📚 IB Chemistry: Redox Reactions Key Points | IB 化学：氧化还原 考点精讲
Redox reactions are at the heart of IB Chemistry, linking topics from bonding to energetics and electrochemistry. A systematic grasp of oxidation numbers, half-equations, electrode potentials, and electrolysis will unlock high marks in both Paper 1 and Paper 2. This revision guide distils every essential concept, common pitfall, and calculation technique you need, presented in clear paired English and Chinese explanations.
氧化还原反应是 IB 化学的核心，将结构、能量学和电化学联系起来。系统地掌握氧化数、半反应方程式、电极电势和电解，能够帮助你在卷一和卷二中取得高分。本复习指南提炼了所有关键概念、常见陷阱和计算技巧，并通过清晰的中英对照讲解，助你全面备考。
1. Oxidation Number Rules | 氧化数规则
The oxidation number (ON) is the hypothetical charge an atom would carry if all its bonds were completely ionic. You must apply these rules in order: free elements have ON = 0; the ON of a monatomic ion equals its charge; oxygen is almost always −2 (except in peroxides where it is −1, and in OF₂ where it is +2); hydrogen is +1 when bonded to non‑metals and −1 when bonded to metals; the sum of ONs in a neutral compound is 0; in a polyatomic ion the sum equals the ion’s charge. For example, in KMnO₄, K is +1, O is −2, so Mn must be +7.
氧化数（ON）是假设所有化学键完全离子化时原子所带的电荷。必须按顺序应用以下规则：游离态元素 ON = 0；单原子离子的 ON 等于其所带电荷；氧通常为 −2（过氧化物中为 −1，OF₂ 中为 +2）；与非金属结合的氢为 +1，与金属结合的氢为 −1；中性分子中 ON 总和为 0；多原子离子中 ON 总和等于离子电荷。例如，在 KMnO₄ 中，K 为 +1，O 为 −2，因此 Mn 必为 +7。
| Species | Oxidation Number | 物种 | 氧化数 |
|---|---|---|---|
| N₂ | 0 | N₂ | 0 |
| H₂O₂ | O: −1 | H₂O₂ | O: −1 |
| Cr₂O₇²⁻ | Cr: +6 | Cr₂O₇²⁻ | Cr: +6 |
2. Identifying Redox Reactions | 识别氧化还原反应
A reaction is redox if any atom’s oxidation number changes. Oxidation means an increase in ON (loss of electrons); reduction means a decrease in ON (gain of electrons). The mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain) helps. For instance, when Zn reacts with Cu²⁺, Zn goes from 0 to +2 (oxidation) and Cu²⁺ goes from +2 to 0 (reduction). A reaction where no ON changes — e.g., precipitation of AgCl from Ag⁺ and Cl⁻ — is not redox.
只要任何原子的氧化数发生变化，该反应就是氧化还原反应。氧化是指 ON 升高（失去电子）；还原是指 ON 降低（得到电子）。助记口诀 OIL RIG（氧化失电子，还原得电子）很有帮助。例如，Zn 与 Cu²⁺ 反应时，Zn 从 0 变为 +2（被氧化），Cu²⁺ 从 +2 变为 0（被还原）。如果 ON 未变，如 Ag⁺ 与 Cl⁻ 生成 AgCl 沉淀，则不是氧化还原反应。
3. Oxidising and Reducing Agents | 氧化剂和还原剂
The oxidising agent (oxidant) is itself reduced; it accepts electrons and causes the oxidation of another species. The reducing agent (reductant) is itself oxidised; it donates electrons and causes reduction. Common oxidising agents include KMnO₄ (Mn⁺⁷ → Mn²⁺), K₂Cr₂O₇ (Cr⁺⁶ → Cr³⁺), and halogens. Common reducing agents include metals like Zn, Fe²⁺ ions, and I⁻ ions. A substance like SO₂ can act as both an oxidising and a reducing agent depending on the reaction partner.
氧化剂本身被还原；它接受电子并使另一种物质被氧化。还原剂本身被氧化；它给出电子并使另一种物质被还原。常见的氧化剂包括 KMnO₄（Mn⁺⁷ → Mn²⁺）、K₂Cr₂O₇（Cr⁺⁶ → Cr³⁺）和卤素。常见的还原剂有 Zn 等金属、Fe²⁺ 离子和 I⁻ 离子。像 SO₂ 这样的物质可根据反应对象同时充当氧化剂和还原剂。
4. Writing Half-Reactions | 书写半反应
Half-reactions show either the oxidation or the reduction process separately. Steps to write a half-reaction in acidic solution: balance the atom undergoing change; add H₂O to balance O; add H⁺ to balance H; finally add electrons (e⁻) to balance charge. Example: MnO₄⁻ → Mn²⁺. Balance Mn, then 4 H₂O on right to balance O, then 8 H⁺ on left, charge ( −1 + 8 = +7 vs +2 ), so add 5e⁻ on left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. In basic solution, after balancing in acid, add OH⁻ to both sides to neutralise H⁺ and simplify.
半反应分别显示氧化过程或还原过程。在酸性溶液中书写半反应的步骤：配平发生变化的原子；加 H₂O 配平 O；加 H⁺ 配平 H；最后加电子（e⁻）配平电荷。示例：MnO₄⁻ → Mn²⁺。先配平 Mn，右边加 4 H₂O 配平 O，左边加 8 H⁺，电荷（-1 +8 = +7 对比 +2），因此左边加 5e⁻：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。在碱性溶液中，先用酸性方法配平，然后两边加 OH⁻ 中和 H⁺ 并化简。
5. Balancing Redox Equations | 配平氧化还原方程式
To balance a full redox equation, split it into two half-reactions. Balance each following the rules above. Multiply each half-reaction by appropriate integers so that the number of electrons lost equals the number gained. Add the half-reactions and cancel electrons and any identical species appearing on both sides. For the reaction of Fe²⁺ with MnO₄⁻ in acid: oxidation half: Fe²⁺ → Fe³⁺ + e⁻; reduction half: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Multiply oxidation by 5, then add: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. Always check that both atoms and charges balance.
配平完整的氧化还原方程式时，先分成两个半反应。按上述规则分别配平。将两个半反应乘以适当的整数，使失去的电子数等于得到的电子数。将半反应相加，消去电子以及两边相同的物种。Fe²⁺ 与 MnO₄⁻ 在酸性条件下的反应：氧化半反应 Fe²⁺ → Fe³⁺ + e⁻；还原半反应 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。将氧化半反应乘以 5 后相加：5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O。务必检查原子和电荷是否均配平。
6. Introduction to Electrochemical Cells | 电化学电池简介
A voltaic (galvanic) cell converts chemical energy into electrical energy via a spontaneous redox reaction. It consists of two half-cells connected by a wire and a salt bridge. The anode is the electrode where oxidation occurs (negative in a voltaic cell); the cathode is where reduction occurs (positive). The salt bridge allows ions to flow and maintain electrical neutrality. Conversely, an electrolytic cell uses an external power source to drive a non‑spontaneous reaction; here the anode is positive and the cathode is negative.
伏打（原）电池通过自发的氧化还原反应将化学能转化为电能。它由两个半电池通过导线和盐桥连接组成。阳极发生氧化反应（在原电池中为负极）；阴极发生还原反应（正极）。盐桥允许离子流动以维持电中性。相反，电解池则使用外部电源驱动非自发反应；此时阳极为正极，阴极为负极。
7. Standard Electrode Potentials | 标准电极电势
The standard electrode potential E° is measured under standard conditions (298 K, 100 kPa, 1 mol dm⁻³ solutions) relative to the standard hydrogen electrode, which is assigned 0 V. All half-cell potentials are listed as reduction potentials. A more positive E° means a greater tendency to be reduced, i.e., a stronger oxidising agent. For example, F₂/F⁻ has E° = +2.87 V; Li⁺/Li has E° = −3.04 V. The table makes it easy to predict the direction of electron flow.
标准电极电势 E° 是在标准条件（298 K、100 kPa、1 mol dm⁻³ 溶液）下相对于标准氢电极（0 V）测定的。所有半电池电势均以还原电势列出。E° 值越正，表示该物种越容易被还原，即氧化性越强。例如，F₂/F⁻ 的 E° = +2.87 V；Li⁺/Li 的 E° = −3.04 V。利用此表可轻松预测电子流动方向。
8. Calculating Cell EMF and Spontaneity | 计算电池电动势和自发性
The electromotive force (EMF) of a cell is given by E°cell = E°cathode − E°anode, where both potentials are reduction potentials. A positive E°cell indicates a spontaneous reaction (ΔG° < 0). The relationship between EMF and Gibbs free energy is:
ΔG° = −nFE°cell
where n is the number of moles of electrons transferred and F is the Faraday constant (96 500 C mol⁻¹). For example, if E°cell = +1.10 V and n = 2, ΔG° = −212 kJ mol⁻¹, confirming spontaneity.
电池的电动势（EMF）由 E°cell = E°cathode − E°anode 求得，其中两个电势均为还原电势。E°cell > 0 表示反应自发（ΔG° < 0）。EMF 与吉布斯自由能的关系为：
ΔG° = −nFE°cell
其中 n 是转移电子的物质的量，F 为法拉第常数（96 500 C mol⁻¹）。例如，若 E°cell = +1.10 V，n = 2，则 ΔG° = −212 kJ mol⁻¹，证实反应自发。
9. Electrolysis | 电解
Electrolysis is the decomposition of an electrolyte by an electric current. For molten ionic compounds, the cation is reduced at the cathode and the anion is oxidised at the anode. In aqueous solutions, water can also be oxidised or reduced. At the cathode, the species with the more positive E° (less reactive metal or H₂O) is reduced. At the anode, if a halide is present it is usually oxidised; otherwise water is oxidised to O₂. Faraday’s laws state that the mass of substance produced is proportional to the quantity of charge passed: m = (M × I × t) / (n × F), where M is molar mass, I is current, t is time, n is number of electrons in the half-equation.
电解是指电流使电解质分解的过程。对于熔融离子化合物，阳离子在阴极还原，阴离子在阳极氧化。在水溶液中，水也可能被氧化或还原。在阴极，E° 较正（较不活泼的金属或水）的物质优先还原。在阳极，若存在卤离子通常先氧化；否则水被氧化生成 O₂。法拉第定律指出，生成物质的质量与通过的电量成正比：m = (M × I × t) / (n × F)，其中 M 为摩尔质量，I 为电流，t 为时间，n 为半反应中的电子数。
10. Redox Titrations | 氧化还原滴定
Redox titrations determine an unknown concentration by reacting it with a standardised oxidising or reducing agent. Potassium manganate(VII), KMnO₄, is a common self‑indicator: its purple colour disappears at the end point. Another important system is iodine‑thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, using starch as an indicator added near the end point. Calculations follow the same mole‑ratio method as acid‑base titrations, using the balanced redox equation to relate moles of titrant to moles of analyte.
氧化还原滴定通过让待测物与标准氧化剂或还原剂反应来确定未知浓度。高锰酸钾 KMnO₄ 是常见的自身指示剂，其紫色在终点消失。另一个重要体系是碘‑硫代硫酸盐：I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻，以淀粉为指示剂并在接近终点时加入。计算方法与酸碱滴定相同，利用配平的氧化还原方程式建立滴定剂与待测物物质的量之间的关系。
11. Common Exam Pitfalls | 常见考试陷阱
Students often lose marks by assigning wrong oxidation numbers — especially in unfamiliar compounds or peroxides. In half-equations, forgetting to balance atoms before electrons or failing to convert to basic conditions when required are frequent errors. A classic mistake is using E°cell = E°anode − E°cathode (incorrect) instead of cathode minus anode. For electrolysis of aqueous solutions, remembering that water competes is vital. Finally, units and sign conventions in ΔG° = −nFE° must be consistent: n in mol, F in C mol⁻¹, E° in V, giving ΔG° in J.
学生常因错误指定氧化数而丢分，尤其是在陌生化合物或过氧化物中。在半方程式中，配平电子前未配平原子，或需要时未转化为碱性条件，都是常见错误。经典错误是用 E°cell = E°anode − E°cathode（错误）而非阴极减阳极。对于水溶液电解，必须记得水也会参与竞争。最后，ΔG° = −nFE° 中的单位和符号须前后一致：n 以 mol 计，F 以 C mol⁻¹ 计，E° 以 V 计，ΔG° 的单位为 J。
12. Summary and Exam Tips | 核心考点总结与应试技巧
Redox mastery requires fluency in oxidation numbers, half-equation balancing, and the language of electrochemistry. Always label which species is oxidised and which is reduced, and identify the oxidising and reducing agents. For calculations, memorise E°cell = E°cathode − E°anode and ΔG° = −nFE°, and practise with past IB questions. Use the electrochemical series to predict whether a reaction is spontaneous, and never forget the competition from water in electrolysis. A structured, step‑by‑step approach will ensure you reach the right answer every time.
掌握氧化还原需要熟练运用氧化数、半反应配平以及电化学术语。始终标明哪种物质被氧化、哪种被还原，并标明氧化剂和还原剂。在计算方面，记住 E°cell = E°cathode − E°anode 和 ΔG° = −nFE°，要多练习 IB 历年真题。利用电化学序预测反应是否自发，并且千万别忽视电解时水的竞争作用。采用系统化的分步解题方法，每次都能得出正确答案。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导