A-Level OCR Chemistry: Redox Exam Insights | A-Level OCR 化学：氧化还原考点精讲

📚 A-Level OCR Chemistry: Redox Exam Insights | A-Level OCR 化学：氧化还原考点精讲

Redox reactions form the backbone of many chemical processes assessed in OCR A-Level Chemistry, from electron transfer and oxidation number calculations to electrochemical cells and disproportionation. Mastering this topic means thinking in terms of electrons gained and lost, and being able to apply standard potentials to predict feasibility.

氧化还原反应是 OCR A-Level 化学中许多考点的基础，涉及电子转移、氧化数计算、电化学电池以及歧化反应等。掌握这一主题意味着能够从电子得失角度思考问题，并运用标准电极电势判断反应发生的可能性。

1. What is Redox? | 什么是氧化还原？

Redox is short for reduction–oxidation. A redox reaction is any chemical process in which electrons are transferred between species, causing a change in oxidation number.

氧化还原是还原和氧化的合称。任何发生电子转移并导致氧化数变化的化学反应都属于氧化还原反应。

Oxidation is defined as the loss of electrons, and reduction is the gain of electrons. The mnemonic OIL RIG helps: Oxidation Is Loss, Reduction Is Gain.

氧化定义为失去电子，还原定义为得到电子。记忆口诀 OIL RIG 代表 Oxidation Is Loss（氧化即失电子）, Reduction Is Gain（还原即得电子）。

Whenever oxidation occurs, reduction must occur simultaneously. The substance that is oxidised is the reducing agent, and the substance that is reduced is the oxidising agent.

氧化与还原总是同时发生。被氧化的物质是还原剂，被还原的物质是氧化剂。

2. Oxidation Number Rules | 氧化数规则

An oxidation number (or oxidation state) is a book-keeping number assigned to an atom in a compound or ion to track electron ownership. The OCR exam expects you to know these key rules:

氧化数（氧化态）是给化合物或离子中的原子指定的一个记账数值，用于追踪电子归属。OCR 考试要求掌握以下关键规则：

Free elements have an oxidation number of 0 (e.g. O₂, Na, Cl₂).

游离态单质的氧化数为 0（如 O₂、Na、Cl₂）。
For simple monatomic ions, the oxidation number equals the charge on the ion (e.g. Na⁺ = +1, Cl⁻ = –1).

简单单原子离子的氧化数等于离子所带电荷数（如 Na⁺: +1，Cl⁻: –1）。
In compounds, hydrogen is usually +1 (except in metal hydrides like NaH, where it is –1).

在化合物中，氢通常为 +1（金属氢化物如 NaH 中为 –1）。
Oxygen is usually –2 (except in peroxides where it is –1, and in OF₂ where it is +2).

氧通常为 –2（过氧化物中为 –1，OF₂ 中为 +2）。
The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion’s charge.

中性分子中所有原子的氧化数之和为 0；多原子离子中所有原子的氧化数之和等于离子所带电荷。
In compounds, Group 1 metals are +1, Group 2 metals are +2, and aluminium is +3.

化合物中，第 1 族金属为 +1，第 2 族金属为 +2，铝为 +3。
In a molecule or ion, the more electronegative element is given the negative oxidation number.

在分子或离子中，电负性较大的元素获得负氧化数。

3. Calculating Oxidation Numbers | 氧化数的计算

Use the rules to work out the oxidation number of an underlined atom in a formula. For example, in H₂SO₄, O is –2, H is +1, so 2(+1) + S + 4(–2) = 0, giving S = +6.

运用上述规则可以计算分子中指定原子的氧化数。例如 H₂SO₄ 中 O 为 –2，H 为 +1，则 2(+1) + S + 4(–2) = 0，解得 S = +6。

In MnO₄⁻, O is –2, total charge is –1: Mn + 4(–2) = –1, so Mn = +7.

在 MnO₄⁻ 中，O 为 –2，离子总电荷 –1：Mn + 4(–2) = –1，故 Mn = +7。

For Fe₃O₄, the iron atoms have a mixed oxidation state: the formula can be viewed as FeO·Fe₂O₃, giving Fe oxidation states of +2 and +3.

Fe₃O₄ 中铁原子具有混合氧化态，可视为 FeO·Fe₂O₃，因此 Fe 的氧化态分别为 +2 和 +3。

In organic molecules, the oxidation number of carbon can be determined by the electronegativity shifts between C–C, C–H, C–O bonds, which helps track oxidation and reduction in organic redox.

在有机分子中，碳的氧化数可通过 C–C、C–H、C–O 等键的电负性差异进行分配，这有助于判断有机反应中的氧化与还原。

4. Oxidising and Reducing Agents | 氧化剂与还原剂

An oxidising agent (oxidant) gains electrons and is itself reduced. A reducing agent (reductant) loses electrons and is itself oxidised.

氧化剂得到电子，本身被还原。还原剂失去电子，本身被氧化。

Common oxidising agents tested by OCR include acidified potassium manganate(VII) (MnO₄⁻/H⁺), potassium dichromate(VI) (Cr₂O₇²⁻/H⁺), oxygen, and halogens.

OCR 常考的氧化剂包括酸化高锰酸钾（MnO₄⁻/H⁺）、重铬酸钾（Cr₂O₇²⁻/H⁺）、氧气以及卤素单质。

Common reducing agents include metals like zinc and iron, sulfite ions (SO₃²⁻), iodide ions (I⁻), and thiosulfate ions (S₂O₃²⁻).

常见还原剂有锌、铁等金属，亚硫酸根离子（SO₃²⁻），碘离子（I⁻）和硫代硫酸根离子（S₂O₃²⁻）。

Recognising the agent: in the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, Zn is oxidised (reducing agent), Cu²⁺ is reduced (oxidising agent).

识别试剂：在反应 Zn + Cu²⁺ → Zn²⁺ + Cu 中，Zn 被氧化，充当还原剂；Cu²⁺ 被还原，充当氧化剂。

5. Half-Equations and the Ion-Electron Method | 半反应与离子电子法

A half-equation shows either the oxidation or reduction part of a redox reaction, with electrons explicitly included. The ion-electron method uses these half-equations to balance the overall equation in acidic or alkaline conditions.

半反应分别展示氧化或还原部分，并明确写出电子。离子电子法运用这两个半反应，在酸性或碱性条件下配平完整的氧化还原方程式。

For example, the oxidation half-equation for Fe²⁺ to Fe³⁺ is: Fe²⁺ → Fe³⁺ + e⁻.

例如，Fe²⁺ 氧化为 Fe³⁺ 的半反应为：Fe²⁺ → Fe³⁺ + e⁻。

The reduction half-equation for MnO₄⁻ in acidic solution is: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

酸性条件下 MnO₄⁻ 的还原半反应为：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。

To combine, multiply each half-equation so that the electrons lost equal the electrons gained. Then add and cancel electrons and any other species appearing on both sides.

合并时，给半反应乘以适当系数使失电子数目与得电子数目相等，然后相加，消去电子和两边都出现的物质。

6. Balancing Redox Equations in Acidic and Alkaline Media | 酸性及碱性介质中氧化还原方程式的配平

In acidic medium, add H⁺ and H₂O to balance oxygen and hydrogen. In alkaline medium, balance as if in acid, then add OH⁻ to both sides to neutralise H⁺, forming water.

在酸性介质中，加入 H⁺ 和 H₂O 来平衡氧和氢。在碱性介质中，可先按酸性条件配平，然后在两边添加 OH⁻ 中和 H⁺，生成水。

Example – acidic: Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺. Step 1: Write half-equations. Step 2: Multiply Fe oxidation by 5 to match 5 electrons from MnO₄⁻ reduction. Overall: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

酸性示例：配平 MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺。步骤：写出半反应，将铁氧化半反应乘以 5 以匹配 MnO₄⁻ 还原的 5 个电子，总反应为 MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。

Example – alkaline: Balance Cr(OH)₃ + ClO⁻ → CrO₄²⁻ + Cl⁻. First, treat as acidic: Cr(OH)₃ + ClO⁻ → CrO₄²⁻ + Cl⁻. Add H₂O, H⁺ as needed, then neutralise with OH⁻. The final balanced equation in base is 2Cr(OH)₃ + 3ClO⁻ + 4OH⁻ → 2CrO₄²⁻ + 3Cl⁻ + 5H₂O.

碱性示例：配平 Cr(OH)₃ + ClO⁻ → CrO₄²⁻ + Cl⁻。先按酸性介质处理，需加 H₂O 和 H⁺，再加 OH⁻ 中和。最终碱性条件下的配平为 2Cr(OH)₃ + 3ClO⁻ + 4OH⁻ → 2CrO₄²⁻ + 3Cl⁻ + 5H₂O。

7. Introduction to Electrochemical Cells | 电化学电池简介

An electrochemical cell consists of two half-cells connected by an external circuit and a salt bridge. Oxidation occurs at the anode (negative electrode in a galvanic cell), and reduction occurs at the cathode (positive electrode).

电化学电池由两个半电池通过外电路和盐桥连接而成。氧化发生在阳极（原电池中的负极），还原发生在阴极（原电池中的正极）。

A half-cell typically contains a metal electrode dipped in a solution of its ions, e.g. Zn(s) | Zn²⁺(aq). A platinum electrode is used when no solid metal is involved, as in Fe³⁺/Fe²⁺ half-cells.

半电池通常包含浸在其离子溶液中的金属电极，如 Zn(s) | Zn²⁺(aq)。当没有固态金属参与时（如 Fe³⁺/Fe²⁺ 半电池），则使用铂电极。

The salt bridge (often a strip of filter paper soaked in KNO₃) completes the circuit by allowing ions to move and maintaining electrical neutrality in the half-cells.

盐桥（常为浸有 KNO₃ 溶液的滤纸条）允许离子迁移，维持半电池的电中性，从而形成完整回路。

The cell potential (EMF) is the voltage measured when no current is flowing, reflecting the difference in reducing power of the two half-cells.

电池电动势（EMF）是在无电流通过时测得的电压，反映了两个半电池还原能力的差异。

8. Standard Hydrogen Electrode and Standard Electrode Potentials | 标准氢电极与标准电极电势

The standard hydrogen electrode (SHE) is the reference against which all other electrode potentials are measured. It consists of a platinum electrode in 1 mol dm⁻³ H⁺ with H₂ gas at 1 atm and 298 K. Its standard electrode potential is defined as 0.00 V.

标准氢电极（SHE）是所有电极电势的测量基准。它由铂电极浸入 1 mol dm⁻³ H⁺ 溶液中，通入 1 atm H₂ 气体，温度为 298 K。其标准电极电势定义为 0.00 V。

The standard electrode potential (E°) of a half-cell is measured by connecting it to the SHE under standard conditions (298 K, 1 mol dm⁻³ ion concentrations, 100 kPa for gases).

半电池的标准电极电势（E°）是在标准条件下（298 K、离子浓度 1 mol dm⁻³、气体压强 100 kPa）与 SHE 连接时测得的电势。

A half-cell with a more negative E° has a greater tendency to undergo oxidation (stronger reducing agent). A more positive E° means a greater tendency to undergo reduction (stronger oxidising agent).

E° 越负的半电池越易发生氧化（还原性更强）；E° 越正则越易发生还原（氧化性更强）。

For example, E° for Zn²⁺/Zn = –0.76 V; E° for Cu²⁺/Cu = +0.34 V. Zinc therefore is a stronger reducing agent than copper.

例如，Zn²⁺/Zn 的 E° = –0.76 V，Cu²⁺/Cu 的 E° = +0.34 V，因此锌作为还原剂比铜更强。

9. Calculating Cell EMF and Predicting Feasibility | 计算电池电动势与预测反应可行性

The cell EMF is calculated as E°(cell) = E°(cathode) – E°(anode), where the cathode is the half-cell with the more positive (or less negative) electrode potential.

电池电动势计算公式：E°(cell) = E°(正极) – E°(负极)，其中正极是电极电势较高（或负值较小）的半电池。

A positive E°(cell) indicates that the reaction is thermodynamically feasible under standard conditions. For example, for the Daniell cell Zn|Zn²⁺||Cu²⁺|Cu: E°(cell) = +0.34 – (–0.76) = +1.10 V, indicating the reaction Zn + Cu²⁺ → Zn²⁺ + Cu is feasible.

E°(cell) 为正，表明反应在标准条件下热力学可行。例如丹尼尔电池 Zn|Zn²⁺||Cu²⁺|Cu：E°(cell) = +0.34 – (–0.76) = +1.10 V，说明 Zn + Cu²⁺ → Zn²⁺ + Cu 可行。

If E°(cell) is negative, the reaction is not feasible in the forward direction; the reverse reaction would be spontaneous.

若 E°(cell) 为负值，则正向反应不可行；逆反应可自发进行。

Under non-standard conditions, the Nernst equation can be used: E = E° + (0.059 / z) log₁₀([oxidised]/[reduced]) at 298 K. Changing ion concentrations shifts the electrode potential and can alter feasibility.

在非标准条件下，可使用 Nernst 方程（298 K 时简化形式）：E = E° + (0.059 / z) log₁₀([氧化型]/[还原型])。离子浓度改变会导致电极电势移动，从而影响反应可行性。

10. Disproportionation Reactions | 歧化反应

A disproportionation reaction occurs when the same element in a single species is simultaneously oxidised and reduced, forming two different products.

歧化反应是指同一物质中某元素同时发生氧化和还原，生成两种不同产物的反应。

Classic OCR examples: the reaction of copper(I) ions in aqueous solution: 2Cu⁺(aq) → Cu(s) + Cu²⁺(aq). Here, Cu⁺ (oxidation state +1) is both oxidised to Cu²⁺ and reduced to Cu.

OCR 经典示例：铜(I)离子在水溶液中发生歧化：2Cu⁺(aq) → Cu(s) + Cu²⁺(aq)。其中 Cu⁺（+1 氧化态）既被氧化为 Cu²⁺，又被还原为 Cu。

Another is the reaction of chlorine with cold dilute sodium hydroxide: Cl₂ + 2NaOH → NaCl + NaOCl + H₂O. Oxidation state of Cl changes from 0 to –1 (in NaCl) and +1 (in NaOCl).

另一个例子是氯气与冷的稀氢氧化钠溶液反应：Cl₂ + 2NaOH → NaCl + NaOCl + H₂O。氯的氧化数从 0 变为 –1（NaCl）和 +1（NaOCl）。

To check if a species is likely to disproportionate, compare the standard electrode potentials for the half-reactions involved. If the potential for the reduction half-step is greater than that for the oxidation half-step, disproportionation is feasible.

判断某物质能否歧化，可比较相关半反应的标准电极电势。若还原步骤的 E° 大于氧化步骤的 E°，则歧化反应可行。

11. Redox Titrations Overview | 氧化还原滴定概述

Redox titrations are a key practical application in OCR, most commonly using manganate(VII) or iodine/thiosulfate.

氧化还原滴定是 OCR 中的重要实验应用，最常见的是高锰酸钾法和碘/硫代硫酸钠法。

In manganate(VII) titrations, acidified KMnO₄ acts as its own indicator – the purple MnO₄⁻ decolourises at the endpoint when reduced to almost colourless Mn²⁺. The half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

高锰酸钾滴定中，酸化的 KMnO₄ 自身作为指示剂——紫色的 MnO₄⁻ 在终点被还原为几乎无色的 Mn²⁺。半反应：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。

In iodine–thiosulfate titrations, I₂ is generated or reacted, and the remaining I₂ is titrated with S₂O₃²⁻ using starch indicator (blue-black to colourless). The relevant half-equation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻.

碘量法中，生成或消耗 I₂，然后用 S₂O₃²⁻ 滴定剩余的 I₂，以淀粉作指示剂（蓝黑色褪至无色）。相关半反应：2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻。

From titration data, you can calculate the amount of oxidising or reducing agent in the sample by using the stoichiometric ratio from the balanced redox equation.

根据滴定数据，可利用配平的氧化还原反应方程式的化学计量比计算样品中氧化剂或还原剂的含量。

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A-Level OCR Chemistry: Redox Exam Insights | A-Level OCR 化学：氧化还原 考点精讲