18.2: Balancing Redox Reactions By Inspection

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Learning Objectives

Learn to balanced simple redox reactions by inspection.
Learn to balance complex redox reactions by the half reaction method.
Use the solvent, or parts of it, as a reactant or a product in balancing a redox reaction.

Balancing simple redox reactions can be a straightforward matter of going back and forth between products and reactants. For example, in the redox reaction of Na and Cl₂:

\[\ce{Na + Cl2 → NaCl}\]

it should be immediately clear that the Cl atoms are not balanced. We can fix this by putting the coefficient 2 in front of the product:

\[\ce{Na + Cl2 → 2NaCl}\]

However, now the sodium is unbalanced. This can be fixed by including the coefficient 2 in front of the Na reactant:

\[\ce{2Na + Cl2 → 2NaCl}\]

This reaction is now balanced. That was fairly straightforward; we say that we are able to balance the reaction by inspection. Many simple redox reactions can be balanced by inspection.

Example \(\PageIndex{1}\):

Balance this redox reaction by inspection.

SO₂ + O₂ → SO₃

Solution

There is one S atom on both sides of the equation, so the sulfur is balanced. However, the reactant side has four O atoms while the product side has three. Clearly we need more O atoms on the product side, so let us start by including the coefficient 2 on the SO₃:

SO₂ + O₂ → 2SO₃

This now gives us six O atoms on the product side, and it also imbalances the S atoms. We can balance both the elements by adding coefficient 2 on the SO₂ on the reactant side:

2SO₂ + O₂ → 2SO₃

This gives us two S atoms on both sides and a total of six O atoms on both sides of the chemical equation. This redox reaction is now balanced.

Exercise \(\PageIndex{1}\)

Balance this redox reaction by inspection.

Al + O₂ → Al₂O₃

Answer

4Al + 3O₂ → 2Al₂O₃

The first thing you should do when encountering an unbalanced redox reaction is to try to balance it by inspection.