Ch-8. Redox Reactions (Exercise) Page 275

Question 27:	Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes (ii) An aqueous solution AgNO3 with platinum electrodes (iii) A dilute solution of H2SO4 with platinum electrodes (iv) An aqueous solution of CuCl2 with platinum electrodes.
Answer :	(i) AgNO3 ionizes in aqueous solutions to form Ag+ and  ions. On electrolysis, either Ag+ ions or H2O molecules can be reduced at the cathode. But the reduction potential of Ag+ ions is higher than that of H2O. Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H2O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H2O molecules. Therefore, Ag metal gets oxidized at the anode. (ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O2. At the cathode, Ag+ ions are reduced and get deposited. (iii) H2SO4 ionizes in aqueous solutions to give H+ and ions. On electrolysis, either of H+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of H+ ions is higher than that of H2O molecules. Hence, at the cathode, H+ ions are reduced to liberate H2 gas. On the other hand, at the anode, either of ions or H2O molecules can get oxidized. But the oxidation of involves breaking of more bonds than that of H2O molecules. Hence, ions have a lower oxidation potential than H2O. Thus, H2O is oxidized at the anode to liberate O2molecules. (iv) In aqueous solutions, CuCl2 ionizes to give Cu2+ and Cl– ions as: On electrolysis, either of Cu2+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of Cu2+ is more than that of H2O molecules. Hence, Cu2+ ions are reduced at the cathode and get deposited. Similarly, at the anode, either of Cl– or H2O is oxidized. The oxidation potential of H2O is higher than that of Cl–. But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl– ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl– ions are oxidized at the anode to liberate Cl2 gas.
Question 28:	Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Answer :	A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt. The order of the increasing reducing power of the given metals is Cu < Fe < Zn < Al < Mg. Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is given below: Mg>Al> Zn> Fe,>Cu
Question 29:	Given the standard electrode potentials, K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –2.37V. Cr3+/Cr = –0.74V Arrange these metals in their increasing order of reducing power.
Answer :	The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is Ag < Hg < Cr < Mg < K.
Question 30:	Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place, further show: (i) which of the electrode is negatively charged, (ii) the carriers of the current in the cell, and (iii) individual reaction at each electrode.
Answer :	The galvanic cell corresponding to the given redox reaction can be represented as: (i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+ and the leaving electrons accumulate on this electrode. (ii) Ions are the carriers of current in the cell. (iii) The reaction taking place at Zn electrode can be represented as: And the reaction taking place at Ag electrode can be represented as:  . (iv) In aqueous solutions, CuCl2 ionizes to give Cu2+ and Cl– ions as: On electrolysis, either of Cu2+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of Cu2+ is more than that of H2O molecules. Hence, Cu2+ ions are reduced at the cathode and get deposited. Similarly, at the anode, either of Cl– or H2O is oxidized. The oxidation potential of H2O is higher than that of Cl–. But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl– ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl– ions are oxidized at the anode to liberate Cl2 gas.

Ch-8. Redox Reactions (Exercise) Page 274

Question 18:	Balance the following redox reactions by ion-electron method: (a)  (aq) + I– (aq) → MnO2 (s) + I2(s) (in basic medium) (b)  (aq) + SO2 (g) → Mn2+ (aq) +  (aq) (in acidic solution) (c) H2O2 (aq) + Fe2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution) (d) + SO2(g) → Cr3+ (aq) + (aq) (in acidic solution)
Answer :	(a) Step 1: The two half reactions involved in the given reaction are: Oxidation half reaction: Reduction half reaction: Step 2: Balancing I in the oxidation half reaction, we have: Now, to balance the charge, we add 2 e– to the RHS of the reaction. Step 3: In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction. Now, to balance the charge, we add 4 OH– ions to the RHS of the reaction as the reaction is taking place in a basic medium. Step 4: In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS. Step 5: Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have: Step 6: Adding the two half reactions, we have the net balanced redox reaction as: (b)Following the steps as in part (a), we have the oxidation half reaction as: And the reduction half reaction as: Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as: (c) Following the steps as in part (a), we have the oxidation half reaction as: And the reduction half reaction as: Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as: (d) Following the steps as in part (a), we have the oxidation half reaction as: And the reduction half reaction as: Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
Question 19:	Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
Answer :	(a) The O.N. (oxidation number) of P decreases from 0 in P4 to – 3 in PH3 and increases from 0 in P4 to + 2 in. Hence, P4 acts both as an oxidizing agent and a reducing agent in this reaction. Ion–electron method: The oxidation half equation is: The P atom is balanced as: The O.N. is balanced by adding 8 electrons as: The charge is balanced by adding 12OH– as: The H and O atoms are balanced by adding 4H2O as: The reduction half equation is: The P atom is balanced as The O.N. is balanced by adding 12 electrons as: The charge is balanced by adding 12OH– as: The O and H atoms are balanced by adding 12H2O as: By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as: (b) The oxidation number of N increases from – 2 in N2H4 to + 2 in NO and the oxidation number of Cl decreases from + 5 in  to – 1 in Cl–. Hence, in this reaction, N2H4 is the reducing agent and  is the oxidizing agent. Ion–electron method: The oxidation half equation is: The N atoms are balanced as: The oxidation number is balanced by adding 8 electrons as: The charge is balanced by adding 8 OH–ions as: The O atoms are balanced by adding 6H2O as: The reduction half equation is: The oxidation number is balanced by adding 6 electrons as: The charge is balanced by adding 6OH– ions as: The O atoms are balanced by adding 3H2O as: The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as: Oxidation number method: Total decrease in oxidation number of N = 2 × 4 = 8 Total increase in oxidation number of Cl = 1 × 6 = 6 On multiplying N2H4 with 3 and  with 4 to balance the increase and decrease in O.N., we get: The N and Cl atoms are balanced as: The O atoms are balanced by adding 6H2O as: This is the required balanced equation. (c) The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in and the oxidation number of O increases from – 1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidizing agent and H2O2 is the reducing agent. Ion–electron method: The oxidation half equation is: The oxidation number is balanced by adding 2 electrons as: The charge is balanced by adding 2OH–ions as: The oxygen atoms are balanced by adding 2H2O as: The reduction half equation is: The Cl atoms are balanced as: The oxidation number is balanced by adding 8 electrons as: The charge is balanced by adding 6OH– as: The oxygen atoms are balanced by adding 3H2O as: The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:  Oxidation number method: Total decrease in oxidation number of Cl2O7 = 4 × 2 = 8 Total increase in oxidation number of H2O2 = 2 × 1 = 2 By multiplying H2O2 and O2 with 4 to balance the increase and decrease in the oxidation number, we get: The Cl atoms are balanced as: The O atoms are balanced by adding 3H2O as: The H atoms are balanced by adding 2OH– and 2H2O as: This is the required balanced equation.
Question 20:	What sorts of informations can you draw from the following reaction ?
Answer :	The oxidation numbers of carbon in (CN)2, CN– and CNO– are +3, +2 and +4 respectively. These are obtained as shown below: Let the oxidation number of C be x. (CN)2 2(x – 3) = 0 ∴x = 3 CN– x – 3 = –1 ∴x = 2 CNO– x – 3 – 2 = –1 ∴x = 4 The oxidation number of carbon in the various species is: It can be easily observed that the same compound is being reduced and oxidised simultaneously in the given equation. Reactions in which the same compound is reduced and oxidised is known as disproportionation reactions. Thus, it can be said that the alkaline decomposition of cyanogen is an example of disproportionation reaction.
Question 21:	The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ion. Write a balanced ionic equation for the reaction.
Answer :	The given reaction can be represented as: The oxidation half equation is: The oxidation number is balanced by adding one electron as: The charge is balanced by adding 4H+ ions as: The O atoms and H+ ions are balanced by adding 2H2O molecules as: The reduction half equation is: The oxidation number is balanced by adding one electron as: The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
Question 22:	Consider the elements: Cs, Ne, I and F (a) Identify the element that exhibits only negative oxidation state. (b) Identify the element that exhibits only postive oxidation state. (c) Identify the element that exhibits both positive and negative oxidation states. (d) Identify the element which exhibits neither the negative nor does the positive oxidation state.
Answer :	(a) F exhibits only negative oxidation state of –1. (b) Cs exhibits positive oxidation state of +1. (c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of – 1, + 1, + 3, + 5, and + 7. (d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.
Question 23:	Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Answer :	The given redox reaction can be represented as: The oxidation half reaction is: The oxidation number is balanced by adding two electrons as: The charge is balanced by adding 4H+ ions as: The O atoms and H+ ions are balanced by adding 2H2O molecules as: The reduction half reaction is: The chlorine atoms are balanced as: The oxidation number is balanced by adding electrons The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
Question 24:	Refer to the periodic table given in your book and now answer the following questions: (a) Select the possible non metals that can show disproportionation reaction. (b) Select three metals that can show disproportionation reaction.
Answer :	In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states. (a) P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states. (b) Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or more oxidation states.
Question 25:	In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?
Answer :	The balanced chemical equation for the given reaction is given as: Thus, 68 g of NH3 reacts with 160 g of O2. Therefore, 10g of NH3 reacts with g of O2, or 23.53 g of O2. But the available amount of O2 is 20 g. Therefore, O2 is the limiting reagent (we have considered the amount of O2 to calculate the weight of nitric oxide obtained in the reaction). Now, 160 g of O2 gives 120g of NO. Therefore, 20 g of O2 givesg of N, or 15 g of NO. Hence, a maximum of 15 g of nitric oxide can be obtained.
Question 26:	Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible: (a) Fe3+(aq) and I–(aq) (b) Ag+(aq) and Cu(s) (c) Fe3+ (aq) and Cu(s) (d) Ag(s) and Fe3+(aq) (e) Br2(aq) and Fe2+(aq)
Answer :	(a) The possible reaction between is given by, E° for the overall reaction is positive. Thus, the reaction between  and  is feasible. (b) The possible reaction between is given by, E° positive for the overall reaction is positive. Hence, the reaction between  and  is feasible. (c) The possible reaction between and is given by, E° positive for the overall reaction is positive. Hence, the reaction between  and  is feasible. (d) The possible reaction betweenand is given by, Here, E° for the overall reaction is negative. Hence, the reaction between and  is not feasible. (e) The possible reaction betweenand is given by, Here, E° for the overall reaction is positive. Hence, the reaction between and  is feasible.