Balance the redox reaction MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic medium

Question

Balance the following redox reaction using the ion-electron (half-reaction) method in acidic medium:

\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}

Solution — Step by Step

Split the reaction into oxidation and reduction half-reactions:

Reduction half-reaction (MnO₄⁻ is reduced — Mn goes from +7 to +2):

\text{MnO}_4^- \rightarrow \text{Mn}^{2+}

Oxidation half-reaction (Fe²⁺ is oxidised — Fe goes from +2 to +3):

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}

For the reduction half-reaction, Mn is already balanced (1 Mn on each side). ✓

For the oxidation half-reaction, Fe is already balanced (1 Fe on each side). ✓

Reduction half-reaction has 4 oxygen atoms on the left and 0 on the right. Add 4 H₂O to the right:

\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Oxidation half-reaction has no oxygen atoms — no H₂O needed. ✓

Reduction half-reaction: Now we have 8 H atoms on the right (from 4 H₂O). Add 8 H⁺ to the left:

8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Oxidation half-reaction: No H to balance. ✓

Reduction half-reaction: Left charge: $8(+1) + (-1) = +7$ Right charge: $+2 + 0 = +2$

To make charges equal: add 5 electrons to the left:

5e^- + 8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Left charge check: $-5 + 8 - 1 = +2$ ✓ Right charge: $+2$ ✓

Oxidation half-reaction: Left charge: $+2$ Right charge: $+3$

Remove 1 electron from right (or add 1e⁻ to right side going to products = subtract from left = add to right as product… simplest: 1 electron released):

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

Left charge: $+2$ , Right charge: $+3 - 1 = +2$ ✓

Reduction gains 5e⁻, oxidation releases 1e⁻. Multiply oxidation by 5:

5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-

Now add the two half-reactions. The 5e⁻ cancels:

5e^- + 8\text{H}^+ + \text{MnO}_4^- + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+} + 5e^-

Cancelling 5e⁻ from both sides:

\boxed{8\text{H}^+ + \text{MnO}_4^- + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}}

Why This Works

The ion-electron method works because redox reactions are fundamentally electron transfer processes. By separating into half-reactions, we:

Track electrons explicitly — we know exactly how many are gained (5) and lost (1 × 5 = 5) — must be equal
Use water and H⁺ (in acid) or OH⁻ and water (in base) to balance O and H without changing the overall charge balance

This method always gives a balanced equation because we enforce both:

Mass balance: atoms of each element equal on both sides
Charge balance: total charge equal on both sides (electrons cancelled out)

Alternative Method — Oxidation Number Method

Change in oxidation number approach:

Mn: $+7 \to +2$ , gain of 5 electrons per Mn
Fe: $+2 \to +3$ , loss of 1 electron per Fe

To balance electrons: 5 Fe atoms needed for every 1 Mn atom (5 × 1 = 1 × 5).

Skeleton equation: $\text{MnO}_4^- + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+}$

Then balance O by adding 4H₂O on right and 8H⁺ on left. Same final result.

KMnO₄ reactions in acidic medium always involve: MnO₄⁻ → Mn²⁺ (gain 5e⁻, add 8H⁺, produce 4H₂O). Memorise this half-reaction — it appears in nearly every JEE Main and NEET redox question involving KMnO₄.

Common Mistake

Students confuse the acidic medium method with the basic medium method. In acidic medium: add H₂O for oxygen balance, add H⁺ for hydrogen balance. In basic medium: add H₂O for hydrogen balance, add OH⁻ for oxygen balance. Using H⁺ in a basic medium problem gives the wrong balanced equation and full marks lost.

Question

Solution — Step by Step

Why This Works

Alternative Method — Oxidation Number Method

Common Mistake

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