Redox Reactions 13 1 b Complex HalfRxtns HalfReaction

Redox Reactions 13. 1 b Complex Half-Rxtns

Half-Reaction Method Ø separate the full equation into 2 half- reactions: oxidation and reduction Ø balance each half-reaction separately Ø add them together to get full balanced equation

Steps for acidic solution Write 2 separate equations for oxidation and reduction 1. l include any compound containing the element involved

Steps for acidic solution For each half-reaction: 2. l l balance all elements but H and O balance O using H 2 O balance H using H+ balance charge using electrons (e-)

Steps for acidic solution 3. If needed, multiply one or both half-reactions by an integer so that number of electrons is equal in both half reactions

Steps for acidic solution 4. 5. Add the half-reactions together and simplify Check to be sure all is balanced.

Steps for basic solution 1. 2. 3. 4. 5. balance using acidic method add OH- ions equal to #H+ ions to both sides form water on the side of equation that contains both ions simplify # of water molecules check for balance

Example 4 x

Example

Acidic Example Mn. O 4–(aq) + Br–(aq) Mn 2+(aq) + Br 2(aq) Ø 1. Determine oxidation and reduction half-reactions: • Oxidation half-reaction: Br–(aq) Br 2(aq) • Reduction half-reaction: Mn. O 4–(aq) Mn 2+(aq)

2. Balance for atoms other than H and O: Oxidation: 2 Br–(aq) Br 2(aq) Reduction: Mn. O 4–(aq) Mn 2+(aq) 3. Balance for oxygen by adding H 2 O: 2 Br–(aq) Br 2(aq) l Oxidation: l Reduction: Mn. O 4–(aq) Mn 2+(aq) + 4 H 2 O(l)

Ø 4. Balance for hydrogen by adding H+: Oxidation: 2 Br–(aq) Br 2(aq) Reduction: Mn. O 4–(aq) + 8 H+(aq) Mn 2+(aq) + 4 H 2 O(l) 5. Balance for charge by adding electrons (e–): Oxidation: 2 Br–(aq) Br 2(aq) + 2 e– Reduction: Mn. O 4–(aq) + 8 H+(aq) + 5 e– Mn 2+(aq) + 4 H 2 O(l)

Ø Ø 6. Balance for numbers of electrons by multiplying: l Oxidation: 5[2 Br–(aq) Br 2(aq) + 2 e–] l Reduction: 2[Mn. O 4–(aq) + 8 H+(aq) + 5 e– Mn 2+(aq) + 4 H 2 O(l)] 7. Combine and cancel to form one equation: l Oxidation: 10 Br–(aq) 5 Br 2(aq) + 10 e– l Reduction: 2 Mn. O 4–(aq) + 16 H+(aq) + 10 e– 2 Mn 2+(aq) + 8 H 2 O(l) 2 Mn. O 4–(aq) + 10 Br–(aq) + 16 H+(aq) 2 Mn 2+(aq) + 5 Br 2(aq) + 8 H 2 O(l)

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