# 8.3.2 Balancing of Redox Reactions

Two methods are used to balance chemical equations for redox processes. One of these methods is based on the change in the oxidation number of reducing agent and the oxidising agent and the other method is based on splitting the redox reaction into two half reactions — one involving oxidation and the other involving reduction. Both these methods are in use and the choice of their use rests with the individual using them.

(a) Oxidation Number Method: In writing equations for oxidation-reduction reactions, just as for other reactions, the compositions and formulas must be known for the substances that react and for the products that are formed.

The oxidation number method is now best illustrated in the following steps:

Step 1: Write the correct formula for each reactant and product.

Step 2: Identify atoms which undergo change in oxidation number in the reaction by assigning the oxidation number to all elements in the reaction.

Step 3: Calculate the increase or decrease in the oxidation number per atom and for the entire molecule/ion in which it occurs. If these are not equal then multiply by suitable number so that these become equal. (If you realise that two substances are reduced and nothing is oxidised or vice-versa, something is wrong. Either the formulas of reactants or products are wrong or the oxidation numbers have not been assigned properly).

Step 4: Ascertain the involvement of ions if the reaction is taking place in water, add H+ or OH– ions to the expression on the appropriate side so that the total ionic charges of reactants and products are equal. If the reaction is carried out in acidic solution, use H+ ions in the equation; if in basic solution, use OH– ions.

Step 5: Make the numbers of hydrogen atoms in the expression on the two sides equal by adding water (H2O) molecules to the reactants or products. Now, also check the number of oxygen atoms. If there are the same number of oxygen atoms in the reactants and products, the equation then represents the balanced redox reaction.

Let us now explain the steps involved in the method with the help of a few problems given below:

(b) Half Reaction Method: In this method, the two half equations are balanced separately and then added together to give balanced equation. Suppose we are to balance the equation showing the oxidation of Fe2+ ions to Fe3+ ions by dichromate ions (Cr2O7)2– in acidic medium, wherein, Cr2O72– ions are reduced to Cr3+ ions. The following steps are involved in this task.

Fe2+ (aq) + Cr2O72– (aq) → Fe3+ (aq) + Cr3+ (aq)

Step 2: Separate the equation into halfreactions:

Oxidation half :

+2                +3

Fe2+ (aq) → Fe3+ (aq)

Reduction half:

+6 -2                 +3

Cr2O72– (aq) → Cr3+ (aq)

Step 3: Balance the atoms other than O and H in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms. For the reduction half reaction, we multiply the Cr3+ by 2 to balance Cr atoms.

Cr2O72– (aq) → 2 Cr3+ (aq)

Step 4: For reactions occurring in acidic medium, add H2O to balance O atoms and H+ to balance H atoms.

Thus we get,

Cr2O72– (aq) + 14H+ (aq)→ 2 Cr3+ (aq) + 7H2O (l)

Step 5: Add electrons to one side of the half reaction to balance the charges. If need be, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate number.

The oxidation half reaction is thus rewritten to balance the charge:

Fe2+ (aq) → Fe3+ (aq) + e-

Now in the reduction half reaction there are net twelve positive charges on the left hand side and only six positive charges on the right hand side. Therefore, we add six electrons on the left side.

Cr2O72– (aq) + 14H+ (aq) + 6 e- → 2 Cr3+ (aq) + 7H2O (l)

To equalise the number of electrons in both the half reactions, we multiply the oxidation half reaction by 6 and write as :

6Fe2+ (aq) + Cr2O72– (aq) + 14H+ (aq) → 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7H2O (l)

Step 7: Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number of atoms and the charges.

For the reaction in a basic medium, first balance the atoms as is done in acidic medium. Then for each H+ ion, add an equal number of OH ions to both sides of the equation.

Where H+ and OH appear on the same side of the equation, combine these to give H2O.