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Balancing redox reactions 2
• Balance oxidation-reduction reactions using redox methods
Include: oxidation number method, and half-reaction method
Additional KEY Terms
Some redox reactions require an acidic or basic solution.
Half-reaction method
It depends on the reaction and substance needing to be oxidized – I will always tell you if it is in an acidic or basic solution
Half-reaction method is used for balancing redox reactions in the presence of acid or base.
• Acid / base not oxidized or reduced in reaction• Usually converted to water
Cr2O72–
(aq) + SO32–
(aq) → Cr3+(aq) + SO4
2–(aq)
Step 1: Assign O#s and write half-reactions
Balancing in Acidic Solutions
+3+4-2+6 -2+6-2
Oxidation: SO32- → SO4
2- + 2e–
Reduction: Cr2O72- + 3e– → Cr3+
Aqueous ions cannot exist by themselves – the spectator ions must have already been removed
Step 2: Balance all elements except H and O
Step 3: Balance oxygen atoms by adding H2O
Oxidation: SO32- → SO4
2- + 2e–
Reduction: Cr2O72- + e– → Cr3+2
Oxidation: SO32- + H2O → SO4
2- + 2e–
Reduction: Cr2O72- + 6e– → 2 Cr3+ + 7 H2O
Be sure to adjust the 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
36
Step 4: Balance hydrogen atoms adding H+ ions
Step 5: Balance the number of electrons between half-reactions
Oxidation: SO32- + H2O → SO4
2- + 2e–
Reduction: Cr2O72- + 6e– → 2 Cr3+ + 7 H2O
+ 2 H+
14 H+ +
Oxidation: 3 x (SO32- + H2O → SO4
2- + 2e– + 2 H+)
Reduction: 14 H+ + Cr2O72- + 6e– → 2 Cr3+ + 7 H2O
3 SO32- + 3 H2O → 3 SO4
2- + 6e– + 6 H+
Multiply the whole equation by the common multiple needed to make the half-reaction electrons equal
6. Add the two half-reactions
Oxidation: 3 SO32- + 3 H2O → 3 SO4
2- + 6e– + 6 H+
Reduction: 14 H+ + Cr2O7
2- + 6e– → 2 Cr3+ + 7 H2O
8 H+ + Cr2O72- + 3 SO3
2- → 2 Cr3+ + 3 SO42- + 4 H2O
Take care here - cancel out what you can and combine the half-reactions into a single equation
8 4
MnO4– + I– → MnO2 + I2
8 H+ + 2 MnO4– + 6 I– → 2 MnO2 + 3 I2 + 4 H2O
Balance the following reaction in a acidic solution.
-1+7 0+4
Oxidation: I- → I2 + e–
Reduction: MnO4- + 3e– → MnO2
2
+ 2 H2O4 H+ +
3 x ()
2x (
)Oxidation: 6 I- → 3 I2 + 6e–Reduction: 8 H+ + 2 MnO4
- + 6e– → 2 MnO2 + 4 H2O
No spectator ion on the reactant side – keep compound together – so you have to keep the product together too
12
Balancing in Basic Solutions• Steps 1-4 are the same as in Acid solutions
MnO4– + C2O4
2– → CO2 + MnO2
+4+3+7 +4
Oxidation: C2O42- → CO2 + e–
Reduction: MnO41- + 3e– → MnO2
2
+ 2 H2O4 H+ +
12
Tip: Take your time and do each step on a new line to avoid any mis-steps
**5b. Eliminate H+ / OH- by forming water
Oxidation C2O42- → 2 CO2 + 2e–
Reduction: 4 H+ + MnO41- + 3e– → MnO2 + 2
H2O+ 4 OH-4 OH- +
**5a. Add the same number of OH- as H+ to BOTH sides of the equation
The point here is to cancel any water to simplify the half-reactions
Oxidation C2O42- → 2 CO2 + 2e–
Reduction: 4 OH- + 4 H+ + MnO4- + 3e– → MnO2 + 2 H2O + 4
OH-
4 H2O2
4 H2O + 2 MnO4– + 3 C2O4
2– → 2 MnO2 + 6 CO2 + 8 OH–
Oxidation 3 C2O42-
→ 6 CO2 + 6e–
Reduction: 4 H2O + 2 MnO41- + 6e– → 2 MnO2 + 8
OH-
Step 7: Add the two half-reactions
Oxidation C2O42- → 2 CO2 + 2e–
Reduction: 2 H2O + MnO41- + 3e– → MnO2 + 4
OH-
3 x ()
2x (
)
Step 6: Balance the number of electrons between half-reactions
N2O + ClO– → NO2– + Cl–
Balance the following reaction in a basic solution.
+3+1+1 -1
Oxidation: N2O → NO2- + e–
Reduction: ClO- + 2e– → Cl-
23 H2O +
2 H+ +
+ 6 H+
+ 1 H2O
• Steps 1-4
24
2 OH– + 2 ClO– + N2O → 2 Cl– + 2 NO2– + H2O
Oxidation: N2O → NO2- + 4e–
Reduction: ClO- + 2e– → Cl-
23 H2O +
2 H+ +
+ 6 H+
+ 1 H2O
+ 6 OH-
2 OH- + 2 H2O
3
+ 2 OH-
6 OH- + 6 H2O
1
Reduction: ClO- + 2e– → Cl-1 H2O + + 2 OH-
Oxidation: N2O → NO2- + 4e–2 + 3 H2O6 OH- +
2x ( )
2 H2O + 2 ClO- + 4e– → 2 Cl- + 4 OH-
12
• Steps 5-7
CAN YOU / HAVE YOU?
• Balance oxidation-reduction reactions using redox methods
Include: oxidation number method, and half-reaction method
Additional KEY Terms