PF1009 2024 12B Balancing Redox Reaction Aqueous Base PDF
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This document presents a detailed procedure for balancing redox reactions in an aqueous base environment. It walks through various steps and concepts, along with example reaction scenarios. The document focuses on balancing chemical equations by using the method of balancing half-reactions in a basic solution.
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### Identify ox/red. species - Zn + NO<sub>3</sub><sup>-</sup> → Zn(OH)<sub>4</sub><sup>2-</sup> + NH<sub>3</sub> ### BASIC RXN AQ 1. Write 2 skeletal equations for 1/2 rxns 2. Balance by inspection all elements in 1/2 rxns except H & charge 3. Zn + 4H<sub>2</sub>O → Zn(OH)<sub>4</sub><sup>2-</s...
### Identify ox/red. species - Zn + NO<sub>3</sub><sup>-</sup> → Zn(OH)<sub>4</sub><sup>2-</sup> + NH<sub>3</sub> ### BASIC RXN AQ 1. Write 2 skeletal equations for 1/2 rxns 2. Balance by inspection all elements in 1/2 rxns except H & charge 3. Zn + 4H<sub>2</sub>O → Zn(OH)<sub>4</sub><sup>2-</sup> NO<sub>3</sub><sup>-</sup> + 4H<sub>2</sub>O → NH<sub>3</sub> + 3H<sub>2</sub>O 4. Zn + 4H<sub>2</sub>O → Zn(OH)<sub>4</sub><sup>2-</sup> + 4H<sup>+</sup> NO<sub>3</sub><sup>-</sup> + 6H<sup>+</sup> + 9H<sub>2</sub>O → NH<sub>3</sub> + 3H<sub>2</sub>O 5. Zn + 4H<sub>2</sub>O + 4OH<sup>- </sup>→ Zn(OH)<sub>4</sub><sup>2-</sup> + 4H<sup>+</sup> + 4OH<sup>-</sup> NO<sub>3</sub><sup>-</sup> + 9H<sup>+</sup> + 9OH<sup>-</sup> → NH<sub>3</sub> + 3H<sub>2</sub>O + 9OH<sup>-</sup> 6. Zn + 4H<sub>2</sub>O + 4OH<sup>-</sup> → Zn(OH)<sub>4</sub><sup>2-</sup> + 4H<sub>2</sub>O NO<sub>3</sub><sup>-</sup> + 9H<sub>2</sub>O → NH<sub>3</sub> + 3H<sub>2</sub>O + 9OH<sup>-</sup> 7. Zn + HOH<sup>-</sup> → Zn(OH)<sub>4</sub><sup>2-</sup> NO<sub>3</sub><sup>-</sup> + 6H<sub>2</sub>O → NH<sub>3</sub> + 9OH<sup>-</sup> ### Balance O by adding H<sub>2</sub>O ### Balance H by adding H<sup>+</sup> ### Convert acidic to basic. Add sufficient OH to all H+ of 1/2 reactions ### Combine H<sup>+</sup> and OH<sup>-</sup> to form H<sub>2</sub>O and place H<sub>2</sub>O to one side of equations - Zn + 4OH<sup>-</sup> → Zn(OH)<sub>4</sub><sup>2-</sup> + 2e<sup>-</sup> - NO<sub>3</sub><sup>-</sup> + 6H<sub>2</sub>O + 8e<sup>-</sup> → NH<sub>3</sub> + 9OH<sup>-</sup> ### Balance charge with electrons - Multiply X4 - 4Zn + 16OH<sup>-</sup> → 4Zn(OH)<sub>4</sub><sup>2-</sup> + 8e<sup>-</sup> - NO<sub>3</sub><sup>-</sup> + 6H<sub>2</sub>O + 8e<sup>-</sup> → NH<sub>3</sub> + 9OH<sup>-</sup> ### Have same elemeber of electrons in each 1/2 RX 2 ### Combine 4O + 2 1/2 rxns as simultaneous equations. ### Combine common species across to balanced reaction. - 4Zn + 7OH<sup>-</sup> + NO<sub>3</sub><sup>-</sup> + 6H<sub>2</sub>O → 4Zn(OH)<sub>4</sub><sup>2-</sup> + NH<sub>3</sub> ### CHECK: - 4xZn → 4xZn - 1xNO<sub>3</sub><sup>-</sup> → 1xNO<sub>3</sub><sup>-</sup> - 16xO → 16xO - 19xH → 19xH - 8x negative charge → 8x negative charge
