Redox Reactions PDF

redox reactions 235 Unit 7 redox reactions...

redox reactions 235 Unit 7 redox reactions Where there is oxidation, there is always reduction – Chemistry is essentially a study of redox systems. After studying this unit you will be able to Chemistry deals with varieties of matter and change of one kind of matter into the other. Transformation of identify redox reactions as a class of reactions in which oxidation matter from one kind into another occurs through the and reduction reactions occur various types of reactions. One important category of such simultaneously; reactions is Redox Reactions. A number of phenomena, define the terms oxidation, both physical as well as biological, are concerned with reduction, oxidant (oxidising redox reactions. These reactions find extensive use in agent) and reductant (reducing pharmaceutical, biological, industrial, metallurgical and agent); agricultural areas. The importance of these reactions is explain mechanism of redox apparent from the fact that burning of different types of reactions by electron transfer fuels for obtaining energy for domestic, transport and process; other commercial purposes, electrochemical processes use the concept of oxidation for extraction of highly reactive metals and non-metals, number to identify oxidant and manufacturing of chemical compounds like caustic reductant in a reaction; soda, operation of dry and wet batteries and corrosion of classify redox reaction into metals fall within the purview of redox processes. Of late, combination (synthesis), decomposition, displacement environmental issues like Hydrogen Economy (use of and disproportionation liquid hydrogen as fuel) and development of ‘Ozone Hole’ reactions; have started figuring under redox phenomenon. suggest a comparative order 7.1 CLASSICAL IDEA OF REDOX REACTIONS – among various reductants and oxidants; OXIDATION AND REDUCTION REACTIONS balance chemical equations Originally, the term oxidation was used to describe the using (i) oxidation number addition of oxygen to an element or a compound. Because (ii) half reaction method; of the presence of dioxygen in the atmosphere (~20%), lear n the concept of redox 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) (7.1) S (s) + O2 (g) → SO2 (g) (7.2) 2024-25 Unit 7.indd 235 10/10/2022 10:37:02 AM  236 chemistry In reactions (7.1) and (7.2), the elements broadened these days to include removal magnesium and sulphur are oxidised on of oxygen/electronegative element from account of addition of oxygen to them. a substance or addition of hydrogen/ Similarly, methane is oxidised owing to the electropositive element to a substance. addition of oxygen to it. According to the definition given above, CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) (7.3) the following are the examples of reduction processes: A careful examination of reaction (7.3) in which hydrogen has been replaced by oxygen 2 HgO (s) 2 Hg (l) + O2 (g) (7.8) prompted chemists to reinterpret oxidation (removal of oxygen from mercuric oxide ) in terms of removal of hydrogen from it and, 2 FeCl3 (aq) + H2 (g) →2 FeCl2 (aq) + 2 HCl(aq) therefore, the scope of term oxidation was (7.9) 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) (7.10) (addition of hydrogen) 2 H2S(g) + O2 (g) → 2 S (s) + 2 H2O (l) (7.4) 2HgCl2 (aq) + SnCl2 (aq) → Hg2Cl2 (s)+SnCl4 (aq) As knowledge of chemists grew, it was natural to extend the term oxidation for (7.11) reactions similar to (7.1 to 7.4), which do (addition of mercury to mercuric chloride) not involve oxygen but other electronegative In reaction (7.11) simultaneous oxidation elements. The oxidation of magnesium with of stannous chloride to stannic chloride is fluorine, chlorine and sulphur etc. occurs also occurring because of the addition of according to the following reactions : electronegative element chlorine to it. It was soon realised that oxidation and reduction Mg (s) + F2 (g) → MgF2 (s) (7.5) always occur simultaneously (as will be Mg (s) + Cl2 (g) → MgCl2 (s) (7.6) apparent by re-examining all the equations given above), hence, the word “redox” was Mg (s) + S (s) → MgS (s) (7.7) coined for this class of chemical reactions. Incorporating the reactions (7.5 to Problem 7.1 7.7) within the fold of oxidation reactions In the reactions given below, identify encouraged chemists to consider not only the species undergoing oxidation and the removal of hydrogen as oxidation, but reduction: also the removal of electropositive elements as oxidation. Thus the reaction : (i) H2S (g) + Cl2 (g) → 2 HCl (g) + S (s) 2K4 [Fe(CN)6](aq) + H2O2 (aq) →2K3[Fe(CN)6](aq) (ii) 3Fe3O4 (s) + 8 Al (s) → 9 Fe (s) + 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) H 2 S is oxidised because a more term “oxidation” is defined as the addition electronegative element, chlorine is added of oxygen/electronegative element to to hydrogen (or a more electropositive a substance or removal of hydrogen/ element, hydrogen has been removed electropositive element from a substance. from S). Chlorine is reduced due to addition of hydrogen to it. In the beginning, reduction was considered as removal of oxygen from a compound. (ii) Aluminium is oxidised because However, the term reduction has been oxygen is added to it. Ferrous ferric oxide 2024-25 Unit 7.indd 236 10/10/2022 10:37:03 AM  redox reactions 237 For convenience, each of the above (Fe3O4) is reduced because oxygen has processes can be considered as two separate been removed from it. steps, one involving the loss of electrons (iii) With the careful application of the and the other the gain of electrons. As an concept of electronegativity only we illustration, we may further elaborate one of may infer that sodium is oxidised and these, say, the formation of sodium chloride. hydrogen is reduced. 2 Na(s) → 2 Na+(g) + 2e– Reaction (iii) chosen here prompts us to think in terms of another way to define Cl2(g) + 2e– → 2 Cl–(g) redox reactions. Each of the above steps is called a half reaction, which explicitly shows involvement 7.2 REDOX REACTIONS IN TERMS OF of electrons. Sum of the half reactions gives ELECTRON TRANSFER REACTIONS the overall reaction : We have already learnt that the reactions 2 Na(s) + Cl2 (g) → 2 Na+ Cl– (s) or 2 NaCl (s) 2Na(s) + Cl2(g) → 2NaCl (s) (7.12) Reactions 7.12 to 7.14 suggest that half 4Na(s) + O2(g) → 2Na2O(s) (7.13) reactions that involve loss of electrons are 2Na(s) + S(s) → Na2S(s) (7.14) called oxidation reactions. Similarly, the are redox reactions because in each of these half reactions that involve gain of electrons reactions sodium is oxidised due to the addition are called reduction reactions. It may not of either oxygen or more electronegative be out of context to mention here that the element to sodium. Simultaneously, chlorine, new way of defining oxidation and reduction oxygen and sulphur are reduced because to has been achieved only by establishing a each of these, the electropositive element correlation between the behaviour of species sodium has been added. From our knowledge as per the classical idea and their interplay of chemical bonding we also know that sodium in electron-transfer change. In reactions (7.12 chloride, sodium oxide and sodium sulphide to 7.14) sodium, which is oxidised, acts as are ionic compounds and perhaps better a reducing agent because it donates electron written as Na+Cl– (s), (Na+)2O2–(s), and (Na+)2 to each of the elements interacting with it and S2–(s). Development of charges on the species thus helps in reducing them. Chlorine, oxygen produced suggests us to rewrite the reactions and sulphur are reduced and act as oxidising (7.12 to 7.14) in the following manner : agents because these accept electrons from sodium. To summarise, we may mention that Oxidation : 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 7.2 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 represented as Na+H– (s), this suggests that one half reaction in this process is : 2 Na (s) → 2 Na+(g) + 2e– 2024-25 Unit 7.indd 237 10/10/2022 10:37:03 AM  238 chemistry 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 (7.15). 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 reactions solution. No visible reaction is noticed and automatically reveals that here sodium attempt to detect the presence of Cu2+ ions is oxidised and hydrogen is reduced, by passing H2S gas through the solution to therefore, the complete reaction is a produce the black colour of cupric sulphide, redox change. CuS, does not succeed. Cupric sulphide has such a low solubility that this is an extremely 7.2.1 Competitive Electron T ransfer sensitive test; yet the amount of Cu2+ formed Reactions cannot be detected. We thus conclude that Place a strip of metallic zinc in an aqueous the state of equilibrium for the reaction (7.15) solution of copper nitrate as shown in Fig. greatly favours the products over the reactants. 7.1, for about one hour. You may notice that the strip becomes coated with reddish Let us extend electron transfer reaction metallic copper and the blue colour of the now to copper metal and silver nitrate solution solution disappears. Formation of Zn2+ ions in water and arrange a set-up as shown in among the products can easily be judged Fig. 7.2. The solution develops blue colour when the blue colour of the solution due to due to the formation of Cu2+ ions on account Cu2+ has disappeared. If hydrogen sulphide of the reaction: gas is passed through the colourless solution containing Zn2+ ions, appearance of white zinc sulphide, ZnS can be seen on making the solution alkaline with ammonia. The reaction between metallic zinc and the (7.16) aqueous solution of copper nitrate is : Here, Cu(s) is oxidised to Cu 2+ (aq) and Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) (7.15) Ag+(aq) is reduced to Ag(s). Equilibrium greatly In reaction (7.15), 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 Evidently, now if zinc is oxidised, releasing the reaction of metallic cobalt placed in nickel electrons, something must be reduced, sulphate solution. The reaction that occurs accepting the electrons lost by zinc. Copper here is : ion is reduced by gaining electrons from the zinc. Reaction (7.15) may be rewritten as : (7.17) Fig. 7.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker. 2024-25 Unit 7.indd 238 11/11/2022 09:48:49  redox reactions 239 Fig. 7.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker. 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 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 [Co2+(aq) and Ni (s)] are greatly favoured. O. What has been said here with respect This competition for release of electrons to equation (7.18) may be true for a good incidently reminds us of the competition for number of other reactions involving covalent 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: which metals and their ions are listed on the H2(s) + Cl2(g) → 2HCl(g) (7.19) basis of their tendency to release electrons and, just as we do in the case of acids to indicate CH 4(g) + 4Cl2(g) → CCl4(l) + 4HCl(g) (7.20) the strength of the acids. As a matter of fact we have already made certain comparisons. In order to keep track of electron shifts By comparison we have come to know that in chemical reactions involving formation zinc releases electrons to copper and copper of covalent compounds, a more practical releases electrons to silver and, therefore, method of using oxidation number has the electron releasing tendency of the metals been developed. In this method, it is always is in the order: Zn>Cu>Ag. We would love to assumed that there is a complete transfer make our list more vast and design a metal of electron from a less electronegative atom activity series or electrochemical series. to a more electonegative atom. For example, The competition for electrons between various we rewrite equations (7.18 to 7.20) to show metals helps us to design a class of cells, charge on each of the atoms forming part of named as Galvanic cells in which the chemical the reaction : reactions become the source of electrical 0 0 +1 –2 energy. We would study more about these 2H2(g) + O2(g) → 2H2O (l) (7.21) cells in Class XII. 0 0 +1 –1 H2 (s) + Cl2(g) → 2HCl(g) (7.22) 7.3 OXIDATION NUMBER –4+1 0 +4 –1 +1 –1 A less obvious example of electron transfer is realised when hydrogen combines with oxygen CH4(g) + 4Cl2(g) → CCl4(l) +4HCl(g) (7.23) to form water by the reaction: It may be emphasised that the assumption 2H2(g) + O2 (g) → 2H2O (l) (7.18) of electron transfer is made for book-keeping Though not simple in its approach, yet purpose only and it will become obvious at we can visualise the H atom as going from a a later stage in this unit that it leads to the neutral (zero) state in H2 to a positive state in simple description of redox reactions. H2O, the O atom goes from a zero state in O2 Oxidation number denotes the oxidation to a dinegative state in H2O. It is assumed that state of an element in a compound there is an electron transfer from H to O and ascertained according to a set of rules consequently H2 is oxidised and O2 is reduced. formulated on the basis that electron pair 2024-25 Unit 7.indd 239 11/10/2022 15:18:03  240 chemistry in a covalent bond belongs entirely to more bonding state of oxygen but this number electronegative element. would now be a positive figure only. It is not always possible to remember or 4. The oxidation number of hydrogen is +1, make out easily in a compound/ion, which except when it is bonded to metals in binary element is more electronegative than the compounds (that is compounds containing other. Therefore, a set of rules has been two elements). For example, in LiH, NaH, formulated to determine the oxidation and CaH2, its oxidation number is –1. number of an element in a compound/ion. 5. In all its compounds, fluorine has an If two or more than two atoms of an element oxidation number of –1. Other halogens (Cl, are present in the molecule/ion such as Br, and I) also have an oxidation number Na2S2O3/Cr2O72–, the oxidation number of the of –1, when they occur as halide ions in atom of that element will then be the average their compounds. Chlorine, bromine and of the oxidation number of all the atoms of iodine when combined with oxygen, for that element. We may at this stage, state the example in oxoacids and oxoanions, have rules for the calculation of oxidation number. positive oxidation numbers. These rules are: 6. The algebraic sum of the oxidation number of all the atoms in a compound must be 1. In elements, in the free or the uncombined zero. In polyatomic ion, the algebraic sum state, each atom bears an oxidation of all the oxidation numbers of atoms of number of zero. Evidently each atom in the ion must equal the charge on the ion. H2, O2, Cl2, O3, P4, S8, Na, Mg, Al has the Thus, the sum of oxidation number of three oxidation number zero. oxygen atoms and one carbon atom in the 2. For ions composed of only one atom, the carbonate ion, (CO3)2– must equal –2. oxidation number is equal to the charge By the application of above rules, we can on the ion. Thus Na+ ion has an oxidation find out the oxidation number of the desired number of +1, Mg2+ ion, +2, Fe3+ ion, +3, element in a molecule or in an ion. It is clear Cl– ion, –1, O2– ion, –2; and so on. In their that the metallic elements have positive compounds all alkali metals have oxidation oxidation number and nonmetallic elements number of +1, and all alkaline earth metals have positive or negative oxidation number. have an oxidation number of +2. Aluminium The atoms of transition elements usually is regarded to have an oxidation number of display several positive oxidation states. The +3 in all its compounds. highest oxidation number of a representative 3. The oxidation number of oxygen in most element is the group number for the first compounds is –2. However, we come two groups and the group number minus 10 across two kinds of exceptions here. (following the long form of periodic table) for One arises in the case of peroxides and the other groups. Thus, it implies that the superoxides, the compounds of oxygen in highest value of oxidation number exhibited which oxygen atoms are directly linked to by an atom of an element generally increases each other. While in peroxides (e.g., H2O2, across the period in the periodic table. In the Na2O2), each oxygen atom is assigned an third period, the highest value of oxidation oxidation number of –1, in superoxides number changes from 1 to 7 as indicated (e.g., KO2, RbO2) each oxygen atom is below in the compounds of the elements. assigned an oxidation number of –(½). A term that is often used interchangeably The second exception appears rarely, i.e. with the oxidation number is the oxidation when oxygen is bonded to fluorine. In state. Thus in CO2, the oxidation state of such compounds e.g., oxygen difluoride carbon is +4, that is also its oxidation number (OF2) and dioxygen difluoride (O2F2), the and similarly the oxidation state as well oxygen is assigned an oxidation number as oxidation number of oxygen is – 2. This of +2 and +1, respectively. The number implies that the oxidation number denotes the assigned to oxygen will depend upon the oxidation state of an element in a compound. 2024-25 Unit 7.indd 240 10/10/2022 10:37:04 AM  redox reactions 241 Group 1 2 13 14 15 16 17 Element Na Mg Al Si P S Cl Compound NaCl MgSO4 AlF3 SiCl4 P4O10 SF6 HClO 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 compound is sometimes presented according invariably applied to define oxidation, to the notation given by German chemist, reduction, oxidising agent (oxidant), reducing Alfred Stock. It is popularly known as Stock 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 increase 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 the molecular formula. Thus aurous chloride Reduction : A decrease in the oxidation 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 reagent which can 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 implies change in oxidation state, which in called as oxidants also. turn helps to identify whether the species Reducing agent: A reagent which lowers the is present in oxidised form or reduced 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 Problem 7.3 reductants. Using Stock notation, represent the Redox reactions: Reactions which involve following compounds :HAuCl4, Tl2O, FeO, change in oxidation number of the interacting Fe2O3, CuI, CuO, MnO and MnO2. species. Solution Problem 7.4 By applying various rules of calculating Justify that the reaction: the oxidation number of the desired 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 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 species 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 therefore, conclude 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, Fe 2(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. 2024-25 Unit 7.indd 241 10/10/2022 10:37:04 AM  242 chemistry here that all decomposition reactions are not Further, Cu2O helps sulphur in Cu2S to redox reactions. For example, decomposition increase its oxidation number, therefore, of calcium carbonate is not a redox reaction. Cu(I) is an oxidant; and sulphur of Cu2S +2 +4 –2 +2 –2 +4 –2 helps copper both in Cu2S itself and Cu2O to decrease its oxidation number; CaCO3 (s) CaO(s) + CO2(g) therefore, sulphur of Cu2S is reductant. 3. Displacement reactions In a displacement reaction, an ion (or an 7.3.1 Types of Redox Reactions atom) in a compound is replaced by an ion (or an atom) of another element. It may be 1. Combination reactions denoted as: A combination reaction may be denoted in X + YZ → XZ + Y the manner: A+B → C Displacement reactions fit into two Either A and B or both A and B must be in categories: metal displacement and non-metal the elemental form for such a reaction to be displacement. a redox reaction. All combustion reactions, (a) Metal displacement: A metal in a which make use of elemental dioxygen, as well compound can be displaced by another metal as other reactions involving elements other in the uncombined state. We have already than dioxygen, are redox reactions. Some discussed about this class of the reactions important examples of this category are: under section 7.2.1. Metal displacement 0 0 +4 –2 reactions find many applications in C(s) + O2 (g) CO2(g) (7.24) metallurgical processes in which pure metals are obtained from their compounds in ores. A 0 0 +2 –3 few such examples are: 3Mg(s) + N2(g) Mg3N2(s) (7.25) +2 +6 –2 0 0 +2 +6 –2 –4+1 0 +4 –2 +1 –2 CuSO4(aq) + Zn (s) → Cu(s) + ZnSO4 (aq) CH4(g) + 2O2(g) CO2(g) + 2H2O (l) (7.29) 2. Decomposition reactions +5 –2 0 0 +2 –2 V2O5 (s) + 5Ca (s) 2V (s) + 5CaO (s) Decomposition reactions are the opposite of combination reactions. Precisely, a (7.30) decomposition reaction leads to the breakdown +4 –1 0 0 +2 –1 of a compound into two or more components TiCl4 (l) + 2Mg (s) Ti (s) + 2 MgCl2 (s) at least one of which must be in the elemental (7.31) state. Examples of this class of reactions are: +1 –2 0 0 +3 –2 0 +3 –2 0 2H2O (l) 2H2 (g) + O2(g) (7.26) Cr2O3 (s) + 2 Al (s) Al2O3 (s) + 2Cr(s) (7.32) +1 –1 0 0 In each case, the reducing metal is a 2NaH (s) 2Na (s) + H2(g) (7.27) better reducing agent than the one that is +1 +5 –2 +1 –1 0 being reduced which evidently shows more 2KClO3 (s) 2KCl (s) + 3O2(g) (7.28) capability to lose electrons as compared to It may carefully be noted that there is no the one that is reduced. change in the oxidation number of hydrogen (b) Non-metal displacement: The non- in methane under combination reactions metal displacement redox reactions include and that of potassium in potassium chlorate hydrogen displacement and a rarely occurring in reaction (7.28). This may also be noted reaction involving oxygen displacement. 2024-25 Unit 7.indd 242 10/10/2022 10:37:04 AM  redox reactions 243 All alkali metals and some alkaline earth Cu>Ag. Like metals, activity series also exists metals (Ca, Sr, and Ba) which are very good for the halogens. The power of these elements reductants, will displace hydrogen from cold as oxidising agents decreases as we move water. down from fluorine to iodine in group 17 of the 0 +1 –2 +1 –2 +1 0 periodic table. This implies that fluorine is so 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) reactive that it can replace chloride, bromide (7.33) and iodide ions in solution. In fact, fluorine is 0 +1 –2 +2 –2 +1 0 so reactive that it attacks water and displaces Ca(s) + 2H2O(l) → Ca(OH)2 (aq) + H2(g) the oxygen of water : (7.34) +1 –2 0 +1 –1 0 Less active metals such as magnesium and 2H2O (l) + 2F2 (g) → 4HF(aq) + O2(g) (7.40) 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 using fluorine are not generally carried out in Mg(s) + 2H2O(l) Mg(OH)2(s) + H2(g) aqueous solution. On the other hand, chlorine (7.35) can displace 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) (7.36) 0 +1 –1 +1 –1 0 Many metals, including those which do not Cl2 (g) + 2KBr (aq) → 2 KCl (aq) + Br2 (l) react with cold water, are capable of displacing (7.41) hydrogen from acids. Dihydrogen from acids 0 +1–1 +1 –1 0 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 (7.42) the examples of such metals. A few examples As Br2 and I2 are coloured and dissolve in CCl4, for the displacement of hydrogen from acids can easily be identified from the colour of the are: solution. The above reactions can be written 0 +1 –1 +2 –1 0 in ionic form as: Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2 (g) 0 –1 –1 0 (7.37) Cl2 (g) + 2Br (aq) → 2Cl (aq) + Br2 (l) (7.41a) – – 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) (7.42b) – – (7.38) Reactions (7.41) and (7.42) 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 (7.39) Test’. It may not be out of place to mention Reactions (7.37 to 7.39) 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 0 –1 –1 0 the rate of hydrogen gas evolution, which is Br2 (l) + 2I – (aq) → 2Br– (aq) + I2 (s) (7.43) the slowest for the least active metal Fe, and The halogen displacement reactions have the fastest for the most reactive metal, Mg. a direct industrial application. The recovery Very less active metals, which may occur in of halogens from their halides requires an the native state such as silver (Ag), and gold (Au) do not react even with hydrochloric acid. oxidation process, which is represented by: 2X– → X2 + 2e– (7.44) In section (7.2.1) we have already discussed that the metals – zinc (Zn), copper (Cu) and here X denotes a halogen element. Whereas silver (Ag) through tendency to lose electrons chemical means are available to oxidise Cl–, show their reducing activity in the order Zn> Br– and I–, as fluorine is the strongest oxidising 2024-25 Unit 7.indd 243 10/10/2022 10:37:04 AM  244 chemistry 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 electrolytically, 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 (7.49) Disproportionation reactions are a special type (It is to be noted with care that fluorine in of redox reactions. In a disproportionation reaction (7.49) will undoubtedly attack water reaction an element in one oxidation state to produce some oxygen also). This departure is simultaneously oxidised and reduced. shown by fluorine is not surprising for us as One of the r eacting substances in a we know the limitation of fluorine that, being disproportionation reaction always contains the most electronegative element, it cannot an element that can exist in at least three exhibit any positive oxidation state. This oxidation states. The element in the form means that among halogens, fluorine does not of reacting substance is in the intermediate show a disproportionation tendency. oxidation state; and both higher and lower oxidation states of that element are formed in Problem 7.5 the reaction. The decomposition of hydrogen Which of the following species, do not peroxide is a familiar example of the reaction, show disproportionation reaction and where oxygen experiences disproportionation. why ? +1 –1 +1 –2 0 ClO–, ClO2–, ClO3– and ClO4– 2H2O2 (aq) → 2H2O(l) + O2(g) (7.45) Also write reaction for each of the species Here the oxygen of peroxide, which is present that disproportionates. in –1 state, is converted to zero oxidation state Solution in O2 and decreases to –2 oxidation state in Among the oxoanions of chlorine listed H2O. above, ClO4– does not disproportionate Phosphorous, sulphur and chlorine because in this oxoanion chlorine is undergo disproportionation in the alkaline present in its highest oxidation state that medium as shown below : is, +7. The disproportionation reactions 0 –3 +1 for the other three oxoanions of chlorine P4(s) + 3OH–(aq)+ 3H2O(l) → PH3(g) + 3H2PO2– are as follows: (aq) +1 –1 +5 (7.46) 3ClO– → 2Cl– + ClO–3 0 –2 +2 +3 +5 –1 S8(s) + 12 OH (aq) → 4S (aq) + 2S2O32–(aq) – 2– 6 ClO2– 4ClO3– + 2Cl– + 6H2O(l) +5 –1 +7 (7.47) 4ClO–3 → Cl– + 3 ClO4– 0 +1 –1 Cl2 (g) + 2 OH– (aq) → ClO– (aq) + Cl– (aq) + Problem 7.6 H2O (l) Suggest a scheme of classification of the (7.48) following redox reactions The reaction (7.48) describes the (a) N2 (g) + O2 (g) → 2 NO (g) formation of household bleaching agents. The hypochlorite ion (ClO –) formed in the (b) 2Pb(NO3)2(s) → 2PbO(s) + 4 NO2 (g) + reaction oxidises the colour-bearing stains O2 (g) of the 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 NO3– (aq)+H2O(l) as exhibited by chlorine in reaction (7.48), 2024-25 Unit 7.indd 244 11/10/2022 15:18:43  redox reactions 245 Solution reaction (c), hydrogen of water has been In reaction (a), the compound nitric displaced by hydride ion into dihydrogen oxide is formed by the combination of gas. Therefore, this may be called as the elemental substances, nitrogen and displacement redox reaction. The reaction oxygen; therefore, this is an example (d) involves disproportionation of NO2 of combination redox reactions. 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 components; therefore, this is categorised under example of disproportionation redox decomposition redox reaction. In reaction. The Paradox of Fractional Oxidation Number Sometimes, we come across with certain compounds in which the oxidation number of a particular element in the compound is in fraction. Examples are: C3O2 [where oxidation number of carbon is (4/3)], Br3O8 [where oxidation number of bromine is (16/3)] and Na2S4O6 (where oxidation number of sulphur is 2.5). We know that the idea of fractional oxidation number is unconvincing to us, because electrons are never shared/transferred in fraction. Actually this fractional oxidation state is the average oxidation state of the element under examination and the structural parameters reveal that the element for whom fractional oxidation state is realised is present in different oxidation states. Structure of the species C3O2, Br3O8 and S4O62– reveal the following bonding situations: +2 0 +2 O = C = C*= C = O Structure of C3O2 (carbon suboxide) Structure of Br3O8 (tribromooctaoxide) Structure of S4O62– (tetrathionate ion) The element marked with asterisk in each species is exhibiting the different oxidation state (oxidation number) from rest of the atoms of the same element in each of the 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 the average is 4/3. However, the realistic picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br3O8, each of the two terminal bromine atoms are present in +6 oxidation state and the middle bromine is present in +4 oxidation state. Once again the average, that is different from reality, is 16/3. In the same fashion, in the species S4O62–, each of the two extreme sulphurs exhibits oxidation state of +5 and the two middle sulphurs as zero. The average of four oxidation numbers of sulphurs of the S4O62– is 2.5, whereas the reality being + 5,0,0 and +5 oxidation number respectively for each sulphur. We may thus, in general, conclude that the idea of fractional oxidation state should be taken with care and the reality is revealed by the structures only. Further, whenever we come across with fractional oxidation state of any particular element in any species, we must understand that this is the average oxidation number only. In reality (revealed by structures only), the element in that particular species is present in more than one whole number oxidation states. Fe3O4, Mn3O4, 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. 2024-25 Unit 7.indd 245 10/10/2022 10:37:04 AM  246 chemistry (a) Oxidation Number Method: In writing Problem 7.7 equations for oxidation-reduction reactions, Why do the following reactions proceed just as for other reactions, the compositions and differently ? formulas must be known for the substances Pb3O4 + 8HCl → 3PbCl2 + Cl2 + 4H2O and that react and for the products that are formed. The oxidation number method is now Pb3O4 + 4HNO3 → 2Pb(NO3)2 + PbO2 + best illustrated in the following steps: 2H2O Solution Step 1: Write the correct formula for each reactant and product. Pb 3 O 4 is actually a stoichiometric Step 2: Identify atoms which undergo mixture of 2 mol of PbO and 1 mol of change in oxidation number in the reaction PbO2. In PbO 2, lead is present in +4 by assigning the oxidation number to all oxidation state, whereas the stable elements in the reaction. oxidation state of lead in PbO is +2. PbO2 thus can act as an oxidant (oxidising Step 3: Calculate the increase or decrease agent) and, therefore, can oxidise Cl– ion in the oxidation number per atom and for of HCl into chlorine. We may also keep in the entire molecule/ion in which it occurs. If these are not equal then multiply by suitable mind that PbO is a basic oxide. Therefore, number so that these become equal. (If you 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 2PbO + 4HCl → 2PbCl2 + 2H2O products are wrong or the oxidation numbers have not been assigned properly). (acid-base reaction) +4 –1 +2 0 Step 4: Ascertain the involvement of ions if PbO2 + 4HCl → PbCl2 + Cl2 +2H2O the reaction is taking place in water, add H+ or OH– ions to the expression on the appropriate (redox reaction) 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 may occur between PbO 2 and HNO 3. the equation; if in basic solution, use OH– ions. However, the acid-base reaction occurs Step 5 : Make the numbers of hydrogen between PbO and HNO3 as: atoms in the expression on the two sides 2PbO + 4HNO3 → 2Pb(NO3)2 + 2H2O equal by adding water (H2O) molecules to the It is the passive nature of PbO2 against reactants or products. Now, also check the HNO3 that makes the reaction different number of oxygen atoms. If there are the same from the one that follows with HCl. number of oxygen atoms in the reactants and products, the equation then represents the 7.3.2 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 based on the change in the given below: oxidation number of reducing agent and the Problem 7.8 oxidising agent and the other method is based Write the net ionic equation for the on splitting the redox reaction into two half reaction of potassium dichromate(VI), reactions — one involving oxidation and the K2Cr2O7 with sodium sulphite, Na2SO3, other involving reduction. Both these methods are in use and the choice of their use rests in an acid solution to give chromium(III) with the individual using them. ion and the sulphate ion. 2024-25 Unit 7.indd 246 10/10/2022 10:37:04 AM  redox reactions 247 Solution Problem 7.9 Step 1: The skeletal ionic equation is: Permanganate ion reacts with bromide Cr2O72–(aq) + SO32–(aq) → Cr3+(aq) ion in basic medium to give manganese dioxide and bromate ion. Write the + SO42–(aq) balanced ionic equation for the reaction. Step 2: Assign oxidation numbers for Cr and S Solution +6 –2 +4 –2 +3 +6 –2 Step 1: The skeletal ionic equation is : Cr2O72–(aq) + SO32–(aq) → Cr(aq)+SO42–(aq) MnO4–(aq) + Br–(aq) → MnO2(s) + BrO3– (aq) This indicates that the dichromate ion is Step 2: Assign oxidation numbers for the oxidant and the sulphite ion is the Mn and Br reductant. +7 –1 +4 +5 Step 3: Calculate the increase and MnO4–(aq) + Br–(aq) →MnO2 (s) + BrO3– (aq) decrease of oxidation number, and make this indicates that permanganate ion them equal: from step-2 we can notice is the oxidant and bromide ion is the that there is change in oxidation state reductant. of chromium and sulphur. Oxidation state of chromium changes form +6 to Step 3: Calculate the increase and +3. There is decrease of +3 in oxidation decrease of oxidation number, and make state of chromium on right hand side of the increase equal to the decrease. +7 –1 +4 +5 the equation. Oxidation state of sulphur changes from +4 to +6. There is an increase 2MnO (aq)+Br (aq) → 2MnO2(s)+BrO3–(aq) – 4 – of +2 in the oxidation state of sulphur on Step 4: As the reaction occurs in the right hand side. To make the increase and basic medium, and the ionic charges are decrease of oxidation state equal, place not equal on both sides, add 2 OH– ions numeral 2 before cromium ion on right on the right to make ionic charges equal. hand side and numeral 3 before sulphate 2MnO4– (aq) + Br– (aq) → 2MnO2(s) + ion on right hand side and balance the BrO3–(aq) + 2OH–(aq) chromium and sulphur atoms on both the Step 5: Finally, count the hydrogen atoms sides of the equation. Thus we get and add appropriate number of water +6 –2 +4 –2 +3 molecules (i.e. one H2O molecule) on the Cr2O72–(aq) + 3SO32– (aq) → 2Cr3+ (aq) + left side to achieve balanced redox change. +6 –2 2MnO4–(aq) + Br–(aq) + H2O(l) → 2MnO2(s) 3SO42– (aq) + BrO3– (aq) + 2OH–(aq) Step 4: As the reaction occurs in the acidic medium, and further the ionic (b) Half Reaction Method: In this method, charges are not equal on both the sides, the two half equations are balanced separately add 8H+ on the left to make ionic charges and then added together to give balanced equal equation. Cr2O72–(aq) + 3SO32–(aq)+ 8H+→ 2Cr3+(aq) Suppose we are to balance the equation + 3SO42– (aq) showing the oxidation of Fe2+ ions to Fe3+ ions by dichromate ions (Cr2O7)2– in acidic medium, Step 5: Finally, count the hydrogen wherein, Cr2O72– ions are reduced to Cr3+ ions. atoms, and add appropriate number of The following steps are involved in this task. water molecules (i.e., 4H2O) on the right to achieve balanced redox change. Step 1: Produce unbalanced equation for the reaction in ionic form : Cr2O72– (aq) + 3SO32– (aq)+ 8H+ (aq) → Fe2+(aq) + Cr2O72– (aq) → Fe3+ (aq) + Cr3+(aq) 2Cr3+ (aq) + 3SO42– (aq) +4H2O (l) (7.50) 2024-25 Unit 7.indd 247 10/10/2022 10:37:05 AM  248 chemistry Step 2: Separate the equation into half- Step 7: Verify that the equation contains the reactions: same type and number of atoms and the same +2 +3 charges on both sides of the equation. This Oxidation half : Fe2+ (aq) → Fe3+(aq) (7.51) last check reveals that the equation is fully +6 –2 +3 balanced with respect to number of atoms Reduction half : Cr2O72–(aq) → Cr3+(aq) and the charges. (7.52) For the reaction in a basic medium, first Step 3: Balance the atoms other than O and balance the atoms as is done in acidic medium. H in each half reaction individually. Here the Then for each H+ ion, add an equal number of oxidation half reaction is already balanced OH– ions to both sides of the equation. Where with respect to Fe atoms. For the reduction H+ and OH– appear on the same side of the half reaction, we multiply the Cr3+ by 2 to equation, combine these to give H2O. balance Cr atoms. Problem 7.10 Cr2O72–(aq) → 2 Cr3+(aq) (7.53) Permanganate(VII) ion, MnO4– in basic Step 4: For reactions occurring in acidic solution oxidises iodide ion, I– to produce medium, add H2O to balance O atoms and H+ molecular iodine (I2) and manganese to balance H atoms. (IV) oxide (MnO2). Write a balanced ionic Thus, we get : equation to represent this redox reaction. Cr2O72– (aq) + 14H+ (aq) → 2 Cr3+(aq) + 7H2O (l) Solution (7.54) Step 1: First we write the skeletal Step 5: Add electrons to one side of the half ionic equation, which is reaction to balance the charges. If need be, MnO4– (aq) + I– (aq) → MnO2(s) + I2(s) make the number of electrons equal in the Step 2: The two half-reactions are: two half reactions by multiplying one or both half reactions by appropriate number. –1 0 Oxidation half : I–(aq) → I2 (s) The oxidation half reaction is thus +7 +4 rewritten to balance the charge: Reduction half: MnO4–(aq) → MnO2(s) Fe2+ (aq) → Fe3+ (aq) + e– (7.55) Step 3: To balance the I atoms in the Now in the reduction half reaction there oxidation half reaction, we rewrite it as: are net twelve positive charges on the left hand 2I– (aq) → I2 (s) side and only six positive charges on the right hand side. Therefore, we add six electrons on Step 4: To balance the O atoms in the the left side. reduction half reaction, we add two water molecules on the right: Cr2O72– (aq) + 14H+ (aq) + 6e– → 2Cr3+(aq) + MnO4– (aq) → MnO2 (s) + 2 H2O (l) 7H2O (l) (7.56) To balance the H atoms, we add four H+ To equalise the number of electrons in both ions on the left: the half reactions, we multiply the oxidation MnO–4 (aq) + 4 H+ (aq) → MnO2(s) + 2H2O (l) half reaction by 6 and write as : As the reaction takes place in a basic 6Fe2+ (aq) → 6Fe3+(aq) + 6e– (7.57) solution, therefore, for four H+ ions, we Step 6: We add the two half reactions to add four OH– ions to both sides of the achieve the overall reaction and cancel the equation: electrons on each side. This gives the net ionic MnO–4 (aq) + 4H+ (aq) + 4OH–(aq) → equation as : MnO2 (s) + 2 H2O(l) + 4OH– (aq) 6Fe2+(aq) + Cr2O72–(aq) + 14H+(aq) → 6 Fe3+(aq) + Replacing the H+ and OH– ions with water, 2Cr (aq) + 7H2O(l) (7.58) 3+ 2024-25 Unit 7.indd 248 10/10/2022 10:37:05 AM  redox reactions 249 the resultant equation is: (ii) If there is no dramatic auto-colour change (as with MnO4– titration), there are indicators MnO4– (aq) + 2H2O (l) → MnO2 (s) + 4 OH– (aq) which are oxidised immediately after the Step 5 : In this step we balance the last bit of the reactant is consumed, charges of the two half-reactions in the producing a dramatic colour change. The manner depicted as: best example is afforded by Cr2O72–, which 2I– (aq) → I2 (s) + 2e– is not a self-indicator, but oxidises the MnO4–(aq) + 2H2O(l) + 3e– → MnO2(s) indicator substance diphenylamine just after the equivalence point to produce an + 4OH–(aq) intense blue colour, thus signalling the Now to equalise the number of electrons, end point. we multiply the oxidation half-reaction by (iii) There is yet another method which is 3 and the reduction half-reaction by 2. interesting and quite common. Its use is 6I–(aq) → 3I2 (s) + 6e– restricted to those reagents which are able 2 MnO4– (aq) + 4H2O (l) +6e– → 2MnO2(s) to oxidise I– ions, say, for example, Cu(II): + 8OH– (aq) 2Cu2+(aq) + 4I–(aq) → Cu2I2(s) + I2(aq) (7.59) Step 6: Add two half-reactions to This method relies on the facts that obtain the net reactions after cancelling iodine itself gives an intense blue colour with electrons on both sides. starch and has a very specific reaction with 6I –(aq) + 2MnO4–(aq) + 4H2O(l) → 3I2(s) + thiosulphate ions (S2O32–), which too is a redox 2MnO2(s) +8 OH–(aq) reaction: Step 7: A final verification shows that I2(aq) + 2 S2O2– 3 (aq)→2I–(aq) + S4O62–(aq) (7.60) the equation is balanced in respect of I2, though insoluble in water, remains in the number of atoms and charges on solution containing KI as KI3. both sides. On addition of starch after the liberation 7.3.3 Redox Reactions as the Basis for of iodine from the reaction of Cu2+ ions on Titrations iodide ions, an intense blue colour appears. In acid-base systems we come across with a This colour disappears as soon as the iodine titration method for finding out the strength is consumed by the thiosulphate ions. Thus, of one solution against the other using a the end-point can easily be tracked and the pH sensitive indicator. Similarly, in redox rest is the stoichiometric calculation only. systems, the titration method can be adopted 7.3.4 Limitations of Concept of Oxidation to determine the strength of a reductant/ Number oxidant using a redox sensitive indicator. The usage of indicators in redox titration is As you have observed in the above discussion, illustrated below: the concept of redox processes has been evolving with time. This process of evolution (i) In one situation, the reagent itself is is continuing. In fact, in recent past the intensely coloured, e.g., permanganate oxidation process is visualised as a decrease ion, MnO4–. Here MnO4– acts as the self in electron density and reduction process as indicator. The visible end point in this case is achieved after the last of the reductant an increase in electron density around the (Fe2+ or C2O42–) is oxidised and the first atom(s) involved in the reaction. lasting tinge of pink colour appears at 7.4 Redox Reactions and Electrode MnO4– concentration as low as 10–6 mol dm–3 Processes (10–6 mol L–1). This ensures a minimal ‘overshoot’ in colour beyond the equivalence The experiment corresponding to reaction point, the point where the reductant and (7.15), can also be observed if zinc rod is the oxidant are equal in terms of their dipped in copper sulphate solution. The mole stoichiometry. redox reaction takes place and during the 2024-25 Unit 7.indd 249 10/10/2022 10:37:05 AM  250 chemistry reaction, zinc is oxidised to zinc ions and are represented as Zn2+/Zn and Cu2+/Cu. copper ions are reduced to metallic copper In both cases, oxidised form is put before due to direct transfer of electrons from zinc the reduced form. Now we put the beaker to copper ion. During this reaction heat is containing copper sulphate solution and the also evolved. Now we modify the experiment beaker containing zinc sulphate solution in such a manner that for the same redox side by side (Fig. 7.3). We connect solutions reaction transfer of electrons takes place in two beakers by a salt bridge (a U-tube indirectly. This necessitates the separation containing a solution of potassium chloride of zinc metal from copper sulphate solution. or ammonium nitrate usually solidified by We take copper sulphate solution in a beaker boiling with agar agar and later cooling to a and put a copper strip or rod in it. We also jelly like substance). This provides an electric take zinc sulphate solution in another beaker contact between the two solutions without and put a zinc rod or strip in it. Now reaction allowing them to mix with each other. The takes place in either of the beakers and at the zinc and copper rods are connected by a interface of the metal and its salt solution in metallic wire with a provision for an ammeter each beaker both the reduced and oxidized and a switch. The set-up as shown in Fig.7.3 forms of the same species are present. These is known as Daniell cell. When the switch is represent the species in the reduction and in the off position, no reaction takes place in oxidation half reactions. A redox couple is either of the beakers and no current flows defined as having together the oxidised and through the metallic wire. As soon as the reduced forms of a substance taking part in switch is in the on position, we make the an oxidation or reduction half reaction. following observations: This is represented by separating the 1. The transfer of electrons now does not oxidised form from the reduced form by take place directly from Zn to Cu2+ but a vertical line or a slash representing an through the metallic wire connecting the interface (e.g. solid/solution). For example two rods as is apparent from the arrow in this experiment the two redox couples which indicates the flow of current. 2. The electricity from solution in one beaker to solution in the other beaker flows by the migration of ions through the salt bridge. We know that the flow of current is possible only if there is a potential difference between the copper and zinc rods known as electrodes here. The potential associated with each electrode is known as electrode potential. If the concentration of each species taking part in the electrode reaction is unity (if any gas appears in the electrode reaction, it is confined to 1 atmospheric pressure) and further the reaction is carried out at 298K, then the potential of each electrode is said Fig.7.3 The set-up for Daniell cell. Electrons to be the Standard Electrode Potential. By produced at the anode due to oxidation convention, the standard electrode potential of Zn travel through the external circuit (E) of hydrogen electrode is 0.00 volts. The to the cathode where these reduce the electrode potential value for each electrode copper ions. The circuit is completed inside the cell by the migration of ions process is a measure of the relative tendency through the salt bridge. It may be noted of the active species in the process to remain that the direction of current is opposite to in the oxidised/reduced form. A negative E the direction of electron flow. means that the redox couple is a stronger 2024-25 Unit 7.indd 250 10/10/2022 10:37:05 AM  redox reactions 251 reducing agent than the H+/H2 couple. A information from them. The values of standard positive E means that the redox couple is a electrode potentials for some selected electrode weaker reducing agent than the H+/H2 couple. processes (reduction reactions) are given in The standard electrode potentials are very Table 7.1. You will learn more about electrode important and we can get a lot of other useful reactions and cells in Class XII. Table 7.1 The Standard Electrode Potentials at 298 K Ions are present as aqueous species and H2O as liquid; gases and solids are shown by g and s respectively. Reaction (Oxidised form + ne– → Reduced form) E / V F2(g) + 2e– → 2F– 2.87 Co 3+ +e – → Co2+ 1.81 H2O2 + 2H+ + 2e– → 2H2O 1.78 MnO4– + 8H+ + 5e– → Mn2+ + 4H2O 1.51 Au3+ + 3e– → Au(s) 1.40 Cl2(g) + 2e– → 2Cl 1.36 – Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O 1.33 O2(g) + 4H+ + 4e– → 2H2O 1.23 MnO2(s) + 4H+ + 2e– → Mn2+ + 2H2O 1.23 Br2 + 2e– → 2Br 1.09 – NO3– + 4H+ + 3e– → NO(g) + 2H2O 0.97 Increasing strength of reducing agent Increasing strength of oxidising agent 2Hg2+ + 2e– → Hg22+ 0.92 Ag+ + e– → Ag(s) 0.80 Fe3+ + e– → Fe2+ 0.77 O2(g) + 2H+ + 2e– → H2O2 0.68 I2(s) + 2e– → 2I 0.54 – Cu+ + e– → Cu(s) 0.52 Cu2+ + 2e– → Cu(s) 0.34 AgCl(s) + e– → Ag(s) + Cl– 0.22 AgBr(s) + e– → Ag(s) + Br– 0.10 2H+ + 2e– → H2(g) 0.00 Pb2+ + 2e– → Pb(s) –0.13 Sn2+ + 2e– → Sn(s) –0.14 Ni2+ + 2e– → Ni(s) –0.25 Fe2+ + 2e– → Fe(s) –0.44 Cr3+ + 3e– → Cr(s) –0.74 Zn2+ + 2e– → Zn(s) –0.76 2H2O + 2e–