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246 CHEMISTRY (a) Oxidation Number Method: In writing Problem 7.7 equations for oxidation-reduction reactions, Why do the following reactions proceed justasforotherreactions,thecompositionsand differently ? formulas must be known for the substances Pb3O 4 8HCl → 3PbCl + 2l + 2HO an2 that react and for the products that are formed. The oxidation number method is now Pb3O 4 4HNO → 23b(NO ) + Pb3 2 2 best illustrated in the following steps: 2H O 2 Step 1: Write the correct formula for each Solution Pb O is actually a stoichiometric reactant and product. 3 4 Step 2: Identify atoms which undergo mixture of 2 mol of PbO and 1 mol of PbO . In PbO , lead is present in +4 change in oxidation number in the reaction 2 2 by assigning the oxidation number to all oxidation state, whereas the stable elements in the reaction. oxidation state of lead in PbO is +2. PbO2 Step 3: Calculate the increase or decrease thus can act as an oxidant (oxidising in the oxidation number per atom and for agent) and, therefore, can oxidise Cl–ion of HCl into chlorine. We may also keep in the entire molecule/ion in which it occurs. If these are not equal then multiply by suitable mind that PbO is a basic oxide. Therefore, the reaction number so that these become equal. (If you realise that two substances are reduced and Pb3O 4 8HCl → 3PbCl + C2 + 4H2O 2 nothing is oxidised or vice-versa, something can be splitted into two reactions namely: is wrong. Either the formulas of reactants or products are wrong or the oxidation numbers 2PbO + 4HCl → 2PbCl + 22 O 2 (acid-base reaction) have not been assigned properly). Step 4: Ascertain the involvement of ions if + –1 +2 0 + PbO + 4HCl → PbCl + Cl +2H O the reaction is taking place in water, add H or 2 2 2 2 OH –ions to the expression on the appropriate (redox reaction) side so that the total ionic charges of reactants Since HNO itself is an oxidising agent 3 and products are equal. If the reaction is therefore, it is unlikely that the reaction carried out in acidic solution, use H + ions in may occur between PbO and HNO . – 2 3 the equation; if in basic solution, use OH ions. However, the acid-base reaction occurs Step 5 : Make the numbers of hydrogen between PbO and HNO 3as: atoms in the expression on the two sides 2PbO + 4HNO → 3Pb(NO ) + 2H3 2 2 equal by adding water (H 2O) molecules to the reactants or products. Now, also check the It is the passive nature of PbO 2gainst HNO 3that makes the reaction different number of oxygen atoms. If there are the same from the one that follows with HCl. number of oxygen atoms in the reactants and products, the equation then represents the 7.3.2 Balancing of Redox Reactions balanced redox reaction. Two methods are used to balance chemical Let us now explain the steps involved in equations for redox processes. One of these the method with the help of a few problems given below: methods is based on the change in the oxidation number of reducing agent and the Problem 7.8 oxidising agent and the other method is based Write the net ionic equation for the on splitting the redox reaction into two half reactions — one involving oxidation and the reaction of potassium dichromate(VI), K Cr O with sodium sulphite, Na SO , other involving reduction. Both these methods 2 2 7 2 3 are in use and the choice of their use rests in an acid solution to give chromium(III) ion and the sulphate ion. with the individual using them. Reprint 2026-27