Ch-8. Redox Reactions (Exercise) Page 273

Question 10:	The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Answer :	The oxidation state of Ag in AgF2 is +2. But, +2 is an unstable oxidation state of Ag. Therefore, whenever AgF2 is formed, silver readily accepts an electron to form Ag+. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, AgF2 acts as a very strong oxidizing agent.
Question 11:	Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
Answer :	Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows: (i)P4 and F2 are reducing and oxidising agents respectively. If an excess of P4 is treated with F2, then PF3 will be produced, wherein the oxidation number (O.N.) of P is +3. However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the O.N. of P is +5. (ii)K acts as a reducing agent, whereas O2 is an oxidising agent. If an excess of K reacts with O2, then K2O will be formed, wherein the O.N. of O is –2. However, if K reacts with an excess of O2, then K2O2 will be formed, wherein the O.N. of O is –1. (iii)C is a reducing agent, while O2 acts as an oxidising agent. If an excess of C is burnt in the presence of insufficient amount of O2, then CO will be produced, wherein the O.N. of C is +2. On the other hand, if C is burnt in an excess of O2, then CO2 will be produced, wherein the O.N. of C is +4.
Question 12:	How do you count for the following observations? (a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction. (b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer :	(a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of the following reasons. (i) In a neutral medium, OH– ions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced. (ii) KMnO4 and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence, in alcohol, KMnO4 and toluene can react at a faster rate. The balanced redox equation for the reaction in a neutral medium is give as below: (b) When conc. H2SO4 is added to an inorganic mixture containing bromide, initially HBr is produced. HBr, being a strong reducing agent reduces H2SO4 to SO2 with the evolution of red vapour of bromine. But, when conc. H2SO4 is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H2SO4 to SO2.
Question 13:	Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions: (a) 2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq) (b) HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → 2Ag(s) + HCOO–(aq) + 4NH3(aq) + 2H2O(l) (c) HCHO (l) + 2Cu2+(aq) + 5 OH–(aq) → Cu2O(s) + HCOO–(aq) + 3H2O(l) (d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l) (e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
Answer :	(a) Oxidised substance → C6H6O2 Reduced substance → AgBr Oxidising agent → AgBr Reducing agent → C6H6O2 (b)Oxidised substance → HCHO Reduced substance →  Oxidising agent →  Reducing agent → HCHO (c) Oxidised substance → HCHO Reduced substance → Cu2+ Oxidising agent → Cu2+ Reducing agent → HCHO (d) Oxidised substance → N2H4 Reduced substance → H2O2 Oxidising agent → H2O2 Reducing agent → N2H4 (e) Oxidised substance → Pb Reduced substance → PbO2 Oxidising agent → PbO2 Reducing agent → Pb
Question 14:	Consider the reactions: (aq) + I2(s) → (aq) + 2I–(aq) aq) + 2Br2(l) + 5 H2O(l) → (aq) + 4Br–(aq) + 10H+(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer :	The average oxidation number (O.N.) of S in is +2. Being a stronger oxidising agent than I2, Br2 oxidises  to , in which the O.N. of S is +6. However, I2 is a weak oxidising agent. Therefore, it oxidises to , in which the average O.N. of S is only +2.5. As a result,reacts differently with iodine and bromine.
Question 15:	Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer :	F2 can oxidize Cl– to Cl2, Br– to Br2, and I– to I2 as: On the other hand, Cl2, Br2, and I2 cannot oxidize F– to F2. The oxidizing power of halogens increases in the order of I2 < Br2 < Cl2 < F2. Hence, fluorine is the best oxidant among halogens. HI and HBr can reduce H2SO4 to SO2, but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF. Again, I– can reduce Cu2+ to Cu+, but Br– cannot. Hence, hydroiodic acid is the best reductant among hydrohalic compounds. Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.
Question 16:	Why does the following reaction occur? (aq) + 2F– (aq) + 6H+(aq) → XeO3(g) + F2(g) + 3H2O(l) What conclusion about the compound Na4XeO6 (of which  is a part) can be drawn from the reaction.
Answer :	The given reaction occurs because oxidises  and reduces. In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in to +6 in XeO3 and the O.N. of F increases from –1 in F– to O in F2. Hence, we can conclude that is a stronger oxidising agent than F–.
Question 17:	Consider the reactions: (a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq) (b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq) (c) C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH–(aq) → C6H5COO–(aq) + 2Ag(s) + 4NH3 (aq) + 2 H2O(l) (d) C6H5CHO(l) + 2Cu2+(aq) + 5OH–(aq) → No change observed. What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?
Answer :	Ag+ and Cu2+ act as oxidising agents in reactions (a) and (b) respectively. In reaction (c), Ag+ oxidises C6H5CHO to C6H5COO–, but in reaction (d), Cu2+ cannot oxidise C6H5CHO. Hence, we can say that Ag+ is a stronger oxidising agent than Cu2+.