1. Balancing redox reactions 2

2. Half-reaction Spectator ion
OUTCOME QUESTION(S):
C REACTIONS AND BALANCING
Balance oxidation-reduction reactions using the oxidation number method
Balance oxidation-reduction reactions either the acid or base half-reaction method
Vocabulary & Concepts
Half-reaction Spectator ion

3. 2 Al(s) + 3 CuSO4(aq) → Al2(SO4)3(aq) + 3 Cu(s)
Half reactions show oxidation and reduction reactions separately as aqueous net ionic equations
*spectator IONS not included* +2 +3
2 Al(s) + 3 CuSO4(aq) → Al2(SO4)3(aq) + 3 Cu(s)
Oxidation: Al → Al3+ + 3e–
(losing 3 electrons)
Reduction: Cu2+ + 2e– → Cu
(gaining 2 electrons)
Spectator ions (SO42-) are just that – ions that don’t change and aren’t really part of what is going on (usually they will have already been removed)

4. Some redox reactions require an acidic or basic solutions.
Half-reaction method
Some redox reactions require an acidic or basic solutions.
It depends on the reaction and substance needing to be oxidized – it will always be clear if it is an acidic or basic solution
Acid / base not oxidized or reduced in reaction
Usually converted to water
Half-reaction method is used for balancing redox reactions in the presence of acid or base.

5. Cr2O72–(aq) + SO32–(aq) → Cr3+(aq) + SO42–(aq)
Balancing in Acidic Solutions +6 -2 +4 -2 +3 +6 -2
Cr2O72–(aq) + SO32–(aq) → Cr3+(aq) + SO42–(aq)
Aqueous ions cannot exist by themselves – the spectator ions must have already been removed
Step 1: Assign O#s and write half-reactions
Oxidation: SO32- → SO e–
Reduction: Cr2O e– → Cr3+

6. Each Cr gains 3e- but there are 2 Cr atoms - so a total of 6e- gained
Step 2: Balance all elements except H and O
Oxidation: SO32- → SO e–
Reduction: Cr2O e– → Cr3+ 6 3 2
Be sure to adjust electrons to the number of atoms for each half-reaction
Each Cr gains 3e- but there are 2 Cr atoms - so a total of 6e- gained
Step 3: Balance oxygen atoms by adding H2O
Oxidation: SO H2O → SO e–
Reduction: Cr2O e– → 2 Cr H2O

7. Oxidation: SO32- + H2O → SO42- + 2e–
Step 4: Balance hydrogen atoms adding H+ ions
Oxidation: SO H2O → SO e–
+ 2 H+
Reduction: Cr2O e– → 2 Cr H2O
14 H+ +
Step 5: Balance the number of electrons between half-reactions
Oxidation: 3 x (SO H2O → SO e– + 2 H+)
3 SO H2O → 3 SO e– + 6 H+
Reduction: 14 H+ + Cr2O e– → 2 Cr H2O
Multiply the whole equation by the common multiple needed to make the half-reaction electrons equal

8. 8 H+ + Cr2O72- + 3 SO32- → 2 Cr3+ + 3 SO42- + 4 H2O
6. Add the two half-reactions
Oxidation: 3 SO H2O → 3 SO e– + 6 H+ 8 4
Reduction: 14 H+ + Cr2O e– → 2 Cr H2O
8 H+ + Cr2O SO32- → 2 Cr SO H2O
Take care here - cancel out what you can and combine the half-reactions into a single equation

9. Balance the following reaction in a acidic solution.+7 -1 +4
MnO4– + I– → MnO2 + I2
Oxidation: I- → I e–
3 x ( ) 2 2 1
Reduction: MnO e– → MnO2
2x ( )
4 H+ +
+ 2 H2O
No spectator ion on the reactant side – keep compound together – so you have to keep the product together too
Oxidation: I- → 3 I2 + 6e–
Reduction: 8 H+ + 2 MnO e– → 2 MnO2 + 4 H2O
8 H+ + 2 MnO4– + 6 I– → 2 MnO2 + 3 I2 + 4 H2O

10. Balancing in Basic Solutions
Steps 1-4 are the same as in Acid solutions +3 +7 +4 +4
MnO4– + C2O42– → CO2 + MnO2
Oxidation: C2O → CO e– 2 2 1
Reduction: MnO e– → MnO2
4 H+ +
+ 2 H2O
Tip: Take your time and do each step on a new line to avoid any mis-steps

11. The point here is to cancel any water to simplify the half-reactions
**5a. Add the same number of OH- as H+ to BOTH sides of the equation
Oxidation: C2O42- → 2 CO2 + 2e–
Reduction: 4 H+ + MnO e– → MnO2 + 2 H2O
4 OH- +
+ 4 OH-
**5b. Eliminate H+ / OH- by forming water
Oxidation: C2O42- → 2 CO2 + 2e– 2
4 H2O
Reduction: 4 OH- + 4 H+ + MnO e– → MnO2 + 2 H2O + 4 OH-
The point here is to cancel any water to simplify the half-reactions

12. 4 H2O + 2 MnO4– + 3 C2O42– → 2 MnO2 + 6 CO2 + 8 OH–
Step 6: Balance the number of electrons between half-reactions
Oxidation: C2O42- → 2 CO2 + 2e–
3 x ( )
Reduction: 2 H2O + MnO e– → MnO2 + 4 OH-
2x ( )
Step 7: Add the two half-reactions
Oxidation: C2O42- → 6 CO2 + 6e–
Reduction: 4 H2O + 2 MnO e– → 2 MnO2 + 8 OH-
4 H2O + 2 MnO4– + 3 C2O42– → 2 MnO2 + 6 CO2 + 8 OH–

13. Balance the following reaction in a basic solution.+1 +1 +3 -1
N2O ClO– → NO2– Cl–
Oxidation: N2O → NO e–
3 H2O + 2 4 2
+ 6 H+
Reduction: ClO e– → Cl-
2 H+ +
+ 1 H2O
Steps 1-4

14. 2 OH– + 2 ClO– + N2O → 2 Cl– + 2 NO2– + H2O3
6 H2O
Oxidation: N2O → NO e–
6 OH- +
3 H2O + 2
+ 6 H+
+ 6 OH-
Reduction: ClO e– → Cl-
2 OH- +
2 H+ +
+ 1 H2O
+ 2 OH- 1
2 H2O 2 1
Oxidation: N2O → NO e– 2
+ 3 H2O
6 OH- +
Reduction: ClO e– → Cl-
1 H2O +
+ 2 OH-
2x ( )
2 H2O ClO- + 4e– → 2 Cl- + 4 OH-
2 OH– + 2 ClO– + N2O → 2 Cl– + 2 NO2– + H2O
Steps 5-7

15. Half-reaction Spectator ion
CAN YOU / HAVE YOU?
C REACTIONS AND BALANCING
Balance oxidation-reduction reactions using the oxidation number method
Balance oxidation-reduction reactions either the acid or base half-reaction method
Vocabulary & Concepts
Half-reaction Spectator ion