Answer: 2 : 5.
- A 1 : 1
- B 2 : 5
- C 5 : 2
- D 3 : 5
Correct answer: B. 2 : 5
Explanation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (reduction, 5e⁻ per Mn). C₂O₄²⁻ → 2CO₂ + 2e⁻ (oxidation, 2e⁻ per oxalate). LCM(5,2) = 10. Multiply: 2 MnO₄⁻ (10e⁻ gained) and 5 C₂O₄²⁻ (10e⁻ lost). Ratio = 2 : 5.
Electron transfer in a redox reaction: zinc loses electrons (oxidised, acts as reducing agent) and copper(II) ions gain those same electrons (reduced, acts as oxidising agent).
Concept context
Master electron transfer in chemistry: assign oxidation states, balance half-reactions, identify oxidising and reducing agents, and connect redox to everyday reactions.