Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

(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: P4(s)→HPO−2(aq)P4s→HPO2-(aq) The P atom is balanced as: P4(s)→4HPO−2(aq)P4s→4HPO2-(aq) The O atom is balanced by adding 8 H2O molecules: P4(s)+8H2O→4HPO−2(aq)P4s+8H2O→4HPO2-(aq)The H atom is balanced by adding 12 H+ ions: P4(s)+8H2O→4HPO−2(aq)+12H+P4s+8H2O→4HPO2-(aq)+12H+ The charge is balanced by adding e– as: P4(s)+8H2O→4HPO−2(aq)+12H++8e−P4s+8H2O→4HPO2-(aq)+12H++8e- ...(i) The reduction half equation is: The P atom is balanced as: P4(s)→4PH3(g)P4(s)→4PH3(g) The H is balanced by adding 12 H+ as: P4(s)+12H+→4PH3(g)P4(s)+12H+→4PH3(g) The charge is balanced by adding 12e– as: P4(s)+12H++12e−→4PH3(g)P4(s)+12H++12e-→4PH3(g) ...(ii) By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as: 5P4(s)+24H2O→12HPO−2+8PH3(g)+12H+5P4(s)+24H2O→12HPO2-+8PH3(g)+12H+As, the medium is basic, add 12OH– both sides as:5P4(s)+12H2O+12OH−→12HPO−2+8PH3(g)5P4(s)+12H2O+12OH-→12HPO2-+8PH3(g)This is the required balanced equation. Oxidation number method: Let, total no of P reduced = x∴∴Total no of P oxidised = 4– xP4(s)+OH−→xPH3(g)+(4−x)HPO−2P4(s)+OH-→xPH3(g)+4-xHPO2- ... (i)Total decrease in oxidation number of P =x × 3 = 3x Total increase in oxidation number of P = (4– x) × 2 = 8 – 2x∵∵3x = 8 – 2x x = 8/5From (i), 5P4(s)+5OH−→8PH3(g)+12HPO−25P4(s)+5OH-→8PH3(g)+12HPO2-Since, reaction occures in basic medium, the charge is balanced by adding 7OH– on LHS as:5P4(s)+12OH−→8PH3(g)+12HPO−25P4(s)+12OH-→8PH3(g)+12HPO2-The O atoms are balanced by adding 12H2O as:5P4(s)+12H2O+12OH−→+12HPO−2+8PH3(g)5P4(s)+12H2O+12OH-→+12HPO2-+8PH3(g)This is the required balanced equation. (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, Cl2O7is 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. read less