Edexcel A-Level Chemistry Topic 14: Redox II & Electrode Potentials | 氧化还原 II 与电极电势

📚 Edexcel A-Level Chemistry Topic 14: Redox II & Electrode Potentials | 氧化还原 II 与电极电势

Welcome to this Edexcel A-Level Chemistry revision guide for Topic 14: Redox II and Electrode Potentials. This topic extends your earlier work on oxidation states and half-equations, linking electron transfer to measurable electrical potentials and using standard electrode potentials to predict whether redox reactions are feasible.

欢迎阅读本篇 Edexcel A-Level 化学复习指南，主题为氧化还原 II 与电极电势。本专题将拓展你此前学习过的氧化数与半反应式知识，把电子转移与可测量的电势联系起来，并利用标准电极电势判断氧化还原反应是否可行。

1. Oxidation States and Redox Review | 氧化数与氧化还原复习

In Topic 3 you met oxidation states and redox reactions. Topic 14 starts by ensuring you can assign oxidation states confidently to atoms in molecules, polyatomic ions and complex ions. Oxidation is an increase in oxidation state and involves loss of electrons; reduction is a decrease in oxidation state and involves gain of electrons.

在 Topic 3 中你已经学过氧化数与氧化还原反应。Topic 14 首先要求你能够熟练判断分子、多原子离子以及配离子中原子的氧化数。氧化是氧化数升高并失去电子；还原是氧化数降低并获得电子。

Useful rules include: elements have an oxidation state of 0; oxygen is usually -2; hydrogen is usually +1; the sum of oxidation states equals the overall charge on the species. In MnO₄⁻, for example, Mn has an oxidation state of +7 because four O atoms contribute -8 and the overall charge is -1.

常用规则包括：单质的氧化数为 0；氧通常为 -2；氢通常为 +1；各原子氧化数的总和等于粒子所带电荷。例如在 MnO₄⁻ 中，四个 O 贡献 -8，整个离子带 -1 电荷，因此 Mn 的氧化数为 +7。

• Oxidation: increase in oxidation state, loss of electrons
• 氧化：氧化数升高，失去电子
• Reduction: decrease in oxidation state, gain of electrons
• 还原：氧化数降低，获得电子

2. Half-Equations and Balancing Redox | 半反应式与氧化还原配平

A redox reaction can be split into two half-equations: an oxidation half-equation and a reduction half-equation. To balance a half-equation under acidic conditions, add H₂O to balance oxygen, add H⁺ to balance hydrogen, and then add electrons to balance charge.

氧化还原反应可以拆分为两个半反应式：氧化半反应式和还原半反应式。在酸性条件下配平半反应式时，通常先加 H₂O 平衡氧，再加 H⁺ 平衡氢，最后加电子平衡电荷。

A common reduction half-equation is the conversion of manganate(VII) to manganese(II) under acidic conditions:

一个常见的还原半反应式是酸性条件下高锰酸根(VII)被还原为锰(II)：

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

An oxidation half-equation can then be combined with it. For oxalic acid, ethanedioate ions are oxidised to carbon dioxide:

氧化半反应式可以与其组合。例如草酸中的乙二酸根被氧化为二氧化碳：

C₂O₄²⁻ → 2CO₂ + 2e⁻

To combine the two half-equations, multiply so the electrons cancel. The overall ionic equation is:

将两个半反应式组合时，需要乘以适当系数使电子抵消。总离子方程式为：

2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

3. Disproportionation | 歧化反应

Disproportionation is a redox reaction in which the same element is both oxidised and reduced simultaneously. A classic example is the disproportionation of copper(I) ions into copper(II) and copper metal:

歧化反应是同一元素在同一步反应中既被氧化又被还原的氧化还原反应。一个典型例子是铜(I)离子歧化为铜(II)离子和铜单质：

2Cu⁺ → Cu²⁺ + Cu

Here the oxidation state of copper changes from +1 to +2 in one product and from +1 to 0 in the other. Disproportionation is also shown by chlorine when it reacts with cold dilute sodium hydroxide:

在该反应中，铜的氧化数从 +1 变为 +2，同时在另一种产物中从 +1 变为 0。氯气与冷的稀氢氧化钠溶液反应也表现出歧化：

Cl₂ + 2NaOH → NaCl + NaClO + H₂O

Chlorine is simultaneously reduced to chloride, Cl⁻, and oxidised to chlorate(I), ClO⁻. Identifying disproportionation requires you to track the oxidation state of one element across both products.

氯气同时被还原为氯离子 Cl⁻，并被氧化为次氯酸根 ClO⁻。判断歧化反应需要追踪同一元素在两种产物中的氧化数变化。

4. Electrochemical Cells Basics | 电化学电池基础

An electrochemical cell consists of two half-cells connected by a salt bridge and an external conducting wire. Each half-cell contains

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