REDOX REACTIONS
REDOX REACTIONS263UNIT 8REDOX REACTIONSWhere there is oxidation, there is always reduction –Chemistry is essentially a study of redox systems.After studying this unit you will beable to identify redox reactions as a classof reactions in which oxidationand reduction reactions occursimultaneously; define the terms oxidation,reduction, oxidant (oxidisingagent) and reductant (reducingagent); explain mechanism of redoxreactions by electron transferprocess; use the concept of oxidationnumber to identify oxidant andreductant in a reaction; classify redox reaction intocombination(synthesis),decomposition, displacementanddisproportionationreactions; suggest a comparative orderamong various reductants andoxidants; balance chemical equationsusing (i) oxidation number(ii) half reaction method; learn the concept of redoxreactions in terms of electrodeprocesses.Chemistry deals with varieties of matter and change of onekind of matter into the other. Transformation of matter fromone kind into another occurs through the various types ofreactions. One important category of such reactions isRedox Reactions. A number of phenomena, both physicalas well as biological, are concerned with redox reactions.These reactions find extensive use in pharmaceutical,biological, industrial, metallurgical and agricultural areas.The importance of these reactions is apparent from the factthat burning of different types of fuels for obtaining energyfor domestic, transport and other commercial purposes,electrochemical processes for extraction of highly reactivemetals and non-metals, manufacturing of chemicalcompounds like caustic soda, operation of dry and wetbatteries and corrosion of metals fall within the purview ofredox processes. Of late, environmental issues likeHydrogen Economy (use of liquid hydrogen as fuel) anddevelopment of ‘Ozone Hole’ have started figuring underredox phenomenon.8.1 CLASSICAL IDEA OF REDOX REACTIONS –OXIDATION AND REDUCTION REACTIONSOriginally, the term oxidation was used to describe theaddition of oxygen to an element or a compound. Becauseof the presence of dioxygen in the atmosphere ( 20%),many elements combine with it and this is the principalreason why they commonly occur on the earth in theform of their oxides. The following reactions representoxidation processes according to the limited definition ofoxidation:(8.1)2 Mg (s) O2 (g) 2 MgO (s)(8.2)S (s) O2 (g) SO2 (g)2019-20
264CHEMISTRYIn reactions (8.1) and (8.2), the elementsmagnesium and sulphur are oxidised onaccount of addition of oxygen to them.Similarly, methane is oxidised owing to theaddition of oxygen to it.CH4 (g) 2O2 (g) CO2 (g) 2H2O (l)(8.3)A careful examination of reaction (8.3) inwhich hydrogen has been replaced by oxygenprompted chemists to reinterpret oxidation interms of removal of hydrogen from it and,therefore, the scope of term oxidation wasbroadened to include the removal of hydrogenfrom a substance. The following illustration isanother reaction where removal of hydrogencan also be cited as an oxidation reaction.(8.4)2 H2S(g) O2 (g) 2 S (s) 2 H2O (l)As knowledge of chemists grew, it wasnatural to extend the term oxidation forreactions similar to (8.1 to 8.4), which do notinvolve oxygen but other electronegativeelements. The oxidation of magnesium withfluorine, chlorine and sulphur etc. occursaccording to the following reactions :Mg (s) F2 (g) MgF2 (s)(8.5)Mg (s) Cl2 (g) MgCl2 (s)(8.6)Mg (s) S (s) MgS (s)(8.7)Incorporating the reactions (8.5 to 8.7)within the fold of oxidation reactionsencouraged chemists to consider not only theremoval of hydrogen as oxidation, but also theremoval of electropositive elements asoxidation. Thus the reaction :been broadened these days to include removalof oxygen/electronegative element from asubstance or addition of hydrogen/electropositive element to a substance.According to the definition given above, thefollowing are the examples of reductionprocesses:2 Hg (l) O2 (g)(8.8)2 HgO (s)(removal of oxygen from mercuric oxide )2 FeCl3 (aq) H2 (g) 2 FeCl2 (aq) 2 HCl(aq)(8.9)(removal of electronegative element, chlorinefrom ferric chloride)CH2 CH2 (g) H2 (g) H3C – CH3 (g) (8.10)(addition of hydrogen)2HgCl2 (aq) SnCl2 (aq) Hg2Cl2 (s) SnCl4 (aq)(8.11)(addition of mercury to mercuric chloride)In reaction (8.11) simultaneous oxidationof stannous chloride to stannic chloride is alsooccurring because of the addition ofelectronegative element chlorine to it. It wassoon realised that oxidation and reductionalways occur simultaneously (as will beapparent by re-examining all the equationsgiven above), hence, the word “redox” wascoined for this class of chemical reactions.2K4 [Fe(CN)6](aq) H2O2 (aq) 2K3[Fe(CN)6](aq) 2 KOH (aq)is interpreted as oxidation due to the removalof electropositive element potassium frompotassium ferrocyanide before it changes topotassium ferricyanide. To summarise, theterm “oxidation” is defined as the additionof oxygen/electronegative element to asubstance or removal of hydrogen/electropositive element from a substance.In the beginning, reduction wasconsidered as removal of oxygen from acompound. However, the term reduction has2019-20Problem 8.1In the reactions given below, identify thespecies undergoing oxidation andreduction:(i) H2S (g) Cl2 (g) 2 HCl (g) S (s)(ii) 3Fe3O4 (s) 8 Al (s) 9 Fe (s) 4Al2O3 (s)(iii) 2 Na (s) H2 (g) 2 NaH (s)Solution(i) H 2 S is oxidised because a moreelectronegative element, chlorine is addedto hydrogen (or a more electropositiveelement, hydrogen has been removedfrom S). Chlorine is reduced due toaddition of hydrogen to it.(ii) Aluminium is oxidised becauseoxygen is added to it. Ferrous ferric oxide
REDOX REACTIONS265For convenience, each of the aboveprocesses can be considered as two separatesteps, one involving the loss of electrons andthe other the gain of electrons. As anillustration, we may further elaborate one ofthese, say, the formation of sodium chloride.(Fe3O4) is reduced because oxygen hasbeen removed from it.(iii) With the careful application of theconcept of electronegativity only we mayinfer that sodium is oxidised andhydrogen is reduced.Reaction (iii) chosen here prompts us tothink in terms of another way to defineredox reactions. 2 Na(s) 2 Na (g) 2e–8.2 REDOX REACTIONS IN TERMS OFELECTRON TRANSFER REACTIONSWe have already learnt that the reactions(8.12)2Na(s) Cl2(g) 2NaCl (s)(8.13)4Na(s) O2(g) 2Na2O(s)(8.14)2Na(s) S(s) Na2S(s)are redox reactions because in each of thesereactions sodium is oxidised due to theaddition of either oxygen or moreelectronegative element to sodium.Simultaneously, chlorine, oxygen and sulphurare reduced because to each of these, theelectropositive element sodium has beenadded. From our knowledge of chemicalbonding we also know that sodium chloride,sodium oxide and sodium sulphide are ioniccompounds and perhaps better written as – 2– 2–Na Cl (s), (Na ) 2O (s), and (Na ) 2 S (s).Development of charges on the speciesproduced suggests us to rewrite the reactions(8.12 to 8.14) in the following manner :––Cl2(g) 2e 2 Cl (g)Each of the above steps is called a halfreaction, which explicitly shows involvementof electrons. Sum of the half reactions givesthe overall reaction : –2 Na(s) Cl2 (g) 2 Na Cl (s) or 2 NaCl (s)Reactions 8.12 to 8.14 suggest that halfreactions that involve loss of electrons arecalled oxidation reactions. Similarly, thehalf reactions that involve gain of electronsare called reduction reactions. It may notbe out of context to mention here that the newway of defining oxidation and reduction hasbeen achieved only by establishing acorrelation between the behaviour of speciesas per the classical idea and their interplay inelectron-transfer change. In reactions (8.12 to8.14) sodium, which is oxidised, acts asa reducing agent because it donates electronto each of the elements interacting with it andthus helps in reducing them. Chlorine, oxygenand sulphur are reduced and act as oxidisingagents because these accept electrons fromsodium. To summarise, we may mention thatOxidation : Loss of electron(s) by any species.Reduction: Gain of electron(s) by any species.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 redoxchange.SolutionSince in the above reaction the compoundformed is an ionic compound, which may –also be represented as Na H (s), thissuggests that one half reaction in thisprocess is : –2 Na (s) 2 Na (g) 2e2019-20
266CHEMISTRYand the other half reaction is:––H2 (g) 2e 2 H (g)This splitting of the reaction underexamination into two half reactionsautomatically reveals that here sodium isoxidised and hydrogen is reduced,therefore, the complete reaction is a redoxchange.8.2.1 Competitive Electron TransferReactionsPlace a strip of metallic zinc in an aqueoussolution of copper nitrate as shown in Fig. 8.1,for about one hour. You may notice that thestrip becomes coated with reddish metalliccopper and the blue colour of the solution2 disappears. Formation of Zn ions among theproducts can easily be judged when the blue2 colour of the solution due to Cu hasdisappeared. If hydrogen sulphide gas ispassed through the colourless solution2 containing Zn ions, appearance of white zincsulphide, ZnS can be seen on making thesolution alkaline with ammonia.The reaction between metallic zinc and theaqueous solution of copper nitrate is :2 2 (8.15)Zn(s) Cu (aq) Zn (aq) Cu(s)In reaction (8.15), zinc has lost electronsto form Zn2 and, therefore, zinc is oxidised.Evidently, now if zinc is oxidised, releasingelectrons, something must be reduced,accepting the electrons lost by zinc. Copperion is reduced by gaining electrons from the zinc.Reaction (8.15) may be rewritten as :At this stage we may investigate the stateof equilibrium for the reaction represented byequation (8.15). For this purpose, let us placea strip of metallic copper in a zinc sulphatesolution. No visible reaction is noticed and2 attempt to detect the presence of Cu ions bypassing H2S gas through the solution toproduce the black colour of cupric sulphide,CuS, does not succeed. Cupric sulphide hassuch a low solubility that this is an extremely2 sensitive test; yet the amount of Cu formedcannot be detected. We thus conclude that thestate of equilibrium for the reaction (8.15)greatly favours the products over the reactants.Let us extend electron transfer reaction nowto copper metal and silver nitrate solution inwater and arrange a set-up as shown inFig. 8.2. The solution develops blue colour due2 to the formation of Cu ions on account of thereaction:(8.16)Here, Cu(s) is oxidised to Cu 2 (aq) and Ag (aq) is reduced to Ag(s). Equilibrium greatlyfavours the products Cu2 (aq) and Ag(s).By way of contrast, let us also compare thereaction of metallic cobalt placed in nickelsulphate solution. The reaction that occurshere is :(8.17)Fig. 8.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker.2019-20
REDOX REACTIONS267Fig. 8.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker.At equilibrium, chemical tests reveal that both2 2 Ni (aq) and Co (aq) are present at moderateconcentrations. In this case, neither the2 reactants [Co(s) and Ni (aq)] nor the products2 [Co (aq) and Ni (s)] are greatly favoured.This competition for release of electronsincidently reminds us of the competition forrelease of protons among acids. The similaritysuggests that we might develop a table inwhich metals and their ions are listed on thebasis of their tendency to release electrons justas we do in the case of acids to indicate thestrength of the acids. As a matter of fact wehave already made certain comparisons. Bycomparison we have come to know that zincreleases electrons to copper and copperreleases electrons to silver and, therefore, theelectron releasing tendency of the metals is inthe order: Zn Cu Ag. We would love to makeour list more vast and design a metal activityseries or electrochemical series. Thecompetition for electrons between variousmetals helps us to design a class of cells,named as Galvanic cells in which the chemicalreactions become the source of electricalenergy. We would study more about these cellsin Class XII.However, as we shall see later, the chargetransfer is only partial and is perhaps betterdescribed as an electron shift rather than acomplete loss of electron by H and gain by O.What has been said here with respect toequation (8.18) may be true for a good numberof other reactions involving covalentcompounds. Two such examples of this classof the reactions are:(8.19)H2(s) Cl2(g) 2HCl(g)and,CH 4(g) 4Cl2(g) CCl4(l) 4HCl(g) (8.20)In order to keep track of electron shifts inchemical reactions involving formation ofcovalent compounds, a more practical methodof using oxidation number has beendeveloped. In this method, it is alwaysassumed that there is a complete transfer ofelectron from a less electronegative atom to amore electonegative atom. For example, werewrite equations (8.18 to 8.20) to showcharge on each of the atoms forming part ofthe reaction :00 1 –22H2(g) O2(g) 2H2O (l)00(8.21) 1 –1H2 (s) Cl2(g) 2HCl(g)8.3 OXIDATION NUMBERA less obvious example of electron transfer isrealised when hydrogen combines with oxygento form water by the reaction:2H2(g) O2 (g) 2H2O (l)(8.18)Though not simple in its approach, yet wecan visualise the H atom as going from aneutral (zero) state in H2 to a positive state inH2O, the O atom goes from a zero state in O2to a dinegative state in H2O. It is assumed thatthere is an electron transfer from H to O andconsequently H2 is oxidised and O2 is reduced.– 4 10 4 –1(8.22) 1 –1CH4(g) 4Cl2(g) CCl4(l) 4HCl(g)(8.23)It may be emphasised that the assumptionof electron transfer is made for book-keepingpurpose only and it will become obvious at alater stage in this unit that it leads to the simpledescription of redox reactions.Oxidation number denotes theoxidation state of an element in acompound ascertained according to a setof rules formulated on the basis that2019-20
268CHEMISTRYelectron pair in a covalent bond belongsentirely to more electronegative element.It is not always possible to remember ormake out easily in a compound/ion, whichelement is more electronegative than the other.Therefore, a set of rules has been formulatedto determine the oxidation number of anelement in a compound/ion. If two or morethan two atoms of an element are present in2–the molecule/ion such as Na2S2O3/Cr2O7 , theoxidation number of the atom of that elementwill then be the average of the oxidationnumber of all the atoms of that element. Wemay at this stage, state the rules for thecalculation of oxidation number. These rules are:1. In elements, in the free or the uncombinedstate, each atom bears an oxidationnumber of zero. Evidently each atom in H2,O2, Cl2, O3, P4, S8, Na, Mg, Al has theoxidation number zero.2. For ions composed of only one atom, theoxidation number is equal to the charge on the ion. Thus Na ion has an oxidation2 3 number of 1, Mg ion, 2, Fe ion, 3,–2–Cl ion, –1, O ion, –2; and so on. In theircompounds all alkali metals haveoxidation number of 1, and all alkalineearth metals have an oxidation number of 2. Aluminium is regarded to have anoxidation number of 3 in all itscompounds.3. The oxidation number of oxygen in mostcompounds is –2. However, we come acrosstwo kinds of exceptions here. One arisesin the case of peroxides and superoxides,the compounds of oxygen in which oxygenatoms are directly linked to each other.While in peroxides (e.g., H2O2, Na2O2), eachoxygen atom is assigned an oxidationnumber of –1, in superoxides (e.g., KO2,RbO2) each oxygen atom is assigned anoxidation number of –(½). The secondexception appears rarely, i.e. when oxygenis bonded to fluorine. In such compoundse.g., oxygen difluoride (OF2) and dioxygendifluoride (O2F2), the oxygen is assignedan oxidation number of 2 and 1,respectively. The number assigned tooxygen will depend upon the bonding stateof oxygen but this number would now bea positive figure only.4. The oxidation number of hydrogen is 1,except when it is bonded to metals in binarycompounds (that is compounds containingtwo elements). For example, in LiH, NaH,and CaH2, its oxidation number is –1.5. In all its compounds, fluorine has anoxidation number of –1. Other halogens (Cl,Br, and I) also have an oxidation numberof –1, when they occur as halide ions intheir compounds. Chlorine, bromine andiodine when combined with oxygen, forexample in oxoacids and oxoanions, havepositive oxidation numbers.6. The algebraic sum of the oxidation numberof all the atoms in a compound must bezero. In polyatomic ion, the algebraic sumof all the oxidation numbers of atoms ofthe ion must equal the charge on the ion.Thus, the sum of oxidation number of threeoxygen atoms and one carbon atom in the2–carbonate ion, (CO3) must equal –2.By the application of above rules, we canfind out the oxidation number of the desiredelement in a molecule or in an ion. It is clearthat the metallic elements have positiveoxidation number and nonmetallic elementshave positive or negative oxidation number.The atoms of transition elements usuallydisplay several positive oxidation states. Thehighest oxidation number of a representativeelement is the group number for the first twogroups and the group number minus 10(following the long form of periodic table) forthe other groups. Thus, it implies that thehighest value of oxidation number exhibitedby an atom of an element generally increasesacross the period in the periodic table. In thethird period, the highest value of oxidationnumber changes from 1 to 7 as indicated belowin the compounds of the elements.A term that is often used interchangeablywith the oxidation number is the oxidationstate. Thus in CO2, the oxidation state ofcarbon is 4, that is also its oxidation numberand similarly the oxidation state as well asoxidation number of oxygen is – 2. This impliesthat the oxidation number denotes theoxidation state of an element in a compound.2019-20
REDOX ClMgSO4AlF3SiCl4P4O10SF6HClO4Highest oxidationnumber state ofthe group element 1 2 3 4 5 6 7The oxidation number/state of a metal in acompound is sometimes presented accordingto the notation given by German chemist,Alfred Stock. It is popularly known as Stocknotation. According to this, the oxidationnumber is expressed by putting a Romannumeral representing the oxidation numberin parenthesis after the symbol of the metal inthe molecular formula. Thus aurous chlorideand auric chloride are written as Au(I)Cl andAu(III)Cl3. Similarly, stannous chloride andstannic chloride are written as Sn(II)Cl2 andSn(IV)Cl4. This change in oxidation numberimplies change in oxidation state, which inturn helps to identify whether the species ispresent in oxidised form or reduced form.Thus, Hg2(I)Cl2 is the reduced form of Hg(II) Cl2.Problem 8.3Using Stock notation, represent thefollowing compounds :HAuCl4, Tl2O, FeO,Fe2O3, CuI, CuO, MnO and MnO2.SolutionBy applying various rules of calculatingthe oxidation number of the desiredelement in a compound, the oxidationnumber of each metallic element in itscompound is as follows: Au has 3HAuCl4Tl2O Tl has 1FeO Fe has 2 Fe has 3Fe2O3CuI Cu has 1CuO Cu has 2MnO Mn has 2 Mn has 4MnO2Therefore, these compounds may berepresented as:HAu(III)Cl4, Tl2(I)O, Fe(II)O, Fe2(III)O3,Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O2.14151617The idea of oxidation number has beeninvariably applied to define oxidation,reduction, oxidising agent (oxidant), reducingagent (reductant) and the redox reaction. Tosummarise, we may say that:Oxidation: An increase in the oxidationnumber of the element in the given substance.Reduction: A decrease in the oxidationnumber of the element in the given substance.Oxidising agent: A reagent which canincrease the oxidation number of an elementin a given substance. These reagents are calledas oxidants also.Reducing agent: A reagent which lowers theoxidation number of an element in a givensubstance. These reagents are also called asreductants.Redox reactions: Reactions which involvechange in oxidation number of the interactingspecies.Problem 8.4Justify that the reaction:2Cu2O(s) Cu2S(s) 6Cu(s) SO2(g)is a redox reaction. Identify the speciesoxidised/reduced, which acts as anoxidant and which acts as a reductant.SolutionLet us assign oxidation number to eachof the species in the reaction underexamination. This results into: 1 –2 1 –20 4 –22Cu2O(s) Cu2S(s) 6Cu(s) SO2We therefore, conclude that in thisreaction copper is reduced from 1 stateto zero oxidation state and sulphur isoxidised from –2 state to 4 state. Theabove reaction is thus a redox reaction.2019-20
270CHEMISTRYthat all decomposition reactions are not redoxreactions. For example, decomposition ofcalcium carbonate is not a redox reaction.Further, Cu2O helps sulphur in Cu2S toincrease its oxidation number, therefore,Cu(I) is an oxidant; and sulphur of Cu2Shelps copper both in Cu2S itself and Cu2Oto decrease its oxidation number;therefore, sulphur of Cu2S is reductant. 2 4 –28.3.1 Types of Redox Reactions1. Combination reactionsA combination reaction may be denoted in themanner:A B CEither A and B or both A and B must be in theelemental form for such a reaction to be a redoxreaction. All combustion reactions, whichmake use of elemental dioxygen, as well asother reactions involving elements other thandioxygen, are redox reactions. Some importantexamples of this category are:00 4 –2C(s) O2 (g)0CO2(g)0 2 –33Mg(s) N2(g)–4 1(8.24)Mg3N2(s)0(8.25) 4 –2CH4(g) 2O2(g)2H2O (l) 1 –12NaH (s) 1 5 –202H2O (l)0(a) Metal displacement: A metal in acompound can be displaced by another metalin the uncombined state. We have alreadydiscussed about this class of the reactionsunder section 8.2.1. Metal displacementreactions find many applications inmetallurgical processes in which pure metalsare obtained from their compounds in ores. Afew such examples are: 5 –2 1 –10 2 6 –200 2 –22V (s) 5CaO (s)(8.30) 4 –100TiCl4 (l) 2Mg (s)0 2 –1Ti (s) 2 MgCl2 (s)(8.31) 3 –20(8.26)Cr2O3 (s) 2 Al (s)(8.27)In each case, the reducing metal is a betterreducing agent than the one that is beingreduced which evidently shows more capabilityto lose electrons as compared to the one thatis reduced.02Na (s) H2(g)0V2O5 (s) 5Ca (s) 3 –22H2 (g) O2(g)0Displacement reactions fit into two categories:metal displacement and non-metaldisplacement.CuSO4(aq) Zn (s) Cu(s) ZnSO4 (aq)(8.29)2. Decomposition reactionsDecomposition reactions are the opposite ofcombination reactions. Precisely, adecomposition reaction leads to the breakdownof a compound into two or more componentsat least one of which must be in the elementalstate. Examples of this class of reactions are: 1 –2 4 –2CaCO3 (s)CaO(s) CO2(g)3. Displacement reactionsIn a displacement reaction, an ion (or an atom)in a compound is replaced by an ion (or anatom) of another element. It may be denotedas:X YZ XZ Y 2 6 –2 1 –2CO2(g) 2 –202KClO3 (s)2KCl (s) 3O2(g)(8.28)It may carefully be noted that there is nochange in the oxidation number of hydrogenin methane under combination reactions andthat of potassium in potassium chlorate inreaction (8.28). This may also be noted hereAl2O3 (s) 2Cr(s)(8.32)(b) Non-metal displacement: The non-metaldisplacement redox reactions includehydrogen displacement and a rarely occurringreaction involving oxygen displacement.2019-20
REDOX REACTIONS271All alkali metals and some alkaline earthmetals (Ca, Sr, and Ba) which are very goodreductants, will displace hydrogen from coldwater.0 1 –22Na(s) 2H2O(l)0 1 –2 1 –2 10 2NaOH(aq) H2(g)(8.33) 2 –2 10Ca(s) 2H2O(l) Ca(OH)2 (aq) H2(g)(8.34)Less active metals such as magnesium andiron react with steam to produce dihydrogen gas:0 1 –2Mg(s) 2H2O(l)0 1 –22Fe(s) 3H2O(l) 2 –2 10Mg(OH)2(s) H2(g)(8.35) 3 –20Fe2O3(s) 3H2(g) (8.36)Many metals, including those which do notreact with cold water, are capable of displacinghydrogen from acids. Dihydrogen from acidsmay even be produced by such metals whichdo not react with steam. Cadmium and tin arethe examples of such metals. A few examplesfor the displacement of hydrogen from acidsare:0 1 –1 2 –10order Zn Cu Ag. Like metals, activity seriesalso exists for the halogens. The power of theseelements as oxidising agents decreases as wemove down from fluorine to iodine in group17 of the periodic table. This implies thatfluorine is so reactive that it can replacechloride, bromide and iodide ions in solution.In fact, fluorine is so reactive that it attackswater and displaces the oxygen of water : 1 –200 1 –1 2 –10 1 –1 2 –1 1 –10Fe(s) 2HCl(aq) FeCl2(aq) H2(g) 1 –1 1–10 1 –10Cl2 (g) 2KI (aq) 2 KCl (aq) I2 (s)(8.42)As Br2 and I2 are coloured and dissolve in CCl4,can easily be identified from the colour of thesolution. The above reactions can be writtenin ionic form as:–1–1–0–Cl2 (g) 2Br (aq) 2Cl (aq) Br2 (l) (8.41a)000(8.41)00Mg (s) 2HCl (aq) MgCl2 (aq) H2 (g)(8.38) 1 –1Cl2 (g) 2KBr (aq) 2 KCl (aq) Br2 (l)Zn(s) 2HCl(aq) ZnCl2 (aq) H2 (g)(8.37)02H2O (l) 2F2 (g) 4HF(aq) O2(g) (8.40)It is for this reason that the displacementreactions of chlorine, bromine and iodineusing fluorine are not generally carried out inaqueous solution. On the other hand, chlorinecan displace bromide and iodide ions in anaqueous solution as shown below:–1–1–0–Cl2 (g) 2I (aq) 2Cl (aq) I2 (s)(8.42b)Reactions (8.41) and (8.42) form the basisof identifying Br– and I – in the laboratorythrough the test popularly known as ‘LayerTest’. It may not be out of place to mentionhere that bromine likewise can displace iodideion in solution:(8.39)Reactions (8.37 to 8.39) are used toprepare dihydrogen gas in the laboratory.Here, the reactivity of metals is reflected in therate of hydrogen gas evolution, which is theslowest for the least active metal Fe, and thefastest for the most reactive metal, Mg. Veryless active metals, which may occur in thenative state such as silver (Ag), and gold (Au)do not react even with hydrochloric acid.The halogen displacement reactions havea direct industrial application. The recoveryof halogens from their halides requires anoxidation process, which is represented by:In section (8.2.1) we have alreadydiscussed that the metals – zinc (Zn), copper(Cu) and silver (Ag) through tendency to loseelectrons show their reducing activity in the2X X2 2e(8.44)here X denotes a halogen element. Whereas–chemical means are available to oxidise Cl ,––Br and I , as fluorine is the strongest oxidising0–1–1–0–Br2 (l) 2I (aq) 2Br (aq) I2 (s)2019-20––(8.43)
272CHEMISTRY–agent; there is no way to convert F ions to F2by chemical means. The only way to achieve–F2 from F is to oxidise electrolytically, thedetails of which you will study at a later stage.4. Disproportionation reactionsDisproportionation reactions are a special typeof redox reactions. In a disproportionationreaction an element in one oxidation state issimultaneously oxidised and reduced. One ofthereactingsubstancesinadisproportionation reaction always containsan element that can exist in at least threeoxidation states. The element in the form ofreacting substance is in the intermediateoxidation state; and both higher and loweroxidation states of that element are formed inthe reaction. The decomposition of hydrogenperoxide is a familiar example of the reaction,where oxygen experiences disproportionation. 1 –1 1 –2fluorine shows deviation from this behaviourwhen it reacts with alkali. The reaction thattakes place in the case of fluorine is as follows:––2 F2(g) 2OH (aq) 2 F (aq) OF2(g) H2O(l)(8.49)(It is to be noted with care that fluorine inreaction (8.49) will undoubtedly attack waterto produce some oxygen also). This departureshown by fluorine is not surprising for us aswe know the limitation of fluorine that, beingthe most electronegative element, it cannotexhibit any positive oxidation state. Thismeans that among halogens, fluorine does notshow a disproportionation tendency.02H2O2 (aq) 2H2O(l) O2(g)(8.45)Here the oxygen of peroxide, which is presentin –1 state, is converted to zero oxidation statein O2 and decreases to –2 oxidation state inH2O.Phosphorous, sulphur and chlorineundergo disproportionation in the alkalinemedium as shown below :0–3 1––P4(s) 3OH (aq) 3H2O(l) PH3(g) 3H2PO2(aq)(8.46)0–2–S8(s) 12 OH (aq) 4S02–– 1–1 32––1––3ClO 2(aq) 2S2O3 (aq) 6H2O(l)(8.47) 1Cl2 (g) 2 OH (aq) Problem 8.5Which of the following species, do notshow disproportionation reaction andwhy ?––––ClO , ClO2 , ClO3 and ClO4Also write reaction for each of the speciesthat disproportionates.SolutionAmong the oxoanions of chlorine listed–above, ClO4 does not disproportionatebecause in this oxoanion chlorine ispresent in its highest oxidation state thatis, 7. The disproportionation reactionsfor the other three oxoanions of chlorineare as follows:–ClO (aq) Cl (aq) H2O (l)(8.48)The reaction (8.48) describes the formationof household bleaching agents. The–hypochlorite ion (ClO ) formed in the reactionoxidises the colour-bearing stains of thesubstances to colourless compounds.It is of interest to mention here that whereasbromine and iodine follow the same trend asexhibited by chlorine in reaction (8.48),2019-206 5– 5–ClO2–4ClO3 5–4ClO 3–2Cl ClO 3–1 ––1 2Cl 7––Cl 3 ClO4Problem 8.6Suggest a scheme of classification of thefollowing redox reactions(a) N2 (g) O2 (g) 2 NO (g)(b) 2Pb(NO3)2(s) 2PbO(s) 4 NO2 (g) O2 (g)(c) NaH(s) H2O(l) NaOH(aq) H2 (g)––(d) 2NO2(g) 2OH (aq) NO2(aq) –NO3 (aq) H2O(l)
REDOX REACTIONS273SolutionIn reaction (a), the compound nitric oxideis formed by the combination of theelemental substances, nitrogen andoxygen; theref
reactions that involve loss of electrons are called oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions. It may not be out of context to mention here that the new way of defining oxidation and reduction has been achieved only by establishing a correlation between the behaviour of species
ii. acid–base neutralization reactions iii. oxidation–reduction or redox reactions. Q.3. What are the important aspects of redox reactions? Ans: Almost every element participate in redox reactions. The important aspects of redox reactions are as follows: i. Large number of natural, biological and industrial processes involve redox reactions .
CHAPTER 12: Redox Reactions . Goals of Chapter: - Understand redox reactions in detail - Review oxidation numbers - Learn electrochemical techniques Application of Redox Chemistry - extracting metals from ores, e.g. Need to learn to balance tricky redox reactions Cu 2CO 3(OH) 2(s) C(s) Æ2Cu(s) 2CO 2(g) H 2O(g .
The Major Classes of Chemical Reactions. 4.6 Elements in Redox Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions 4.4 Acid -Base Reactions. 4.5 Oxidation -Reduction (Redox) Reactions 4.7
The above example described a single-replacement reaction. All reactions of this kind are redox . A series of reactions are attempted. For each attempt, it is recorded whether a reaction occurs or not. . Worksheet 1. The following reactions were performed. Construct a redox table.
Special Topic 6.1: Oxidizing Agents and Aging 6.2 Oxidation Numbers Internet: Balancing Redox Reactions 6.3 Types of Chemical Reactions Combination Reactions Decomposition Reactions Combustion Reactions Special Topic 6.2: Air Pollution and Catalytic Converters Single-Displacement Reactions Internet: Single-Displacement Reaction 6.4 Voltaic Cells
Scaffold : Redox Reactions Example Worksheet ATTENTION: This worksheet will be used to play the Redox relay and is modified into cards To determine if a reaction is a REDOX reaction: 1. Assign oxidation numbers to each element in the reaction. 2. If any of the oxidation numbers change in the reaction then the reaction is REDOX. Practice:
Chemical Reactions Slide 3 / 142 Table of Contents: Chemical Reactions · Balancing Equations Click on the topic to go to that section · Types of Chemical Reactions · Oxidation-Reduction Reactions · Chemical Equations · Net Ionic Equations · Types of Oxidation-Reduction Reactions · Acid-Base Reactions · Precipitation Reactions
Most of these reactions can be classified into one of three main types of chemical reactions: precipitation reactions, acid-base neutralization reactions, and oxidation-reduction (also called “redox”) reactions. Aqueous Solutions(aq) Many reactions occur in an aqueous environment (i.e.,
