Number of moles of H⁺ ions required by 1 mole of MnO₄⁻ to oxidise oxalate ion to CO₂ is _________.

Solution

Chemistry ma’am · 4 steps

All the H+ goes to permanganate's half-reaction, so balance that half and read off the 8.

1. Reduction half. In acid, Mn goes from +7 in MnO₄⁻ to +2 in Mn²⁺ and gains 5 electrons. The 4 oxygens leave as 4 water molecules, and those need 8 H⁺.

   MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

2. Oxidation half. Each C goes from +3 to +4: C₂O₄²⁻ → 2CO₂ + 2e⁻. This half uses no H⁺ at all.
3. Full equation. Multiply the halves by 2 and 5 so the electrons cancel: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. That is 16 H⁺ for 2 MnO₄⁻.
4. So. 1 mole of MnO₄⁻ needs 16 ÷ 2 = 8 moles of H⁺.

The common slip: 16. Reading the 16 from the full balanced equation and forgetting it is for 2 moles of MnO₄⁻ gives 16, not 8.

The idea to keep. In acidic medium, acidified permanganate always needs 8 H+ per MnO₄⁻ to reach Mn²⁺. If the other half uses no H+, that 8 is your answer straight away.

Answer from the NTA final answer key. Solution by PrepWiser.