Edexcel A Level Chemistry Topic 3: Redox I | Edexcel A Level化学 主题3：氧化还原基础

📚 Edexcel A Level Chemistry Topic 3: Redox I | Edexcel A Level化学 主题3：氧化还原基础

Welcome to this TutorHao revision guide for Edexcel A Level Chemistry Topic 3: Redox I. Redox reactions are fundamental across the whole A Level specification: they connect atomic structure, bonding, inorganic chemistry, and later electrode potentials. Mastering oxidation states and half-equations now will save you time in organic, transition metal and energy topics.

欢迎阅读 TutorHao 针对 Edexcel A Level 化学主题 3 '氧化还原基础' 的复习指南。氧化还原反应贯穿 A Level 全部知识体系：它连接原子结构、化学键、无机化学以及后续的电极电势。现在掌握氧化态和半反应方程式，可以为你今后学习有机化学、过渡金属和能量主题节省大量时间。

1. Oxidation states and rules | 氧化态及其规则

The oxidation state, also called oxidation number, is a bookkeeping number assigned to an atom in a substance. It tells you how many electrons an atom has lost or gained relative to the free element. Edexcel requires you to apply a standard set of rules rather than memorise every compound.

氧化态，也叫氧化数，是分配给物质中原子的 '记账数字'。它表示一个原子相对于单质失去或得到了多少个电子。Edexcel 考试要求你掌握一套标准规则，而不是死记每个化合物。

Rule	氧化数规则
Uncombined element has oxidation state 0.	单质原子的氧化态为零。
Simple ion has oxidation state equal to its charge.	单原子离子的氧化态等于其所带电荷。
Hydrogen is +1, except in metal hydrides where it is -1.	氢通常为 +1，但金属氢化物中为 -1。
Oxygen is -2, except in peroxides where it is -1 and in OF₂ where it is +2.	氧通常为 -2，但过氧化物中为 -1，OF₂ 中为 +2。
The sum of oxidation states equals the total charge on the species.	化合物或离子中各原子氧化态之和等于总电荷。

Sum of oxidation states = overall charge

2. Worked examples for common compounds | 常见化合物氧化态示例

so 2 + x - 8 = 0, giving x = +6. The sulfur in sulfuric acid has an oxidation state of +6.

计算 H₂SO₄ 中 S 的氧化态。设 S = x。氢为 +1×2 = +2；氧为 -2×4 = -8。总和为 0，所以 2 + x - 8 = 0，解得 x = +6。硫酸中硫的氧化态为 +6。

-1, so x - 8 = -1, giving x = +7. This is why Mn in MnO₄⁻ is described as manganese(VII).

对于高锰酸根离子 MnO₄⁻，设 Mn = x。氧为 -2×4 = -8。离子电荷为 -1，所以 x - 8 = -1，解得 x = +7。这就是 MnO₄⁻ 中 Mn 被称为锰(VII) 的原因。

is -2 × 7 = -14. Let Cr = x, so 2 + 2x - 14 = 0, x = +6. Chromium is +6 in dichromate(VI).

在 K₂Cr₂O₇ 中，钾为 +1×2 = +2；氧为 -2×7 = -14。设 Cr = x，则 2 + 2x - 14 = 0，x = +6。重铬酸根中的铬为 +6。

3. Oxidation and reduction definitions | 氧化和还原的定义

In A Level chemistry, oxidation is defined as an increase in oxidation state, and reduction is a decrease in oxidation state. This electronic definition is much more useful than the old oxygen/hydrogen definitions because it works for reactions with no oxygen present.

在 A Level 化学中，氧化被定义为氧化态升高，还原被定义为氧化态降低。这种电子定义比旧的 '加氧去氢' 定义更有用，因为它适用于没有氧参与的许多反应。

The mnemonic OIL RIG is often used: Oxidation Is Loss of electrons, Reduction Is Gain of electrons. A species that is oxidised loses electrons and its oxidation number increases; a species that is reduced gains electrons and its oxidation number decreases.

常用助记口诀 OIL RIG：氧化是失去电子，还原是得到电子。被氧化的物种失去电子，氧化数升高；被还原的物种得到电子，氧化数降低。

Zn → Zn²⁺ + 2e⁻   |   Cu²⁺ + 2e⁻ → Cu

4. Half-equations | 半反应方程式

ge on both sides must balance. For example, Fe³⁺ + e⁻ → Fe²⁺ is a reduction half-equation.

半反应方程式单独表示氧化过程或还原过程，并把电子明确写出来。电子带电荷，因此方程式两边的总电荷必须守恒。例如 Fe³⁺ + e⁻ → Fe²⁺ 是一个还原半反应。

and finally add electrons to balance charge. For example: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

对于酸性条件下的复杂离子，写半反应方程式时先配平发生氧化还原的原子，再用 H₂O 配平氧，用 H⁺ 配平氢，最后加电子配平电荷。例如：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

5. Combining half-equations | 合并半反应方程式

To construct an overall redox equation, combine the oxidation and reduction half-equations so that the number of electrons lost equals the number gained. Electrons then cancel and do not appear in the final equation.

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要写出完整的氧化还原方程式，需要合并氧化半反应和还原半反应，使失去的电子数等于得到的电子数。电子随后相互抵消，不再出现在总方程式中。

+ 2e⁻. Multiply the Fe³⁺ half-equation by 2 so both involve 2e⁻. The overall equation is:

例如：Fe³⁺ 与 I⁻ 反应，还原半反应为 Fe³⁺ + e⁻ → Fe²⁺，氧化半反应为 2I⁻ → I₂ + 2e⁻。将 Fe³⁺ 半反应乘以 2，使两个半反应都涉及 2e⁻。总方程式为：

2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

the left side has total charge 2(3+) + 2(1-) = +4, and the right side has 2(2+) + 0 = +4.

务必检查质量守恒和电荷守恒。上述方程中，左边总电荷为 2(3+) + 2(1-) = +4，右边为 2(2+) + 0 = +4。

6. Oxidising and reducing agents | 氧化剂与还原剂

An oxidising agent is a species that accepts electrons and is itself reduced in the reaction. A reducing agent is a species that donates electrons and is itself oxidised. Do not confuse the action with the species: the oxidising agent causes oxidation, so it gains electrons.

氧化剂是接受电子、本身被还原的物质。还原剂是给出电子、本身被氧化的物质。不要混淆作用与物种：氧化剂使别的物质氧化，因此它本身得到电子。

s the oxidising agent because Cu²⁺ is reduced to Cu, and H₂ is the reducing agent because H₂ is oxidised to H₂O.

在反应 CuO + H₂ → Cu + H₂O 中，CuO 是氧化剂，因为 Cu²⁺ 被还原为 Cu；H₂ 是还原剂，因为 H₂ 被氧化为 H₂O。

CuO + H₂ → Cu + H₂O

7. Disproportionation | 歧化反应

Disproportionation is a redox reaction in which one element is simultaneously oxidised and reduced, meaning the same species forms products containing that element in higher and lower oxidation states. This is a favourite Edexcel exam concept.

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被还原，也就是同一物种生成含有该元素更高和更低氧化态的产物。这是 Edexcel 考试中常见的重点概念。

₂ + 2NaOH → NaCl + NaClO + H₂O. Chlorine goes from 0 in Cl₂ to -1 in NaCl and +1 in NaClO.

典型例子是氯气与冷的氢氧化钠溶液反应：Cl₂ + 2NaOH → NaCl + NaClO + H₂O。氯的氧化态从 Cl₂ 中的 0 变为 NaCl 中的 -1 和 NaClO 中的 +1。

Cl₂ + 2NaOH → NaCl + NaClO + H₂O

oportionation: 2H₂O₂ → 2H₂O + O₂. Oxygen changes from -1 in H₂O₂ to -2 in H₂O and 0 in O₂.

过氧化氢分解也是歧化反应：2H₂O₂ → 2H₂O + O₂。氧的氧化态从 H₂O₂ 中的 -1 变为 H₂O 中的 -2 和 O₂ 中的 0。

8. Metal–acid and metal–salt redox | 金属与酸、金属与盐的氧化还原

Reactive metals are reducing agents because they readily lose electrons to form positive ions. When magnesium reacts with hydrochloric acid, Mg is oxidised from 0 to +2, while H⁺ is reduced from +1 to 0 in H₂.

活泼金属是还原剂，因为它们容易失去电子形成阳离子。镁与盐酸反应时，Mg 从 0 被氧化到 +2，而 H⁺ 从 +1 被还原为 H₂ 中的 0。

Mg + 2HCl → MgCl₂ + H₂

The ionic equation shows electron transfer more clearly:

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