📚 IB Chemistry: Concepts and Identification of Redox Reactions | IB化学：氧化还原反应的概念与判断
Redox reactions are among the most important processes in chemistry. They explain rusting, combustion, respiration, photosynthesis, batteries, and countless industrial reactions. In IB Chemistry, mastering the concepts of oxidation and reduction and knowing how to identify redox reactions are essential skills for Paper 1 and Paper 2.
氧化还原反应是化学中最重要的过程之一。它可以解释铁的生锈、燃烧、呼吸作用、光合作用、电池以及无数工业反应。在IB化学中，掌握氧化与还原的概念并学会判断氧化还原反应，是Paper 1和Paper 2必备的核心技能。
1. Oxidation and Reduction: Historical and Modern Definitions | 氧化与还原：历史与现代定义
iginally the removal of oxygen from a compound. These definitions were useful but limited.
从历史上看，氧化被定义为物质与氧结合的过程。例如，镁在空气中燃烧生成氧化镁：2Mg + O₂ → 2MgO。还原最初定义为从化合物中去除氧。这些定义虽然有用，但适用范围有限。
The modern definition is based on electron transfer. Oxidation is the loss of electrons, while reduction is the gain of electrons. A useful mnemonic is OIL RIG: Oxidation Is Loss, Reduction Is Gain.
现代定义基于电子转移。氧化是失去电子，还原是得到电子。一个有用的助记口诀是OIL RIG：Oxidation Is Loss（氧化即失电子），Reduction Is Gain（还原即得电子）。
ons to become Zn²⁺, so zinc is oxidised. The Cu²⁺ ions gain these two electrons to become copper atoms, so copper ions are reduced.
例如，在反Zn + Cu²⁺ → Zn²⁺ + Cu中，锌原子失去两个电子变成Zn²⁺，因此锌被氧化。Cu²⁺离子得到这些电子变成铜原子，因此铜离子被还原。
2. Oxidation Numbers | 氧化数（氧化态）
Oxidation number (or oxidation state) is a bookkeeping tool used to track electron transfer in chemical reactions. It is the charge an atom would have if electrons were transferred completely, even in compounds with covalent bonds.
氧化数（或氧化态）是一种用于追踪化学反应中电子转移的记账工具。它表示假设电子完全转移后原子所带的电荷，即使对于共价化合物也是如此。
By comparing the oxidation numbers of each element before and after a reaction, we can determine whether oxidation or reduction has occurred. An increase in oxidation number means oxidation; a decrease means reduction.
通过比较反应前后各元素的氧化数，可以判断是否发生了氧化或还原。氧化数升高意味着氧化；氧化数降低意味着还原。
2FeCl₃. The oxidation number of iron increases from +2 to +3, so iron is oxidised. Chlorine in Cl₂ has an oxidation number of 0 and becomes -1 in FeCl₃, so chlorine is reduced.
考虑反应：2FeCl₂ + Cl₂ → 2FeCl₃。铁的氧化数从+2升高到+3，因此铁被氧化。Cl₂中氯的氧化数为0，在FeCl₃中变为-1，因此氯被还原。
3. Rules for Assigning Oxidation Numbers | 氧化数确定规则
The following rules are essential for correctly assigning oxidation numbers. These rules are applied in order of priority.
以下规则是正确确定氧化数的关键。这些规则按优先顺序应用。
- An element in its free state has an oxidation number of 0. Example: Na, O₂, S₈, H₂.
- 游离态元素的氧化数为0。例如：Na、O₂、S₈、H₂。
- For a monatomic ion, the oxidation number equals its ionic charge. Example: Na⁺ is +1, Cl⁻ is -1.
- 对于单原子离子，氧化数等于其离子电荷。例如：Na⁺为+1，Cl⁻为-1。
- Fluorine always has an oxidation number of -1 in compounds.
- 氟在化合物中始终为-1。
- Hydrogen is usually +1, but in metal hydrides such as NaH it is -1.
- 氢通常为+1，但在NaH等金属氢化物中为-1。
- Oxygen is usually -2, but in peroxides such as H₂O₂ it is -1, and in OF₂ it is +2.
- 氧通常为-2，但在H₂O₂等过氧化物中为-1，在OF₂中为+2。
- The sum of oxidation numbers in a neutral molecule is 0; in a polyatomic ion it equals the ion charge.
- 中性分子中所有原子氧化数之和为0；多原子离子中其和等于离子电荷。
The table below shows examples of applying these rules.
下表展示了应用这些规则的示例。
| Species | Element | Oxidation Number |
| H₂SO₄ | S | +6 |
| NO₃⁻ | N | +5 |
| Cr₂O₇²⁻ | Cr | +6 |
| MnO₄⁻ | Mn | +7 |
| S₂O₃²⁻ | S (average) | +2 |
4. Using Oxidation Numbers to Identify Redox Reactions | 用氧化数判断氧化还原反应
A reaction is classified as redox if there is a change in oxidation number for at least one element. If no oxidation numbers change, the reaction is not a redox reaction.
如果至少一种元素的氧化数发生变化，该反应就是氧化还原反应。如果没有任何氧化数改变，则该反应不是氧化还原反应。
mbers are Ag +1, N +5, O -2, Na +1, Cl -1 on both sides. Because no oxidation number changes, this is a double displacement reaction, not a redox reaction.
考虑硝酸银与氯化钠的反应：AgNO₃ + NaCl → AgCl + NaNO₃。反应前后Ag为+1、N为+5、O为-2、Na为+1、Cl为-1。因为氧化数没有变化，所以这是一个复分解反应，而不是氧化还原反应。
from +3 to 0, so it is reduced. Aluminium changes from 0 to +3, so it is oxidised. This is definitely a redox reaction.
现在考虑铝热反应：Fe₂O₃ + 2Al → Al₂O₃ + 2Fe。铁从+3变为0，被还原；铝从0变为+3，被氧化。这显然是一个氧化还原反应。
Many acid-base reactions, precipitation reactions, and simple ion combination reactions are not redox reactions because the oxidation states of all atoms remain unchanged.
许多酸碱反应、沉淀反应和简单的离子结合反应都不是氧化还原反应，因为所有原子的氧化态保持不变。
5. Half Equations and Electron Transfer | 半方程与电子转移
Redox reactions can be written as two half equations: one for oxidation and one for reduction. Each half equation shows the electron transfer explicitly.
氧化还原反应可以写成两个半方程：一个表示氧化，一个表示还原。每个半方程都明确显示电子转移。
For the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, the half equations are:
对于反应Zn + Cu²⁺ → Zn²⁺ + Cu，半方程为：
Oxidation: Zn → Zn²⁺ + 2e⁻
Reduction: Cu²⁺ + 2e⁻ → Cu
Notice that the number of electrons lost in the oxidation half equation equals the number of electrons gained in the reduction half equation. This is why no free electrons appear in the overall balanced equation.
注意：氧化半方程中失去的电子数等于还原半方程中得到的电子数。因此，在总平衡方程中不会出现自由电子。
+ 2e⁻ → 2Cl⁻. To balance the electrons, the sodium half equation must be multiplied by 2.
在反应2Na + Cl₂ → 2NaCl中，钠被氧化：Na → Na⁺ + e⁻；氯被还原：Cl₂ + 2e⁻ → 2Cl⁻。为了平衡电子数，钠的半方程需要乘以2。
6. Oxidizing and Reducing Agents | 氧化剂与还原剂
An oxidising agent (oxidant) is a substance that causes oxidation by accepting electrons. Because it accepts electrons, the oxidising agent itself is reduced. A reducing agent (reductant) is a substance that causes reduction by donating electrons, and is itself oxidised.
氧化剂是接受电子而使他物氧化的物质。因为氧化剂接受电子，所以它自身被还原。还原剂是提供电子而使他物还原的物质，它自身被氧化。
e Cu²⁺ ion is the oxidising agent because it accepts electrons; Zn is the reducing agent because it donates electrons.
在反应Zn + Cu²⁺ → Zn²⁺ + Cu中，Cu²⁺是氧化剂，因为它接受电子；Zn是还原剂，因为它提供电子。
The table below lists common oxidising and reducing agents in IB Chemistry.
下表列出了IB化学中常见的氧化剂和还原剂。
| Oxidising Agents | Product After Reduction | Reducing Agents | Product After Oxidation |
| KMnO₄ (acidic) | Mn²⁺ | KI | I₂ |
| K₂Cr₂O₇ (acidic) | Cr³⁺ | Fe²⁺ | Fe³⁺ |
| H₂O₂ | H₂O | H₂C₂O₄ (oxalic acid) | CO₂ |
| Halogens (Cl₂, Br₂) | Cl⁻, Br⁻ | S₂O₃²⁻ | S₄O₆²⁻ |
7. Disproportionation Reactions | 歧化反应
A disproportionation reaction is a special redox reaction in which a single element is simultaneously oxidised and reduced. The element must have at least three oxidation states: one intermediate state and one lower and one higher state.
歧化反应是一种特殊的氧化还原反应，其中同一种元素同时被氧化和还原。该元素必须至少具有三种氧化态：一种中间氧化态、一种较低氧化态和一种较高氧化态。
A classic example is the reaction of chlorine with cold dilute sodium hydroxide:
一个经典例子是氯气与冷稀氢氧化钠溶液的反应：
Cl₂ + 2NaOH → NaCl + NaClO + H₂O
Chlorine in Cl₂ has an oxidation number of 0. In NaCl, chlorine is -1, so it is reduced. In NaClO, chlorine is +1, so it is oxidised. Thus, chlorine is both oxidised and reduced in the same reaction.
Cl₂中氯的氧化数为0。在NaCl中氯为-1，因此被还原；在NaClO中氯为+1，因此被氧化。因此，氯在同一反应中既被氧化又被还原。
Hydrogen peroxide also undergoes disproportionation when it decomposes:
过氧化氢分解时也会发生歧化反应：
2H₂O₂ → 2H₂O + O₂
Oxygen in H₂O₂ has an oxidation number of -1. In H₂O it is -2 (reduction), and in O₂ it is 0 (oxidation).
H₂O₂中氧的氧化数为-1。在H₂O中为-2（还原），在O₂中为0（氧化）。
8. Common Redox Reactions in IB Chemistry | IB化学中的常见氧化还原反应
Redox reactions appear throughout the IB Chemistry syllabus. Recognising common types helps you respond quickly in exams.
氧化还原反应贯穿IB化学教学大纲。识别常见类型有助于你在考试中快速作答。
- Metal displacement: Zn + CuSO₄ → ZnSO₄ + Cu. Zinc displaces copper from its salt solution.
- 金属置换反应：Zn + CuSO₄ → ZnSO₄ + Cu。锌将铜从其盐溶液中置换出来。
- logen displacement: Cl₂ + 2KBr → 2KCl + Br₂. Chlorine displaces the less reactive bromine.
- 卤素置换反应：Cl₂ + 2KBr → 2KCl + Br₂。氯置换出活泼性较弱的溴。
- CH₄ + 2O₂ → CO₂ + 2H₂O. Carbon is oxidised from -4 to +4; oxygen is reduced from 0 to -2.
- 燃烧反应：CH₄ + 2O₂ → CO₂ + 2H₂O。碳从-4被氧化到+4；氧从0被还原到-2。
- Corrosion: 2Fe + O₂ + 2H₂O → 2Fe(OH)₂. Iron is oxidised to Fe²⁺.
- 腐蚀反应：2Fe + O₂ + 2H₂O → 2Fe(OH)₂。铁被氧化为Fe²⁺。
- Respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. This is essentially a slow combustion.
- 呼吸作用：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O。本质上是一种缓慢燃烧。
In metal displacement reactions, the more reactive metal acts as the reducing agent. In halogen displacement, a more reactive halogen oxidises halide ions to produce a less reactive halogen.
在金属置换反应中，较活泼的金属作为还原剂。在卤素置换反应中，较活泼的卤素氧化卤离子，生成较不活泼的卤素。
9. Balancing Redox Equations in Acidic and Basic Media | 酸碱介质中氧化还原方程的配平
In IB Chemistry, you may be asked to balance redox equations using the half-equation method. This is especially important for reactions involving permanganate, dichromate, or hydrogen peroxide.
在IB化学中，你可能会被要求使用半方程法配平氧化还原方程。这对于涉及高锰酸盐、重铬酸盐或过氧化氢的反应尤为重要。
The following steps apply to acidic solutions.
以下步骤适用于酸性溶液。
- Write the two half equations for the oxidising and reducing agents.
- 写出氧化剂和还原剂的两个半方程。
- Balance atoms other than O and H.
- 平衡除O和H以外的原子。
- Balance oxygen atoms by adding H₂O.
- 通过添加H₂O来平衡氧原子。
- Balance hydrogen atoms by adding H⁺.
- 通过添加H⁺来平衡氢原子。
- Balance charge by adding electrons e⁻.
- 通过添加电子e⁻来平衡电荷。
- Multiply each half equation so that the electrons cancel, then add the half equations.
- 将两个半方程乘以适当系数使电子相等，然后相加。
For the reaction between acidic permanganate and iron(II):
对于酸性高锰酸盐与铁(II)的反应：
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Fe²⁺ → Fe³⁺ + e⁻
Multiplying the iron half equation by 5 and adding gives:
将铁半方程乘以5后相加，得到：
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
For basic solutions, the same method is used, but after balancing with H⁺, add OH⁻ to neutralise the H⁺ and form water. For example, in basic media, permanganate is reduced to MnO₂.
对于碱性溶液，使用相同方法，但在用H⁺配平后，添加OH⁻中和H⁺并形成水。例如，在碱性介质中，高锰酸盐被还原为MnO₂。
10. Redox Titration and Applications | 氧化还原滴定与应用
Redox titrations are a practical application of oxidation-reduction reactions. In IB Chemistry, potassium permanganate and iodine are commonly used as titrants.
氧化还原滴定是氧化还原反应的一种实际应用。在IB化学中，高锰酸钾和碘常用作滴定剂。
Potassium permanganate, KMnO₄, is a strong oxidising agent. In acidic solution, its intense purple colour disappears as it is reduced to colourless Mn²⁺. This means no separate indicator is needed; the endpoint is shown by the first permanent pale pink colour.
高锰酸钾（KMnO₄）是强氧化剂。在酸性溶液中，其深紫色会随着被还原为无色的Mn²⁺而消失。因此无需额外指示剂；终点由第一次出现持久淡粉色表示。
Iodine redox titrations use starch as an indicator. Starch forms a deep blue-black complex with iodine. In the titration of thiosulfate with iodine, the reaction is:
碘的氧化还原滴定使用淀粉作为指示剂。淀粉与碘形成深蓝黑色配合物。在硫代硫酸盐与碘的滴定中，反应为：
2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻
Redox reactions are also the basis of batteries, fuel cells, and electrochemical cells. In a voltaic cell, the oxidising agent and reducing agent are separated so that the electron transfer occurs through an external circuit, producing electrical energy.
氧化还原反应还是电池、燃料电池和电化学电池的基础。在原电池中，氧化剂和还原剂被分开，使电子转移通过外电路进行，从而产生电能。
11. Common Mistakes and Exam Tips | 常见错误与考试提示
Students often confuse the terms oxidising agent and reducing agent. Remember: the oxidising agent oxidises something else and is itself reduced. The reducing agent reduces something else and is itself oxidised.
学生经常混淆氧化剂和还原剂这两个术语。记住：氧化剂使他物氧化，自身被还原；还原剂使他物还原，自身被氧化。
Another common mistake is assigning oxidation numbers incorrectly in compounds such as H₂O₂, OF₂, and metal hydrides. Always apply the special rules before the general rules.
另一个常见错误是在H₂O₂、OF₂和金属氢化物等化合物中错误分配氧化数。务必先应用特殊规则，再应用一般规则。
- Always check whether oxidation numbers have changed before classifying a reaction as redox.
- 在判断一个反应是否为氧化还原反应之前，务必检查氧化数是否发生变化。
- If an element appears as a free element on one side of the equation and in a compound on the other, the reaction is usually redox.
- 如果某元素在方程一侧以游离态存在，而在另一侧以化合物形式存在，该反应通常是氧化还原反应。
- In half equations, the total charge must balance as well as the atoms.
- 在半方程中，总电荷和原子数都必须平衡。
- Don't use curly quotes or LaTeX in your exam paper; write equations clearly with correct superscripts and subscripts.
- 考试中不要使用弯引号或LaTeX；应清晰书写方程，正确使用上标和下标。
12. Summary | 总结
Redox reactions involve the transfer of electrons. Oxidation is the loss of electrons; reduction is the gain of electrons. Oxidation numbers are the key tool for identifying redox reactions and balancing equations.
氧化还原反应涉及电子转移。氧化是失去电子；还原是得到电子。氧化数是判断氧化还原反应和配平方程的关键工具。
An oxidising agent accepts electrons and is reduced; a reducing agent donates electrons and is oxidised. Disproportionation reactions feature one element that is both oxidised and reduced. Mastering half equations and the rules for acidic and basic media will help you solve complex redox problems with confidence.
氧化剂接受电子并自身被还原；还原剂提供电子并自身被氧化。歧化反应的特征是同一种元素既被氧化又被还原。掌握半方程法以及酸碱介质中的配平规则，能帮助你自信地解决复杂的氧化还原问题。
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