Redox Reaction and Equivalent Concept Exercise # 3

Exercise # 3 1. Part # I [Matrix Match Type Questions] Column-I Column-II (A) Eq. wt. = Molecular weight 33 (p) When CrI3 oxidises into Cr2O72– and IO4– (B) Eq. wt. = Molecular weight 27 (q) When Fe(SCN)2 oxidises into Fe3+, SO42–, CO32– and NO3– (C) Eq. wt. = Molecular weight 28 (r) When NH4SCN oxidizes into SO42–, CO32– and NO3– (D) 2. (A) (B) (C) (D) 3. Eq. wt. = Molecular weight 24 Column-I When Bi2S3 converted into Bi5+ and S When Al2(Cr2O7)3 reduced into Cr3+ in acidic medium When FeS2 converted into Fe2O3 and SO2 When Mn(NO3)2 converted into MnO42– and NO (s) When As2S3 oxidises into AsO3– and SO42– (p) (q) Column-II 18 11 (r) 2 (s) 10 Column-I Column-II (A) P2H4 PH3 + P4H2 (p) E= 3M 4 (B) I2 I – + IO3– (q) E= 3M 5 (C) MnO4– + Mn2+ + H2O Mn3O4 + H+ (r) E= 15M 26 (D) H3PO2 PH3 + H3PO3 (s) E= 5M 6 (A) Column-I Molarity (p) Column-II Dependent on temperature (B) Molality (q) MA × n A n A M A + n BM B (C) Mole fraction (r) Independent of temperature (D) Mass % (s) XA XBMB × 1000 4. x 100 Where MA , MB are molar masses, nA , nB are no of moles & XA , XB are mole fractions of solute and solvent respectively. REDOX REACTION AND EQUIVALENT CONCEPT 5. Column-I (A) 100 ml of 0.2 M AlCl3 solution + 400 ml (B) 50 ml of 0.4 M KCl + 50 ml H2O (C) 30 ml of 0.2 M K2SO4 + 70 ml H2O (D) 200 ml 24.5% (w/v) H2SO4 6. Column-II (p) Total concentration of cation(s) = 0.12 M of 0.1 M HCl solution (q) [SO42–] = 0.06 M (r) [SO42–] = 2.5 M (s) [Cl¯] = 0.2 M Column-I Column-II (A) 4.1 g H2SO3 (B) 4.9 g H3PO4 (C) 4.5 g oxalic acid (H2C2O4) (D) 5.3 g Na2CO3 7. (p) 200 mL of 0.5 N base is used for complete neutralization (q) 200 millimoles of oxygen atoms (r) Central atom is in its highest oxidation number (s) May react with an oxidising agent Column-I Column-II (A) Sn+2 + MnO4– (acidic) 3.5 mole (p) Amount of oxidant available decides the number of electrons transfer 1.2 mole (B) H2C2O4 + MnO4– (acidic) 8.4 mole 3.6 mole –2 (C) S2O3 + I2 7.2 mole 3.6 mole (D) Fe+2 + Cr2O7–2 (acidic) 9.2 mole 1.6 mole Part # II (q) Amount of reductant available decides the number of electrons transfer (r) Number of electrons involved per mole of oxidant > Number of electrons involved per mole of reductant (s) Number of electrons involved per mole of oxidant < Number of electrons involved per mole of reductant. [Comprehension Type Questions] Comprehension # 1 The strength of H2O2 is expressed in several ways like molarity, normality, % (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H2O2 on decomposition gives 10 volumes of oxygen at STP or 1 litre of H2O2 gives 10 litre of O2 at STP. The decomposition of H2O2 is shown as under : 1 O (g) 2 2 H2O2 can acts as oxidising as well as reducing agent, as oxidizing agent H 2O2 converted into H2O and as reducing agent H2O2 converted into O2, both cases it's n-factor is 2.  Normality of H2O2 solution = 2 × Molarity of H2O2 solution H2O2 (aq) H2O () + 1. What is the molarity of "11.2 V" of H2O2 ? (A) 1 M 2. (C) 5.6 M (D) 11.2 M What is the percentage strength (% w/V) of "11.2 V" H2O2 ? (A) 1.7 3. (B) 2 M (B) 3.4 (C) 34 (D) none of these 20 mL of H2O2 solution is reacted with 80 mL of 0.05 M KMnO4 in acidic medium then what is the volume strength of H2O2 ? (A) 2.8 (B) 5.6 (C) 11.2 (D) none of these 4. 40 g Ba(MnO4)2 (mol. wt. = 375) sample containing some inert impurities in acidic medium is completely reacted with 125 mL of "33.6 V" of H2O2. What is the percentage purity of the sample ? (A) 28.12 % (B) 70.31 % (C) 85 % (D) none of these Comprehension # 2 Oleum is considered as a solution of SO3 in H2SO4, which is obtained by passing SO3 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H2O then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum. For example, a oleum bottle labelled as '109% H2SO4' means the 109 g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of H2O which combines with all the free SO3 present in oleum to form H2SO4 as SO3 + H2O  H2SO4 1. What is the % of free SO3 in an oleum that is labelled as '104.5 % H2SO4'? (A) 10 2. (B) 20 (B) 7.46 L at STP (C) 3.73 L at STP (D) 11.2 L at STP If excess water is added into a bottle sample labelled as '112 % H2SO4' and is reacted with 5.3 g Na2CO3, then find the volume of CO2 evolved at 1 atm pressure and 300 K temperature after the completion of the reaction : (A) 2.46 L 4. (D) none of these 9.0 g water is added into oleum sample labelled as '112% H2SO4' then the amount of free SO3 remaining in the solution is : (A) 14.93 L at STP 3. (C) 40 (B) 24.6 L (C) 1.23 L (D) 12.3 L 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % of free SO3 in the sample is : (A) 74 (B) 26 (C) 20 (D) none of these Comprehension # 3 Equivalent weight = Molecular weight / Atomic weight n – factor n-factor is very important in redox as well as non-redox reactions. With the help of n-factor we can predicts the molar ratio of the reactant species taking part in reactions. The reaciprocal of n-factor's ratio of the reactants is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H + /OH – furnished per mole of acid/base. n-factor of a reactant is no. of moles of electrons lost or gained per mole of reactant. Example 1 1. In acidic medium : KMnO4 (n = 5)  Mn2+ 2. In basic medium : KMnO4 (n = 3)  Mn2+ 3. In neutral medium : KMnO4 (n = 1)  Mn6+ Example 2 FeC2O4  Fe3+ + 2CO2 Total no. of moles of e– lost by 1 mole of FeC2O4 = 1 + 1 × 2  3  1. n-factor of FeC2O4 = 3 n-factor of Ba(MnO4)2 in acidic medium is : (A) 2 (B) 6 (C) 10 (D) none of these REDOX REACTION AND EQUIVALENT CONCEPT 2. For the reaction, H3PO2 + NaOH  NaH2PO2 + H2O What is the equivalent weight of H3PO2 ? (mol. wt. is M) (A) M 3. (C) M/3 (D) none of these For the reaction, Fe0.95O (molar mass : M)  Fe2O3. What is the eq. wt. of Fe0.95O ? (A) 4. (B) M/2 M 0.85 (B) M 0.95 (C) M 0.8075 (D) none of these In the reaction, xVO + yFe2O3 FeO + V2O5. What is the value of x and y respectively ? (A) 1, 1 (B) 2, 3 (C) 3, 2 (D) none of these Comprehension # 4 Some amount of “20V” H2O2 is mixed with excess of acidified solution of KI. The iodine so liberated required 200 mL of 0.1 N Na2S2O3 for titration. 1. 2. 3. The volume of H2O2 solution is : (A) 11.2 mL (B) 37.2 mL (C) 5.6 mL (D) 22.4 mL The mass of K2Cr2O7 needed to oxidise the above volume of H2O2 solution is : (A) 3.6 g (B) 0.8 g (C) 4.2 g (D) 0.98 g The volume of O2 at STP that would be liberated by above H2O2 solution is : (A) 56 mL (B) 112 mL (C) 168 mL (D) 224 mL Comprehension # 5 Equivalent Mass The equivalent mass of a substance is defined as the number of parts by mass of it which combine with or displace 1.0078 parts by mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of chlorine. The equivalent mass of a substance expressed in grams is called gram equivalent mass. The equivalent mass of a substance is not constant. It depends upon the reaction in which the substance is participating. A compound may have different equivalent mass in different chemical reactions and under different experimental conditions. (A) Equivalent mass of an acid It is the mass of an acid in grams which contains 1.0078 g of replaceable H+ ions or it is the mass of acid which contains one mole of replaceable H+ ions. It may be calculated as : Equivalent mass of acid = Molecular mass of acid Basicity of acid Basicity of acid = number of replaceable hydrogen atoms present in one molecule of acid (B) Equivalent mass of a base It is the mass of the base which contains one mole of replaceable OH– ions in molecule. Equivalent mass of base = Molecular mass of base Aciditiy of base Acidity of base = Number of replaceable OH– ions present in one molecule of the base Equivalent mass of an oxidising agent (a) Electron concept : Equivalent mass of oxidising agent = Molecular mass of oxidising agent Number of electrons gained by one molecule (b) Oxidation number concept : Equivalent mass of oxidising agent = Molecular mass of oxidising agent Total change in oxidation number per molecule of oxidising agent 1. Equivalent mass of Ba(MnO4)2 in acidic medium is : (where M stands for molar mass) (A) M/5 (B) M/6 (C) M/10 (D) M/2 2. Equivalent mass of Fe0.9 O in reaction with acidic K2Cr2O7 is : (M = Molar mass) (A) 7 M/10 (B)10 M/7 (C) 7 M/9 (D) 9 M/7 3. Equivalent weight of oxalic acid salt in following reaction is : (Atomic masses : O = 16, C = 12, K = 39) H2C2O4 + Ca(OH)2  CaC2O4 + H2O (A) 90 (B) 45 (C) 64 (D) 128 EXERCISE - 3 : PART # I 1. 3. 5. 7. A (q), B (p), C (s), D (r) 2. A (s), B (q), C (r), D (p) 4. A (p, s), B (s), C (p, q), D (r) 6. A (p, r), B (q, r), C (p, q, r), D (q, r) A (s), B (p), C (q), D (r) A (p), B (r,s ), C (r), D (r, q) A (p, s), B (q, r), C (p, q, s), D (r) PART # II Comprehension # 1 : 1. A 2. B 3. B 4. B Comprehension # 2 : 1. B 2. C 3. C 4. B Comprehension # 3 : 1. C 2. A 3. C 4. B Comprehension # 4 : 1. C 2. D 3. B Comprehension # 5 : 1. C 2. B 3. C