📚 Redox and Oxidation Number | 氧化还原与氧化数
Redox reactions are a fundamental class of chemical reactions involving the transfer of electrons between species. They are essential in both biological systems and industrial processes, such as respiration, combustion, and electrochemistry.
氧化还原反应是一类基本的化学反应，涉及物质之间的电子转移。它们在生物系统和工业过程中都至关重要，如呼吸作用、燃烧和电化学。
The term ‘redox’ is a combination of reduction and oxidation, two complementary processes that always occur simultaneously. When one substance is oxidised, another must be reduced.
“氧化还原”一词是还原和氧化的结合，这两个互补过程总是同时发生。当一种物质被氧化时，另一种物质必然被还原。
1. Early Definitions: Oxygen and Hydrogen Transfer | 早期定义：氧和氢的转移
Historically, oxidation was defined as the gain of oxygen by an element or compound. For example, when magnesium burns in air to form MgO, magnesium is oxidised because it gains oxygen.
早期定义中，氧化被定义为元素或化合物获得氧。例如，镁在空气中燃烧生成 MgO，镁获得氧，因此被氧化。
Reduction was seen as the loss of oxygen. In the reaction CuO + H₂ → Cu + H₂O, the copper(II) oxide loses oxygen to form copper, so it is reduced.
还原则被视为失去氧。在反应 CuO + H₂ → Cu + H₂O 中，氧化铜失去氧生成铜，因此被还原。
An alternative early definition used hydrogen transfer: oxidation is the loss of hydrogen, while reduction is the gain of hydrogen. For instance, ethanol (CH₃CH₂OH) is oxidised to ethanal (CH₃CHO) by losing two hydrogen atoms.
另一种早期定义使用氢的转移：氧化是失去氢，还原则是获得氢。例如，乙醇 (CH₃CH₂OH) 失去两个氢原子被氧化成乙醛 (CH₃CHO)。
These definitions are limited in scope and cannot explain many reactions, such as the formation of sodium chloride from its elements, where no oxygen or hydrogen is transferred.
这些定义适用范围有限，无法解释许多反应，例如氯化钠由单质生成时，并没有氧或氢的转移。
2. The Modern Definition: Electron Transfer | 现代定义：电子转移
The modern definition of redox reactions is based on the transfer of electrons. Oxidation is the loss of electrons, and reduction is the gain of electrons. A helpful mnemonic is OIL RIG (Oxidation Is Loss, Reduction Is Gain).
氧化还原反应的现代定义基于电子转移。氧化是失去电子，还原是获得电子。一个有用的记忆法是 OIL RIG（氧化失电子，还原得电子）。
In the reaction between zinc and copper(II) ions, Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), each zinc atom loses two electrons to form Zn²⁺, so zinc is oxidised. Each Cu²⁺ ion gains two electrons to form Cu, so copper(II) ions are reduced.
在锌与铜(II)离子的反应 Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) 中，每个锌原子失去两个电子形成 Zn²⁺，锌被氧化。每个 Cu²⁺ 离子获得两个电子形成 Cu，铜(II)离子被还原。
This electron-transfer model can be applied to any redox reaction, including those in electrochemical cells, and it directly links to the concept of oxidation number.
这种电子转移模型可应用于任何氧化还原反应，包括电化学电池中的反应，并直接与氧化数的概念相联系。
3. What is an Oxidation Number? | 什么是氧化数？
An oxidation number (also called oxidation state) is the hypothetical charge an atom would have if all bonds in a compound were completely ionic. It is a bookkeeping tool used to track electron distribution in reactions.
氧化数（也称氧化态）是假设化合物中所有键均为完全离子键时，原子将带有的形式电荷。它是一种用于跟踪反应中电子分布的簿记工具。
Unlike ionic charges, which are real, oxidation numbers can be assigned even to atoms in covalent molecules, making it possible to decide whether a redox process has occurred without having to identify electron transfer directly.
与实际存在的离子电荷不同，氧化数甚至可以分配给共价分子中的原子，从而无需直接确认电子转移即可判断是否发生了氧化还原过程。
4. Rules for Assigning Oxidation Numbers | 分配氧化数的规则
Rule 1: The oxidation number of an atom in its elemental form is zero. For example, in O₂, N₂, Cl₂, S₈, and metallic zinc, each atom has an oxidation number of 0.
规则1：元素单质中原子的氧化数为零。例如 O₂, N₂, Cl₂, S₈ 和金属锌中，每个原子的氧化数均为 0。
Rule 2: For a monatomic ion, the oxidation number equals the charge on the ion. Thus, Na⁺ has an oxidation number of +1, Cl⁻ is –1, and Al³⁺ is +3.
规则2：对于单原子离子，氧化数等于离子所带的电荷。因此 Na⁺ 的氧化数为 +1，Cl⁻ 为 –1，Al³⁺ 为 +3。
Rule 3: In most compounds, oxygen has an oxidation number of –2. Exceptions include peroxides (e.g., H₂O₂, where oxygen is –1), superoxides (KO₂, where oxygen is –½), and when bonded to fluorine (e.g., OF₂, where oxygen is +2).
规则3：在大多数化合物中，氧的氧化数为 –2。例外包括过氧化物（如 H₂O₂ 中氧为 –1）、超氧化物（KO₂ 中氧为 –½）以及与氟成键时（如 OF₂ 中氧为 +2）。
Rule 4: Hydrogen usually has an oxidation number of +1 when bonded to non-metals. However, in metal hydrides such as NaH and CaH₂, hydrogen is present as H⁻ with an oxidation number of –1.
规则4：氢与非金属成键时，氧化数通常为 +1。但在金属氢化物如 NaH 和 CaH₂ 中，氢以 H⁻ 形式存在，氧化数为 –1。
Rule 5: Fluorine always has an oxidation number of –1 in its compounds. Other halogens generally have –1 unless bonded to oxygen or a more electronegative halogen, in which case they can have positive oxidation numbers.
规则5：氟在其化合物中始终为 –1 氧化数。其他卤素通常为 –1，除非与氧或电负性更大的卤素成键，此时它们可能具有正氧化数。
Rule 6: The sum of oxidation numbers in a neutral compound is zero; in a polyatomic ion, the sum equals the ion’s charge. For example, in SO₄²⁻, the sum of the oxidation numbers of one sulfur and four oxygen atoms must equal –2.
规则6：中性化合物中所有原子的氧化数总和为零；在多原子离子中，总和等于离子所带电荷。例如在 SO₄²⁻ 中，一个硫和四个氧的氧化数总和必须等于 –2。
Rule 7: In compounds, Group 1 metals (Li, Na, K, etc.) always have an oxidation number of +1, and Group 2 metals (Be, Mg, Ca, etc.) always have +2. Aluminium is always +3 in its compounds.
规则7：在化合物中，第1族金属（Li, Na, K 等）始终为 +1 氧化数，第2族金属（Be, Mg, Ca 等）始终为 +2。铝在其化合物中始终为 +3。
5. Applying Oxidation Number Rules: Examples | 应用氧化数规则：示例
Let us determine the oxidation number of manganese in KMnO₄. Potassium is +1, oxygen is –2 (Rule 3). Four oxygen atoms give a total of –8. The sum is +1 + x + 4(–2) = 0, so x = +7. Manganese is in the +7 oxidation state.
我们来计算 KMnO₄ 中锰的氧化数。钾为 +1，氧为 –2（规则3）。四个氧原子共提供 –8。总和为 +1 + x + 4(–2) = 0，因此 x = +7。锰处于 +7 氧化态。
For the dichromate ion, Cr₂O₇²⁻, oxygen is –2. Seven O atoms contribute –14. Let x be the oxidation number of each Cr. Then 2x + (–14) = –2, giving 2x = +12, so x = +6. Chromium is +6 in dichromate.
对于重铬酸根离子 Cr₂O₇²⁻，氧为 –2。七个氧原子贡献 –14。设每个 Cr 的氧化数为 x，则 2x + (–14) = –2，得出 2x = +12，x = +6。重铬酸根中铬为 +6。
In hydrogen peroxide, H₂O₂, hydrogen is +1 (Rule 4). Two H atoms give +2, so both oxygen atoms together must total –2, giving each oxygen an oxidation number of –1, not the usual –2.
在过氧化氢 H₂O₂ 中，氢为 +1（规则4）。两个 H 原子提供 +2，因此两个氧原子总和必须为 –2，每个氧的氧化数为 –1，而非通常的 –2。
6. Using Oxidation Numbers to Identify Redox | 利用氧化数判断氧化还原反应
A reaction is classified as a redox reaction if the oxidation numbers of some elements change from reactants to products. An increase in oxidation number indicates oxidation; a decrease indicates reduction.
如果某些元素的氧化数从反应物到生成物发生变化，则该反应属于氧化还原反应。氧化数升高表示氧化，氧化数降低表示还原。
Consider the reaction: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq). In Fe³⁺, the oxidation number is +3; in Fe²⁺ it is +2 — a decrease, so iron(III) is reduced. Iodide, I⁻, has an oxidation number of –1, which increases to 0 in I₂, so iodide is oxidised.
考虑反应：2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq)。在 Fe³⁺ 中，氧化数为 +3；在 Fe²⁺ 中为 +2，降低了，因此铁(III)被还原。碘离子 I⁻ 的氧化数为 –1，在 I₂ 中升至 0，因此碘离子被氧化。
If there is no change in oxidation numbers, the reaction is not a redox reaction. For example, in the neutralisation reaction NaOH + HCl → NaCl + H₂O, the oxidation numbers remain: Na (+1), O (–2), H (+1), Cl (–1) throughout. No electron transfer occurs.
如果氧化数没有变化，则该反应不是氧化还原反应。例如，在中和反应 NaOH + HCl → NaCl + H₂O 中，各元素氧化数始终为：Na (+1)、O (–2)、H (+1)、Cl (–1)，没有电子转移发生。
7. Oxidising and Reducing Agents | 氧化剂与还原剂
An oxidising agent (or oxidant) is a substance that causes another substance to be oxidised, while itself being reduced. It accepts electrons from the species it oxidises. Common oxidising agents include potassium manganate(VII), potassium dichromate(VI), and halogens.
氧化剂（或氧化试剂）是使其他物质被氧化而自身被还原的物质。它从被氧化的物种接受电子。常见氧化剂包括高锰酸钾、重铬酸钾和卤素。
A reducing agent (or reductant) is a substance that causes another substance to be reduced, and in doing so it is itself oxidised. It donates electrons. Examples are reactive metals like zinc and magnesium, hydrogen, and iron(II) salts.
还原剂（或还原试剂）是使其他物质被还原而自身被氧化的物质。它提供电子。例子有活泼金属如锌和镁、氢气以及铁(II)盐。
In the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), zinc is the reducing agent (it reduces Cu²⁺ to Cu and is oxidised to Zn²⁺), while Cu²⁺ acts as the oxidising agent (it oxidises Zn to Zn²⁺ and is reduced to Cu).
在反应 Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) 中，锌是还原剂（它将 Cu²⁺ 还原为 Cu，自身被氧化为 Zn²⁺），而 Cu²⁺ 充当氧化剂（它将 Zn 氧化为 Zn²⁺，自身被还原为 Cu）。
8. Balancing Redox Equations: Half-Reaction Method | 氧化还原方程式配平：半反应法
Many redox equations, especially those involving ions in aqueous solution, are balanced using the half-reaction method. The overall reaction is split into two half-equations: one for oxidation and one for reduction.
许多氧化还原方程式，尤其是涉及水溶液中离子的，都使用半反应法进行配平。总反应被拆分成两个半反应方程式：一个氧化半反应，一个还原半反应。
For each half-equation, balance all atoms except O and H first. Then add H₂O to balance oxygen atoms, add H⁺ to balance hydrogen atoms (for acidic conditions), and finally add electrons (e⁻) to balance the charge.
每个半反应方程式，首先平衡除氧和氢以外的所有原子。然后加入 H₂O 来平衡氧原子，加入 H⁺ 来平衡氢原子（酸性条件下），最后加入电子 (e⁻) 来平衡电荷。
Example: balancing the reaction between MnO₄⁻ and Fe²⁺ in acid. Reduction half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxidation half-equation: Fe²⁺ → Fe³⁺ + e⁻. Multiply the oxidation half-equation by 5, then add to cancel electrons, giving: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
示例：平衡酸性溶液中 MnO₄⁻ 与 Fe²⁺ 的反应。还原半反应：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。氧化半反应：Fe²⁺ → Fe³⁺ + e⁻。将氧化半反应乘以5，然后相加消去电子，得到：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。
For alkaline conditions, instead of H⁺, use H₂O and OH⁻ to balance hydrogen and charge as appropriate.
对于碱性条件，适当使用 H₂O 和 OH⁻ 而不是 H⁺ 来平衡氢和电荷。
9. Disproportionation Reactions | 歧化反应
A disproportionation reaction is a special type of redox reaction in which a single substance is simultaneously oxidised and reduced, with the same element undergoing both an increase and a decrease in oxidation number.
歧化反应是一种特殊类型的氧化还原反应，其中一种物质同时被氧化和还原，同一元素的氧化数既有升高也有降低。
A classic example is the reaction of chlorine with water: Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HClO(aq). Here, chlorine (0) is both reduced to chloride, Cl⁻ (–1), and oxidised to chlorate(I), ClO⁻ (where Cl is +1).
一个经典例子是氯气与水的反应：Cl₂(g) + H₂O(l) ⇌ HCl(aq) + HClO(aq)。这里，氯 (0) 既被还原为氯离子 Cl⁻ (–1)，又被氧化为次氯酸根 ClO⁻（其中 Cl 为 +1）。
Another important disproportionation is the decomposition of hydrogen peroxide: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). Oxygen in H₂O₂ has an oxidation number of –1; in H₂O it becomes –2 (reduction) and in O₂ it becomes 0 (oxidation).
另一个重要的歧化反应是过氧化氢的分解：2H₂O₂(aq) → 2H₂O(l) + O₂(g)。H₂O₂ 中氧的氧化数为 –1；在 H₂O 中变为 –2（还原），在 O₂ 中变为 0（氧化）。
Disproportionation is thermodynamically favourable for species that are unstable with respect to their higher and lower oxidation states, often occurring in the absence of other redox partners.
对于那些相对于其较高和较低氧化态不稳定的物种，歧化在热力学上是有利的，通常在没有其他氧化还原伙伴的情况下发生。
10. Redox Titrations | 氧化还原滴定
Redox titrations are widely used in quantitative analysis to determine the concentration of an unknown reducing or oxidising agent. A standard solution of known concentration is used to react with the analyte in a controlled manner.
氧化还原滴定广泛用于定量分析，以确定未知还原剂或氧化剂的浓度。使用已知浓度的标准溶液，以可控方式与分析物反应。
Potassium manganate(VII), KMnO₄, is a common titrant because it acts as its own indicator — the intense purple colour of MnO₄⁻ disappears as it is reduced to nearly colourless Mn²⁺ in acidic solution. The endpoint is signalled by the first permanent pink colour.
高锰酸钾 KMnO₄ 是常见的滴定剂，因为它本身可作为指示剂——MnO₄⁻ 的浓紫色在酸性溶液中被还原为几乎无色的 Mn²⁺ 时消失。终点由第一次持久的粉红色指示。
Iodine-thiosulfate titrations are also important. Iodine (I₂) can be generated by an oxidising agent and then titrated with sodium thiosulfate (Na₂S₂O₃), where iodine is reduced to I⁻ and thiosulfate is oxidised to tetrathionate (S₄O₆²⁻).
碘-硫代硫酸盐滴定也很重要。碘 (I₂) 可由氧化剂产生，然后用硫代硫酸钠 (Na₂S₂O₃) 滴定，碘被还原为 I⁻，硫代硫酸根被氧化为连四硫酸根 (S₄O₆²⁻)。
Understanding oxidation numbers helps deduce the stoichiometry of these titrations without memorising full equations.
理解氧化数有助于推导这些滴定的化学计量关系，而无需记忆完整方程式。
11. Applications of Redox and Oxidation Numbers | 氧化还原与氧化数的应用
Redox chemistry underpins energy production in batteries and fuel cells. In a lithium-ion battery, lithium atoms are oxidised to Li⁺ at the anode, releasing electrons that travel through an external circuit, while at the cathode a transition metal oxide is reduced.
氧化还原化学是电池和燃料电池中能量生产的基础。在锂离子电池中，锂原子在负极被氧化为 Li⁺，释放电子流经外电路，而在正极，过渡金属氧化物被还原。
In extractive metallurgy, reduction processes are used to obtain metals from their ores. For instance, iron is extracted from iron(III) oxide in a blast furnace using carbon monoxide as the reducing agent: Fe₂O₃ + 3CO → 2Fe + 3CO₂.
在提取冶金中，还原过程用于从矿石中获取金属。例如，铁在高炉中用一氧化碳作为还原剂从氧化铁中提取：Fe₂O₃ + 3CO → 2Fe + 3CO₂。
Biological systems rely heavily on redox reactions, such as the electron transport chain in respiration, where NADH and FADH₂ are oxidised, driving the production of ATP.
生物系统严重依赖氧化还原反应，例如呼吸作用中的电子传递链，其中 NADH 和 FADH₂ 被氧化，驱动 ATP 的生成。
Assigning oxidation numbers is also crucial in environmental chemistry, for example, to understand the speciation of nitrogen in the nitrogen cycle (NH₃: –3, NO₂⁻: +3, NO₃⁻: +5).
分配氧化数在环境化学中也至关重要，例如，理解氮循环中氮的形态（NH₃：–3，NO₂⁻：+3，NO₃⁻：+5）。
12. Summary and Key Points | 总结与要点
Redox reactions involve electron transfer; oxidation is loss of electrons (OIL), reduction is gain of electrons (RIG). Oxidation numbers are assigned according to a set of rules and help track electron flow.
氧化还原反应涉及电子转移；氧化是失去电子 (OIL)，还原是获得电子 (RIG)。氧化数根据一系列规则分配，有助于跟踪电子流向。
An increase in oxidation number indicates oxidation, while a decrease indicates reduction. An oxidising agent is reduced, and a reducing agent is oxidised. Balancing redox equations uses half-reactions, and disproportionation involves a single element being both oxidised and reduced.
氧化数升高表示氧化，降低表示还原。氧化剂本身被还原，还原剂本身被氧化。氧化还原方程式的配平使用半反应法，而歧化反应涉及同一元素同时被氧化和还原。
Memorising the priority rules, especially for O (–2 except peroxides), H (+1 except hydrides), and recognising common oxidising agents (MnO₄⁻, Cr₂O₇²⁻) will help you solve exam problems with confidence.
记住优先规则，特别是氧（除过氧化物外 –2）、氢（除氢化物外 +1），并识别常见氧化剂（MnO₄⁻, Cr₂O₇²⁻），将有助你自信地解决考试问题。
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