SUBJECTIVE. O Thus, evaluation of ox. no. of S should be made as a + 3 ( 2) + 2 ( 1) = 0 a = +6

Size: px

Start display at page:

Download "SUBJECTIVE. O Thus, evaluation of ox. no. of S should be made as a + 3 ( 2) + 2 ( 1) = 0 a = +6"

Rodney Nash
6 years ago
Views:

1 Solved Problems Chemistry: Oxidation and Reduction-1 SUBJECTIVE Problem 1: Determine the oxidation no. of following elements underlined: a) K Cr O 7 b) H SO 5 c) H S O 8 d) NH NO e) Fe O f) KI g) Na S O 6 h) Na S O Solution: a) Ox. no. of K = +1 Ox. no. of O = Ox. no. of Cr = x x (+1) + x + 7x ( ) = 0 x = +1 x = +6 b) Ox. no. of H = +1 Ox. no. of O = Ox. no. of S = a 1+ a + 5 ( ) = 0 a = +8 But this can not be true as maximum ox. no. for S (VI gp) stands +6. The exceptional value is due to the fact that two O atoms in HSO5 show peroxide linkage, i.e., O H O O S O H O Thus, evaluation of ox. no. of S should be made as 1 + a + ( ) + ( 1) = 0 a = +6 c) Two O atoms form per-oxide linkage. i.e., O O H O S O O S O H O O 1 + a + 6 ( ) + ( 1) = 0 a = + 6 d) a ( ) = 0 by rules a = + 1 (wrong)

2 Chemistry: Oxidation and Reduction- No doubt there are two N atoms in NHNO, but one N atom has negative ox. no. (attached to H) and the other has positive ox. no. (attached to O). Therefore, evaluation should be made separately as NO Ox. no. of N in NH and Ox. no. of N in a + (+1) = +1 a + ( ) = 1 a = a = +5 e) a + ( ) = 0 a = + 8 or FeO is a mixed oxide of FeO. FeO f) 1 + (a) = 0 a = 1 or KI is KI + I I has two oxidation no. 1 and 0 respectively. However, factually speaking ox. no. of I in KI is an average of two values 1 and O. 1 (0) 1 Average ox. no. = g) (+1) + a + 6 ( ) = 0 a = + 5 h) 1 + a + ( ) = 0 a = + Problem : a) One mole of N H loses 10 mole electrons to form a new compound Y. Assuming that all the N appears in new compound. What is oxidation state of N in Y? There is no change in oxidation state of H. b) The composition of a sample of Wustite is Fe 0.9 O What percentage of iron is present in the form of Fe (III )? Solution: a) NH (Y) + 10e Y contains all N atoms N (N) a + 10e Therefore, a ( ) = 10 a = + 00 b) Oxidation no. of Fe in Wustite is =.15 9 It is an intermediate value in between two oxidation state of Fe as, Fe (II) and (III), Let percentage of Fe (III) be a, then (100 a) + a = or a = Percentage of Fe (III) = %

3 Chemistry: Oxidation and Reduction- Problem : Select the species acting as reductant and oxidant in the reaction given below : i) PCl + Cl PCl 5 ii) AlCl + K Al + KCl iii) SO + H S S + H O iv) BaCl + Na SO BaSO + NaCl v) I + 6NaOH NalO + 5Nal + H O Solution: In a conjugate pair, oxidant has higher ox. no. i) P + P +5 + e 0 e + Cl Cl 1 PCl is reductant and Cl is oxidant. In a conjugate pair of redox, the one having higher ox. no. is oxidant. ii) For AlCl : Al + + e Al 0 : For K : K 0 K +1 + e Oxidant is AlCl and reductant is K. iii) For SO : S + + e S 0 ; For HS : S e SO is oxidant HS is reductant iv) No change in ox. no. of either of the conjugate pair. None is oxidant or reductant. 0 v) For I : I I e and 0 I + e I 1 I acts as oxidant and reductant both. Problem : Balance the following redox equation, MnO AsO MnO AsO using ion-electron method (alkaline medium). Solution: i) Identify the oxidation & reduction halves Reduction half reaction : MnO MnO Oxidation half reaction: AsO AsO ii) Atoms of the element undergoing oxidation and reduction are already balanced. iii) Balancing O atoms, Reduction half reaction: HO + MnO MnO + OH Oxidation half reaction: OH + AsO + HO AsO iv) Balancing H atoms, H atoms are already balanced in both the half-reactions. v) Balancing charge, Reduction half reaction: e + HO + MnO MnO + OH (B) Oxidation half reaction: OH + AsO AsO + HO + e (A) vi) Multiply equation (A) by and equation (B) by and then add (A) and (B). e + HO + MnO MnO + OH ] OH + AsO AsO + HO + e ] AsO + MnO + HO AsO + MnO + OH

4 Chemistry: Oxidation and Reduction- Problem 5: Balance the following reactions by oxidation number method: Zn + N O Zn + +N O + H O (Acidic Medium) Solution: Find the oxidation states of atoms undergoing redox changes Z n N O Zn N O Increase in O.S = decrease in O.S = 8 Zn + N O Zn + + NO Charge on LHS = Charge on RHS = + 8 Difference in charge = +8 ( ) = 10 As the medium of reaction is acidic, adding 10H + on LHS to equalize the charges Zn + N O + 10H + Zn + + NO To equalize the O and H atoms, add 5HO on RHS to get the balanced equation. Zn + N O + 10H + Zn + + NO + 5HO

5 Assignments (New Pattern) Chemistry: Oxidation and Reduction-5 SECTION I Single Choice Questions 1. Which of the following reaction involves oxidation and reduction? (a) NaBr + HCl NaCl + HBr (b) HBr + AgNO AgBr + HNO (c) H + Br HBr (d) NaO + HSO NaSO + HO. In which one of the following, hydrogen is acting as an oxidizing agent? (a) With Li to give LiH (b) With I to give HI (c) With N to give NH (d) With S to give HS. Which of the following equation is a balanced one? (a) 5 BiO + H + + Mn + 5Bi + + 7HO + MnO (b) 5 BiO + 1H + + Mn + 5Bi + +7HO + MnO (c) BiO (d) 6BiO + H + Mn + Bi + + HO + MnO + 1H + + Mn + 6Bi + + 6HO + MnO. The number of electrons required to balance the following equation NO + H + + e HO + NO are (a) 5 (b) (c) (d) 5. Which of the following statement is correct about the oxidation number? (a) The oxidation number of all atoms in elementary state is 0. (b) The sum of oxidation number of all the atoms in the formula of a compound is always zero. (c) Alkali and alkaline earth metals have +1 and + oxidation states respectively. (d) All of the above. 6. True statement for the reaction [Fe(CN)6] + OH + HO [Fe(CN)6] + HO+ O is (a) HO is an oxidant and reductant (b) CN undergoes neither oxidation nor reduction (c) [Fe(CN)6] gains electrons (d) All the above are correct 7. In the conversion NHOH NO, the equivalent weight of NHOH is M M (a) (b) M (c) (d) M 5 8. Oxidation number of S in S O is (a) (b) + (c) +6 (d) 0

6 Chemistry: Oxidation and Reduction-6 9. The oxidation state of iron in Fe(CO)9 is (a) (b) 0 (c) (d) The oxidation states of the most electronegative element in the products of the reaction, BaO with dilute HSO are (a) 0 and 1 (b) 1 and (c) and 0 (d) and In which of the following compound, iron has the lowest oxidation state? (a) FeSO(NH)SO 6HO (b) K[Fe(CN)6] (c) Fe(CO)5 (d) FeO 1. Oxidation number of iron in Fe0.9O is (a) + (b) + (c) 00/9 (d) 8/ 1. In the reaction, Fe + O Fe + + 6O, which of the following statement is incorrect? (a) It is a redox reaction (b) Metallic iron is a reducing agent (c) Fe + is an oxidising agent (d) Metallic iron is reduced to Fe + 1. For the redox reaction Cr O 7 + H + + Ni Cr + + Ni + + HO the correct coefficients of the reactants for the balanced reaction are Cr O 7 Ni H + (a) 1 1 (b) 1 (c) (d) The compound having the lowest oxidation state of iron is (a) KFe(CN)6 (b) KFeO (c) FeO (d) Fe(CO)5 16. The oxidation state of nickel in Ni(CO) is (a) 0 (b) + (c) + (d) 17. Oxidation state of nitrogen is incorrectly given for Compound Oxidation state (a) [CO(NH)5Cl]Cl (b) NHOH 1 (c) (NH5)SO + (d) MgN 18. A compound contains atoms X, Y, Z. The oxidation number of X is +, Y is +5 and Z is. The possible form of the compound is (a) XY1Z (b) Y(XZ) (c) X(YZ) (d) X(YZ)

7 Chemistry: Oxidation and Reduction The two sulphur atoms in NaSO have (a) + and + oxidation state (b) and +6 oxidation state (c) + and +6 oxidation state (d) same oxidation state 0. In the presence of H + ions, which of the following species can act as very strong oxidizing agent? (a) ClO (b) ClO (c) ClO (d) All 1. Which of the following species can oxidize Mn + to MnO ion (a) IO (b) H IO 6 (c) H I O 9 (d) All. Which of the following has been arranged in order of decreasing oxidation number of sulphur (a) HSO7 > NaSO6 > NaSO > S8 (b) FeS > SO SO > HSO (c) HSO5 > HSO > SCl > HSO (d) HSO > SO > HS > HSO8. Oxidation number of Fe in FeO is fractional because (a) it is a mixed [Fe(+) Fe(+)] oxide (b) it is a non-stoichoiometric compound (c) it is a mixed [Fe (+) Fe(+)] oxide (d) None of the above. A compound containing only sodium, nitrogen and oxygen has.% by weight of sodium. What is the possible oxidation number of nitrogen in the compound? (a) (b) + (c) (d) When one mole NO is converted into 1 mole NO, 0.5 mole N and 0.5 mole NO respectively. It accepts x, y and z mole of electrons, x, y and z are respectively (a) 1, 5, (b) 1,, (c), 1, (d),, SECTION II May be more than one choice 1. NH reacts with BF to form the adduct HN BF. In doing so the hybridisation of (a) B and N both change (b) Only N changes (c) Only B changes (d) Neither B nor N changes. HBr and HI can reduce sulphuric acid, HCl can reduce KMnO and HF can reduce (a) HSO (b) KMnO (c) KCrO7 (d) none of these. A solution of sodium metal in liquid ammonia is strongly reducing agent due to the presence of (a) sodium atoms (b) sodium hydride (c) sodium amide (d) solvated electrons

8 Chemistry: Oxidation and Reduction-8. The oxidation number of cabron in CHO is (a) (b) + (c) 0 (d) + 5. The brown ring complex compound is formulated as [Fe(HO)5(NO) + ]SO. The oxidation state of iron is (a) 1 (b) (c) (d) 0 6. The equivalent mass of MnSO is half of its molar mass when it is converted to (a) MnO (b) MnO (c) MnO (d) MnO 7. The oxidation number of phosphorus in Ba(HPO) is (a) + (b) + (c) +1 (d) 1 8. The pair of compounds having metals in their highest oxidation state is (a) MnO, FeCl (b) MnO, CrOCl (c) [Fe(CN)6], [Co(CN)] (d) [NiCl], [CoCl] 9. In a reaction FeS + KMnO + H + Fe + + SO + Mn + + HO the equivalent mass of FeS would be equal to (a) molar mass molar mass (b) 10 molar mass molar mass (c) (d) The oxidation numbers of sulphur in S8, SF, HS respectively, are (a) 0, +1 and (b) +, +1 and (c) 0, +1 and + (d), +1 and 11. The number of moles of KMnO that will be needed to react with one mole of sulphite ion in acidic solution is (a) /5 (b) /5 (c) /5 (d) 1 1. Amongst the following, identify the species with an atom in +6 oxidation state (a) MnO (b) Cr(CN) 6 (c) (d) CrOCl NiF 6 1. White phosphorus reacts with caustic soda. The products are PH and NaHPO. This reaction is an example of (a) oxidation (b) reduction (c) oxidation and reduction (d) neutralization 1. Oxidation state of oxygen atom in potassium superoxide is (a) 1/ (b) 1 (c) (d) 0

9 Chemistry: Oxidation and Reduction Which of the following are not redox reactions (a) Mg + N MgN (b) K[Fe(CN)6] + HSO + HO KSO + CO + FeSO + (NH)SO (c) I + Cl ICl (d) CuSO + NH [Cu(NH)]SO 16. In which of the following co-ordination compounds, the transition metals have an oxidation number of +. (a) [Cr(HO)Cl]Cl. HO (b) [Fe(CO)5] (c) [(HO)5 Cr O Cr (HO)5] + (d) K[Fe (CN)6] 17. Which of the following reactions is a redox reaction? (a) CrO + 6HCl CrCl + HO (b) CrO + NaOH NaCrO + HO (c) CrO H CrO 7 OH (d) Cr O 6I 1H Cr I 7H O The compound having + as the oxidation state of oxygen is (a) HO (b) CO (c) FO (d) MnO 19. The oxidation state of oxygen in O is (a) +1 and 1 (b) + and (c) 0.5 and 0.5 (d) zero 0. A compound contains atoms of three elements A, B and C. If the oxidation number of A is +, B is +5 and that of C is, the possible formula of the compound is (a) A(BC) (b) A(BC) (c) ABC (d) A(BC) 1. Which of the following is not a reducing agent? (a) SO (c) CO (d) NO (b) HO. Equivalent mass of oxidizing agent in the reaction, SO + HS S + HO is (a) (b) 6 (c) 16 (d) 8. The oxidation state of chromium in [Cr(PPh) (CO) ] is (a) + (b) +8 (c) zero (d) +5. The reaction, ClO (aq) ClO (aq) + Cl (aq) is an example of (a) Oxidation reaction (b) Reduction reaction (c) disproportionation reaction (d) Decomposition reaction

10 Chemistry: Oxidation and Reduction Which of the following is a redox reaction? (a) NaCl + KNO NaNO + KCl (b) CaCO + HCl CaCl + HCO (c) Mg(OH) + NHCl MgCl + NHOH (d) Zn + AgCN Ag + Zn(CN) SECTION III 1. Write complete balanced equation for the following in acidic medium by ion-electron method: (a) Cl O + Fe + Cl + Fe + + HO (b) CuS + N O (c) S O Cu + + S8 + NO + HO + SbO5 SbO + HSO (d) HCl + KMnO Cl + KCl + MnCl + HO (e) KClO + HSO KHSO + HClO + ClO + HO (f) HNO + HBr NO + Br + HO (g) I O + I + H + I + HO (h) Br + Br O + H + Br + HO (i) HS + Cr O 7 + H + CrO + S8 + HO (j) Au + N O + Cl + H + AuC l + NO + HO (k) CuO + H + + N O Cu + + NO + HO (l) Mn Mn + MnO (m) Cu + + SO Cu + + S O (n) Cl + I I O + Cl O 1 O (o) Fe(CN ) 6 + Mn O 7 Fe + + CO + N O + Mn + (p) CuP + Cr Cu + + HPO + Cr +. Write complete balanced equation for the following in basic medium by ion-electron method: (a) S O + AgO Ag + S O O (b) Cl + OH Cl + ClO (c) H + Re O ReO + HO (d) ClO + Sb O Cl O + Sb(OH ) 6 (e) I + OH I + I O (f) Mn O + Fe + Mn + + Fe + (g) Cu + + I Cu + + I (h) FeO + Mn 1 O FeO + MnO (i) CH5OH + Mn O CHO + MnO(s) + HO (j) CrI + HO + OH Cr O + I O + HO (k) KOH + KFe(CN)6 + Ce(NO) Fe(OH) + Ce(OH) + KCO + KNO + HO

11 . Balance the following equations by oxidation number method: (a) Cu + N O + Cu + + NO + (Acid medium) (b) Cl + I O + OH I O + + HO (Basic medium) (c) HS + KCrO + HSO (Acid medium) (d) Fe + + Mn O Fe + + Mn + +. (Acid medium) Chemistry: Oxidation and Reduction-11 (e) KMnO + HSO + HO KSO + MnSO + HO + (Acid medium) (f) MnO + HO Mn O + HO (Basic Medium) (g) I + HO HO + I (Acid medium) (h) Cu + + I Cu + + I (i) CuO + NH Cu + N + HO (j) HSO + Cr (k) Cr O 7 7 O HSO + Cr + + HO (Acid medium) + CHO + H + CHO + Cr + (Acid medium) (l) SbCl + KIO + HCl SbCl5 + ICl + HO + KCl (Acid medium) (m) AsS5 + HNO HAsO + HSO + NO (Acid medium). Fill in the blanks: (a) In the conversion of Br to BrO, the oxidation state of bromine changes from. to.. (b) Oxidation state of chlorine in HClO is... (c) The sum of the oxidation numbes of all the atoms in C6H5CHO is.. (d) In Ba(HPO), the oxidation number of phosphorus is... (e) Among SO, HSO and NaSO, sulphur has the highest oxidation state in.. 5. State, whether the following statements are True or False. (a) The oxidation number of nitrogen in both NH and NH is. (b) The reaction between NaOH and HSO is a redox reaction. (c) The element which shows highest oxidation number of +8 is Os in OsO. (d) Both in CH and CO, carbon has the same oxidation number. (e) Decolourisation of KMnO in acidic medium when it acts as an oxidizing agent is due to formation of manganous salt. 6. Which of the following half reaction is an oxidation and which one is a reduction one? (a) Zn Zn + + e (b) Cl + e Cl (c) Sn + + e Sn + (d) Hg + + e (e) Fe Fe + + e Hg 7. Indicate which of the substance/ion in the following reactions is an oxidizing agent and which is a reducing agent? (a) FeCl + SnCl FeCl + SnCl (b) Mg + SO MgO + S (c) SO + O SO (d) Ca + Cl CaCl (e) Sn + + Hg + Hg Sn

12 Chemistry: Oxidation and Reduction-1 8. Which substance/ion is oxidized and which substance/ion is reduced in the following reactions? (a) PbS + HO PbSO + HO (b) HS + FeCl FeCl + HCl + S (c) MnO + HCl MnCl + HO + Cl (d) SnCl + FeCl SnCl + FeCl (e) MnO 16H 5C O Mn 8H O 10CO 9. Write down the molecular and ionic equations for the following chemical changes: (a) When mercuric chloride reacts with stannous chloride (b) when potassium dichromate reacts with ferrous sulphate in presence of dilute sulphuric acid (c) When zinc is added to copper sulphate solution (d) When chlorine is passed through potassium iodide solution. 10. Complete and balance the following equations: (a) NaCO +.. NaSiO +.. (b) HNO (conc.) + HCl (conc.). + HO + Cl ZnO / CrO (c) H + CO.. 00C,Pr essure (d) Cu + HNO (dil.) Cu(NO) + HO + SECTION IV Subjective Questions LEVEL I 1. Indicate the oxidation number of underlined in each case : (a) NaNO (b) H (c) ClO7 (d) KCrOCl (e) BaCl (f) ICl (g) KCrO7 (h) CHO (i) Ni(CO) (j) NHOH (k) (NH5)SO (l) MgN (m) [Co(NH)5Cl]Cl (n) KFeO (o) Ba(HPO) (p) HSO (q) CS (r) S (s) NaSO6 (t) SCl (u) RNO (v) PbO (w) S O (x) C6H1O6 8 (y) MgPO7 (z) KClO. Indicate in each reaction which of the reactant is oxidized or reduced if any : (a) CuSO + KI CuI + I + KSO (b) NaS + HCl + SO NaCl + S + HO (c) NH NO N + HO

13 . Calculate the number of electrons lost or gained during the changes : (a) Fe + HO FeO + H (b) AlCl + K Al + KCl Chemistry: Oxidation and Reduction-1. Explain, why? (a) HS acts as reductant whereas, SO acts as reductant and oxidant both. (b) HO acts as reductant and oxidant both. 5. MnO can oxidize NO to NO in basic medium. How many mol of NO by 1 mol of MnO? 6. Which is stronger base in each pair? a) HSO ; HSO - (b) NO, NO ; (c) Cl, ClO are oxidized 7. Fill in the blanks and balance the following equations: (a) Zn + HNO.. + NO +.. (b) HI + HNO + NO + HO 8. Can the reaction, Cr O H O CrO H be regarded as a redox reaction? 7 9. Identify the oxidants and reductants in the following reactions: (a) CH(g) + Cl(g) CCl(g) + HCl(g) (b) CHO(aq) + H + (aq) + MnO (s) Mn + (aq) + CO(g) + HO(l) 10. Nitric acid acts only as an oxidizing agent while nitrous acid acts both as an oxidizing as well as a reducing agent. Why? LEVEL II (Judge yourself at JEE level) 1. a) For the redox reaction, MnO CO H Mn CO HO, the correct coefficients of the reactants are. [199] b) The compound YBaCuO7, which shows super conductivity, has copper in oxidation state.. Assume that the rare earth element yttrium is in its usual + oxidation state. [199]. Arrange the following in the order of: [1986] (a) increasing oxidation number of iodine: I, HI, HIO, ICl (b) increasing oxidation number of chlorine: ClO7, ClO, HCl, ClF, Cl (c) increasing oxidation number of nitrogen: NH, NH, NO, NO, NO5. Give proper reasoning for the question: [199] [CuCl] is formed but [CuI] is not?. Balance the equation by oxidation number method. [199] MnO H MnO MnO H O

14 Chemistry: Oxidation and Reduction-1 5. Calculate the oxidation state of underlined: [1975] (a) BaXeO (b) BaCl (c) C1HO11 (d) IF7 (e) Na[Fe(CN)5NO] (f) RuO (g) KTaF7 (h) NaMoO (i) UO 7 (j) C in diamond Answers to Assignments SECTION I 1. (c). (a). (b). (c) 5. (d) 6. (d) 7. (b) 8. (b) 9. (b) 10. (b) 11. (c) 1. (c) 1. (d) 1. (a) 15. (d) 16. (a) 17. (c) 18. (c) 19. (b) 0. (d) 1. (d). (a). (c). (b) 5. (a) SECTION II 1. (c). (d). (d). (c) 5. (a) 6. (b) 7. (c) 8. (b) 9. (c) 10. (a) 11. (a) 1. (d) 1. (c) 1. (a) 15. (b), (d) 16. (a), (c) 17. (d) 18. (c) 19. (d) 0. (a) 1. (c). (c). (c). (a), (c) 5. (d) SECTION III 1. (a) 6H+ + Cl O + 6Fe + Cl + 6Fe + + HO (b) CuS + 16N O + 6H + Cu + + S8 + 16NO + HO (c) S O + SbO5 + 6H + + HO SbO + 6HSO (d) 16HCl + KMnO 5Cl + KCl + MnCl + 8HO (e) KClO + HSO KHSO + HClO + ClO + HO (f) HNO + 6HBr NO + Br + HO (g) I O + 7I + 8H + I + HO (h) 5Br + Br O + 6H + Br + HO (i) HS + 8Cr O H + 8CrO + S8 + HO

15 (j) Au + N O + Cl + 6H + AuC I + NO + HO (k) CuO + 1H + + N O 6Cu + + NO + 7HO (l) Mn O + H + Mn + MnO + HO (m) Cu + + SO + HO Cu + + H + + S (n) 5Cl + I 6HO I O + 10Cl + 1H + 1 O (o) 5Fe(CN ) H Mn O (p) 6CuP + 1H Cr O7. (a) S + AgO + OH Ag + S (b) Cl + OH Cl + ClO + HO O (c) H + Re O O Chemistry: Oxidation and Reduction-15 5Fe + + 0CO + 0N O + 61Mn + + 9HO 18Cu + + 6HPO + Cr + + 5HO O ReO + HO + OH + HO (d) ClO + Sb O + OH + H O Cl O + Sb(OH ) 6 1 (e) 6I + 1OH 10I + I O + 6HO (f) Mn O + 5Fe + + HO Mn + + 5Fe + + 8OH (g) Cu + + I Cu + + I 1 (h) 6FeO + Mn O + HO 9FeO + MnO + OH (i) CH5OH + Mn O + OH CHO + MnO(s) + 5HO (j) CrI + 7HO + 10OH Cr O + 6I O + HO (k) 58KOH + KFe(CN)6 + 61Ce(NO) 61Ce(OH) + Fe(OH) + 6HO + 6KCO + 50 KNO. (a) Cu + H + + N O Cu + + NO + HO (b) Cl + I O + OH I O + Cl + HO (c) HS + KCrO + 5HSO Cr(SO) + KSO + 8HO + S (d) 5Fe + + Mn O + 8H + 5Fe + + Mn + + HO (e) KMnO + HSO + 5HO KSO + MnSO + 8HO + 5O (f) HO + MnO + OH Mn O + HO (g) HO + I + H + HO + I (h) Cu + + I Cu + + I (i) CuO + NH Cu + N + HO 7 (j) HSO + Cr +8H + HSO + Cr + + HO (k) Cr O7 + CHO + 8H + CHO + Cr + + HO (l) SbCl + KIO + 6HCl SbCl5 + ICl + HO + KCl O (m) AsS5 + HNO 5HSO + 0NO + HAsO + 1HO. (a) 0, +5 (b) +7 (c) zero (d) +1 (e) HSO

16 Chemistry: Oxidation and Reduction (a) False Ox. no. of nitrogen in NH is but in NH it is 1/. (b) False Neutralisation reaction (c) True (d) False in CH, Ox. no. is While in CO it is +. (e) True 6. (a) Oxidation (b) Reduction (c) Reduction (d) Reduction (e) Oxidation 7. Oxidising Agent Reducing Agent (a) FeCl SnCl (b) SO Mg (c) O SO (d) Cl Ca (e) Hg + Sn + 8. Oxidised Reduced (a) PbS HO (b) HS FeCl (c) HCl MnO (d) SnCl FeCl (e) MnO CO 9. (a) Molecular equation: HgCl + SnCl HgCl + SnCl Ionic equation: Hg + + Sn + Hg Sn (b) Molecular equation: KCrO7 + 6FeSO + 7HSO KSO + Fe(SO) + 7HO + Cr(SO) Ionic equation : Cr O 6Fe 1H Cr 6Fe 7H O 7 (c) Molecular equation: Zn + CuSO ZnSO + Cu Ionic equation: Zn + Cu + Zn + + Cu (d) Molecular equation: KI + Cl KCl + I Ionic equation: I + Cl Cl + I 10. (a) NaCO + SiO NaSiO + CO (b) HNO + HCl NOCl + HO + Cl (c) H + CO CHOH (d) Cu + 10HNO Cu (NO) + 5HO + NO SECTION IV LEVEL - I 1. (a) + (b) 0 (c) +7 (d) +6 (e) + (f) +

17 (g) +6 (h) 0 (i) 0 (j) 1 (k) (l) + (m) + (n) +6 (o) +1 (p) +6 (q) (r) (s) +5/ (t) +1 (u) + (v) +8/ (w) +6 (x) 0 (y) +5 (z) +5 Chemistry: Oxidation and Reduction-17. Oxidized : KI, NaS, N H Reduced : CuSO, SO, N O. (a) 8 electrons (b) electrons. a) Oxidation number of sulphur in HS and SO are respectively and +. 1 b) HO HO + O Oxidation number of O changes from 1 (HO) to (HO) and zero (O) respectively. 5. NO is oxidsed to NO by MnO (in basic medium) which is reduced to MnO. MnO NO NO MnO 7 5 oxidation reduction Thus, MnO MnO oxidation number decreases by -units NO NO oxidation number increases by units Thus, MnO NO MnO NO = 1.5 mol NO 6. a) HSO ; b) NO ; c) Cl 7. (a) Zn + 10HNO Zn(NO) + NO + 5HO (b) 6HI + HNO I + NO + HO 8. Oxidation number of Cr in CrO 7 6 Oxidation number of Cr in CrO 6 Since during this reaction, the oxidation number of Cr has neither decreased nor increased, therefore, the above reaction cannot be regarded as a redox reaction. 9. Writing the O.N. of all the atoms above their symbols, we have, (a) CH (g) Cl (g) CCl (g) H Cl(g)

18 Chemistry: Oxidation and Reduction-18 (b) O.N. of C increases from - in CH to + in CCl and that of Cl decreases from 0 in Cl to 1 in CCl or HCl, therefore, Cl acts as the oxidant and CH acts as the reductant. 1 C H O (aq) H (aq) Mn O (s) Mn(aq) CO (g) H O(l) O.N. of C increases from in CHO to + in CO and that of Mn decreases from + in MnO to + in Mn +, therefore, MnO acts as the oxidant while CHO acts as the reductant. 10. i) HNO: Oxidation number of N is HNO = +5 Maximum oxidation number of N = +5 Minimum oxidation number of N = Since the oxidation number of N is HNO is maximum (+5), therefore, it can only decrease. Hence HNO acts only as an oxidizing agent. ii) HNO: Oxidation number of N in HNO = + Maximum oxidation number of N = +5 Minimum oxidation number of N = Thus, the oxidation number of N can increase by losing electrons or can decrease by accepting electrons. Therefore, HNO acts both as an oxidizing as well as a reducing agent. 1. (a), 5, 16 (b) +7/ LEVEL - II. (a) HI ( 1), I(0), ICl (+1), HIO (+7) (b) HCl ( 1), Cl(0), ClO(+1), ClF (+), ClO7 (+7) (c) NH ( ), NH ( 1/), NO(+1), NO(+), NO5(+5). [Hint: I ion reduces Cu + to CuI and itself undergoes oxidation to form I. However, Cl does not reduce Cu + ].. VI Mn O VI IV Mn O VII [Mn O Mn O ] MnO H MnO MnO H O 5. (a) 0 (b) + (c) 0 (d) +7 (e) (f) +8 (g) +5 (h) + (i) +5 (j) 0

19 Chemistry: Oxidation and Reduction-19 NOTE

7-OXIDATION, REDUCTION, RED-OX REACTIONS, TYPES

7-OXIDATION, REDUCTION, RED-OX REACTIONS, TYPES 1) In the reaction CuO + NH Cu + N + HO, the oxidation number of N changes from 1) to 0 ) 0 to + ) to + 4) to 0 ) In the reaction MnO 4 MnO, the number of