1. Balancing redox reactions 2

2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional KEY Terms

3. 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

4. Cr 2 O 7 2– (aq) + SO 3 2– (aq) → Cr 3+ (aq) + SO 4 2– (aq) Step 1: Assign O#s and write half-reactions Balancing in Acidic Solutions +3 +4 -2 +6 -2 +6 -2 Oxidation:SO 3 2- → SO 4 2- + 2e – Reduction:Cr 2 O 7 2- + 3e – → Cr 3+ Aqueous ions cannot exist by themselves – the spectator ions must have already been removed

5. Step 2: Balance all elements except H and O Step 3: Balance oxygen atoms by adding H 2 O Oxidation:SO 3 2- → SO 4 2- + 2e – Reduction:Cr 2 O 7 2- + e – → Cr 3+ 2 Oxidation:SO 3 2- + H 2 O → SO 4 2- + 2e – Reduction:Cr 2 O 7 2- + 6e – → 2 Cr 3+ + 7 H 2 O 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 3 6

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

7. 6. Add the two half-reactions Oxidation: 3 SO 3 2- + 3 H 2 O → 3 SO 4 2- + 6e – + 6 H + Reduction: 14 H + + Cr 2 O 7 2- + 6e – → 2 Cr 3+ + 7 H 2 O 8 H + + Cr 2 O 7 2- + 3 SO 3 2- → 2 Cr 3+ + 3 SO 4 2- + 4 H 2 O Take care here - cancel out what you can and combine the half- reactions into a single equation 8 4

8. MnO 4 – + I – → MnO 2 + I 2 8 H + + 2 MnO 4 – + 6 I – → 2 MnO 2 + 3 I 2 + 4 H 2 O Balance the following reaction in a acidic solution. +7 0 +4 Oxidation: I - → I 2 + e – Reduction: MnO 4 - + 3e – → MnO 2 2 + 2 H 2 O 4 H + + 3 x () 2x ( ) Oxidation: 6 I - → 3 I 2 + 6e – Reduction: 8 H + + 2 MnO 4 - + 6e – → 2 MnO 2 + 4 H 2 O No spectator ion on the reactant side – keep compound together – so you have to keep the product together too 1 2

9. Balancing in Basic Solutions Steps 1-4 are the same as in Acid solutions MnO 4 – + C 2 O 4 2– → CO 2 + MnO 2 +4 +3 +7 +4 Oxidation: C 2 O 4 2- → CO 2 + e – Reduction: MnO 4 1- + 3e – → MnO 2 2 + 2 H 2 O4 H + + 1 2 Tip: Take your time and do each step on a new line to avoid any mis-steps

10. **5b. Eliminate H + / OH - by forming water Oxidation C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 4 H + + MnO 4 1- + 3e – → MnO 2 + 2 H 2 O + 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 C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 4 OH - + 4 H + + MnO 4 - + 3e – → MnO 2 + 2 H 2 O + 4 OH - 4 H 2 O 2

11. 4 H 2 O + 2 MnO 4 – + 3 C 2 O 4 2– → 2 MnO 2 + 6 CO 2 + 8 OH – Oxidation 3 C 2 O 4 2- → 6 CO 2 + 6e – Reduction: 4 H 2 O + 2 MnO 4 1- + 6e – → 2 MnO 2 + 8 OH - Step 7: Add the two half-reactions Oxidation C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 2 H 2 O + MnO 4 1- + 3e – → MnO 2 + 4 OH - 3 x () 2x ( ) Step 6: Balance the number of electrons between half-reactions

12. N 2 O + ClO – → NO 2 – + Cl – Balance the following reaction in a basic solution. +3+1 Oxidation: N 2 O → NO 2 - + e – Reduction: ClO - + 2e – → Cl - 2 3 H 2 O + 2 H + + + 6 H + + 1 H 2 O Steps 1-4 2 4

13. 2 OH – + 2 ClO – + N 2 O → 2 Cl – + 2 NO 2 – + H 2 O Oxidation: N 2 O → NO 2 - + 4e – Reduction: ClO - + 2e – → Cl - 23 H 2 O + 2 H + + + 6 H + + 1 H 2 O + 6 OH - 2 OH - + 2 H 2 O 3 + 2 OH - 6 OH - + 6 H 2 O 1 Reduction: ClO - + 2e – → Cl - 1 H 2 O + + 2 OH - Oxidation: N 2 O → NO 2 - + 4e – 2 + 3 H 2 O 6 OH - + 2x ( ) 2 H 2 O + 2 ClO - + 4e – → 2 Cl - + 4 OH - 12 Steps 5-7

14. CAN YOU / HAVE YOU? Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional KEY Terms