Answer: Cu 2+ + Cu → 2Cu+.
- A 2H<sub>2</sub>O<sub>2</sub> → 2H<sub>2</sub>O + O<sub>2</sub>
- B Cu<sup>2+</sup> + Cu → 2Cu+
- C Cl<sub>2</sub> + 2NaOH → NaCl + NaOCl + H<sub>2</sub>O
- D 2MnO<sub>4</sub><sup>-</sup> + 5H<sub>2</sub>C<sub>2</sub>O<sub>4</sub> → 2Mn<sup>2+</sup> + 10CO<sub>2</sub> + 8H<sub>2</sub>O
Correct answer: B. Cu<sup>2+</sup> + Cu → 2Cu+
Explanation: Cu²⁺ (OS +2) + Cu (OS 0) → 2Cu⁺ (OS +1). The higher and lower oxidation state of the same element combine to give an intermediate oxidation state. This is comproportionation.
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.