Redox Reactions | 氧化还原反应

📚 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

Historically, oxidation was defined as gain of oxygen and reduction as loss of oxygen. The modern definition, however, is based on electron transfer: 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

An oxidising agent (oxidant) causes oxidation by accepting electrons; it is itself reduced in the process. A reducing agent (reductant) donates electrons and is itself oxidised. Strong oxidising agents have a high tendency to gain electrons – examples include F₂, acidic MnO₄⁻, and Cr₂O₇²⁻. Strong reducing agents, such as Na, Mg and Al, readily lose electrons. In any redox reaction, the substance 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

Redox reactions can be split into two half-equations: one representing oxidation (loss of electrons) and the other representing reduction (gain 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.

Example: MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ in acid. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxidation: C₂O₄²⁻ → 2CO₂ + 2e⁻. Multiply oxidation 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

For reactions in alkaline solution, a convenient approach is to first balance the equation as if it were in acidic medium, and then add OH⁻ ions to both 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Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

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Redox Reactions | 氧化还原反应