📚 Redox Reactions | 氧化还原反应
Redox reactions, short for reduction-oxidation reactions, are fundamental processes in chemistry where electrons are transferred between species. These reactions are central to energy conversion, electrochemistry, and many biological processes. In A-Level Chemistry, a thorough understanding of redox concepts is essential for tackling topics such as electrode potentials, electrolysis, and titration calculations.
氧化还原反应（氧化-还原反应）是化学中电子在物种间转移的基本过程。这些反应是能量转换、电化学和许多生物过程的核心。在A-Level化学中，深入理解氧化还原概念对于掌握电极电势、电解和滴定计算等主题至关重要。
1. Oxidation and Reduction: Two Definitions | 氧化与还原：两种定义
er: oxidation is the loss of electrons, while reduction is the gain of electrons. Another useful perspective uses oxidation numbers: oxidation corresponds to an increase in oxidation number, and reduction to a decrease in oxidation number. For example, in the reaction Mg + Cu²⁺ → Mg²⁺ + Cu, magnesium is oxidised (loses electrons, oxidation number increases from 0 to +2), while copper(II) ions are reduced (gain electrons, oxidation number decreases from +2 to 0). This dual definition allows us to analyse both ionic and molecular redox processes.
传统上，氧化被定义为得氧，还原为失氧。而现代定义基于电子转移：氧化是失去电子，还原是得到电子。另一种有用的视角是运用氧化数：氧化意味着氧化数升高，还原意味着氧化数降低。例如在反应 Mg + Cu²⁺ → Mg²⁺ + Cu 中，镁被氧化（失去电子，氧化数由0升至+2），铜(II)离子被还原（得到电子，氧化数由+2降至0）。这种双重定义为分析离子型和分子型氧化还原过程提供了便利。
2. Oxidation Numbers: Rules and Calculation | 氧化数：规则与计算
Assigning oxidation numbers is a systematic way to track electron transfer. The following rules are applied in order of priority:
- Free elements have an oxidation number of 0 (e.g., O₂, Na, Fe).
- For monatomic ions, the oxidation number equals the charge (e.g., Cl⁻ = –1, Al³⁺ = +3).
- In compounds, fluorine always has oxidation number –1.
- Oxygen usually has oxidation number –2, except in peroxides (–1) and with fluorine (+2).
- Hydrogen is +1 when bonded to nonmetals, –1 when bonded to metals (hydrides).
- The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion it equals the ion's charge.
- In a compound, the more electronegative element is assigned the negative oxidation number.
指定氧化数是追踪电子转移的系统方法。以下规则按优先顺序应用：
- 游离态元素的氧化数为0（如 O₂、Na、Fe）。
- 单原子离子的氧化数等于其所带电荷（如 Cl⁻ = –1，Al³⁺ = +3）。
- 化合物中氟的氧化数总为 –1。
- 氧的氧化数通常为 –2，但在过氧化物中为 –1，与氟结合时为 +2。
- 氢与非金属结合时为 +1，与金属结合时为 –1（氢化物）。
- 中性化合物中各元素氧化数之和为0；多原子离子中氧化数之和等于离子电荷。
- 化合物中电负性较大的元素被赋予负氧化数。
For example, in Fe₃O₄, assuming O is –2, total negative charge is –8, so the three Fe atoms must sum to +8, giving an average oxidation number of +8/3 (actually a mixture of +2 and +3). In H₂O₂ (hydrogen peroxide), oxygen is –1, not –2.
例如在 Fe₃O₄ 中，假设 O 为 –2，总负电荷为 –8，因此三个 Fe 原子总的氧化数为 +8，平均氧化数为 +8/3（实际上是 +2 与 +3 的混合）。在 H₂O₂（过氧化氢）中，氧的氧化数为 –1 而非 –2。
3. Identifying Oxidising and Reducing Agents | 识别氧化剂和还原剂
lectrons and is itself oxidised. Strong oxidising agents have a high tendency to gain electrons – bstance that is oxidised acts as the reducing agent, while the substance that is reduced acts as the oxidising agent. For instance, in the reaction 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, Fe³⁺ is the oxidising agent (it gains an electron to become Fe²⁺) and I⁻ is the reducing agent (it loses an electron to form I₂).
氧化剂（oxidant）通过接受电子引起氧化，自身被还原。还原剂（reductant）提供电子，自身被氧化。强氧化剂具有很高的得电子倾向，例如 F₂、酸性 MnO₄⁻ 和 Cr₂O₇²⁻。强还原剂如 Na、Mg、Al 则易于失去电子。在任何氧化还原反应中，被氧化的物质充当还原剂，被还原的物质充当氧化剂。例如在反应 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ 中，Fe³⁺ 是氧化剂（它获得一个电子变成 Fe²⁺），I⁻ 是还原剂（它失去一个电子形成 I₂）。
4. Half-Equations and Electron Transfer | 半反应与电子转移
of electrons). For the reaction Zn + Cu²⁺ → Zn²⁺ + Cu, the oxidation half-equation is Zn → Zn²⁺ + 2e⁻, and the reduction half-equation is Cu²⁺ + 2e⁻ → Cu. When combining half-equations, the electrons must cancel. Half-equations are essential for understanding electrochemical cells and for balancing complex redox equations. For example, in the reaction of acidified permanganate with iron(II) ions:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Fe²⁺ → Fe³⁺ + e⁻
Multiplying the iron half-equation by 5 and adding gives the overall ionic equation.
氧化还原反应可拆分为两个半反应：一个表示氧化（失去电子），另一个表示还原（得到电子）。对于反应 Zn + Cu²⁺ → Zn²⁺ + Cu，氧化半反应为 Zn → Zn²⁺ + 2e⁻，还原半反应为 Cu²⁺ + 2e⁻ → Cu。合并半反应时，电子必须相互抵消。半反应对于理解电化学电池和配平复杂氧化还原方程式至关重要。例如在酸性高锰酸钾与铁(II)离子的反应中，半反应如上所示，将铁半反应乘以5后相加即可得到总离子方程式。
5. Balancing Redox Equations in Acidic Medium | 酸性介质中配平氧化还原方程式
The half-equation method is the most reliable way to balance redox equations in acid. The steps are:
- Assign oxidation numbers and identify what is oxidised and reduced.
- Write separate half-equations for oxidation and reduction.
- Balance all atoms except O and H.
- Balance oxygen by adding H₂O molecules.
- Balance hydrogen by adding H⁺ ions.
- Balance charge by adding electrons (e⁻).
- Multiply the half-equations so that the electrons are equal, then add them together and cancel common species.
idation by 5 and reduction by 2, then add: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.
半反应法是配平酸性条件下氧化还原方程式最可靠的方法。步骤为：分配氧化数，确定氧化与还原部分；写出各自的半反应；先配平氧和氢以外的原子；用 H₂O 配平氧；用 H⁺ 配平氢；用电子配平电荷；使电子数相等后将两个半反应相加，消去相同物质。例如 MnO₄⁻ 与 C₂O₄²⁻ 在酸中的反应：还原半反应为 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O，氧化半反应为 C₂O₄²⁻ → 2CO₂ + 2e⁻，经过倍数调整后相加即可得到配平方程式。
6. Balancing Redox Equations in Alkaline Medium | 碱性介质中配平氧化还原方程式
h sides to neutralise the H⁺ (forming water). Alternatively, one can use OH⁻ and H₂O directly to balance O and H. For example, the oxidation of Cr³⁺ to CrO₄²⁻ in alkaline conditions: first in acid, Cr³⁺ + 4H₂O → CrO₄²⁻ + 8H⁺ + 3e⁻. Then add 8 OH⁻ to both sides: Cr³⁺ + 4H₂O + 8OH⁻ → CrO₄²⁻ + 8H₂O + 3e⁻, which simplifies to Cr³⁺ + 8OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻.
对于碱性溶液中的反应，一种便捷的方法是先按酸性介质配平，然后在等式两边同时加入 OH⁻ 以中和 H⁺，生成水。也可以直接使用 OH⁻ 和 H₂O 来配平氧和氢。例如在碱性条件下 Cr³⁺ 被氧化为 CrO₄²⁻：先在酸性条件下得到 Cr³⁺ + 4H₂O → CrO₄²⁻ + 8H⁺ + 3e⁻，两边加 8OH⁻ 后化简即得 Cr³⁺ + 8OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻。
7. Standard Electrode Potentials and the Electrochemical Series | 标准电极电势与电化学序
The standard electrode potential (E°) of a half-cell measures its tendency to be reduced, under standard conditions (298 K, 1 mol dm⁻³ for ions, 100 kPa for gases). The standard hydrogen electrode (SHE: 2H⁺ + 2e⁻ ⇌ H₂) is assigned a potential of exactly 0 V. The electrochemical series lists half-reactions with their E° values; a more positive E° indicates a stronger oxidising agent. Selected values are shown below:
| Half-reaction | E° / V |
| F₂ + 2e⁻ ⇌ 2F⁻ | +2.87 |
| MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O | +1.51 |
| Cu²⁺ + 2e⁻ ⇌ Cu | +0.34 |
| 2H⁺ + 2e⁻ ⇌ H₂ | 0.00 |
| Zn²⁺ + 2e⁻ ⇌ Zn | –0.76 |
| Na⁺ + e⁻ ⇌ Na | –2.71 |
The cell EMF under standard conditions is given by E°(cell) = E°(reduction) – E°(oxidation) when using reduction potentials, or E°(cell) = E°(right) – E°(left) for a conventional cell diagram.
标准电极电势（E°）衡量半电池在标准条件（298 K，离子浓度1 mol dm⁻³，气体100 kPa）下被还原的趋势。标准氢电极（SHE: 2H⁺ + 2e⁻ ⇌ H₂）被指定为0 V。电化学序列将半反应及其 E° 列出；E° 越正，氧化剂越强。上表列出了一些典型值。标准条件下电池电动势由 E°(电池) = E°(还原) – E°(氧化) 计算，或根据电池图式用右侧减左侧。
8. Predicting Feasibility of Redox Reactions | 预测氧化还原反应的自发性
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