Balancing Redox Reactions (Half-Reaction Method) | Class 11 Chemistry

Balancing Redox Reactions

Half-Reaction (Ion-Electron) Method | Class 11 Chemistry

1. The Half-Reaction Method

Also known as the Ion-Electron Method, this approach splits the overall redox reaction into two separate half-reactions: the Oxidation Half and the Reduction Half. Each half is balanced individually and then added together.

2. Steps to Balance (Acidic Medium)

1. Write the skeletal equation in ionic form.
2. Split into two half-reactions: Oxidation (Increase in ON) and Reduction (Decrease in ON).
3. Balance atoms other than Oxygen and Hydrogen.
4. Balance Oxygen by adding $H_2O$ molecules to the side deficient in oxygen.
5. Balance Hydrogen by adding $H^+$ ions to the side deficient in hydrogen.
6. Balance Charge by adding electrons ($e^-$) to the more positive side.
7. Equalize electrons in both half-reactions by multiplying with suitable integers.
8. Add the two half-reactions and cancel out common species (electrons, $H_2O, H^+$).

3. Worked Example (Acidic Medium)

Reaction: Dichromate ion oxidizes Ferrous to Ferric in acid solution.

$Cr_2O_7^{2-} + Fe^{2+} \rightarrow Cr^{3+} + Fe^{3+}$

Step A: Oxidation Half (Fe)

1. Skeletal: $Fe^{2+} \rightarrow Fe^{3+}$

2. Balance Charge: Add $1e^-$ to RHS.

Oxidation Half: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ ... (i)

Step B: Reduction Half (Cr)

1. Skeletal: $Cr_2O_7^{2-} \rightarrow Cr^{3+}$

2. Balance Cr: $Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$

3. Balance O (7 atoms): Add $7H_2O$ to RHS.
$Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$

4. Balance H (14 atoms): Add $14H^+$ to LHS.
$Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O$

5. Balance Charge: LHS = +12, RHS = +6. Add $6e^-$ to LHS.

Reduction Half: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ ... (ii)

Step C: Combine

Multiply Eq (i) by 6 to match electrons.

$6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$

Add to Eq (ii):

$Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$

4. Balancing in Basic Medium

The steps are the same as Acidic Medium up to balancing Hydrogen. After obtaining the final balanced equation with $H^+$:

1. Add $OH^-$ ions to both sides equal to the number of $H^+$ ions present.
2. Combine $H^+$ and $OH^-$ to form $H_2O$.
3. Cancel out excess water molecules from both sides.

Alternative Rule for Basic H-Balance: Add $H_2O$ to the side with excess Hydrogen, and add equal $OH^-$ to the other side.

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