Redox Reactions

Redox Reactions OXIDATION AND REDUCTION 1. (b) Oxygen is lost. For example Fe2O3 + 2Al   2Fe + Al2O3 Cr2O3 + 2Al  2Cr + Al2O3 CLASSICAL IDEA OF OXIDATION (a) Oxidation is a chemical reaction in which oxygen is added (c) Electropositive element is added. For example 2HgCl2 + SnCl2   Hg2Cl2 + SnCl4 CuC l 2 + Cu   Cu2Cl2 (d) Electronegative element is removed. For example (b) Hydrogen is removed i.e. hydrogen becomes less Zn + 2HCl Cu + 4HNO 3 ZnCl 2 +H2 ; Cu(NO 3 )2 + 2NO 2 + 2H2O (c) Electronegative element is added 2FeCl2 + Cl2 2Sb + 3Cl2 2FeCl3 + H2   2FeCl2 + 2HCl PbS + H2   Pb + H2S (e) Valency of electropositive element decreases. For example CuSO 4 + Fe 2FeCl3 ; 2SbCl3 (Cu 0 ) (Cu ) FeCl 3 + H 2 S (d) Electropositive element is removed 2NaI+ H2 O 2  2NaOH+ I2 ; FeSO 4 + C u +2 FeCl 2 + 2HCl + S +3 (Fe ) (Fe +2 ) 3. MODERN CONCEPT OF OXIDATION , (e) Valency of electropositive element increases SnCl2 + Cl2 2. SnCl4 CLASSICAL IDEA OF REDUCTION (a) Hydrogen is added. For example H2 + Cl2   2HCl N2 + 3H2   2NH3 The reaction in which an element or an atom or an ion or molecule loses electron is called oxidation. de electronation is oxidation. (a) Neutral atom : When a neutral atom loses electron, it gets converted to a positive ion. Na   Na+1 + e– Al   Al+3 + 3e– 153 (b) Cation : When a cation loses electron, there is an increase in its positive charge. Sn+2   Sn+4 + 2e– Hg+1   Hg+2 + e– O2 + 4e–   2O–2 I2 + 2e–   2I–1 (c) Anion : When an anion loses electron equal to its negative charge, it gets converted to a neutral atom. –2 2O –   O2 + 4e 2N–3   N2 + 6e– (d) Complex Anion : When a complex anion loses electron, its negative charge decreases. [Fe(CN)6]–4   [Fe(CN)6] –3 + e– (e) Molecule : When a molecule loses electrons, it breaks up into it constituents. +1 Redox Reactions (e) Molecule : When a molecule accepts electron, it is a reduction reaction. – H2O2   2H + O2 + 2e Therefore in oxidation reactions– Therefore in reduction reactions– (i) Positive charge decreases and negative charge increases (ii) Oxidation number decreases OXIDANTS (i) Molecules of most electronegative elements e.g. O2, O3, halogens (ii) Compounds having either of an element (under lined) in their highest oxidation state e.g. KMnO4, K2Cr2O7, H2SO4, HNO3, FeCl3, HgCl2, KClO3, NaNO3 etc. (i) Positive charge increases and negative charge decreases (iii) Oxides of metals and non metals e.g. MgO, CaO, (ii) Oxidation number increases CrO3, H2O2, CO2, SO3, etc. 4. MODERN CONCEPT OF REDUCTION REDUCTANTS The reaction in which an element or an atom or an ion (positive or negative) or a molecule accepts (i) All metals e.g. Na, Al, Zn etc. electron, is called reduction. Electronation is (ii) Some non metals e.g. C, S, P, H2 etc. reduction. (iii) Halogen acids e.g. HI, HBr, HCl. (a) Neutral Atom :When a neutral element or atom accepts electron, it get converted into an anion. (iv) Metallic hydrides e.g. NaH, LiH, CaH2 etc. (v) Compounds having either of an element (under lined) N + 3e–   N–3 in their lowest oxidation state e.g. FeCl2, FeSO4, S + 2e–   S–2 Hg2Cl2, (b) Cation : When a cation accepts electron equal to SnCl2, Cu2O etc. its charge, it gets converted into a neutral atom. Mg+2 + 2e–   Mgº +3 – Al + 3e   Alº (c) Similarly, when a cation accepts less electrons than its charge, its positive charge decreases. For example Cu+2 + e–   Cu+1 Fe+3 + e–   Fe+2 (d) Anion : When an anion accepts electron, its negative charge increases. MnO4–1 + e–   MnO4–2 [Fe(CN)6]–3 + e–   [Fe(CN)6]–4 (vi) Some organic compounds e.g. HCOOH, Aldehydes, Oxalic acid, Tartaric acid etc. REDOX REACTIONS Redox reactions are the chemical reactions which involve both oxidation as well as reduction simultaneously. In fact, oxidation and reduction go hand in hand. The redox reactions are of two types: (i) Direct redox and (ii) Indirect redox reactions. 154 Redox Reactions When chemical reactions are carried out then some OXIDATION NUMBER of the species may lose electrons whereas some other may gain electrons. The concept of electron transfer 1. The definition : Oxidation number of an element in a particular compound represents the number of can easily explain in the redox reactions in the case electrons lost or gained by an element during its of ionic substances. However, for covalent change from free state into that compound or compounds we use a new term oxidation number Oxidation number of an element in a particular to explain oxidation and reduction or redox reactions. compound represents the extent of oxidation or Before discussing in detail, some other terms reduction of an element during its change from free frequently being used are: state into that compound. SPECTATOR IONS 2. Oxidation number is given positive sign if electrons are lost. Species that are present in the solution but not take Oxidation number is given negative sign if electrons are gained. part in the reaction and are also omitted while writing the net ionic reaction are called spectator ions or bystander ions. Zn + 2H+ + 2Cl–   Zn+2 + 2Cl– + H2 In this reaction ions are omitted and are called as spectator ions and appear on the reactant as well as product side. TYPES OF REDOX REACTION 1. Oxidation number represents real charge in case of ionic compounds, however, in covalent compounds it represents for imaginary charge. 4. It is the residual charge which an atom appears to have when other atom are withdrawn from the molecules as ions by containing electrons with more electronegative atoms. AUTOXIDATION (A) THE RULE FOR DERIVING OXIDATION NUMBER Turpentine, Phosphorous and metals like Zn and Pb can absorb oxygen from air in the presence of water. Following arbitrary rules have been adopted to derive Oxidation Number of elements on the basis of The water oxidized to hydrogen peroxide. The periodic properties of elements. phenomena of formation of H SO by the oxidation 2 4 of H2O is (i) In uncombined state or free state, Oxidation Number of an element is zero. known as autoxidation. Pb + O2   PbO2 ; 2. 3. (ii) In combined state Oxidation Number of ....... PbO2 + H2O   PbO + H2 O2 DISPROPORTIONATION a. ........ F is always – 1. One and the same substance may act simultaneously as an oxidising and as a reducing agent. As a result a part of it gets oxidised to higher state and rest of it is reduced to a lower state of oxidation. Such as reaction, in which the substance undergoes simultaneous oxidation and reduction is called disproportionation. –1 2 2 2 –1 2 –2 2 0 2 b. ........ O is –2; In peroxides (–O–O–) it is 1 and in superoxide – 1/2 .However in F2O, it is +2. c. ........ H is 1; In ionic hydrides it is –1. d. ........ metals is always positive. e. ........ alkali metals (IA e.g. Li, Na, K, Rb, Cs, Fr) is always + 1. f. ........ alkaline earth metals (IIA e.g. Be, Mg, Ca, Sr, Ba, Ra) is always +2. g. ........ halogens in halides is always – 1. h. ........ sulphur in sulphides in always –2. 155 Redox Reactions (iii) The algebraic sum of all the Oxidation Number of EXCEPTIONS elements in a compound is equal to zero. e.g. (i) Oxidation Number of Cl in Cl2O is +1, because Cl K2MnO4 acts as an electropositive element in this. 2 × Oxidation Number of K + Oxidation Number (ii) Oxidation Number of Cl in ClF3 = +3 of Mn + 4 (Oxidation Number of O) = 0 (iii) Oxidation Number of Cl in KClO = +5 3 (iv) The algebraic sum of all the Oxidation Numbers of (iv) Oxidation Number of I in IF = +7 7 elements in a radical is equal to net charge on that (v) Oxidation Number of I in IF = +5 5 radical e.g.C2O42–. 2 × Oxidation Number of C + (1) OXIDATION NUMBER OF RADICALS 4 (Oxidation Number of O) = – 2. Oxidation Number of radicals is equal to charge (v) Oxidation Number can be zero, +ve, – ve, integer present on them. For example, or fraction. (vi) Maximum Oxidation Number of an element is (except O & F) = Group Number. Minimum Oxidation Number of an element is (except metals) = Group Number – 8. (i) Oxidation Number of sulphite (SO3–2), sulphate (SO4–2), thiosulphate (S2O3–2), oxalate (C2O4–2), carbonate (CO3–2), sulphide (S–2) is equal to charge (–2) present on each of them. Note : Group number in Mendeleef’s modern periodic (ii) Oxidation Number of each of the anions, Cl–1, Br– 1 table. , I–1, NO3–1, CN–1, OH–1,SCN–1,CH3COO–1 and HCO3–1 is –1. < (iii) Oxidation Number of each of the anions. PO 4–3, BO3–3, AsO4–3. (Arsenate) and AsO3–3 (Arsenite) is –3. (viii)Variable oxidation number is most commonly shown by transition elements as well as by p-block elements. (iv) Oxidation Number of each of the cations, CH3+, NH4+, Na+, K+ is +1. Transition elements : Fe (+2 & +3), Cu (+1 & (vii) The most common oxidation states of some representative elements are given below. +2), Mn (+7, +6, +5, +4, +3, +2, +1) etc. p-block elements : As (+3 & +5); Sb (+3 & +5), Sn (+2 & +4) etc. Group Outer shell Common Oxidation configuration Number I gp ns1 0, +1 II gp ns2 0, +2 III gp ns2 np1 2 2 (v) Oxidation Number of each of the cations, Ca+2, Mg+2, Sr+2 and Fe+2 is +2. (vi) Oxidation Number of Al in [Al(H2O)6]+3 is +3. 0, +1, +3 IV gp ns np 0, ±1, ±2, ±3, ±4 V gp ns2 np3 0, ±1, ±3, +5 VI gp ns2 np4 0, ±2, +4, +6 VII gp ns2 np5 0, ±1, +3, +5, +7 Zero gp ns2 np6 0 (usually) 156 Redox Reactions (2) S ELEMENT (i) S in H2S 2(1) + x = 0 +2 + x = 0 x=–2 (ii) S in SO2 x + 2(–2) = 0 x–4=0 x=+4 (iii) S in SO4–2 x + 4(–2) = –2 x–8=–2 x=+6 (iv) S in SO3–2 x + 3(–2) = –2 x–6=–2 x=+4 (v) S in SF6 x + 6(–1) = 0 x–6=0 x=+6 (vi) S in H2SO3 2(–1) + x + 3(–2) = 0 +2 + x – 6 = 0 x=+4 (vii) S in As2S3 2(3) + 3x = 0 6 + 3x = 0 x=–2 (3) P ELEMENT (i) Oxidation number of P in P4 = 0 (ii) P in PO4–3 : x + 4 (–2) = –3 x – 8 = – 3, x = + 5 (iii) P in NaHPO2 : 1(1) + 1(1) + + 2(–2) = 0 +1 +1 + x – 4 = 0, x = +2 (iv) P in H3PO3 : 3(+1) + x + 3(–2) = 0 + 3 + x – 6 = 0, x=+3 (v) P in Na2HPO4 : 2(1) + 1(1) + x + 4(–2) = 0 + 2 + 1 + x – 8 = 0, x=+5 (vi) P in Mg2P2O7 : 2(2) + 2x + 7(–2) = 0 + 4 + 2x – 14 = 0, 2x = 10, x = + 5 (4) OXIDATION NUMBER OF Cr IN ITS VARIOUS COMPOUNDS (i) Cr in CrO : x – 2 = 0, x=+2 (ii) Cr in Cr2O3 : 2x – 6 = 0, x=+3 (iii) Cr in CrSO4 : x – 2 = 0, x=+2 (iv) Cr in Cr2(SO4)3 : 2x – 6 = 0, x=+3 (v) Cr in CrO2Cl2 : 2x – 6 = 0, x=+3 (vi) Cr in K2Cr2O7 : 2 + 2x – 14 = 0, x=+6 (vii) Cr in K2CrO4 : 2 + x – 8 = 0, x=+6 (viii)Cr in Cr2O7–2 : 2x – 14 = –2, 2x = 12 (ix) Cr in CrO4–2 : x – 8 = –2, x=+6 (x) Cr in Cr (NH3)4SO4 : x – 2 = 0, x=+2 x=+6 (Here, Oxidation Number of NH3 is zero) (xi) Oxidation Number of Cr in [Cr(NH3)4]+2 : (xii) Oxidation Number of Cr in Na2CrO4 : (xiii)Oxidation Number of Cr in Cr(CO)6 : x=+2 +2 + x – 8 = 0, x=+6 x=0 (Oxidation Number of Cr = 0) 157 Redox Reactions (5) OXIDATION NUMBER OF Mn IN ITS COMPOUNDS (i) Mn in MnO : x – 2 = 0, x=+2 (ii) Mn in Mn2O3 : 2x – 6 = 0, x=+3 (iii) Mn in MnSO4 : x – 2 = 0, x=+2 (iv) Mn in Mn2(SO4)3 : 2x – 6 = 0, x=+3 (v) Mn in K2MnO4 : +2 + x – 8 = 0, x=+6 (vi) Mn in KMnO4 : +1 + x – 8 = 0, x=+7 (vii) Mn in Mn(CO)10 : x + 10(0) = 0 x=0 (viii)Mn in MnO4– : x–8=–1 x=+7 (ix) Mn in Mn (C2O4)2.2H2O : x – 4 = 0, x=+4 element. The valency of an element is numerically equal to the number of hydrogen atoms or chlorine atoms or twice the number of oxygen atoms that combine with one atom of that element. It is also equal to the number of electrons lost or accepted or shared by the atoms of an element. (6) OXIDATION STATE Oxidation state of an atom is defined as oxidation number per atom. e.g. In K2MnO4, Oxidation number of Mn = +6 Oxidation state of Mn = Mn6+ However, for all practical purposes oxidation state is often expressed as oxidation number. In some cases (mainly in the case of electrovalent compounds), valency and oxidation number are the same but in other cases they may have different values. The difference between the two have been tabulated. (B) VALENCY AND OXIDATION NUMBER Valency of an element represents the power or capacity of the element to combine with the other S.No. 1. Valency Oxidation number (State) It is the combining capacity of the element. Oxidation number is the charge (real or imaginary) No plus or minus sign is attached to it. present on the atom of the element when it is in combination. It may have plus or minus sign. 2. Valency of an element is usually fixed. Oxidation number of an element may have different values. It depends on the nature of compound in which it is present. 3. Valency is always a whole number. Oxidation number of the element may be a whole number or fractional. 4. Valency of the element is never zero except Oxidation number of the element may be zero. in noble gases. For example, in the following compounds of carbon, the oxidation number varies from – 4 to +4 but valency of carbon is 4 in all the compounds. Compound CH4 CH3Cl CH2Cl2 CHCl3 CCl4 Oxidation number of carbon –4 –2 0 +2 +4 158 EVALUATION OF OXIDATION NUMBER Determine Oxidation number of the element underlined in each of the following : O || H- O - O - S- O -H (a) H2SO5 :  2 × 1 + x + 5 × (–2) = 0  || O Redox Reactions (iii) In case of coordinate bond assign +2 value for Oxidation Number to atom from which coordinate bond is directed to other a more electronegative atom and – 2 value to more electronegative atom. (iv) If coordinate bond is directed from more electronegative atom to less electronegative atom, then neglect contribution for coordinate bond. Thus for H – C  N. x = +8 (wrong) 1 + x + 3 × (– 1) = 0;  x = + 2 But this can not be true as maximum oxidation number for S can not exceed +6. The exceptional Note :  N has three covalent bonds and more value is due to the fact that O atom in H2SO5 show electronegative than carbon. peroxide linkage. Therefore evaluation of oxidation  Oxidation Number of N = – 3 number should be made as : 2 × (+1) + x + 3 × (–2) + 2 × (–1) = 0 (for H) (for S) (for O) (for O – O)  a = + 6 (b) NH4NO3 : 2 × x + 4 × 1 + 3 (–2) = 0  x = +1 (wrong) No doubt NH4NO3 has two N atoms but one N atom has negative Oxidation Number (attached to H) and the other has +ve Oxidation Number (attached to O). Therefore, evaluation should be made separately for NH4+ & NO3–. NH4+  x + 4 × (+1) = + 1; x = – 3 (Oxidation Number of N in NH4+) NO 3 – x + 3 × (–2) = – 1;  x = + 5 (Oxidation Number of N in NO 3–) (d) H – N  C : 1 + (–3) + x = 0;  x=+2 [The contribution of coordinate bond is neglected because the bond is directed from more electronegative to less electronegative carbon atom.] (e) Fe3O4 : 3 × x + 4 × (– 2) = 0;  x = + (8/3) or  Fe3O4 is a mixed oxide of FeO. Fe2O3  Fe has two Oxidation Numbers +2 and +3. However factually speaking Oxidation Numbers of Fe in Fe3O4 is an average value of these two (i.e. +2 & +3) (c) HCN : The evaluation can not be made directly by Average Oxidation Number using rules since we have no standard rule for 1²(+2) + 2²(+3) 8 oxidation number of N and C i.e. two values are = =+ 3 3 unknown. In all such cases evaluation of oxidation number should be made by indirect concept or by (f) FeSO4 (NH4)2SO4 . 6H2O : the original concepts of bonding. Put sum of Oxidation Numbers of SO4 = – 2 (i) Each covalent bond contributes for one unit value Sum of Oxidation Numbers in (NH4)2SO4 = 0 for oxidation number. [(NH4)2SO4 is a complete molecule] (ii) Covalently bonded atom with less electronegativity Sum of Oxidation Numbers in H2O = 0 acquires +ve Oxidation Number whereas other with [H2O is complete molecule] more electronegativity acquires – ve Oxidation x + (–2) + 0 + 0 = 0; x = + 2 number. 159 Redox Reactions (g) Fe0.94O : x × 0.94 + (–2) = 0; x = 200/94 BALANCING OF EQUATIONS : (h) Na2[Fe(CN)5NO] : + NO in iron complexes has NO nature. Thus 2 × 1 + x + 5 × (–1) + 1 = 0 (for Na) (for Fe)(for CN)(for NO);  x=+2 (i) FeNO(H2O)5SO4 : x + 1 + 5 × 0 + (–2) = 0;  x = + 1 (j) Na2S4O6 : 2 × (+1) + 4x + 6 × (–2) = 0;  x = + 5/2 Here also this value is the average oxidation Number of S. The structure of Na2S4O6 may be written as: 1. Two methods are generally used to balance a redox equation. BY OXIDATION STATE METHOD : Step I & II of ion electron methods should be changed accordingly a shown below in each case (i.e. neutral, acidic or alkaline) medium. The other steps to be followed as usual. Example : KMnO4 + H2C2O4   CO2 + K2O + MnO + H2O Find the oxidation numbers of elements undergoing oxidation reduction Mn7+   Mn2+ i.e. change in Oxidation Number of Mn (+7   +2) = 5 3+ C2   2C4+ i.e. change in Oxidation Number of C (+6   +8) = 2 O O     || ||      N a   O  S  S  S  S  O  N a   || ||     O O Thus 2Mn7+ + 5C23+   2Mn2+ + 10C4+ (a) Acidic Medium : The side which has one, oxygen less is to be provided with 1 H2O and opposite side by 2H+. Oxidation Number of each S atom in S – S atom (b) Basic Medium : The side which has one oxygen involved in pure covalent bond is zero. extra is to be provided with one H2O and opposite +5 + 5 + 0 + 0 5 side by 2(OH–) ions. The side which has one Average Oxidation Number = =+ hydrogen extra is to be provided with 1(OH–) and 4 2 opposite by 1H2O. (k) Dimethyl sulphoxide or (CH3)2SO : BALANCING OF HALF REACTIONS Oxidation Number of CH3 = 1 : Oxidation Number Example 1 : I2   IO3– (acidic medium) of O = – 2 Balance atoms other than O & H if  2 × (+1) + x + (–2) = 0; x=0 needed I2   2IO3– (l) CrO5 : Balance O atoms using H+ & H2O as reported in step 4 of acidic medium earlier The structure of CrO5 has two peroxide bonds I2 + 6H2O   2IO3– + 12H+ O O \ / | / Cr \ | Balance charge by electrons O || O (A butterfly structure) I2 + 6H2O   2IO3– + 12H+ + 10e– O Example 2 : S2O32–   SO2 (basic medium)  x + 4 × (–1) + 1 × (–2) = 0; x = +6 2– S 2 O3   2SO2 (m) Na2S3O6 : 2– S2O3 + 2OH–   2SO2 + H2O – 2 × 1 + 3 × (x) + 6 × (–2) = 0; (By H2O & OH ) 10 1 S2O32– + 2OH–   2SO2 + H2O +  x = 3 – 3 3 4e 160 2. ION ELECTRON METHOD : This method involves three sets of rules depending upon the nature of equation to be balanced in neutral, acidic or alkaline medium. (i) Divide the overall reaction into oxidation half and reduction half reactions. (ii) Balance the half reactions w.r.t. charges and electrons. (iii) Equalize the electrons lost and gained by multiplying the half reactions with suitable integers. Simultaneously oxygen and Hydrogen will also be balanced. (iv) Add the two half reactions. Ex. MnO4— + Fe+2   Mn+2 + Fe+3 + H+ Balancing in acidic medium First half reaction Second half reaction Redox Reactions Multiplying equation 2 with 5 and adding with equation 1 5Fe+2   5Fe+3 + 5e– 5e– + 8H+ + MnO4–   Mn+2 + 4H2O 5Fe+2 + 8H+ + MnO4—   5Fe+3 + Mn+2 + 4H2O Hence equation balanced EQUIVALENT WEIGHT OF OXIDANTS AND REDUCTANTS By using oxidation number, equivalent weight of oxidising and reducing substance can be determined as follows Equivalent weight of a oxidant = Molecular weight of molecule or ion Electrons accepted by one molecule or ion = Molecular weight of molecule or ion Total change in oxidation number MnO4—   Mn+2 Fe+2   Fe+3 Equivalent weight of a reductant Molecular weight of molecule or ion MnO4—   Mn+2 + 4H2O Fe+2   Fe+3 + e– eq.....2 = Electrons released by one molecule or ion 8H+ + MnO4—   Mn+2 + 4H2O 5e– + 8H+ + MnO4—   Mn+2 + 4H2O eq.....1 = Molecular weight of molecule or ion Total change in oxidation number 161