2.2 Balancing of Redox Equations | Electrochemistry | Chemistry 12

Balancing redox (oxidation-reduction) equations requires ensuring both atom conservation and charge conservation. Two systematic methods are used in FBISE/FSc chemistry.

Why Special Methods Are Needed

Simple inspection balancing fails for redox reactions because electrons are transferred between species. We must explicitly account for electron gain and loss.

Two fundamental rules:

Total electrons lost by the reducing agent = Total electrons gained by the oxidizing agent
Net charge must be equal on both sides

Method 1: Oxidation Number Method

Steps

Assign oxidation numbers to all atoms in the equation.
Identify which atoms are oxidized (increase in oxidation number) and which are reduced (decrease).
Calculate the total increase and total decrease in oxidation numbers.
Multiply the oxidized and reduced species by appropriate coefficients so that total increase = total decrease.
Balance the remaining atoms (H and O) and charges by inspection.

Example

Balance: $K M n O_{4} + F e S O_{4} + H_{2} S O_{4} \to M n S O_{4} + F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + H_{2} O$

Species	Oxidation State Change	Change per atom
$M n$ in $K M n O_{4}$	+7 → +2	decrease of 5 (reduction)
$F e$ in $F e S O_{4}$	+2 → +3	increase of 1 (oxidation)

To equalize: multiply Mn species by 1 and Fe species by 5:

$2 K M n O_{4} + 10 F e S O_{4} + 8 H_{2} S O_{4} \to 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 8 H_{2} O$

Method 2: Ion-Electron (Half-Reaction) Method

This method splits the overall reaction into two half-equations — one for oxidation and one for reduction — balances each separately, then combines them.

Steps for Acidic Medium

Write the unbalanced ionic equation.
Separate into oxidation and reduction half-reactions.
Balance atoms other than O and H first.
Balance O by adding $H_{2} O$ to the deficient side.
Balance H by adding $H^{+}$ to the deficient side.
Balance charge by adding electrons ( $e^{-}$ ) to the more positive side.
Multiply each half-reaction by a factor so electrons cancel.
Add the two half-reactions and simplify.

Steps for Basic Medium

Follow steps 1–8 for acidic medium, then:

For every $H^{+}$ present, add one $O H^{-}$ to both sides.
Combine $H^{+}$ + $O H^{-}$ → $H_{2} O$ on the same side; simplify.

Example: Acidic Medium

Balance: $M n O_{4}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +}$ (acidic)

Reduction half-reaction: $M n O_{4}^{-} \to M n^{2 +}$

Balance Mn: already balanced
Balance O: add $4 H_{2} O$ to right $M n O_{4}^{-} \to M n^{2 +} + 4 H_{2} O$
Balance H: add $8 H^{+}$ to left $M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$
Balance charge: left = $- 1 + 8 = + 7$ ; right = $+ 2$ ; add $5 e^{-}$ to left $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Oxidation half-reaction: $F e^{2 +} \to F e^{3 +} + e^{-}$

Multiply oxidation half by 5: $5 F e^{2 +} \to 5 F e^{3 +} + 5 e^{-}$

Add the two half-reactions: $M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 4 H_{2} O + 5 F e^{3 +}$

Comparison of Methods

Feature	Oxidation Number Method	Ion-Electron Method
Best for	Molecular equations	Ionic equations
Balances charge via	Electron count from oxidation numbers	Explicit $e^{-}$ in half-equations
Medium adjustment	Less explicit	Explicit $H^{+}$ / $O H^{-}$ / $H_{2} O$ steps

Key Points to Remember

Oxidizing agent: gains electrons → oxidation number decreases → undergoes reduction
Reducing agent: loses electrons → oxidation number increases → undergoes oxidation
In acidic medium: use $H^{+}$ and $H_{2} O$ to balance H and O
In basic medium: use $O H^{-}$ and $H_{2} O$ to balance H and O
Always verify: same number of each atom AND same net charge on both sides

Balancing redox (oxidation-reduction) equations requires ensuring both atom conservation and charge conservation. Two systematic methods are used in FBISE/FSc chemistry.

Why Special Methods Are Needed

Simple inspection balancing fails for redox reactions because electrons are transferred between species. We must explicitly account for electron gain and loss.

Two fundamental rules:

Total electrons lost by the reducing agent = Total electrons gained by the oxidizing agent
Net charge must be equal on both sides

Method 1: Oxidation Number Method

Steps

Assign oxidation numbers to all atoms in the equation.
Identify which atoms are oxidized (increase in oxidation number) and which are reduced (decrease).
Calculate the total increase and total decrease in oxidation numbers.
Multiply the oxidized and reduced species by appropriate coefficients so that total increase = total decrease.
Balance the remaining atoms (H and O) and charges by inspection.

Example

Balance: $K M n O_{4} + F e S O_{4} + H_{2} S O_{4} \to M n S O_{4} + F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + H_{2} O$

Species	Oxidation State Change	Change per atom
$M n$ in $K M n O_{4}$	+7 → +2	decrease of 5 (reduction)
$F e$ in $F e S O_{4}$	+2 → +3	increase of 1 (oxidation)

To equalize: multiply Mn species by 1 and Fe species by 5:

$2 K M n O_{4} + 10 F e S O_{4} + 8 H_{2} S O_{4} \to 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 8 H_{2} O$

Method 2: Ion-Electron (Half-Reaction) Method

This method splits the overall reaction into two half-equations — one for oxidation and one for reduction — balances each separately, then combines them.

Steps for Acidic Medium

Write the unbalanced ionic equation.
Separate into oxidation and reduction half-reactions.
Balance atoms other than O and H first.
Balance O by adding $H_{2} O$ to the deficient side.
Balance H by adding $H^{+}$ to the deficient side.
Balance charge by adding electrons ( $e^{-}$ ) to the more positive side.
Multiply each half-reaction by a factor so electrons cancel.
Add the two half-reactions and simplify.

Steps for Basic Medium

Follow steps 1–8 for acidic medium, then:

For every $H^{+}$ present, add one $O H^{-}$ to both sides.
Combine $H^{+}$ + $O H^{-}$ → $H_{2} O$ on the same side; simplify.

Example: Acidic Medium

Balance: $M n O_{4}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +}$ (acidic)

Reduction half-reaction: $M n O_{4}^{-} \to M n^{2 +}$

Balance Mn: already balanced
Balance O: add $4 H_{2} O$ to right $M n O_{4}^{-} \to M n^{2 +} + 4 H_{2} O$
Balance H: add $8 H^{+}$ to left $M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$
Balance charge: left = $- 1 + 8 = + 7$ ; right = $+ 2$ ; add $5 e^{-}$ to left $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Oxidation half-reaction: $F e^{2 +} \to F e^{3 +} + e^{-}$

Multiply oxidation half by 5: $5 F e^{2 +} \to 5 F e^{3 +} + 5 e^{-}$

Add the two half-reactions: $M n O_{4}^{-} + 8 H^{+} + 5 F e^{2 +} \to M n^{2 +} + 4 H_{2} O + 5 F e^{3 +}$

Comparison of Methods

Feature	Oxidation Number Method	Ion-Electron Method
Best for	Molecular equations	Ionic equations
Balances charge via	Electron count from oxidation numbers	Explicit $e^{-}$ in half-equations
Medium adjustment	Less explicit	Explicit $H^{+}$ / $O H^{-}$ / $H_{2} O$ steps

Key Points to Remember

Oxidizing agent: gains electrons → oxidation number decreases → undergoes reduction
Reducing agent: loses electrons → oxidation number increases → undergoes oxidation
In acidic medium: use $H^{+}$ and $H_{2} O$ to balance H and O
In basic medium: use $O H^{-}$ and $H_{2} O$ to balance H and O
Always verify: same number of each atom AND same net charge on both sides