Balancing of Redox Reactions

When we write a reaction, then we have to balance it with some numeric number to respective element. Sometimes we do in simply manner, but there are some method from which we can balance an equation in an easy way.

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 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:

As the name suggests, in writing equations for oxidation – reduction reactions, just as for other reactions, the composition and formulas must be known for the substances that react and for the products that are formed.

This method is followed by some step by step which are defined below:

Step 1: Write the correct formula for each reactant and product and also write down the oxidation number of each element in the reaction.

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: After identifying the element which undergo change in oxidation number, 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.

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 carries out in acidic medium, use H⁺ ions in the equation, if in basic solution, use OH^- ions.

Step 5: Make the number of hydrogen atoms in the expression on the two sides equal by adding water 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.

Example- Write the net ionic equation for the reaction of potassium dichromate(6), K₂Cr₂O₇ with sodium sulphate, Na₂SO₃, in an acid solution to give chromium(3) ion and the sulphate ion.

Solution

Step 1: The skeletal ionic equation is:

Cr₂O₇^2–(aq) + SO₃^2-(aq) → Cr³⁺(aq)+ SO₄^2–(aq)

Step 2: Assign oxidation numbers for Cr and S

+6 –2 +4 –2 +3 +6 –2

Cr₂O₇^2–(aq) + SO₃^2–(aq) → Cr(aq)+SO₄^2–(aq)

This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.

Step 3: Calculate the increase and decrease of oxidation number, and make them equal:

+6 –2 +4 –2 +3 +6 -2

Cr₂O₇^2–(aq) + 3SO₃^2– (aq) → 2Cr³⁺ (aq) + 3SO₄^2– (aq)

Step 4: As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add 8H⁺ on the left to make ionic charges equal.

Cr₂O₇^2–(aq) + 3SO₃^2–(aq)+ 8H⁺→ 2Cr³⁺(aq)+ 3SO₄^2–(aq)

Step 5: Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., 4H₂O) on the right to achieve balanced redox change.

Cr₂O₇^2–(aq) + 3SO₃^2–(aq)+ 8H⁺(aq) →2Cr³⁺(aq) + 3SO₄^2–(aq) +4H₂O (l)

(b) Half Reaction Method:

In this method, the two half equations are balanced separately and then added together to give balanced equation.

For example, we are to balance the equation showing the oxidation of Fe²⁺ ions to Fe³⁺ by dichromate ions in acidic medium, wherein, Cr₂O₇^2- ions are reduced to Cr³⁺ ions, The following steps are involved in this task:

Step 1: Produce unbalanced equation for the reaction in ionic form:

Fe²⁺⁽aq) + Cr₂O₇^2–(aq) → Fe³⁺(aq) + Cr³⁺(aq)

Step 2: Separate the equation into half reactions:

+2 +3

Oxidation half: Fe²⁺ (aq) → Fe³⁺(aq)

+6 –2 +3

Reduction half: Cr₂O₇^2–(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 of Fe atoms. For the reduction half reaction, we multiply the Cr³⁺ by 2 to balance Cr atoms.

Cr₂O₇^2–(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

Cr₂O₇^2–(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.

Cr₂O₇^2–(aq) + 14H⁺(aq) + 6e^–→ 2Cr³⁺(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) → 6Fe³⁺(aq) + 6e^–

Step 6: We add the two half reactions to achieve the overall reaction and cancel the electrons on each side. This gives the net ionic equation as:

6Fe²⁺(aq) + Cr₂O₇^2–(aq) + 14H⁺(aq) → 6 Fe³⁺(aq) +2Cr³⁺(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.

Note- For the reaction in 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.