Electrochemistry

Redox Equations

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Write balanced half-equations for the two reacting species.
Multiply each half-equation to make the number of electrons equal.
Combine the two half-equations. Simplify if necessary.
Change to alkaline medium by neutralising H⁺ with OH^– if required.

How to balance half-equations:

Steps

Half-equation

1. Write skeletal half-equations and balance the atoms being oxidised /reduced.

Do not include spectator ions.

MnO₄^– → Mn²⁺

2. Balance O atoms by adding H₂O

MnO₄^– → Mn²⁺ + 4H₂O

3. Balance H atoms by adding H⁺

MnO₄^– + 8H⁺ → Mn²⁺ + 4H₂O

4. Balance charges by adding e^–

MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O

Example:

MnO₄^– + Br^– → MnO₂ + BrO₃^– (under alkaline conditions)

Steps

Half-equation/ Overall equation

1. Write balanced half-equations for the two reacting species.

MnO₄^– + 4H⁺ + 3e^– → MnO₂+ 2H₂O

Br^– + 3H₂O → BrO₃^– + 6H⁺ + 6e^–

2. Multiply each half-equation to make the number of electrons equal.

2MnO₄^– + 8H⁺ + 6e^– → 2MnO₂+ 4H₂O

Br^– + 3H₂O → BrO₃^– + 6H⁺ + 6e^–

3. Combine the two half-equations. Simplify if necessary.

2MnO₄^– + 8H⁺ + Br^– + 3H₂O → 2MnO₂ + 4H₂O + BrO₃^– + 6H⁺

2MnO₄^– + 2H⁺ + Br^– → 2MnO₂+ H₂O + BrO₃^–

4. Neutralise H⁺ by adding OH^–.

Add 2OH^– to both sides to neutralise the 2H⁺ on the LHS

2MnO₄^– + 2H⁺ + 2OH^– + Br^– → 2MnO₂+ H₂O + BrO₃^– + 2OH^–

2MnO₄^– + H₂O + Br^– → 2MnO₂ + BrO₃^– + 2OH^–

Practice:

Balance the following equation under acidic and alkaline conditions:

MnO₄^– + C₂O₄^2– → Mn²⁺ + CO₂

Answer:

Acidic: 16H⁺ + 2MnO₄^– + 5C₂O₄^2– → 2Mn²⁺ + 8H₂O + 10CO₂

Alkaline: 8H₂O + 2MnO₄^– + 5C₂O₄^2– → 2Mn²⁺ + 16OH^– + 10CO₂

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Electrochemistry

Steps	Half-equation
1. Write skeletal half-equations and balance the atoms being oxidised /reduced. Do not include spectator ions.	MnO₄^– → Mn²⁺
2. Balance O atoms by adding H₂O	MnO₄^– → Mn²⁺ + 4H₂O
3. Balance H atoms by adding H⁺	MnO₄^– + 8H⁺ → Mn²⁺ + 4H₂O
4. Balance charges by adding e^–	MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O

Steps	Half-equation/ Overall equation
1. Write balanced half-equations for the two reacting species.	MnO₄^– + 4H⁺ + 3e^– → MnO₂+ 2H₂O Br^– + 3H₂O → BrO₃^– + 6H⁺ + 6e^–
2. Multiply each half-equation to make the number of electrons equal.	2MnO₄^– + 8H⁺ + 6e^– → 2MnO₂+ 4H₂O Br^– + 3H₂O → BrO₃^– + 6H⁺ + 6e^–
3. Combine the two half-equations. Simplify if necessary.	2MnO₄^– + 8H⁺ + Br^– + 3H₂O → 2MnO₂ + 4H₂O + BrO₃^– + 6H⁺ 2MnO₄^– + 2H⁺ + Br^– → 2MnO₂+ H₂O + BrO₃^–
4. Neutralise H⁺ by adding OH^–.	Add 2OH^– to both sides to neutralise the 2H⁺ on the LHS 2MnO₄^– + 2H⁺ + 2OH^– + Br^– → 2MnO₂+ H₂O + BrO₃^– + 2OH^– 2MnO₄^– + H₂O + Br^– → 2MnO₂ + BrO₃^– + 2OH^–