📚 What Is a Redox Reaction? | 什么是氧化还原反应？
Redox reactions drive respiration, combustion, photosynthesis, corrosion, and the operation of batteries. In A-level chemistry, a redox reaction is defined as a reaction in which both reduction and oxidation occur simultaneously through electron transfer and changes in oxidation state.
氧化还原反应驱动着呼吸作用、燃烧、光合作用、腐蚀以及电池的工作。在 A-level 化学中，氧化还原反应被定义为通过电子转移和氧化数变化、还原与氧化同时发生的反应。
1. Oxidation and Reduction: The Electron Transfer View | 氧化与还原：电子转移视角
At A-level, oxidation is defined as loss of electrons, and reduction is defined as gain of electrons. A useful mnemonic is OIL RIG: Oxidation Is Loss, Reduction Is Gain. Because electrons are negatively charged, losing electrons makes an atom more positive in oxidation state, while gaining electrons makes it more negative.
在 A-level 阶段，氧化被定义为失去电子，还原被定义为得到电子。一个常用的记忆方法是 OIL RIG：氧化是失去电子，还原是得到电子。由于电子带负电，失去电子会使原子的氧化数更正，得到电子则使氧化数更负。
Consider the reaction between zinc and copper(II) ions:
考虑锌与铜(II)离子之间的反应：
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Each zinc atom loses two electrons to form a Zn²⁺ ion, so zinc is oxidised. Each copper(II) ion gains two electrons to form a copper atom, so Cu²⁺ is reduced. Oxidation and reduction always occur together because the electrons lost by one species must be gained by another.
每个锌原子失去两个电子形成 Zn²⁺ 离子，因此锌被氧化。每个铜(II)离子得到两个电子形成铜原子，因此 Cu²⁺ 被还原。氧化与还原总是同时发生，因为一种粒子失去的电子必须被另一种粒子得到。
2. Oxidation States: Rules and Calculation | 氧化数：规则与计算
An oxidation state, also called an oxidation number, is a bookkeeping number assigned to an atom in a compound or ion. It helps identify which species is oxidised and which is reduced without having to draw full electron transfer.
氧化数，也称为氧化态，是分配给化合物或离子中某个原子的记账数值。它有助于判断哪种粒子被氧化、哪种被还原，而无需画出完整的电子转移过程。
| Rule | Example |
|---|---|
| Free element has oxidation state 0 | O₂, Na, Cl₂ |
| Monatomic ion equals its charge | Na⁺ = +1, Cl⁻ = -1 |
| Oxygen is usually -2 | H₂O, CO₂; but -1 in peroxides, +2 in OF₂ |
| Hydrogen is usually +1 | H₂O, HCl; but -1 in metal hydrides |
| Sum equals charge on ion or 0 for neutral compound | SO₄²⁻ has total -2 |
For example, in H₂SO₄, hydrogen is +1 each, oxygen is -2 each, and the total is 2(+1) + S + 4(-2) = 0. Therefore the oxidation state of sulfur is +6.
例如，在 H₂SO₄ 中，每个氢的氧化数为 +1，每个氧为 -2，总氧化数为 2(+1) + S + 4(-2) = 0。因此硫的氧化数为 +6。
3. Using Oxidation States to Identify Redox | 利用氧化数判断氧化还原反应
A redox reaction must have at least one element whose oxidation state increases, meaning oxidation, and at least one element whose oxidation state decreases, meaning reduction. If no oxidation states change, the reaction is not a redox reaction. Acid-base neutralisation and precipitation reactions are usually not redox.
氧化还原反应中至少有一种元素的氧化数升高，即发生氧化；同时至少有一种元素的氧化数降低，即发生还原。如果所有氧化数都不变，该反应就不是氧化还原反应。酸碱中和反应和沉淀反应通常不属于氧化还原反应。
2Mg(s) + O₂(g) → 2MgO(s)
Magnesium changes from oxidation state 0 to +2, so it is oxidised. Oxygen changes from 0 to -2, so it is reduced. This confirms the reaction is redox.
镁的氧化数从 0 变为 +2，因此镁被氧化。氧的氧化数从 0 变为 -2，因此氧被还原。这证实该反应属于氧化还原反应。
By contrast, in H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, the oxidation states of H, S, O, and Na do not change. This is an acid-base reaction, not a redox reaction.
相比之下，在 H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O 中，H、S、O 和 Na 的氧化数均未改变。这是一个酸碱反应，不是氧化还原反应。
4. Oxidising and Reducing Agents | 氧化剂与还原剂
An oxidising agent, also called an oxidant, accepts electrons and is itself reduced during the reaction. A reducing agent, also called a reductant, donates electrons and is itself oxidised. The agent causes the opposite change in the other reactant.
氧化剂又称氧化剂，它在反应中得到电子，自身被还原。还原剂又称还原剂，它在反应中失去电子，自身被氧化。试剂使另一种反应物发生相反的变化。
2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq)
Iodide ions, I⁻, change from -1 to 0 in I₂, so they lose electrons and are oxidised. Therefore I⁻ acts as the reducing agent. Iron(III) ions, Fe³⁺, change from +3 to +2, so they gain electrons and are reduced. Therefore Fe³⁺ acts as the oxidising agent.
碘离子 I⁻ 的氧化数从 -1 变为 I₂ 中的 0，因此 I⁻ 失去电子，被氧化。所以 I⁻ 是还原剂。铁(III)离子 Fe³⁺ 的氧化数从 +3 变为 +2，因此 Fe³⁺ 得到电子，被还原。所以 Fe³⁺ 是氧化剂。
5. Half-Equations: Showing Electron Transfer | 半反应方程式：表示电子转移
Half-equations separate the oxidation process from the reduction process. In acidic solution, balance oxygen atoms by adding H₂O, balance hydrogen atoms by adding H⁺, and balance charge by adding electrons.
半反应方程式将氧化过程与还原过程分开表示。在酸性溶液中，通过加 H₂O 平衡氧原子，通过加 H⁺ 平衡氢原子，通过加电子平衡电荷。
A common reduction half-equation involving manganate(VII) is:
一个常见的涉及高锰酸根离子的还原半反应为：
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
Manganese changes from +7 in MnO₄⁻ to +2 in Mn²⁺, so it gains five electrons. This is reduction. A corresponding oxidation half-equation is Fe²⁺ → Fe³⁺ + e⁻.
锰的氧化数从 MnO₄⁻ 中的 +7 变为 Mn²⁺ 中的 +2，因此它得到五个电子。这是还原反应。相应的氧化半反应为 Fe²⁺ → Fe³⁺ + e⁻。
6. Balancing Redox Equations | 配平氧化还原方程式
To construct an overall redox equation, first write the two half-equations. Then multiply each half-equation by a suitable factor so that the electrons cancel. Finally, add the half-equations and simplify species that appear on both sides.
要写出完整的氧化还原方程式，首先写出两个半反应方程式。然后将各半反应乘以适当的系数，使电子数相互抵消。最后将两个半反应相加，并消去两侧都出现的粒子。
For manganate(VII) and iron(II) in acid, the reduction half-equation has 5e⁻. Multiply the oxidation half-equation Fe²⁺ → Fe³⁺ + e⁻ by 5, then add:
对于酸性条件下的高锰酸根和铁(II)，还原半反应有 5 个电子。将氧化半反应 Fe²⁺ → Fe³⁺ + e⁻ 乘以 5，然后相加：
MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)
Check that atoms and charge balance: one Mn, four O, eight H, five Fe on each side. The total charge is +17 on both sides, confirming the equation is balanced.
检查原子和电荷是否守恒：两侧各有 1 个 Mn、4 个 O、8 个 H 和 5 个 Fe。两侧总电荷均为 +17，证明方程式已配平。
7. Disproportionation Reactions | 歧化反应
Disproportionation is a special type of redox reaction in which the same element in a single species is both oxidised and reduced. The reactant contains an element in an intermediate oxidation state that splits into two different products.
歧化反应是一种特殊的氧化还原反应，其中同一种元素在同一物质中既被氧化又被还原。反应物中某元素处于中间氧化态，反应后转变成两种不同的产物。
Cl₂(aq) + 2OH⁻(aq) → Cl⁻(aq) + ClO⁻(aq) + H₂O(l)
Chlorine starts at oxidation state 0. In Cl⁻, chlorine becomes -1, so it is reduced. In ClO⁻, with oxygen as -2, chlorine becomes +1, so it is oxidised. One chlorine molecule contains two atoms that follow different paths, making this a disproportionation reaction.
氯的起始氧化数为 0。在 Cl⁻ 中，氯的氧化数变为 -1，因此氯被还原。在 ClO⁻ 中，氧的氧化数为 -2，氯的氧化数为 +1，因此氯被氧化。一个氯分子中的两个氯原子走上了不同的变化路径，因此这是一个歧化反应。
Another example is the decomposition of hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂. Oxygen changes from -1 in H₂O₂ to -2 in H₂O and 0 in O₂, so the same element is both reduced and oxidised.
另一个例子是过氧化氢的分解：2H₂O₂ → 2H₂O + O₂。氧的氧化数从 H₂O₂ 中的 -1 变为 H₂O 中的 -2 和 O₂ 中的 0，因此同一种元素既被还原又被氧化。
8. Common Oxidising and Reducing Agents | 常见氧化剂与还原剂
Recognising common oxidising and reducing agents helps predict the direction of electron transfer in unfamiliar reactions. The following table lists typical examples tested at A-level.
识别常见的氧化剂和还原剂有助于预测陌生反应中的电子转移方向。下表列出了 A-level 考试中常见的典型例子。
| Oxidising agents | Reducing agents |
|---|---|
| MnO₄⁻ in acidic solution | Metals such as Zn and Mg |
| Cr₂O₇²⁻ in acidic solution | Fe²⁺ ions |
| Halogens Cl₂, Br₂, I₂ | I⁻ ions |
| H₂O₂ in acidic solution | S₂O₃²⁻ ions |
| Concentrated HNO₃ and H₂SO₄ | H₂O₂ in some conditions |
Hydrogen peroxide is especially interesting because it can act as either an oxid
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