Chemistry by Inam Jazbi: ION ELECTRON METHOD for Balancing of Redox Reactions

Rules for Balancing Redox Reactions by Ion Electron Method (Half-Reaction Method)

Jette and LaMev developed the method for balancing redox-reactions by ion electron method in 1927.

1. Write a skeleton equation which includes those reactants and products that contain the elements undergoing a change in oxidation state.

2. Convert molecular skeleton equation into ionic equation. (If the given equation is already in the iionci form, then this steop is omitted).

3. Split ionic equation into two half-reactions or partial equations i.e. the oxidation and reduction equation. (one for the species that is oxidized and its products and for the species that is reduced and its products).

4. Balance the atoms other than O and H by inserting suitable coefficients in each half reaction separately.

5. Balance each partial equation in terms of atoms.

(a). For reactions in an acidic or neutral medium, add one H₂O for each O atom to balance O and one H⁺ for each H atom to balance H. The O atoms are balanced first.

(b) In case of basic medium, OH^– and H₂O are added to balance oxygen and hydrogen respectively. For reactions in a basic medium, to balance O atom add two OH^– per O atom to the side that is deficient in O while add one H₂O on the other side at the same time. To balance H atom, add one H₂O per H atom to the side where H is less while add one OH^– on the other side at the same time.

For reactions in a basic medium, balance O and H atoms as balanced in acidic medium. Then add OH^– ions to both sides of the equation equal to the number of H⁺ ions. (If there is no H⁺ then do not add OH^– ions). Where H⁺ and OH^– appear on the same side of the equation, combine the ions to give H₂O.

6. Balance the charge in each half equation by inserting e^– (electrons) as a reactant or product on either left or right side. Electrons are added to the left in the partial equation for the reduction equation and to the right in the partial equation for oxidation reaction.

7. Equalize the loss and gain of electrons in both half equations by multiplying one or both half reactions by appropriate co-efficients.

8. Add the two half equations after canceling the electrons. In the sum equation, cancel out any species common to both sides. If H⁺, OH^– or H₂O appears on both sides of the final equation, cancel out the duplications.

(To verify equation is atomically and electrically balanced, count net charges on both sides of the equation, if charges are same then equation is balanced).

Balancing Redox Equations in Acidic medium by Ion Electron Method

Solution

First molecular skeleton equation splits into ionic equation and then oxidation number of only those species have been written who have undergone any change (the species that do not undergo any change in oxidation number do not appear in equation).

HNO₃ being a strong electrolyte is ionized into H⁺ and NO₃^– ions. Here the nitrate ion (NO₃^–) is the oxidizing agent as its N atom undergoes a decrease in oxidation state. H₂S being a weak electrolyte is preferred to be in unionized form due to its slight degree of ionization. Here H₂S is a reducing agent since its sulphur atom undergoes an increase in oxidation state.

Splitting of Ionic Equation into two half-reactions

Overall Redox Reaction

To balance e^–, oxidation half-equation is multiplied by 2 (so that 6e^– are produced) and the reduction half-equation is multiplied by 3 (so that the same 6e^– are consumed) and then add resultant equations after canceling e^–. The equation may be converted back to molecular form by combining ions (NO₃^– and H⁺) together.

Solution

First molecular equation splits into ionic form and then oxidation number of only those species have been written who have undergone any change (the species that do not undergo any change in oxidation number do not appear in equation).

HNO₃ being a strong electrolyte is ionized into H⁺ and NO₃^– ions. Here the nitrate ion (NO₃^–) is the oxidizing agent as its N atom undergoes a decrease in oxidation state. CuS is ionized to give Cu²⁺ and S^2– ion. Here CuS is a reducing agent since its sulphur atom undergoes an increase in oxidation state.

Splitting of ionic equation into two half-reactions

Balancing of Equation No. I (Oxidation Half Reaction)

Balancing of Equation No. II (Reduction half Reaction)

Overall Redox Reaction

To balance e^–, oxidation half-equation is multiplied by 3 (so that 24e^– are produced) and the reduction half-equation is multiplied by 8 (so that the same 24e^– are consumed) and then add resultant equations after canceling e^–.

Solution

Splitting of ionic equation into two half-reactions

After writing oxidation number of those species that undergo any change, split ionic equation into two half reactions:

Balancing of Equation No. I

Balancing of Equation No. II

Overall Redox Reaction

To balance e⁻, oxidation half-equation is multiplied by 5 (so that 10e⁻ are produced) and the reduction half-equation is multiplied by 2 (so that the same 10e⁻ are consumed) and then add resultant equations after canceling e⁻.

Solution

Splitting of ionic equation into two half-reactions

After writing oxidation number of those species that undergo any change, split ionic equation into two half reaction:

Balancing of Equation No. I

Balancing of Equation No. II

Overall Redox Reaction

Since electrons are same in both equations, so they are added together

Solution

Splitting of ionic equation into two half-reactions

After writing oxidation number of those species that undergo any change, split ionic equation into two half reaction:

Balancing of Equation No. I (oxidation half-reaction)

As the reaction occurs in basic medium, and there are 5H⁺ ions, add 5OH^– ions to both sides of the equation. Combining the H⁺ and OH^– ions to form H₂O and subtracting 1 H₂O from both sides, we write:

Balancing of Equation No. II (Reduction Half Equation)

As the reactions occurs in basic medium, and there are 2H⁺ ion, add 2OH^– ion to both sides of the equation. Combining the H⁺ and OH^– ions to form H₂O and subtracting 1H₂O from both sides, we write:

Overall Redox Reaction

To equalize the number of electrons, eq. I is multiplied by 2 and eq. II by 3 and then resultant equations are added.

Solution

Splitting of ionic equation into two half-reactions

After writing oxidation number of those species that undergo any change, split ionic equation into two half reaction:

Balancing of Equation No. I (oxidation half-reaction)

Balancing of Equation No. II

As the reactions occurs in basic medium, and there are 1H⁺ ion, add 1OH^– ion to both sides of the equation. Combining the H⁺ and OH^– ions to form H₂O and subtracting 1H₂O from both sides, we write:

Overall Redox Reaction

To equalize the number of electrons, eq. I is multiplied by 2 and eq. II by 3 and then resultant equations are added.

Solution

Splitting of ionic equation into two half-reactions

After writing oxidation number of those species that undergo any change, split ionic equation into two half reaction:

Balancing of Equation No. I

Balancing of Equation No. II

Overall Redox Reaction

To equalize the number of electrons, eq. I is multiplied by 2 and eq. II by 3 and then resultant equations are added.

Solution

Splitting of ionic equation into two half-reactions

After writing oxidation number of those species that undergo any change, split ionic equation into two half reaction:

Note; This is an auto-redox reaction in which a single specie undergoes simultaneous oxidation and reduction.

Balancing of Equation No. I (reduction half-reaction)

Balancing of Equation No. II

As the reactions occurs in basic medium, and there are 12H⁺ ion, add 12OH⁻ ion to both sides of the equation. Combining the H⁺ and OH⁻ ions to form H₂O and subtracting 1H₂O from both sides, we write: