📚 Disproportionation Reactions | 歧化反应原理与实例
Disproportionation is one of the most distinctive patterns of redox behaviour in A-Level chemistry. In this type of reaction, a single element in a single reactant species is simultaneously oxidised and reduced. Understanding this concept is essential for questions on chlorine chemistry, sulfur chemistry, hydrogen peroxide, and transition-metal ions.
歧化反应是 A-Level 化学中最具特色的氧化还原反应类型之一。在这类反应中，同一种元素在同一反应物中同时被氧化又被还原。理解这一概念，对于回答有关氯化学、硫化学、过氧化氢以及过渡金属离子的题目至关重要。
1. Definition and Key Ideas | 歧化反应的定义与核心要点
A disproportionation reaction is a redox reaction in which the element in one oxidation state is converted into two different oxidation states: one higher and one lower than the original state. The element therefore acts as both the oxidising agent and the reducing agent.
歧化反应是一种氧化还原反应，其中处于某一氧化态的元素被转化为两个不同的氧化态：一个高于原来，一个低于原来。因此，该元素同时充当氧化剂和还原剂。
The reverse process is called comproportionation, in which two different oxidation states of the same element combine to form a single intermediate oxidation state.
相反的过程称为归中反应（comproportionation），即同一种元素的两个不同氧化态结合，生成一个单一的中间氧化态。
General form: A → A(oxidised) + A(reduced)
For example, chlorine molecules can react with cold dilute sodium hydroxide to give sodium chloride and sodium chlorate(I). Chlorine is the only element changing oxidation state, so it must have been both oxidised and reduced.
例如，氯分子能与冷的稀氢氧化钠反应生成氯化钠和次氯酸钠。其中只有氯元素的氧化态发生改变，因此氯必然既被氧化又被还原。
2. Assigning Oxidation States | 氧化态的判断
To identify a disproportionation reaction, the oxidation state of each element must be assigned correctly. The usual rules apply: the sum of oxidation states in a neutral molecule is zero, and in a polyatomic ion the sum equals the charge on the ion.
要判断歧化反应，必须先正确标出各元素的氧化态。常用规则仍然适用：中性分子中所有原子的氧化态之和为零；多原子离子中所有原子的氧化态之和等于该离子所带的电荷。
| Species | 粒子 | Oxidation state of the highlighted element | 目标元素的氧化态 |
| Cl₂ | 0 |
| Cl⁻ | -1 |
| ClO⁻ | +1 |
| ClO₃⁻ | +5 |
| S₂O₃²⁻ | 平均 | +2 |
| H₂O₂ | O 的氧化态 | -1 |
In this table, chlorine moves from 0 to -1 in one product and to +1 or +5 in another product. The change in oxidation state is the evidence for disproportionation.
在上表中，氯从 0 变为其中一个产物中的 -1，同时变为另一个产物中的 +1 或 +5。氧化态的变化就是发生歧化反应的证据。
3. Why Do Disproportionation Reactions Occur? | 歧化反应为什么会发生？
An element in an intermediate oxidation state can act as both an oxidising agent and a reducing agent. If the oxidation and reduction products are thermodynamically more stable than the starting material, the disproportionation is favourable.
处于中间氧化态的元素既可以是氧化剂，也可以是还原剂。如果氧化产物和还原产物在热力学上比反应物更稳定，歧化反应便容易发生。
Reaction conditions are very important. Changing pH, temperature, or concentration can change the stability of the intermediate state and therefore change the products formed. The chemistry of chlorine shows this beautifully: cold dilute alkali gives chlorate(I), whereas hot concentrated alkali gives chlorate(V).
反应条件非常重要。改变 pH、温度或浓度，会改变中间氧化态的稳定性，从而改变生成的产物。氯的化学就清楚地说明了这一点：冷的稀碱生成氯酸盐(I)，而热浓碱生成氯酸盐(V)。
4. Chlorine with Water | 氯气与水的反应
When chlorine dissolves in water, it undergoes a reversible disproportionation reaction:
当氯气溶于水时，会发生可逆的歧化反应：
Cl₂ + H₂O ⇌ HCl + HClO
In this equilibrium, chlorine atoms are reduced from 0 to -1 in hydrogen chloride, and oxidised from 0 to +1 in chloric(I) acid, HClO.
在此平衡中，氯原子在氯化氢中从 0 被还原为 -1，在次氯酸 HClO 中从 0 被氧化为 +1。
This reaction is the basis of using chlorine to treat drinking water and swimming pools. The HClO produced is a weak acid and a powerful oxidising agent that kills bacteria.
该反应是氯气用于饮用水和泳池消毒的基础。生成的 HClO 是弱酸，也是强氧化剂，可以杀灭细菌。
5. Chlorine with Alkali: Cold Dilute versus Hot Concentrated | 氯气与碱：冷稀碱与热浓碱
Chlorine reacts more completely with alkali than with water. The products depend on the temperature and concentration of the alkali.
氯与碱的反应比与水更完全。产物取决于碱的温度和浓度。
Cold dilute alkali | 冷的稀碱
With cold dilute sodium hydroxide, chlorine disproportionates to chloride and chlorate(I):
氯气与冷的稀氢氧化钠反应，歧化生成氯化物和氯酸盐(I)：
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
The half-equations are:
对应的半反应为：
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O + 2e⁻
Reduction: Cl₂ + 2e⁻ → 2Cl⁻
Hot concentrated alkali | 热的浓碱
With hot concentrated alkali, the chlorate(I) ion itself disproportionates further into chloride and chlorate(V):
与热的浓碱反应时，氯酸盐(I)离子本身会进一步歧化为氯化物和氯酸盐(V)：
3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O
The oxidation state of chlorine changes from 0 to -1 in chloride and from 0 to +5 in chlorate(V). This is why the stoichiometric ratio of chloride to chlorate(V) is 5:1.
氯的氧化态在氯化物中从 0 变为 -1，在氯酸盐(V)中从 0 变为 +5。因此氯化物与氯酸盐(V)的化学计量比为 5:1。
6. Thiosulfate Ion in Acid | 硫代硫酸根离子与酸的反应
Thiosulfate is a classic example of a sulfur species undergoing disproportionation. In aqueous acid it decomposes rapidly:
硫代硫酸盐是硫元素发生歧化的经典例子。在酸性水溶液中它会迅速分解：
S₂O₃²⁻ + 2H⁺ → S + SO₂ + H₂O
A yellow precipitate of sulfur forms, and sulfur dioxide gas with a pungent smell is evolved. This reaction is used in the laboratory as a simple test for thiosulfate ions.
反应中生成黄色硫沉淀，同时放出具有刺激性气味的二氧化硫气体。该反应在实验室中可用来检验硫代硫酸根离子。
In S₂O₃²⁻, the average oxidation state of sulfur is +2. In the products, one sulfur atom is reduced from +6 to +4 in SO₂, while the other sulfur atom is oxidised from -2 to 0 in S. The same element therefore undergoes both oxidation and reduction.
在 S₂O₃²⁻ 中，硫的平均氧化态为 +2。产物中，一个硫原子在 SO₂ 中从 +6 被还原为 +4，另一个硫原子在 S 中从 -2 被氧化为 0。因此，同一种元素同时发生了氧化和还原。
7. Hydrogen Peroxide Decomposition | 过氧化氢的分解
Hydrogen peroxide contains oxygen in the intermediate oxidation state -1. It can quietly decompose, especially in the presence of light, transition-metal ions, or manganese(IV) oxide:
过氧化氢中的氧处于中间氧化态 -1。在光照、过渡金属离子或二氧化锰存在下，它会缓慢分解：
2H₂O₂ → 2H₂O + O₂
Oxygen is reduced from -1 to -2 in water and oxidised from -1 to 0 in dioxygen. The reaction is disproportionation because the same element, oxygen, is both oxidised and reduced.
氧在水分子中从 -1 被还原为 -2，在氧气中从 -1 被氧化为 0。由于同一种元素氧既被氧化又被还原，所以这是一个歧化反应。
The decomposition explains why hydrogen peroxide must be stored in a cool, dark place in a suitable container, and why a catalyst such as manganese(IV) oxide produces oxygen gas rapidly.
这一分解反应解释了为什么过氧化氢需要避光、低温储存，也解释了二氧化锰等催化剂为何能使其快速放出氧气。
8. Further Examples: Copper(I) and Nitrogen Dioxide | 更多实例：铜(I)与二氧化氮
Copper(I) compounds are not stable in aqueous acid. Copper(I) disproportionates to copper metal and copper(II):
铜(I)化合物在酸性水溶液中不稳定。铜(I)会歧化为铜单质和铜(II)：
2Cu⁺ → Cu + Cu²⁺
Cu₂O + 2H⁺ → Cu + Cu²⁺ + H₂O
Another familiar example is nitrogen dioxide reacting with water. Nitrogen in NO₂ has oxidation state +4, and it disproportionates to +3 in nitrous acid and +5 in nitric acid:
另一个常见实例是二氧化氮与水的反应。NO₂ 中氮的氧化态为 +4，它歧化为亚硝酸中的 +3 和硝酸中的 +5：
2NO₂ + H₂O → HNO₂ + HNO₃
These examples show that disproportionation is not limited to chlorine and sulfur. Any element
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