Oxidation, Reduction and Redox Equations | 氧化、还原与氧化还原方程式

📚 Oxidation, Reduction and Redox Equations | 氧化、还原与氧化还原方程式

In AQA A-Level Chemistry, oxidation and reduction are not limited to gaining or losing oxygen. The modern definitions use electron transfer and oxidation states, which allow chemists to analyse reactions even when no oxygen is present. This topic underpins half-equations, redox titrations, electrochemical cells and many industrial processes.

在AQA A-Level化学中，氧化与还原不仅限于得氧或失氧。现代定义使用电子转移和氧化数，使化学家即使在没有氧参与的反应中也能进行分析。本主题是半反应方程式、氧化还原滴定、电化学电池和许多工业过程的基础。

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1. What Oxidation and Reduction Mean | 氧化与还原的含义

Oxidation is the loss of electrons; reduction is the gain of electrons. A useful mnemonic is OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.

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：氧化是失电子，还原是得电子。

Oxidation and reduction always occur together because electrons lost by one species must be gained by another. A reaction in which both processes happen is called a redox reaction.

氧化和还原总是同时发生，因为一个物种失去的电子必须被另一个物种获得。同时发生这两个过程的反应称为氧化还原反应。

In terms of oxidation state, oxidation is an increase in oxidation number, and reduction is a decrease in oxidation number.

从氧化数角度看，氧化是氧化数升高，还原是氧化数降低。

Zn → Zn²⁺ + 2e⁻   (oxidation)

Cu²⁺ + 2e⁻ → Cu   (reduction)

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2. Oxidation States and the Rules | 氧化数及其规则

Oxidation state is the charge an atom would have if all bonds were treated as ionic. It is written with the sign before the number, for example +2 or -3.

氧化数是假设所有化学键均为离子键时原子所带的电荷。书写时符号在数字之前，例如 +2 或 -3。

The rules for assigning oxidation states are as follows:

确定氧化数的规则如下：

A pure element has an oxidation state of 0, for example Na(s), O₂(g) and Fe(s).

纯单质的氧化数为 0，例如 Na(s)、O₂(g) 和 Fe(s)。

For a simple monatomic ion, the oxidation state equals the charge on the ion, so Cl⁻ is -1 and Mg²⁺ is +2.

对于简单单原子离子，氧化数等于离子所带的电荷，因此 Cl⁻ 为 -1，Mg²⁺ 为 +2。

In compounds, fluorine is always -1. Oxygen is usually -2, except in peroxides where it is -1, and in OF₂ where it is +2. Hydrogen is +1 except in metal hydrides such as NaH, where it is -1.

在化合物中，氟总是 -1。氧通常为 -2，但在过氧化物中为 -1，在 OF₂ 中为 +2。氢为 +1，但在金属氢化物如 NaH 中为 -1。

The sum of oxidation states in a neutral compound is 0; in a polyatomic ion, the sum equals the overall charge on the ion.

中性化合物中各元素氧化数之和为 0；在多原子离子中，氧化数之和等于离子所带的总电荷。

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3. Calculating Oxidation States | 计算氧化数

Use the sum rule and the fixed values for hydrogen and oxygen to work out an unknown oxidation state in a compound or ion.

利用加和规则以及氢和氧的固定值来计算化合物或离子中未知元素的氧化数。

Example 1: In H₂SO₄, hydrogen is +1 and oxygen is -2. Let sulfur be x.

示例 1：在 H₂SO₄ 中，氢为 +1，氧为 -2。设硫的氧化数为 x。

2(+1) + x + 4(-2) = 0   →   x = +6

Example 2: In MnO₄⁻, oxygen is -2. Let manganese be x.

示例 2：在 MnO₄⁻ 中，氧为 -2。设锰的氧化数为 x。

x + 4(-2) = -1   →   x = +7

Example 3: In Cr₂O₇²⁻, oxygen is -2. Let chromium be x.

示例 3：在 Cr₂O₇²⁻ 中，氧为 -2。设铬的氧化数为 x。

2x + 7(-2) = -2   →   2x = +12   →   x = +6

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4. Recognising Redox Reactions | 识别氧化还原反应

A reaction is redox if any atom changes oxidation state. If no oxidation states change, the reaction is not redox, for example acid-base neutralisation or precipitation.

如果任何原子的氧化数发生变化，反应就是氧化还原反应。如果没有氧化数变化，则反应不是氧化还原反应，例如酸碱中和或沉淀反应。

Cu. Zinc goes from 0 to +2, so it is oxidised; copper goes from +2 to 0, so it is reduced.

例如：Zn + CuSO₄ → ZnSO₄ + Cu。锌的氧化数从 0 变为 +2，因此被氧化；铜的氧化数从 +2 变为 0，因此被还原。

In some reactions, one element is both oxidised and reduced at the same time. This is called disproportionation and is covered in a later section.

在某些反应中，一种元素同时被氧化和被还原。这称为歧化反应，将在后面章节讨论。

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

An oxidising agent accepts electrons and is itself reduced. A reducing agent donates electrons and is itself oxidised.

氧化剂接受电子，本身被还原。还原剂提供电子，本身被氧化。

Common oxidising agents include acidified manganate(VII), acidified dichromate(VI), halogens and oxygen. Common reducing agents include metals, hydrogen, iodide ions and iron(II) compounds.

常见氧化剂包括酸化高锰酸根、酸化重铬酸根、卤素和氧气。常见还原剂包括金属、氢气、碘离子和铁(II)化合物。

In an exam question, identify the species whose oxidation state decreases as the oxidising agent. The species whose oxidation state increases is the reducing agent.

在考试题目中，氧化数降低的物质是氧化剂；氧化数升高的物质是还原剂。

l₂, magnesium is the reducing agent because Mg is oxidised from 0 to +2, and chlorine is the oxidising agent because Cl₂ is reduced from 0 to -1.

示例：在反应 Mg + Cl₂ → MgCl₂ 中，镁是还原剂，因为 Mg 从 0 被氧化为 +2；氯是氧化剂，因为 Cl₂ 从 0 被还原为 -1。

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6. Writing Half-Equations | 书写半反应方程式

Half-equations show either oxidation or reduction separately, including electrons. The key steps are: write the species containing the element, balance atoms other than O and H, add H₂O to balance oxygen, add H⁺ to balance hydrogen, then add electrons to balance charge.

半反应方程式分别表示氧化或还原过程，并包含电子。关键步骤是：写出含该元素的物种，平衡除 O 和 H 以外的原子，加入 H₂O 平衡氧，加入 H⁺ 平衡氢，然后加入电子平衡电荷。

For acidic conditions this method works directly. For alkaline conditions, you can write the acidic half-equation first and then add OH⁻ to both sides to neutralise H⁺.

在酸性条件下此方法直接适用。对于碱性条件，可先写出酸性半反应方程式，然后在两边加入 OH⁻ 中和 H⁺。

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O   (reduction)Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

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