📚 IB Chemistry: Distinguishing Oxidation and Reduction | IB化学：氧化与还原的辨析
Redox reactions are among the most important concepts in IB Chemistry. They appear in bonding, electrochemistry, organic reactions, and industrial processes. To score well, you must be able to distinguish oxidation from reduction through multiple definitions and apply them confidently to half-equations and full equations.
氧化还原反应是IB化学中最重要的概念之一，它贯穿化学键、电化学、有机反应和工业流程。要想取得高分，你必须能够通过多种定义准确辨析氧化与还原，并熟练运用于半方程式和完整方程式的书写。
1. The Historical Definitions: Oxygen and Hydrogen | 历史定义：氧与氢的得失
The earliest definition of oxidation was the combination of a substance with oxygen. Reduction was originally the removal of oxygen from a compound, such as when metal oxides are heated with carbon or hydrogen to give the pure metal.
最早对“氧化”的定义是物质与氧结合；“还原”最初指从化合物中除去氧，例如用碳或氢气加热金属氧化物得到纯金属的过程。
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Combustion of magnesium: 2Mg + O2 → 2MgO. Magnesium is oxidised because it gains oxygen.
镁的燃烧：2Mg + O2 → 2MgO。镁因获得氧而被氧化。
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Reduction of copper(II) oxide: CuO + H2 → Cu + H2O. Copper(II) oxide loses oxygen, so it is reduced; hydrogen gains oxygen and is oxidised.
氧化铜的还原：CuO + H2 → Cu + H2O。氧化铜失去氧，因此被还原；氢气得到氧，因此被氧化。
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In organic chemistry, oxidation can also mean loss of hydrogen, while reduction can mean gain of hydrogen. For example, ethanol → ethanal involves loss of hydrogen.
在有机化学中，氧化也可以指失去氢，而还原可以指得到氢。例如，乙醇转化为乙醛就涉及失去氢。
These definitions are useful but limited. Many redox reactions, such as Na + Cl → NaCl, do not involve oxygen or hydrogen at all.
这些定义虽然有用，但并不完整。许多氧化还原反应，如 Na + Cl → NaCl，完全不涉及氧或氢。
2. The Electron Transfer Definition | 电子转移定义
The modern definition focuses on electron movement. Oxidation is the loss of electrons, and reduction is the gain of electrons. The memory aid OIL RIG is widely used: Oxidation Is Loss, Reduction Is Gain.
现代定义聚焦于电子转移：氧化是失去电子，还原是获得电子。常用记忆口诀是 OIL RIG：Oxidation Is Loss（氧化是失电子），Reduction Is Gain（还原是得电子）。
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Oxidation: a species loses one or more electrons. Oxidation number increases.
氧化：某物质失去一个或多个电子，氧化数升高。
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Reduction: a species gains one or more electrons. Oxidation number decreases.
还原：某物质获得一个或多个电子，氧化数降低。
Consider the reaction between zinc and copper(II) sulfate:
以锌与硫酸铜的反应为例：
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Zinc atoms lose two electrons to form Zn²⁺, so zinc is oxidised. Copper(II) ions gain two electrons to become copper atoms, so Cu²⁺ is reduced.
锌原子失去两个电子形成 Zn²⁺，因此锌被氧化；铜(II)离子得到两个电子变成铜原子，因此 Cu²⁺ 被还原。
The two half-reactions are:
两个半反应为：
Zn → Zn²⁺ + 2e⁻
Cu²⁺ + 2e⁻ → Cu
3. Oxidation Numbers | 氧化数
Oxidation number is a bookkeeping tool used to track electron transfer. It represents the apparent charge an atom would have if all bonds were fully ionic.
氧化数是一种用于追踪电子转移的记数工具，它表示如果所有化学键都完全离子化时，该原子所带有的表观电荷。
The key rules for assigning oxidation numbers are:
确定氧化数的关键规则如下：
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A free element in its standard state has an oxidation number of zero. Example: O2, Cl2, Fe.
处于标准状态的单质，其氧化数为零。例如：O2、Cl2、Fe。
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The oxidation number of a monatomic ion equals its charge. Example: Na⁺ is +1; Cl⁻ is −1.
单原子离子的氧化数等于其电荷。例如：Na⁺ 为 +1；Cl⁻ 为 −1。
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Oxygen is usually −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。
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Hydrogen is usually +1, except in metal hydrides such as NaH, where it is −1.
氢通常为 +1，但在金属氢化物如 NaH 中为 −1。
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Fluorine is always −1 in compounds. Other halogens are usually −1, but can be positive when bonded to oxygen.
氟在化合物中始终为 −1；其他卤素通常为 −1，但与氧成键时可以为正。
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The sum of oxidation numbers in a neutral compound is zero; in a polyatomic ion, it equals the ion’s charge.
中性化合物中所有原子的氧化数总和为零；多原子离子中，所有原子氧化数之和等于该离子所带电荷。
Example: in the reaction 2FeCl2 + Cl2 → 2FeCl3, chlorine gas has oxidation number 0, Fe in FeCl2 is +2, and Fe in FeCl3 is +3. Fe is oxidised because its oxidation number increases; Cl2 is reduced because its oxidation number decreases.
例如，在反应 2FeCl2 + Cl2 → 2FeCl3 中，氯气的氧化数为 0，FeCl2 中的 Fe 为 +2，FeCl3 中的 Fe 为 +3。铁因氧化数升高而被氧化；氯气因氧化数降低而被还原。
4. Identifying Oxidising and Reducing Agents | 氧化剂与还原剂的辨析
An oxidising agent accepts electrons and is itself reduced. A reducing agent donates electrons and is itself oxidised. This is a common source of confusion, so remember: the oxidising agent causes oxidation, but undergoes reduction.
氧化剂接受电子，自身被还原；还原剂给出电子，自身被氧化。这是常见的易混点，务必记住：氧化剂促使别的东西被氧化，但自身发生还原。
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In MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, MnO₄⁻ is the oxidising agent because it accepts five electrons and is reduced to Mn²⁺.
在 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O 中，MnO₄⁻ 是氧化剂，因为它接受五个电子并被还原为 Mn²⁺。
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In Zn → Zn²⁺ + 2e⁻, Zn is the reducing agent because it donates electrons and is oxidised to Zn²⁺.
在 Zn → Zn²⁺ + 2e⁻ 中，Zn 是还原剂，因为它提供电子并被氧化为 Zn²⁺。
| Oxidising Agent | 氧化剂 | Reduced to | 被还原为 | Condition | 条件 |
| KMnO₄ / MnO₄⁻ | Mn²⁺ | Acidic solution | 酸性溶液 |
| K₂Cr₂O₇ / Cr₂O₇²⁻ | Cr³⁺ | Acidic solution | 酸性溶液 |
| Cl₂ | Cl⁻ | Aqueous | 水溶液 |
| Fe³⁺ | Fe²⁺ | Aqueous | 水溶液 |
A strong oxidising agent has a strong tendency to gain electrons and is easily reduced. Conversely, a strong reducing agent readily gives up electrons and is easily oxidised.
强氧化剂具有强烈的得电子趋势，容易被还原；反之，强还原剂容易给出电子，容易被氧化。
5. Writing Half-Equations | 半方程式的书写
Half-equations show either oxidation or reduction separately and must balance both atoms and charge. In acidic solution, follow this order:
半方程式单独表示氧化反应或还原反应，必须同时满足原子守恒和电荷守恒。在酸性溶液中，按以下顺序配平：
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Balance all atoms except oxygen and hydrogen.
配平除氧和氢以外的所有原子。
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Balance oxygen atoms by adding H₂O molecules.
通过添加 H₂O 分子配平氧原子。
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Balance hydrogen atoms by adding H⁺ ions.
通过添加 H⁺ 离子配平氢原子。
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Balance charge by adding electrons, e⁻, to the more positive side.
通过在更正电荷的一侧添加电子 e⁻ 来配平电荷。
Example: convert Cr₂O₇²⁻ to Cr³⁺ in acidic solution.
示例：在酸性溶液中将 Cr₂O₇²⁻ 转化为 Cr³⁺。
Cr₂O₇²⁻ → 2Cr³⁺
Balance oxygen with water:
用水配平氧：
Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
Balance hydrogen with H⁺:
用 H⁺ 配平氢：
14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
Balance charge with electrons. The left side has total charge +12, and the right side has +6, so add six electrons to the left:
用电子配平电荷：左边总电荷为 +12，右边为 +6，因此在左边加入六个电子：
14H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O
For basic solutions, balance as if acidic, then add OH⁻ to neutralise H⁺ on both sides. This converts H⁺ to water.
对于碱性溶液，可先按酸性配平，然后在两边加入 OH⁻ 中和 H⁺，使 H⁺ 转化为水。
6. Balancing Redox Equations | 氧化还原方程式的配平
Full redox equations are obtained by combining the balanced oxidation and reduction half-equations. The key is to make the number of electrons lost equal the number of electrons gained.
完整的氧化还原方程式通过将配平后的氧化半反应和还原半反应相加得到，关键在于使失去的电子数等于得到的电子数。
Example: acidified MnO₄⁻ reacts with Fe²⁺. Write the half-equations:
示例：酸化后的 MnO₄⁻ 与 Fe²⁺ 反应。先写出两个半反应：
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Fe²⁺ → Fe³⁺ + e⁻
Multiply the oxidation half-equation by 5 so that the electrons match:
将氧化半反应乘以 5，使电子数相等：
5Fe²⁺ → 5Fe³⁺ + 5e⁻
Add the two half-equations and cancel the electrons:
将两个半反应相加并消去电子：
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
Finally, check that atoms and charge balance. Here the total charge on each side is +17, so the equation is correct.
最后检查原子和电荷是否守恒。该方程式两边总电荷均为 +17，因此配平正确。
7. Disproportion
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