MnO-4(aq) + Y + 5Fe2+(aq) → Mn2+(aq) + 5Fe2+(aq) + 4H2O(l) In the equation above, Y is

Analyze the question: This is a redox equation in acidic medium. Permanganate ion (MnO₄⁻) oxidises Fe²⁺ to Fe³⁺. The question has a typo: the product side should show 5Fe³⁺, not 5Fe²⁺. We need to find Y, the species that balances the equation.

Concept: In acidic solution, MnO₄⁻ is reduced to Mn²⁺. The oxygen atoms end up as water, and H⁺ ions are needed to balance the hydrogen and the charge.

Step-by-step solution:

1. Half-reaction for reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

• Oxygen: 4 O on the left, so 4H₂O on the right.
• Hydrogen: 4H₂O contains 8 H, so we need 8H⁺ on the left.
• Charge: left = (−1) + 8 + (−5) = +2, right = +2 ✓

2. Half-reaction for oxidation (×5): 5Fe²⁺ → 5Fe³⁺ + 5e⁻

3. Combine: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Check the balance:

• Charge on left: −1 + 8 + 10 = +17
• Charge on right: +2 + 15 = +17 ✓
• H: 8 on each side ✓
• O: 4 on each side ✓

So Y = 8H⁺(aq).

Tip: For MnO₄⁻ → Mn²⁺ in acid, remember the pattern "1, 8, 5, 4": 1 MnO₄⁻ takes 8H⁺ and 5e⁻ to give 4H₂O. The 4 oxygens always give 4 waters, which need 8 H⁺.