Balancing Redox Reactions Using the Ion-Electron Method: The Example of Cr2O72- H I2 → Cr3 IO3- H2O

The ion-electron method, also known as the half-reaction method, is a systematic approach for balancing redox reactions. This article will take you through the step-by-step process of balancing the redox reaction Cr2O72- H I2 → Cr3 IO3- H2O. Understanding this process is essential for mastering redox chemistry.

Step 1: Identify the Oxidation and Reduction Half-Reactions

The first step is to identify which species is being oxidized and which is being reduced.

Oxidation Half-Reaction

Iodine in I2 is oxidized to IO3-. The half-reaction is:

I2 → IO3-

Reduction Half-Reaction

Chromium in Cr2O72- is reduced to Cr3 . The half-reaction is:

Cr2O72- → Cr3

Step 2: Balance Each Half-Reaction

Balance the half-reactions in terms of atoms, charges, and water molecules.

Oxidation Half-Reaction

Balance the iodine, oxygen, hydrogen, and charges:

I2 → 2IO3-

I2 6H2O → 2IO3- 12H

I2 6H2O 8e- → 2IO3- 12H

Reduction Half-Reaction

Balance the chromium, oxygen, hydrogen, and charges:

Cr2O72- → 2Cr3

Cr2O72- → 2Cr3 7H2O

14H Cr2O72-u2192 2Cr3 7H2O

14H Cr2O72- 6e-u2192 2Cr3 7H2O

Step 3: Equalize the Number of Electrons

To ensure the reactants and products have the same number of electrons, we multiply the half-reactions by appropriate factors:

Oxidation Half-Reaction × 3:

3I2 18H2O 24e-u2192 6IO3- 36H

Reduction Half-Reaction × 4:

4Cr2O72- 56H 24e-u2192 8Cr3 28H2O

Step 4: Combine the Half-Reactions

Combine the two balanced half-reactions:

3I2 4Cr2O72- 56H 24e-u2192 6IO3- 8Cr3 28H2O 18H2O

3I2 4Cr2O72- 56H u2192 6IO3- 8Cr3 46H2O

Step 5: Simplify

Combine and simplify the water molecules:

3I2 4Cr2O72- 56H u2192 6IO3- 8Cr3 46H2O

The final balanced equation is:

3I2 4Cr2O72- 56H u2192 6IO3- 8Cr3 46H2O

This equation is now balanced in terms of both mass and charge.