📚 Redox and Oxidation Number | 氧化还原与氧化数
Redox reactions are at the heart of A-Level Chemistry, linking electron transfer, oxidation states and practical titration techniques. This article explains how to assign oxidation numbers, use them to identify oxidation and reduction, balance half-equations and apply redox ideas to titrations and organic reactions.
氧化还原反应是A-Level化学的核心内容之一，它将电子转移、氧化态和滴定实验技术联系在一起。本文将讲解如何确定氧化数、用氧化数判断氧化与还原、配平半反应方程式，以及将氧化还原概念应用于滴定和有机反应。
1. What Is a Redox Reaction? | 什么是氧化还原反应？
A redox reaction is any reaction in which electrons are transferred from one species to another. Oxidation is loss of electrons; reduction is gain of electrons. A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain.
氧化还原反应是指物质之间发生电子转移的反应。氧化是失去电子，还原是得到电子。常用助记口诀是 OIL RIG：氧化是失电子，还原是得电子。
Oxidation and reduction always occur together. One species cannot lose electrons unless another species gains them. The species that loses electrons is oxidised, and the species that gains electrons is reduced.
氧化与还原总是同时发生。一种物质失去电子，必然有另一种物质得到电子。失去电子的物质被氧化，得到电子的物质被还原。
Earlier definitions also describe oxidation as gain of oxygen or loss of hydrogen, and reduction as loss of oxygen or gain of hydrogen. These definitions are still useful in organic chemistry.
早期定义还把氧化描述为加氧或失氢，把还原描述为失氧或加氢。这些定义在有机化学中仍然很有用。
Zn + Cu²⁺ → Zn²⁺ + Cu
2. Oxidation Number: Definition and Rules | 氧化数的定义与规则
The oxidation number, or oxidation state, is the charge an atom would have if all bonding electrons were assigned to the more electronegative atom in the bond. It is a bookkeeping tool, not always a real charge.
氧化数（又称氧化态）是假设成键电子完全归属于电负性较大的原子时，该原子所带的电荷。它是一种记账工具，并不总是真实电荷。
The following rules are used to assign oxidation numbers in molecules and ions.
以下规则用于确定分子和离子中各原子的氧化数。
| Rule | 规则 |
|---|---|
| Free element oxidation number is 0 | 单质氧化数为0 |
| Simple ion oxidation number equals its charge | 简单离子氧化数等于其电荷 |
| Hydrogen is +1, except in metal hydrides where it is -1 | 氢通常为+1，但在金属氢化物中为-1 |
| Oxygen is -2, except in peroxides where it is -1 and in OF₂ where it is +2 | 氧通常为-2，但在过氧化物中为-1，在OF₂中为+2 |
| Fluorine is always -1 in compounds | 化合物中氟始终为-1 |
| Sum of oxidation numbers equals the overall charge | 所有原子氧化数之和等于总电荷 |
3. Calculating Oxidation Numbers | 计算氧化数
To calculate an unknown oxidation number, write an algebraic equation using the known values and the total charge. The following worked examples show the method.
计算未知氧化数时，用已知氧化数和总电荷建立代数方程。以下示例展示该方法。
In H₂SO₄, let the oxidation number of S be x: 2(+1) + x + 4(-2) = 0, so x = +6.
在H₂SO₄中，设S的氧化数为x：2(+1) + x + 4(-2) = 0，因此x = +6。
In MnO₄⁻, let the oxidation number of Mn be x: x + 4(-2) = -1, so x = +7.
在MnO₄⁻中，设Mn的氧化数为x：x + 4(-2) = -1，因此x = +7。
In Cr₂O₇²⁻, let the oxidation number of Cr be x: 2x + 7(-2) = -2, so x = +6.
在Cr₂O₇²⁻中，设Cr的氧化数为x：2x + 7(-2) = -2，因此x = +6。
In NH₄⁺, let the oxidation number of N be x: x + 4(+1) = +1, so x = -3.
在NH₄⁺中，设N的氧化数为x：x + 4(+1) = +1，因此x = -3。
Some compounds, such as Fe₃O₄, appear to have fractional average oxidation numbers. In Fe₃O₄ the three Fe atoms have a total oxidation number of +8, giving an average of +8/3; this is acceptable because Fe₃O₄ contains both Fe²⁺ and Fe³⁺.
有些化合物如Fe₃O₄会出现分数形式的平均氧化数。在Fe₃O₄中三个Fe原子的氧化数总和为+8，因此平均值为+8/3；这是允许的，因为Fe₃O₄同时含有Fe²⁺和Fe³⁺。
4. Using Oxidation Numbers to Identify Redox | 用氧化数判断氧化还原反应
If the oxidation number of an atom increases during a reaction, that atom has been oxidised. If the oxidation number decreases, that atom has been reduced. If no atoms change oxidation number, the reaction is not redox.
若某原子的氧化数在反应中升高，则该原子被氧化；若氧化数降低，则该原子被还原。若所有原子的氧化数都未改变，则该反应不是氧化还原反应。
Consider the reaction Zn + CuSO₄ → ZnSO₄ + Cu. Zinc changes from 0 to +2, so Zn is oxidised. Copper changes from +2 to 0, so Cu²⁺ is reduced.
以反应 Zn + CuSO₄ → ZnSO₄ + Cu 为例。锌从0变为+2，因此Zn被氧化；铜从+2变为0，因此Cu²⁺被还原。
In contrast, neutralisation and precipitation reactions usually do not involve oxidation number changes. For example, Ag⁺ + Cl⁻ → AgCl is not redox because Ag remains +1 and Cl remains -1.
相比之下，中和反应和沉淀反应通常不涉及氧化数变化。例如 Ag⁺ + Cl⁻ → AgCl 不是氧化还原反应，因为Ag仍为+1，Cl仍为-1。
5. Oxidising and Reducing Agents | 氧化剂与还原剂
An oxidising agent, or oxidant, accepts electrons and is itself reduced. A reducing agent, or reductant, donates electrons and is itself oxidised.
氧化剂接受电子，本身被还原；还原剂给出电子，本身被氧化。
In the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, Zn is the reducing agent because it donates electrons and is oxidised. Cu²⁺ is the oxidising agent because it accepts electrons and is reduced.
在反应 Zn + Cu²⁺ → Zn²⁺ + Cu 中，Zn是还原剂，因为它给出电子并被氧化；Cu²⁺是氧化剂，因为它接受电子并被还原。
Common oxidising agents include acidified KMnO₄, K₂Cr₂O₇, O₂, H₂O₂ and halogens. Common reducing agents include reactive metals, I⁻, Fe²⁺, S₂O₃²⁻ and H₂.
常见氧化剂包括酸化KMnO₄、K₂Cr₂O₇、O₂、H₂O₂和卤素。常见还原剂包括活泼金属、I⁻、Fe²⁺、S₂O₃²⁻和H₂。
6. Writing and Combining Half-Equations | 书写与合并半反应方程式
Half-equations show either oxidation or reduction alone. In acidic solution, follow these steps: balance the atoms that change; balance O by adding H₂O; balance H by adding H⁺; then balance charge by adding electrons.
半反应方程式单独表示氧化或还原过程。在酸性溶液中，按以下步骤配平：先配平发生变化的原子；用H₂O配平O；用H⁺配平H；最后用电子配平电荷。
For the reduction of manganate(VII) to Mn²⁺, the half-equation is:
高锰酸根离子还原为Mn²⁺的半反应方程式为：
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
For the oxidation of Fe²⁺ to Fe³⁺, the half-equation is:
Fe²⁺氧化为Fe³⁺的半反应方程式为：
Fe²⁺ → Fe³⁺ + e⁻
To combine the two half-equations, multiply the iron half-equation by 5 so that the electrons cancel. The overall ionic equation is:
合并两个半反应时，将铁的半反应乘以5使电子数相等并抵消。总离子方程式为：
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
7. Disproportionation and Comproportionation | 歧化反应与归中反应
Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced. Its oxidation number both increases and decreases.
歧化反应是同一元素同时被氧化和还原的反应，其氧化数既升高又降低。
A classic example is the reaction of chlorine with cold dilute sodium hydroxide: Cl₂ + 2NaOH → NaCl + NaClO + H₂O. Chlorine changes from 0 in Cl₂ to -1 in NaCl and +1 in NaClO.
一个经典例子是氯与冷的稀氢氧化钠溶液反应：Cl₂ + 2NaOH → NaCl + NaClO + H₂O。氯从Cl₂中的0变为NaCl中的-1和NaClO中的+1。
Hydrogen peroxide can also disproportionate: 2H₂O₂ → 2H₂O + O₂. Oxygen changes from -1 in H₂O₂ to -2 in H₂O and 0 in O₂.
过氧化氢也可以发生歧化：2H₂O₂ → 2H₂O + O₂。氧从H₂O₂中的-1变为H₂O中的-2和O₂中的0。
Comproportionation is the reverse idea: two species containing the same element in different oxidation states react to form a species in an intermediate oxidation state.
归中反应则相反：同一元素的两种不同氧化态物质反应，生成中间氧化态的物质。
8. Redox Titrations | 氧化还原滴定
Redox titrations are used to determine the concentration of oxidising or reducing agents. A key Cambridge example is the titration of Fe²⁺ with acidified potassium manganate(VII).
氧化还原滴定用于测定氧化剂或还原剂的浓度。剑桥考试中的一个重要例子是用酸化高锰酸钾滴定Fe²⁺。
The reaction is MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺. The endpoint is detected because MnO₄⁻ is purple and Mn²⁺ is almost colourless; excess MnO₄⁻ gives a persistent pink colour.
反应为 MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺。终点可通过颜色判断，因为MnO₄⁻为紫色，Mn²⁺几乎无色；过量的MnO₄⁻会使溶液呈持久的粉红色。
Iodine-thiosulfate titrations are also common. The reaction I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻ is used to determine oxidising agents that liberate iodine. Starch is added near the endpoint and forms a blue-black complex with iodine; the endpoint is the disappearance of the blue colour.
碘-硫代硫酸钠滴定也很常见。反应 I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻ 用于测定能释放碘的氧化剂。接近终点时加入淀粉，淀粉与碘形成蓝黑色配合物；终点为蓝色消失。
9. Common Redox Reagents and Colour Changes | 常见氧化还原试剂与颜色变化
The table below summarises colour changes that are frequently tested in Cambridge A-Level Chemistry redox questions.
下表总结了剑桥A-Level化学氧化还原题中常考的颜色变化。
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