Redox Reactions: IB & AQA Chemistry Exam Essentials | IB AQA 化学：氧化还原 考点精讲

📚 Redox Reactions: IB & AQA Chemistry Exam Essentials | IB AQA 化学：氧化还原 考点精讲

Welcome to this focused revision guide on redox chemistry, covering the essential concepts needed for both IB (Standard and Higher Level) and AQA A-level Chemistry. Whether you are analysing oxidation numbers, balancing half-equations, or predicting cell potentials, the principles explained here will help you master the exam-required skills.

欢迎阅读这篇关于氧化还原化学的专项复习指南，涵盖IB（标准级别和高级别）和AQA A-level化学所需的核心概念。无论你是在分析氧化数、配平半反应方程式，还是预测电池电势，这里讲解的原理都将帮助你掌握考试需要的技能。

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1. What is Oxidation and Reduction? | 氧化与还原的定义

In modern chemistry, oxidation is the loss of electrons, and reduction is the gain of electrons. This electron-transfer definition is the most fundamental one used in both IB and AQA specifications.

在现代化学中，氧化是失去电子，还原是得到电子。这种基于电子转移的定义是IB和AQA大纲中最基本的定义。

Oxidation also corresponds to an increase in oxidation number, while reduction involves a decrease in oxidation number. For example, when magnesium reacts with oxygen to form MgO, Mg changes from oxidation number 0 to +2 (oxidation), and O changes from 0 to -2 (reduction).

氧化也对应氧化数的升高，而还原则对应氧化数的降低。例如，当镁与氧气反应生成MgO时，Mg的氧化数从0变为+2（被氧化），O的氧化数从0变为-2（被还原）。

A classic mnemonic is ‘OIL RIG’: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

经典的助记符是”OIL RIG”：氧化是失电子（Oxidation Is Loss），还原是得电子（Reduction Is Gain）。

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2. Oxidation Numbers: Rules and Assigning | 氧化数：规则与确定

Assigning oxidation numbers (also called oxidation states) allows us to track how electrons are redistributed in a reaction. The rules below are essential and are tested regularly in both IB and AQA exams.

确定氧化数（也称氧化态）使我们能够追踪电子在反应中如何重新分布。以下规则至关重要，在IB和AQA考试中经常被考查。

Rule (English)	规则 (中文)
1. The oxidation number of an atom in its elemental form is 0.	1. 元素单质中原子的氧化数为0。
2. For a simple monatomic ion, the oxidation number equals the charge on the ion.	2. 对于简单单原子离子，氧化数等于离子所带的电荷。
3. Fluorine is always -1 in compounds.	3. 化合物中氟总是 -1 价。
4. Oxygen is usually -2, except in peroxides (where it is -1) or when bonded to fluorine (where it can be positive).	4. 氧通常为 -2 价，但在过氧化物中为 -1，与氟结合时可为正值。
5. Hydrogen is +1 when bonded to non-metals, but -1 when bonded to metals (hydrides).	5. 氢与非金属结合时为 +1，与金属结合生成氢化物时为 -1。
6. The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion’s overall charge.	6. 中性分子中氧化数之和为0；多原子离子中氧化数之和等于离子所带电荷。

Example: In KMnO₄, K is +1, O is -2, therefore Mn must be x: (+1) + x + 4(-2) = 0 → x = +7.

示例：在 KMnO₄ 中，K 为 +1，O 为 -2，因此 Mn 必须满足 (+1) + x + 4(-2) = 0，解得 x = +7。

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3. Identifying Oxidizing and Reducing Agents | 识别氧化剂与还原剂

In a redox reaction, the species that accepts electrons (is reduced) is called the oxidizing agent (or oxidant). The species that donates electrons (is oxidized) is called the reducing agent (or reductant).

在氧化还原反应中，接受电子（被还原）的物质称为氧化剂，提供电子（被氧化）的物质称为还原剂。

For example, in the displacement reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), Zn loses electrons (oxidation), so Zn is the reducing agent. Cu²⁺ gains electrons (reduction), so Cu²⁺ is the oxidizing agent.

例如，在置换反应 Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) 中，Zn 失去电子（被氧化），所以 Zn 是还原剂；Cu²⁺ 得到电子（被还原），所以 Cu²⁺ 是氧化剂。

Note that the oxidizing agent always gets reduced, and the reducing agent always gets oxidized.

注意，氧化剂总是被还原，还原剂总是被氧化。

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4. Balancing Redox Reactions: Half-Equation Method | 氧化还原反应配平：半反应法

Under acidic conditions, which are typical in both IB and AQA paper questions, the half-equation method follows these systematic steps:

在酸性条件下（IB和AQA试题中的典型情形），半反应配平法遵循以下系统步骤：

Step 1: Write the unbalanced skeleton half-equations for oxidation and reduction.

步骤1： 分别写出氧化和还原的未配平骨架半反应。

Step 2: Balance all atoms except O and H.

步骤2： 配平除 O 和 H 以外的所有原子。

Step 3: Balance oxygen atoms by adding H₂O molecules.

步骤3： 通过添加 H₂O 分子配平氧原子。

Step 4: Balance hydrogen atoms by adding H⁺ ions (since the medium is acidic).

步骤4： 通过添加 H⁺ 离子配平氢原子（由于介质为酸性）。

Step 5: Balance the charge by adding electrons (e⁻) to the more positive side.

步骤5： 通过在电荷更正的一侧添加电子 (e⁻) 来配平电荷。

Step 6: Multiply the half-equations so that the number of electrons lost equals electrons gained, then add the half-equations and cancel common species.

步骤6： 将半反应乘以适当的系数使失电子数与得电子数相等，然后相加并消去共同物种。

Worked example: Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acid.

示例：配平酸性条件下 MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺。

Oxidation half-equation: Fe²⁺ → Fe³⁺ + e⁻

氧化半反应：Fe²⁺ → Fe³⁺ + e⁻

Reduction half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

还原半反应：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Multiply oxidation by 5 to balance electrons: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O.

将氧化半反应乘以5以平衡电子：5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O。

For alkaline conditions, after balancing as if in acid, add OH⁻ to both sides to neutralise H⁺, forming water.

对于碱性条件，先按酸性条件配平，然后向两边添加 OH⁻ 中和 H⁺，生成水。

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5. Redox Titrations: Principles and Calculations | 氧化还原滴定：原理与计算

Redox titrations such as manganate(VII) with iron(II) or iodine/thiosulfate are common in both IB and AQA papers. The equivalence point is detected either by a colour change of the titrant itself or by using a starch indicator.

高锰酸根(VII)与铁(II)的滴定、碘与硫代硫酸盐的滴定等氧化还原滴定在IB和AQA试卷中都很常见。终点可通过滴定剂自身的颜色变化或使用淀粉指示剂来检测。

Example calculation: A 25.0 cm³ portion of acidified Fe²⁺ solution requires 20.0 cm³ of 0.0200 mol dm⁻³ KMnO₄ to reach the permanent pink endpoint. Find the mass of iron in the sample.

计算示例：一份 25.0 cm³ 酸化后的 Fe²⁺ 溶液需要 20.0 cm³ 0.0200 mol dm⁻³ KMnO₄ 滴定至持续粉红色终点。求样品中铁的质量。

Equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

方程式：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Moles of MnO₄⁻ = concentration × volume = 0.0200 × (20.0/1000) = 4.00 × 10⁻⁴ mol

MnO₄⁻ 的物质的量 = 浓度 × 体积 = 0.0200 × (20.0/1000) = 4.00 × 10⁻⁴ mol

From stoichiometry, moles of Fe²⁺ = 5 × 4.00 × 10⁻⁴ = 2.00 × 10⁻³ mol

根据化学计量比，Fe²⁺ 的物质的量 = 5 × 4.00 × 10⁻⁴ = 2.00 × 10⁻³ mol

Mass of iron = 2.00 × 10⁻³ mol × 55.8 g mol⁻¹ ≈ 0.112 g (present in the 25.0 cm³ aliquot).

铁的质量 = 2.00 × 10⁻³ mol × 55

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