How to Balance Redox Reactions in Acidic Solutions: A Comprehensive Guide

Introduction

Redox reactions, or oxidation-reduction reactions, are fundamental processes in chemistry involving the transfer of electrons between reactants. In acidic solutions, such reactions can be particularly challenging to balance due to the presence of hydrogen ions (H ) and water (H2O). This guide provides a detailed step-by-step method to balance a specific redox reaction in an acidic solution, using the example of MnO-4 S2O3-2 → S4O6-2 Mn2 .

Step-by-Step Guide to Balancing Redox Reactions in Acidic Solutions

Step 1: Identify the Half-Reactions

The first step involves identifying the oxidation and reduction half-reactions.

Oxidation Half-Reaction

The thiosulfate ion (S2O3-2) is oxidized to tetrathionate (S4O6-2).

2 S2O3-2 → S4O6-2

Reduction Half-Reaction

The permanganate ion (MnO-4) is reduced to manganese ion (Mn2 ).

MnO-4 → Mn2

Step 2: Balance Each Half-Reaction

Oxidation Half-Reaction

To balance the sulfur atoms, we multiply the equation by a factor to ensure the correct number of sulfur atoms.

2 S2O3-2 → S4O6-2

To balance the oxygen atoms, add water (H2O).

2 S2O3-2 → S4O6-2 2 H2O

To balance the hydrogen atoms, add H .

2 S2O3-2 4 H → S4O6-2 2 H2O

To balance the charge, add electrons (e-) to the left side.

2 S2O3-2 4 H 2 e- → S4O6-2 2 H2O

Reduction Half-Reaction

To balance the manganese, the equation already has 1 manganese atom on both sides.

MnO-4 → Mn2

To balance the oxygen atoms, add water (H2O).

MnO-4 8 H → Mn2 4 H2O

To balance the charge, add electrons (e-) to the left side.

MnO-4 8 H 5 e- → Mn2 4 H2O

Step 3: Combine the Half-Reactions

To combine the half-reactions, we need to equalize the number of electrons by multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2.

Oxidation half-reaction (multiplied by 5):

10 S2O3-2 20 H 10 e- → 5 S4O6-2 10 H2O

Reduction half-reaction (multiplied by 2):

2 MnO-4 16 H 10 e- → 2 Mn2 8 H2O

Adding the half-reactions together:

10 S2O3-2 20 H 2 MnO-4 16 H 10 e- → 5 S4O6-2 10 H2O 2 Mn2 8 H2O 10 e-

Simplify by removing the electrons (e-) and combining the water (H2O) terms.

10 S2O3-2 2 MnO-4 36 H → 5 S4O6-2 2 Mn2 18 H2O

Step 4: Simplify the Equation

Combine and simplify the hydrogen (H ) and water (H2O) terms to get the final balanced equation.

10 S2O3-2 2 MnO-4 16 H → 5 S4O6-2 2 Mn2 8 H2O

Conclusion

The balanced redox reaction in an acidic solution is:

10 S2O3-2 2 MnO-4 16 H → 5 S4O6-2 2 Mn2 8 H2O

This approach ensures the reaction is balanced by both mass and charge.