📚 Disproportionation | 歧化反应
In A-Level Chemistry, disproportionation is one of the most frequently tested redox concepts, particularly within the halogen and transition metal topics. Understanding how a single species can simultaneously act as both oxidising agent and reducing agent is essential for predicting reaction products and interpreting oxidation states correctly.
在 A-Level 化学中，歧化反应是氧化还原部分最常见的考点之一，尤其是在卤素和过渡金属专题中。理解同一种物质如何同时充当氧化剂和还原剂，对于正确预测反应产物和理解氧化态至关重要。
1. What Is Disproportionation? | 什么是歧化反应？
Disproportionation is a redox reaction in which the same element in one oxidation state is simultaneously oxidised and reduced, forming two different products containing that element in different oxidation states.
歧化反应是指同一元素在同一个氧化态下同时被氧化和还原，生成该元素处于不同氧化态的两种产物的氧化还原反应。
For example, in the reaction of chlorine with water, chlorine atoms (oxidation state 0) are converted into chloride ions (oxidation state −1) and chloric(I) acid (oxidation state +1). The chlorine is both reduced and oxidised.
例如，在氯气与水的反应中，氯原子（氧化态 0）转化为氯离子（氧化态 −1）和次氯酸（氧化态 +1）。氯气同时被还原和氧化。
The reverse process, in which two different species containing the same element in different oxidation states react to form a single product in an intermediate oxidation state, is called comproportionation (or synproportionation).
逆过程称为归中反应（或同化反应），即该元素处于不同氧化态的两种物质反应生成氧化态居中的单一产物。
2. Oxidation States: The Key to Spotting Disproportionation | 氧化态：识别歧化反应的关键
To identify a disproportionation reaction, you must be able to calculate oxidation states confidently. Recall the key rules: the oxidation state of an element in its free state is 0; for a monatomic ion it equals the ionic charge; oxygen is usually −2 (except in peroxides where it is −1); hydrogen is usually +1; and the sum of oxidation states in a neutral compound is 0, while for a polyatomic ion it equals the ionic charge.
要识别歧化反应，必须熟练掌握氧化态的计算。回顾关键规则：游离态元素氧化态为 0；单原子离子的氧化态等于其电荷；氧通常为 −2（过氧化物中为 −1）；氢通常为 +1；中性化合物中各原子氧化态之和为 0，多原子离子中各原子氧化态之和等于离子电荷。
In a disproportionation reaction, the key observation is that a single element appears with three different oxidation states across the reactants and products: one oxidation state in the reactant, and two different oxidation states — one higher and one lower — in the products.
在歧化反应中，关键观察点是一个元素在反应物和产物中出现三种不同的氧化态：反应物中一种氧化态，产物中两种不同的氧化态——一种升高，一种降低。
Element in oxidation state X → Element in oxidation state (X + n) + Element in oxidation state (X − m)
where n and m are positive integers. The element with the intermediate oxidation state X is the species that undergoes disproportionation.
其中 n 和 m 为正整数。氧化态为 X 的中间态元素即为发生歧化的物种。
3. Chlorine with Water | 氯气与水的反应
Chlorine dissolves in water and undergoes a disproportionation reaction to form hydrochloric acid and chloric(I) acid (hypochlorous acid):
氯气溶于水后发生歧化反应，生成盐酸和次氯酸：
Cl₂ + H₂O ⇌ HCl + HClO
In this reaction, the oxidation state of chlorine changes from 0 in Cl₂ to −1 in HCl (reduction) and +1 in HClO (oxidation). Therefore, chlorine is both the oxidising agent and the reducing agent.
在该反应中，氯的氧化态从 Cl₂ 中的 0 变为 HCl 中的 −1（被还原）和 HClO 中的 +1（被氧化）。因此，氯气既是氧化剂又是还原剂。
The equilibrium lies to the left, meaning that only a small proportion of the chlorine reacts with water. This equilibrium is important in water treatment, where chlorine is used to kill bacteria. The HClO (chloric(I) acid) is the active disinfecting species.
该平衡向左移动，意味着只有少量氯气与水反应。这一平衡在水处理中很重要，氯气用于杀菌消毒。次氯酸（HClO）是起消毒作用的活性物种。
4. Chlorine with Cold, Dilute Alkali | 氯气与冷稀碱的反应
When chlorine reacts with cold, dilute sodium hydroxide solution, it disproportionates to form sodium chloride and sodium chlorate(I) (sodium hypochlorite):
氯气与冷稀氢氧化钠溶液反应时，歧化生成氯化钠和次氯酸钠：
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
Again, chlorine changes from oxidation state 0 to −1 (in NaCl) and +1 (in NaClO). This reaction is the basis of the industrial production of bleach. Sodium chlorate(I) is the active ingredient in household bleach.
同样，氯的氧化态从 0 变为 NaCl 中的 −1 和 NaClO 中的 +1。该反应是工业制备漂白剂的基础，次氯酸钠是家用漂白剂中的有效成分。
It is worth noting the ionic equation for this reaction: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. The hydroxide ions provide the alkaline medium required for the disproportionation to proceed readily.
值得注意该反应的离子方程式：Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O。氢氧根离子提供了碱性环境，使歧化反应能够顺利进行。
5. Chlorine with Hot, Concentrated Alkali | 氯气与热浓碱的反应
The conditions of the reaction determine the extent of disproportionation. When chlorine is passed into hot, concentrated sodium hydroxide solution, a different disproportionation product is formed:
反应条件决定歧化的程度。当氯气通入热浓氢氧化钠溶液时，生成不同的歧化产物：
3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O
In this reaction, chlorine is disproportionated from oxidation state 0 to −1 (in NaCl) and +5 (in NaClO₃, sodium chlorate(V)). The higher temperature and concentration provide more energy, allowing the chlorine to be oxidised to a higher oxidation state.
在该反应中，氯从氧化态 0 歧化为 NaCl 中的 −1 和 NaClO₃（氯酸钠(V)）中的 +5。更高的温度和浓度提供更多能量，使氯能够被氧化到更高的氧化态。
This demonstrates an important exam point: the same reactant can undergo disproportionation to different products depending on reaction conditions. You must be able to write and balance both equations.
这揭示了一个重要考点：同一反应物在不同条件下可以发生歧化反应生成不同产物。你必须能够写出并配平这两个方程式。
6. Comparing the Chlorine Reactions | 氯气反应对比
| Condition 条件 | Products 产物 | Oxidation States of Cl 氯的氧化态 |
|---|---|---|
| Water 水 | HCl + HClO | −1 and +1 |
| Cold, dilute NaOH 冷稀 NaOH | NaCl + NaClO | −1 and +1 |
| Hot, concentrated NaOH 热浓 NaOH | 5NaCl + NaClO₃ | −1 and +5 |
The table shows that the oxidation state change is greater under hot, concentrated conditions, illustrating how reaction conditions control the thermodynamics and kinetics of disproportionation.
上表显示，在热浓条件下氧化态变化更大，说明反应条件如何控制歧化反应的热力学和动力学。
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