (aq) + 2OH (aq) them equal: Step 5: Finally, count the hydrogen is +6 –2 +4 –2 +3 atoms and add appr opriate number of Cr2O72– (aq) + 3SO32– (aq) ® 2Cr3+ (aq) + water molecules (i.e. one H2O molecule) +6 –2 on the left side to achieve balanced redox bl 3SO42– (aq) change. – – Step 4: As the reaction occurs in the 2MnO4 (aq) + Br (aq) + H2O(l) ® 2MnO2(s) – – acidic me dium, and further the ionic + BrO3 (aq) + 2OH (aq) pu charges are not equal on both the sides, (b) Half Reaction Method: In this method, add 8H+ on the left to make ionic charges the two half equations are balanced separately be T equal and then added together to giv e balanced 2– 2– + 3+ Cr2O7 (aq) + 3SO3 (aq)+ 8H ® 2Cr (aq) equation. re 2– + 3SO4 (aq) o R Suppose we are to b alance the equation Step 5: Finally, count the hydrogen showing the oxidation of Fe2+ ions to Fe3+ions atoms, and add appropriate number of by dichromate ions (Cr2O7)2– in acidic medium, tt E water molecules (i.e., 4H2O) on the right wherein, Cr2O72– ions are reduced to Cr3+ ions. to achieve balanced redox change. The following steps are involved in this task. 2– 2– + Cr2O7 (aq) + 3SO3 (aq)+ 8H (aq) ® Step 1: Produce unbalanced equation for the C 3+ 2– 2Cr (aq) + 3SO4 (aq) +4H2O (l) reaction in ionic form : 2+ 2– 3+ 3+ Problem 8.9 Fe (aq) + Cr2O7 (aq) ® Fe (aq) + Cr (aq) no N (8.50) Permanganate ion reacts with bromide ion Step 2: Se parate the equation into half- in basic medium to give manganese reactions: dioxide and br omate ion. Write the +2 +3 balanced ionic equation for the reaction. © 2+ Oxidation half : Fe (aq) ® Fe3+(aq) (8.51) Solution +6 –2 +3 Step 1 : The skeletal ionic equation is : 2– Reduction half : Cr2O7 (aq) ® Cr (aq) 3+ – – – MnO4 (aq) + Br (aq) ® MnO2(s) + BrO3 (aq) (8.52) Step 2 : Assign oxidation numbers for Step 3: Balance the atoms other than O and Mn and Br H in each half reaction individually. Here the oxidation half reaction is already balanced with +7 –1 +4 +5 – – – respect to Fe atoms. For the red uction half MnO4 (aq) + Br (aq) ®MnO2 (s) + BrO3 (aq) reaction, we multiply the Cr3+ by 2 to balance this indicates that permanganate ion is Cr atoms.  26 8 CHE MIST RY 2– 3+ Cr2O7 (aq) ® 2 Cr (aq) (8.53) Problem 8.10 Step 4: For re actions occ urring in acidic – Permanganate(V II) ion, MnO4 in basic medium, add H2O to balance O atoms and H+ solution oxidises iodide ion, I– to produce to balance H atoms. molecular iodine (I2) and manganese (IV) Thus, we get : oxide (MnO2). Write a balanced ionic 2– + 3+ Cr2O7 (aq) + 14H (aq) ® 2 Cr (aq) + 7H2O (l) equation to represent this redox reaction. (8.54) Solution Step 1: First we write the skeletal ionic d Step 5: Add electrons to one side of the half reaction to balance the charges. If need be, equation, which is – make the number of electrons equal in the two MnO4 (aq) + I – (aq) ® MnO2(s) + I2(s) he half reactions by multiplying one or both half Step 2: The two half-reactions are: reactions by appropriate number. –1 0 The oxidation half reaction is thus rewritten Oxidation half : I – (aq) ® I2 (s) to balance the charge: is +7 +4 Fe2+ (aq) ® Fe3+ (aq) + e– (8.55) – Reduction half: MnO4 (aq) ® MnO2(s) Now in the reduction half reaction there are Step 3: To balance the I atoms in the bl net twelve positive charges on the left hand side oxidation half reaction, we rewrite it as: and only six positive charges on the right hand 2I – (aq) ® I2 (s) side. Therefore, we add six electrons on the left pu side. Step 4: To balance the O atoms in the 2– + – 3+ reduction half reaction, we add two water Cr2O7 (aq) + 14H (aq) + 6e ® 2Cr (aq) + molecules on the right: be T 7H2O (l) (8.56) MnO4– (aq) ® MnO2 (s) + 2 H2O (l) To equalise the number of electrons in both To balance the H atoms, we add four H+ re o R the half reactions, we multiply the oxidation ions on the left: half reaction by 6 and write as : – MnO4 (aq) + 4 H+ (aq) ® MnO2(s) + 2H2O (l) 6Fe2+ (aq) ® 6Fe3+(aq) + 6e– (8.57) As the reaction takes place in a basic tt E Step 6: We add the two half reactions to solution, therefore, for four H+ ions, we achieve the overall reaction and cancel the add four OH – ions to both sides of the C electrons on each side. This gives the net ionic equation: equation as : – MnO4 (aq) + 4H+ (aq) + 4OH– (aq) ® 2+ 2– + 3+ 6Fe (aq) + Cr2O7 (aq) + 14H (aq) ® 6 Fe (aq) + MnO2 (s) + 2 H2O(l) + 4OH – (aq) no N 3+ 2Cr (aq) + 7H2O(l) (8.58) Replacing the H+ and OH– ions with water, Step 7: Verify that the equation contains the the resultant equation is: same type and number of atoms and the same MnO4– (aq) + 2H2O (l) ® MnO2 (s) + 4 OH– (aq) charges on both sides of the equation. This last © Step 5 : In this step we balance the che ck rev eals that the equation is fully charges of the two half-reactions in the balanced with respect to number of atoms and manner depicted as: the charges. – 2I (aq) ® I2 (s) + 2e– For the reaction in a basic medium, first – MnO4 (aq) + 2H2O(l) + 3e– ® MnO2(s) balance the atoms as is done in acidic medium. – Then for each H+ ion, add an equal number of + 4OH (aq) – OH ions to both sides of the equation. Where Now to equalise the number of electrons, – H+ and OH appear on the s ame side of the we multiply the oxidation half-reaction by equation, combine these to give H2O. 3 and the reduction half-reaction by 2.  REDOX RE ACTIONS 26 9 – – 6I (aq) ® 3I2 (s) + 6e (iii) There is yet another method which is – – interesting and quite common. Its use is 2 MnO4 (aq) + 4H2O (l) +6e ® 2MnO2(s) – restricted to– those reagents which are able + 8OH (aq) to oxidise I ions, say, for example, Cu(II): Step 6: Add two half-reactions to – 2Cu2+(aq) + 4I (aq) ® Cu2I2(s) + I2(aq) (8.59) obtain the net reactions after cancelling electrons on both sides. This method relies on the facts that iodine – – itself gives an intense blue colour with starch 6I (aq) + 2MnO4(aq) + 4H2O(l) ® 3I2(s) + – and has a ve ry s pec if ic r eaction with 2MnO2(s) +8 OH (aq) thiosulphate ions (S2O32– ), which too is a redox d Step 7: A final verification shows that reaction: the equation is balanced in respect of the 2– I2(aq) + 2 S2O3 (aq)®2I–(aq) + S4O62– (aq) (8.60) he number of atoms and charges on both sides. I2, though insoluble in water, remains in solution containing KI as KI3. 8.3.3 Redox Reactions as the Basis for On addition of starch after the liberation of Titrations iodine from the reaction of Cu2+ ions on iodide is In acid-base systems we come across with a ions, an intense blue colour ap pears. This titration method for finding out the strength colour disappears as soon as the iodine is bl of one solution against the other using a pH consumed by the thiosulphate ions. Thus, the sensitive indicator. Similarly, in redox systems, end-point can easily be tracked and the rest the titration method can b e adopted to is the stoichiometric calculation only. pu determine the strength of a reductant/oxidant 8.3.4 Limitations of Concept of Oxidation using a redox sensitive indicator. The usage Number of indicator s in redox titration is illustrated As you have observed in the above discussion, be T below: the concep t of redox processe s has been (i) In one situation, the reagent itself is re evolving with time. This process of evolution o R intensely coloured, e.g., permanganate ion, is continuing. In fact, in rece nt past the MnO–4. Here MnO4– acts as the self indicator. oxidation process is visualised as a decrease The v isible end point in this case is in electron density and reduction process as achieved after the last of the reductant (Fe2+ tt E an increase in electron density around the or C2O42– ) is oxidised and the first lasting atom(s) involved in the reaction. tinge of pink colour appear s at MnO4– C concentration as low as 10–6 mol dm–3 8.4 REDOX REACTIONS AND ELECTRODE (10 –6 mol L –1). This ensur es a minimal PROCESSES ‘ov e r s hoot’ in c olour b e yond the The experiment corresponding to reaction no N equivalenc e point, the point where the (8.15), can also be observed if zinc rod is reductant and the oxidant ar e equal in dipped in copper sulphate solution. The redox terms of their mole stoichiometry. reaction takes place and during the reaction, (ii) If there is no dramatic auto-colour change zinc is oxidised to zinc ions and copper ions © (as with MnO –4 titr ation), ther e ar e are reduced to metallic copper due to direct indicators which are oxidised immediately transfer of electrons from zinc to copper ion. af ter the las t b it of the r eac tant is During this reaction heat is also evolved. Now consumed, producing a dramatic colour we modify the experiment in such a manner change. The best example is afforded by that for the same redox reaction transfer of – Cr2O27 , which is not a self-indicator, but e le ctrons take s p lace indire ctly. T his oxid is e s the indic ator s ub s tanc e necessitates the separation of zinc metal from diphenylamine just after the equivalence copper sulp hate solution. We take copper point to produce an intense blue colour, sulphate solution in a beaker and put a copper thus signalling the end point. strip or rod in it. We also take zinc sulphate  27 0 CHE MIST RY solution in another beaker and put a zinc rod jelly like substance). This provides an electric or strip in it. Now reaction takes place in either contact betwe en the two solutions without of the beakers and at the interface of the metal allowing them to mix w ith each other. The and its salt solution in each beaker both the zinc and copper rods are connected by a metallic reduced and oxidized forms of the same wire with a provision for an ammeter and a species are present. These rep resent the switch. The set-up as shown in Fig.8.3 is known species in the reduction and oxidation half as Daniell cell. When the switch is in the off reactions. A redox couple is defined as having position, no reaction takes place in either of together the oxidised and reduced forms of a the beakers and no current flows through the d substance taking part in an oxidation or metallic wire. As soon as the switch is in the reduction half reaction. on p os ition, w e mak e the f ollow ing he This is re presented by separ ating the observations: oxidised f orm fr om the reduced form by a 1. The transfer of electrons now does not take ver tical line or a slash re prese nting an place directly from Zn to Cu2+ but through interface (e.g. solid/solution). For example the metallic wire connecting the two rods as is apparent from the arrow which is in this experiment the two redox couples are represented as Zn2+/Zn and Cu2+/Cu. In both indicates the flow of current. cases, oxidised form is put before the reduced 2. The electricity from solution in one beaker bl form. Now we put the b eaker containing to solution in the other beaker flows by the copp er sulphate s olution and the be aker migration of ions through the salt bridge. containing zinc sulphate solution side by side We know that the flow of current is possible (Fig. 8.3). We connect s olutions in two only if there is a potential difference pu beakers by a salt bridge (a U-tube containing between the copper and zinc rods known a s olution of p otas s ium c hlorid e or as electrodes here. be T ammonium nitrate usually solidifie d b y The potential associated with each boiling with agar agar and later cooling to a electrode is known as electrode potential. If re the concentration of each species taking part o R in the electrode reaction is unity (if any gas appears in the electrode reaction, it is confined to 1 atmospheric pres sure) and further the tt E reaction is car ried out at 298K, then the potential of each electrode is said to be the C Standard El ectr o de Po tential. By convention, the standard electrode potential 0 (E ) of hydrogen electrode is 0.00 volts. The no N electrode potential value for each electrode process is a measure of the relative tendency of the active species in the process to remain 0 in the oxidised/reduced form. A negative E means that the redox couple is a stronger © reducing agent than the H+/H2 couple. A 0 positive E means that the redox couple is a Fig.8.3 The set-up for Daniell cell. El ectrons weaker reducing agent than the H+/H2 couple. produced at the anode due to oxidation The standar d electrode potentials are very of Zn travel through the external circuit to the cathode where these reduce the important and we can get a lot of other useful copp er ions. The circuit is completed information from them. The values of standard inside the cell by the migrati on of ions electrode potentials for some selected electrode through the salt bridge. It may be noted processes (reduction reactions) are given in that the di rection of current i s opposite Table 8.1. You will learn more about electrode to the di rection of electron flow. reactions and cells in Class XII. C:\Chemistry XI\Unit-8\Unit-8(5)(reprint).pmd 27.7.6, 16.10.6 (reprint)  REDOX RE ACTIONS 27 1 Table 8.1 The Standard Ele ctrode Potentials at 298 K Ions are present as aqueous species and H2O as liquid; gases and solids are shown by g and s respectively. 0 ® Reduced f orm) – Reaction (Oxidised form + ne E /V F2(g) + 2e– ® 2F – 2.87 Co 3+ + e– ® Co 2+ 1.81 H2O2 + 2H+ + 2e– ® 2H2O 1.78 ® Mn 2+ + 4H2O d MnO4– + 8H+ + 5e– 1.51 Au3+ + 3e– ® Au(s) 1.40 he Cl2(g) + 2e– ® 2Cl– 1.36 Cr2O72– + 14H+ + 6e– ® 2Cr3+ + 7H2O 1.33 O2(g) + 4H+ + 4e– ® 2H2O 1.23 MnO2(s) + 4H+ + 2e– ® Mn 2+ + 2H2O 1.23 is Br2 + 2e– ® 2Br– 1.09 Increasing strength of reducing agent Increasing strength of oxidising agent NO3– + 4H+ + 3e– ® NO(g) + 2H2O 0.97 2Hg2+ + 2e– ® Hg22+ 0.92 bl Ag+ + e– ® Ag(s) 0.80 Fe3+ + e– ® Fe2+ 0.77 ® H2O2 pu O2(g) + 2H+ + 2e– 0.68 I2(s) + 2e– ® 2I– 0.54 Cu+ + e– ® Cu(s) 0.52 be T Cu2+ + 2e– ® Cu(s) 0.34 ® Ag(s) + Cl – re AgCl(s) + e– 0.22 o R AgBr(s) + e– ® Ag(s) + Br – 0.10 2H+ + 2e– ® H2(g) 0.00 Pb2+ + 2e– ® Pb(s) –0.13 tt E Sn 2+ + 2e– ® Sn(s) –0.14 Ni2+ + 2e– ® Ni(s) –0.25 C Fe2+ + 2e– ® Fe(s) –0.44 Cr3+ + 3e– ® Cr(s) –0.74 Zn 2+ + 2e– ® Zn(s) –0.76 no N – 2H2O + 2e– ® H2(g) + 2OH –0.83 Al3+ + 3e– ® Al(s) –1.66 Mg2+ + 2e– ® Mg(s) –2.36 ® Na(s) © Na+ + e– –2.71 Ca2+ + 2e– ® Ca(s) –2.87 K+ + e– ® K(s) –2.93 Li+ + e– ® Li(s) –3.05 + 1. A negative E 0 means that the redox couple is a stronger reducing agent than the +H /H2 couple. 2. A positive E 0 means that the redox couple is a weaker reducing agent than the H /H2 couple.  27 2 CHE MIST RY SUMMARY Redox reactions form an important class of reactions in which oxidation and reduction occur simultaneously. Three tier conceptualisation viz, classical, electronic and oxidation number, which is usually available in the texts, has been presented in detail. Oxidation, reductio n, oxidising a gent (oxidant) and reducing agent (reduc tant) have been viewed according to each conceptualisation. Oxidation numbers are assigned in accordance with a co nsistent set of rules. Oxidation number and ion-electron method both are d useful means in writing equations for the redox reactions. Redox reactions are classified into four categories: combination, decomposition displacement a nd disproportionation reactio ns. The concept of redox c ouple and electrode processes is introduced here. he The redox reactions find wide applications in the study of electrode processes a nd cells. is EXERCISES 8.1 Assign oxidation number to t he underlined elements in each of the following bl species: (a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4 (e) CaO2 (f) NaBH4 (g) H2S 2O7 (h) KAl(SO4)2.12 H2O pu 8.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ? be T (a) KI3 (b) H2S 4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH 8.3 Justify th at the following reactions are redox reactions: re (a) CuO(s) + H2(g) ® Cu(s) + H2O(g) o R (b) Fe2O3(s) + 3CO(g) ® 2Fe(s) + 3CO2(g) (c) 4BCl3(g) + 3LiAlH4(s) ® 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s) tt E + – (d) 2K(s) + F2(g) ® 2K F (s) (e) 4 NH3(g) + 5 O2(g) ® 4NO(g) + 6H2O(g) C 8.4 Fluorine reacts with ice a nd results in th e change: H2O(s) + F2(g) ® HF(g) + HOF(g) no N Justify that this reaction is a redox reaction. 8.5 Calculate the oxidation nu mber of sulphur, chromium and nitro gen in H2SO5, 2– – Cr2O7 and NO3. Suggest structure of these compounds. Count for the fallacy. 8.6 Write formulas for the following compounds: © (a) Mercury(II) chloride (b) Nickel(II) sulphate (c) Tin(IV) oxide (d) Thallium(I) sulphate (e) Iron(III) sulphate (f) Chromiu m(III) oxide 8.7 Suggest a list of th e substances where carbon can exhibit o xidation states from –4 to +4 and nitro gen from –3 to +5. 8.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozon e and nitric acid act only as o xidants. Why ? 8.9 Consider the reactions: (a) 6 CO2(g) + 6H2O(l) ® C6 H12 O6(aq) + 6O2(g)  REDOX RE ACTIONS 27 3 (b) O3(g) + H2O2(l) ® H2O(l) + 2O2(g) Why it is more appropriate t o write these reactions as : (a) 6CO2(g) + 12H2O(l) ® C6 H12 O6(aq) + 6H2O(l) + 6O2(g) (b) O3(g) + H2O2 (l) ® H2O(l) + O2(g) + O2(g) Also suggest a technique to investigate the path of the above (a) and (b) redox react ions. 8.10 The compo und AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agen t. Why ? d 8.11 Whenever a reaction between an oxidisin g agent and a reducing agent is carried out, a compound of lower oxidation stat e is formed if the reducing agent is in he excess and a compound of higher oxidation state is formed if the oxidising agent is in ex cess. Justify th is statement giving three illust rations. 8.12 How do you co unt for the following observations ? (a) Tho u gh a lka line pot a ssiu m perma n ga n a te a nd a cidic po ta ssiu m is permangana te both are used a s oxidants, yet in the manufacture o f benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Writ e a balanced redox equation for the reaction. bl (b) Wh en con cen tra ted su lph uric a cid is added to an ino rga nic mixtu re containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromin e. Why ? 8.13 Identify the substance oxidised reduced, oxidising agent and reducing agent for pu each of the following reactions: (a) 2AgBr (s) + C6H6O2(aq) ® 2Ag(s) + 2HBr (aq) + C6H4O2(aq) be T + – – (b) HCHO(l) + 2[Ag (NH3)2] (aq) + 3OH (aq) ® 2Ag(s) + HCOO (aq) + 4NH3(aq) re + 2H2O(l) o R 2+ – – (c) HCHO (l) + 2 Cu (aq) + 5 OH (aq) ® Cu2O(s) + HCOO (aq) + 3H2O(l) (d) N2H4(l) + 2H2O2(l) ?