Balancing a redox reaction with MnO₄⁻ and Fe²⁺, illustrating oxidation states, half-equations, and electron transfer.

Let’s solve this step by step.

We are given the unbalanced redox reaction:

MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O

Our goal is to balance it using the half-reaction method in acidic solution.

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Step 1: Assign oxidation states (already done in image, but let’s verify)

- In MnO₄⁻: Oxygen is -2 each → 4 × (-2) = -8. Total charge is -1 → Mn must be +7
- Wait! The image says -7 for Mn? That’s a mistake. Let me correct that.
- Actually: Mn + 4×(-2) = -1 → Mn - 8 = -1 → Mn = +7
- So Mn goes from +7 in MnO₄⁻ to +2 in Mn²⁺ → reduction (gains electrons)
- Fe²⁺ → Fe³⁺: Iron goes from +2 to +3 → oxidation (loses electron)
- H⁺ is +1, H in H₂O is +1, O is -2 — no change.

✔ Correction: Mn starts at +7, not -7. The image has a typo there. But the rest of the balancing is still correct because they used the right number of electrons later.

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Step 2: Write half-reactions

#### Reduction half: MnO₄⁻ → Mn²⁺

Balance atoms other than H and O:
→ Mn is already balanced.

Balance O by adding H₂O:
→ Add 4 H₂O on right to balance 4 O on left:
MnO₄⁻ → Mn²⁺ + 4H₂O

Balance H by adding H⁺:
→ Right has 8 H (from 4 H₂O), so add 8 H⁺ on left:
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balance charge by adding electrons:
Left side charge: MnO₄⁻ (-1) + 8H⁺ (+8) = +7
Right side: Mn²⁺ (+2) + 4H₂O (0) = +2
To go from +7 to +2, we need to add 5e⁻ to the left:
✔ MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

#### Oxidation half: Fe²⁺ → Fe³⁺

Already balanced for atoms.

Charge: Left = +2, Right = +3 → add 1e⁻ to right:
Fe²⁺ → Fe³⁺ + e⁻

But we need to match electrons with the reduction half (which uses 5e⁻). So multiply entire oxidation half by 5:
✔ 5Fe²⁺ → 5Fe³⁺ + 5e⁻

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Step 3: Combine half-reactions

Add them together:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
5Fe²⁺ → 5Fe³⁺ + 5e⁻
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Cancel out the 5e⁻ on both sides:

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

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Step 4: Check atom and charge balance

Atoms:
- Mn: 1 = 1 ✔
- O: 4 = 4 (in 4H₂O) ✔
- H: 8 = 8 (in 4H₂O) ✔
- Fe: 5 = 5 ✔

Charge:
Left: MnO₄⁻ (-1) + 8H⁺ (+8) + 5Fe²⁺ (+10) = +17
Right: Mn²⁺ (+2) + 5Fe³⁺ (+15) + 4H₂O (0) = +17 ✔

Perfectly balanced!

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Final Answer:
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O