1. 3.3 - Balancing Redox Reactions

2. Half-Reaction Method There are many ways to balance a redox reaction
We are going to learn the most effective and easiest way to do it
This is called the half-reaction method

3. Steps for Balancing Redox Reaction
Assign oxidation numbers
Split your reaction into 2 half-reactions
One for the oxidation
One for the reduction
Balance mass of the equation
In an acid:
To balance for H, use H+
To balance for O, use H2O
In a base:
To balance for H, use H2O
To balance for O, use OH-
Balance charges with electrons
Make electrons in each half-reaction equal by multiplying
Add half-reactions together; if necessary, reform compound/return spectator ions

4. Example 1 Al(s) + Cu2+(aq) → Al3+(aq) + Cu(s) Al(s) → Al3+(aq)
The mass is balanced because there are no O’s or H’s
Al(s) → Al3+(aq) + 3 e-
[Al(s) → Al3+(aq) + 3 e-]*2
2 Al(s) → 2 Al3+(aq) + 6 e-
Cu2+(aq) → Cu(s)
The mass is balanced because there are no O’s or H’s
Cu2+(aq) + 2 e-→ Cu(s)
[Cu2+(aq) + 2 e-→ Cu(s)]*3
3 Cu2+(aq) + 6 e-→ 3 Cu(s)
2 Al(s) → 2 Al3+(aq) + 6 e-
3 Cu2+(aq) + 6 e- → 3 Cu(s)
3 Cu2+(aq) + 2 Al(s) → 2 Al3+(aq) + 3 Cu(s)

5. Cu(s) + AgNO3 (aq) → Cu(NO3)2 (aq) + Ag(s)
Example 2
Next example:
Cu(s) + AgNO3 (aq) → Cu(NO3)2 (aq) + Ag(s)
Identify the elements undergoing oxidation (Cu) and reduction (Ag). The nitrate group (NO3) is a spectator ion which we will not include in our half-reactions.
oxidation
Cu → Cu e-
reduction
Ag+ + 1 e- → Ag

6. Example 2 Write the balanced half-reactions Balance electrons
Step 1
Step 2
Step 3
Write the balanced half-reactions
Balance electrons
Add half-reactions
Cu → Cu e-
Ag+ + 1 e- → Ag
× 2
2 Ag+ + 2e- → 2 Ag
Add equations together
Cu + 2 Ag+ → Cu Ag
Reform compound/return spectator ions
Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

7. MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O
Example 3
Here is a reaction occurring in an acid solution, which accounts for the presence of the H+ions. This example adds a little more complexity to our problem.
MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O
In this example, spectator ions have already been removed. Even though hydrogen and oxygen do not undergo changes in oxidation number they are not spectators and we need to work with them in our half-reactions.

8. Example 3
We determine that Mn undergoes reduction (+7 to +2) while Fe undergoes oxidation (+2 to +3). The iron half-reaction is straight forward, but the manganese reaction is more complex - we must include hydrogen and oxygen in its half-reaction:
oxidation
Fe2+→ Fe e-
reduction
MnO H+ + 5 e- → Mn H2O
To balance the manganese half-reaction - first balance for Mn and O atoms. Next balance the H atoms, and finally add enough electrons to balance the charge on both sides of the equation. Be sure you see what has been done so you can do it on your own.

9. Example 3 Write the balanced half-reactions Add half-reactions
Step 1
Step 2
Step 3
Write the balanced half-reactions
Add half-reactions
Fe2+→ Fe e-
× 5
5Fe2+→ 5Fe e-
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
MnO H+ + 5e- → Mn H2O
Add equations together
MnO Fe H+ → Mn Fe3+ + 4H2O

10. Ce4+ + I– + OH–  Ce3+ + IO3– + H2O
Example 4
Here is a reaction occurring in a basic solution, which accounts for the presence of the OH-ions. This example requires a similar approach as our previous example, just in a base instead of an acid.
Ce4+ + I– + OH–  Ce3+ + IO3– + H2O
In this example, spectator ions have already been removed. Even though hydrogen and oxygen do not undergo changes in oxidation number they are not spectators and we need to work with them in our half-reactions.

11. Example 4 oxidation I– + 6 OH–  IO3– + 3 H2O + 6 e– reduction
Ce e–  Ce3+

12. Example 4
Step 1
Step 2
Step 3
Write the two balanced half-reactions, removing any spectator ions:
Balance electrons
Add the half-reactions, replacing any spectator ions that were removed and/or recombining compounds
Ce e–  Ce3+
 6
6Ce e–  6Ce3+
I– + 6 OH–  IO3– + 3 H2O + 6 e–
added together:
6 Ce4+ + I– + 6 OH–  6 Ce3+ + IO3– + 3 H2O
Reform compound/return spectator ions
Reform compound/return spectator ions