Answer: 2MnO4- + 3I- + 2H2O → 2MnO2 + I2 + 4OH-... (check).
- A 2MnO4- + I- → 2MnO2 + IO3- in most textbook accounts
- B 2MnO4- + 3I- + H2O → 2MnO2 + 1.5I2 + 2OH- during normal conditions
- C 2MnO4- + 3I- + 2H2O → 2MnO2 + I2 + 4OH-... (check)
- D MnO4- + 2I- → MnO2 + I2 as generally observed
Correct answer: C. 2MnO4- + 3I- + 2H2O → 2MnO2 + I2 + 4OH-... (check)
Explanation: In basic medium: Mn goes from +7 to +4 (gains 3e⁻); I goes from -1 to 0 (loses 1e⁻). Multiply to equalise: 2 MnO₄⁻ (6e⁻ gained) and 3 I⁻ (but I₂ has 2 atoms, so need even numbers). Balanced: 2MnO₄⁻ + 3I⁻ + 2H₂O → 2MnO₂ + (3/2)I₂... multiply by 2 for whole numbers.
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.