Balancing redox equations — half-reaction method step by step

Question

Balance the equation: $\text{MnO}_4^- + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+}$ in acidic medium using the half-reaction method.

(CBSE 11 + JEE Main)

Solution — Step by Step

flowchart TD
    A["Unbalanced Redox Equation"] --> B["Step 1: Split into oxidation + reduction half-reactions"]
    B --> C["Step 2: Balance all atoms except O and H"]
    C --> D["Step 3: Balance O using H₂O"]
    D --> E["Step 4: Balance H using H⁺ (acidic) or OH⁻ (basic)"]
    E --> F["Step 5: Balance charge using electrons"]
    F --> G["Step 6: Multiply to equalise electrons"]
    G --> H["Step 7: Add half-reactions, cancel electrons"]
    H --> I["Step 8: Verify atoms and charges"]

Why This Works

In a redox reaction, electrons are transferred from the reducing agent to the oxidising agent. The half-reaction method tracks these electrons explicitly. By balancing each half separately and then combining them so that electrons cancel, we ensure both mass and charge conservation.

The order matters: balance atoms first, then charge. In acidic medium we use $\text{H}^+$ and $\text{H}_2\text{O}$; in basic medium we add $\text{OH}^-$ after balancing in acidic medium (convert all $\text{H}^+$ to $\text{H}_2\text{O}$ by adding $\text{OH}^-$ to both sides).

Alternative Method

The oxidation number method is quicker for simpler equations: find the change in oxidation number, multiply to equalise, then balance by inspection. But for complex ions like $\text{MnO}_4^-$ and $\text{Cr}_2\text{O}_7^{2-}$, the half-reaction method is more systematic and less error-prone.

Common Mistake