歧化反应Disproportionation

氯气与冷的稀 NaOH 溶液反应：Cl₂ + 2NaOH → NaCl + NaClO + H₂O，氯由 0 价变为 −1 价和 +1 价，用于制备漂白液。与热的浓 NaOH 溶液反应：3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O，氯由 0 价变为 −1 价和 +5 价。氯气与水反应 Cl₂ + H₂O ⇌ HCl + HClO 也是歧化反应。

Chlorine with cold dilute NaOH: Cl₂ + 2NaOH → NaCl + NaClO + H₂O, chlorine going from 0 to −1 and +1; this makes bleach. With hot concentrated NaOH: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O, chlorine going from 0 to −1 and +5. The reaction of chlorine with water, Cl₂ + H₂O ⇌ HCl + HClO, is also a disproportionation.

过氧化氢分解：2H₂O₂ → 2H₂O + O₂，氧由 −1 价变为 −2 价和 0 价。Cu⁺ 在水溶液中歧化：2Cu⁺ → Cu + Cu²⁺，所以水溶液中难以存在稳定的 Cu⁺ 化合物。NO₂ 溶于水：3NO₂ + H₂O → 2HNO₃ + NO。

Decomposition of hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂, oxygen going from −1 to −2 and 0. Cu⁺ disproportionates in water, 2Cu⁺ → Cu + Cu²⁺, which is why simple copper(I) compounds are unstable in aqueous solution. NO₂ in water: 3NO₂ + H₂O → 2HNO₃ + NO.

某中间氧化态能否歧化，可比较它作为氧化剂和还原剂的两个电对：E°(Cu⁺/Cu) = +0.52 V，E°(Cu²⁺/Cu⁺) = +0.15 V，前者更正，E°cell = 0.52 − 0.15 = +0.37 V > 0，所以 Cu⁺ 的歧化在标准状态下可自发进行。

Whether an intermediate oxidation state disproportionates can be judged by comparing the couple in which it is reduced with the couple in which it is oxidised: E°(Cu⁺/Cu) = +0.52 V and E°(Cu²⁺/Cu⁺) = +0.15 V; the first is more positive, E°cell = 0.52 − 0.15 = +0.37 V > 0, so disproportionation of Cu⁺ is feasible under standard conditions.