📚 A-Level Chemistry: Using Oxidation Numbers to Determine Redox Reactions | A-Level 化学：利用氧化数判断氧化还原
Redox reactions are among the most important processes in chemistry, from respiration and photosynthesis to batteries and industrial synthesis. A reliable way to identify redox reactions is by tracking oxidation numbers, which allow us to see which atoms lose or gain electrons even when no formal electron transfer is obvious.
氧化还原反应是化学中最重要的过程之一，从呼吸作用和光合作用到电池与工业合成，无处不在。判断氧化还原反应的一个可靠方法就是追踪氧化数，通过氧化数的变化，我们能够看出哪些原子失去或获得电子，即使电子转移并不一目了然。
1. Definition and Rules of Oxidation Numbers | 氧化数的定义与规则
An oxidation number is a bookkeeping value assigned to an atom in a chemical species. It represents the hypothetical charge the atom would have if all bonds were treated as fully ionic, with electrons assigned to the more electronegative atom.
氧化数是分配给化学物种中某个原子的记账数值，它表示若所有化学键都视为完全离子键、电子归属于电负性较大的原子时，该原子所应具有的“假想电荷”。
The following rules are essential for A-Level CIE Chemistry. First, the oxidation number of an element in its free state is zero, for example Na, O₂, and P₄ all have oxidation number 0. Second, for a monatomic ion, the oxidation number equals the charge on the ion, so Mg²⁺ is +2 and Cl⁻ is −1.
以下规则对 CIE A-Level 化学至关重要。第一，游离态元素原子的氧化数为零，例如 Na、O₂、P₄ 的氧化数均为 0。第二，对于单原子离子，氧化数等于离子所带电荷，因此 Mg²⁺ 为 +2，Cl⁻ 为 −1。
Third, in most compounds, hydrogen has oxidation number +1, except in metal hydrides such as NaH where it is −1. Fourth, oxygen usually has oxidation number −2, except in peroxides where it is −1, and in OF₂ where it is +2. Fifth, the sum of oxidation numbers in a neutral compound is zero, and in a polyatomic ion it equals the ion charge.
第三，在大多数化合物中，氢的氧化数为 +1，但在金属氢化物如 NaH 中为 −1。第四，氧的氧化数通常为 −2，但在过氧化物中为 −1，在 OF₂ 中为 +2。第五，中性化合物中所有原子的氧化数之和为零，在多原子离子中则等于离子电荷。
2. Oxidation Number vs Formal Charge | 氧化数与形式电荷的区别
Oxidation numbers are often confused with formal charges, but they are fundamentally different. Formal charges are derived from a Lewis structure by assuming that bonding electrons are shared equally between bonded atoms, even if their electronegativities differ.
氧化数常与形式电荷混淆，但二者有本质区别。形式电荷基于路易斯结构，假设成键电子在成键原子之间均等共享，无论它们的电负性是否不同。
In contrast, oxidation numbers assign both electrons in a bond to the more electronegative atom, making them a measure of ionic character. For example, in carbon dioxide, the formal charge on each oxygen in one resonance form is 0, but the oxidation number of oxygen is −2 and that of carbon is +4.
相比之下，氧化数将共价键中的两个电子都分配给电负性较大的原子，因此它反映的是化学键的离子性程度。例如在二氧化碳中，某一共振结构中氧的形式电荷为 0，但氧的氧化数为 −2，碳的氧化数为 +4。
Oxidation numbers are not physical charges and cannot be measured directly; they are merely a useful accounting tool. However, they are incredibly powerful for balancing redox equations and identifying electron transfer.
氧化数并不是真实存在的电荷，无法直接测量，它们只是一种方便的记账工具。不过，它们在配平氧化还原方程式和判断电子转移方面极为有用。
3. Calculating Oxidation Numbers | 计算氧化数
To calculate the oxidation number of an unknown atom, set up an algebraic equation using the known rules. For example, determine the oxidation number of sulfur in SO₄²⁻. Since each oxygen is −2 and there are four oxygens, the total from oxygen is −8. The sum must equal the ion charge, −2, so sulfur must be +6.
要计算某一未知原子的氧化数，可以利用已知规则建立代数方程。例如，求 SO₄²⁻ 中硫的氧化数。每个氧为 −2，四个氧合计 −8。氧化数总和必须等于离子电荷 −2，因此硫的氧化数为 +6。
Let x = oxidation number of S, then x + 4(−2) = −2, so x = +6
Now consider the dichromate ion, Cr₂O₇²⁻. Each oxygen is −2, giving a total of −14 from seven oxygens. The total charge is −2, so the two chromium atoms must contribute +12, meaning each chromium has oxidation number +6.
再来看重铬酸根离子 Cr₂O₇²⁻。每个氧为 −2，七个氧合计 −14。离子总电荷为 −2，因此两个铬原子合计贡献 +12，即每个铬的氧化数为 +6。
2(x) + 7(−2) = −2, so 2x − 14 = −2, hence x = +6
Always check: in a neutral compound, the sum of all oxidation numbers must be zero; in an ion, it must equal the charge on the ion.
务必检查：对于中性化合物，所有氧化数之和必须为零；对于离子，则必须等于离子所带电荷。
4. Identifying Oxidation and Reduction | 识别氧化与还原
Oxidation is defined as an increase in oxidation number, which corresponds to a loss of electrons. Reduction is a decrease in oxidation number, corresponding to a gain of electrons. These two processes always occur together, hence the term redox.
氧化定义为氧化数的升高，对应失去电子；还原定义为氧化数的降低，对应获得电子。这两个过程总是同时发生，因此称为氧化还原。
Consider the reaction between iron(III) and iodide ions: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. The oxidation number of iron decreases from +3 to +2, so iron is reduced. The oxidation number of iodine increases from −1 to 0, so iodine is oxidised.
以铁(III)与碘离子的反应为例：2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂。铁的氧化数从 +3 降低到 +2，因此铁被还原；碘的氧化数从 −1 升高到 0，因此碘被氧化。
When identifying redox reactions, examine every element in the reaction. If no element changes oxidation number, the reaction is not redox. Many acid–base reactions and precipitation reactions involve no oxidation number change, so they are not classified as redox reactions.
判断氧化还原反应时，应检查反应中的每一种元素。如果没有任何元素的氧化数发生变化，该反应就不是氧化还原反应。许多酸碱反应和沉淀反应不涉及氧化数变化，因此不属于氧化还原反应。
5. Oxidising and Reducing Agents | 氧化剂与还原剂
An oxidising agent is a substance that causes oxidation by removing electrons from another species; it is itself reduced. A reducing agent is a substance that causes reduction by donating electrons; it is itself oxidised.
氧化剂是通过从其他物质夺取电子而引起氧化的物质，自身被还原；还原剂是通过提供电子而引起还原的物质，自身被氧化。
For example, in the reaction of potassium manganate(VII), KMnO₄, with iron(II) ions in acidic solution, the Mn atom is reduced from +7 to +2, so KMnO₄ is the oxidising agent. The Fe²⁺ ion is oxidised from +2 to +3, so Fe²⁺ is the reducing agent.
例如，在酸性溶液中高锰酸钾 KMnO₄ 与铁(II)离子的反应中，Mn 原子从 +7 被还原到 +2，因此 KMnO₄ 是氧化剂；Fe²⁺ 从 +2 被氧化到 +3，因此 Fe²⁺ 是还原剂。
Common oxidising agents include acidified KMnO₄, K₂Cr₂O₇, and acidified hydrogen peroxide. Common reducing agents include metals, iodide ions, and sulfur dioxide. In each case, the oxidising agent contains an element whose oxidation number decreases.
常见的氧化剂包括酸化的 KMnO₄、K₂Cr₂O₇ 以及酸化过氧化氢。常见的还原剂包括金属、碘离子和二氧化硫。在每一种情况中，氧化剂都含有氧化数降低的元素。
6. Disproportionation | 歧化反应
Disproportionation is a special type of redox reaction in which the same element in one oxidation state is simultaneously oxidised and reduced into two different products. This means the element must have oxidation numbers both above and below its original value.
歧化反应是一种特殊的氧化还原反应，同一元素从一种氧化态同时被氧化和被还原，生成两种不同产物。这意味着该元素在产物中的氧化数必须既有高于原值又有低于原值的情况。
A classic CIE example is the reaction of chlorine with cold dilute sodium hydroxide: Cl₂ + 2NaOH → NaCl + NaClO + H₂O. The chlorine in Cl₂ has oxidation number 0. In NaCl, chlorine is −1, so it is reduced. In NaClO, chlorine is +1, so it is oxidised.
CIE 的经典例子是氯气与冷稀氢氧化钠的反应：Cl₂ + 2NaOH → NaCl + NaClO + H₂O。Cl₂ 中氯的氧化数为 0。在 NaCl 中氯为 −1，因此被还原；在 NaClO 中氯为 +1，因此被氧化。
Other examples include the decomposition of hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂. Oxygen changes from −1 in H₂O₂ to −2 in H₂O and 0 in O₂, so H₂O₂ disproportionates. Recognising disproportionation is a common question in A-Level examinations.
其他例子包括过氧化氢的分解：2H₂O₂ → 2H₂O + O₂。氧从 H₂O₂ 中的 −1 变为 H₂O 中的 −2 和 O₂ 中的 0，因此 H₂O₂ 发生歧化。识别歧化反应是 A-Level 考试中的常见题型。
7. Special Case: Peroxides | 特殊情况：过氧化物
Peroxides provide a classic trap in oxidation number calculations. In hydrogen peroxide, H₂O₂, the oxidation number of oxygen is not −2 but −1, because the two oxygen atoms are bonded to each other and share the negative charge equally.
过氧化物是氧化数计算中的经典陷阱。在过氧化氢 H₂O₂ 中，氧的氧化数不是 −2 而是 −1，因为两个氧原子彼此成键，负电荷由它们均等共享。
When H₂O₂ acts as an oxidising agent, oxygen is reduced from −1 to −2, producing water. When H₂O₂ acts as a reducing agent, oxygen is oxidised from −1 to 0, producing oxygen gas. This dual behaviour makes H₂O₂ both an oxidising and a reducing agent.
当 H₂O₂ 作为氧化剂时，氧从 −1 被还原为 −2，生成水；当 H₂O₂ 作为还原剂时，氧从 −1 被氧化为 0，生成氧气。这种双重行为使 H₂O₂ 既是氧化剂又是还原剂。
Similarly, compounds such as Na₂O₂ and BaO₂ contain the peroxide ion, O₂²⁻, in which each oxygen atom has oxidation number −1. In contrast, the oxide ion O²⁻ has oxidation number −2, and the superoxide ion, O₂⁻, has oxygen in oxidation state −½.
类似地，Na₂O₂ 和 BaO₂ 等化合物含有过氧根离子 O₂²⁻，其中每个氧原子的氧化数为 −1。相比之下，氧化物离子 O²⁻ 的氧化数为 −2，超氧根离子 O₂⁻ 中氧的氧化态为 −½。
8. Oxidation Numbers in Organic Compounds | 有机物中的氧化数
Oxidation numbers are not limited to inorganic compounds; they can also be applied to organic molecules to determine whether a reaction is oxidation or reduction. In organic chemistry, oxidation often involves the gain of oxygen or loss of hydrogen, but calculating oxidation numbers gives a more rigorous definition.
氧化数不仅适用于无机化合物，也可以应用于有机分子来判断反应是氧化还是还原。在有机化学中，氧化常表现为得氧或失氢，但通过计算氧化数可以给出更严格的定义。
For example, in the oxidation of ethanol, CH₃CH₂OH, to ethanal, CH₃CHO, the carbon of the alcohol group changes from −1 to +1. This is an increase in oxidation number, so the conversion is an oxidation. Further oxidation to ethanoic acid, CH₃COOH, raises the oxidation number of that carbon to +3.
例如，在乙醇 CH₃CH₂OH 氧化为乙醛 CH₃CHO 的反应中，羟基所连碳的氧化数从 −1 变为 +1。氧化数升高，因此该转化是氧化。继续氧化为乙酸 CH₃COOH 时，该碳的氧化数升至 +3。
Oxidation numbers can also be assigned to each carbon atom using the rule that hydrogen contributes +1 and bonds to more electronegative atoms such as oxygen or halogens contribute negatively to carbon. This approach is particularly useful for predicting whether a reaction involves electron transfer.
计算有机物中各碳原子的氧化数时，可以规定每个 H 对碳贡献 +1，而碳与更电负性原子（如氧或卤素）成键时，碳获得负的贡献。这种方法对于判断反应是否涉及电子转移特别有用。
9. Oxidation Number Changes and Half-Equations | 氧化数变化与半反应
Oxidation numbers provide a systematic method for balancing redox equations. First, write the unbalanced equation and assign oxidation numbers. Identify the element that is oxidised and the element that is reduced, then calculate the total change in oxidation number for each atom.
氧化数为配平氧化还原方程式提供了系统方法。首先写出未配平的方程式并给各原子指定氧化数。找出被氧化的元素和被还原的元素，然后计算每个原子的氧化数总变化量。
When balancing redox equations, the total increase in oxidation number must equal the total decrease, because the number of electrons lost equals the number gained. This allows us to determine the stoichiometric coefficients required for the oxidising and reducing agents.
配平氧化还原方程式时，氧化数总升高量必须等于总降低量，因为失去的电子数等于获得的电子数。由此可以确定氧化剂和还原剂所需的化学计量系数。
For ionic half-equations, the ion-electron method is often preferred. For example, in acidic solution, the reduction of MnO₄⁻ to Mn²⁺ is written as MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The manganese oxidation number changes from +7 to +2, so five electrons are needed.
对于离子半反应，通常优先使用离子–电子法。例如在酸性溶液中，MnO₄⁻ 还原为 Mn²⁺ 的半反应写作 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。锰的氧化数从 +7 变为 +2，因此需要五个电子。
10. Common Pitfalls and Exam Tips | 常见误区与应试技巧
One common mistake is forgetting that the oxidation number of an element in its standard state is zero. Another is incorrectly treating the charge of a polyatomic ion as the overall oxidation number, rather than assigning oxidation numbers to individual atoms.
一个常见错误是忘记元素的单质标准态氧化数为零。另一个错误是将多原子离子的整体电荷误认为总氧化数，而不是分别计算每个原子的氧化数。
A further pitfall is assuming that oxygen always has oxidation number −2, even in peroxides and superoxides. Always check the bonding environment before assigning values. Also remember that hydrogen in metal hydrides has oxidation number −1, not +1.
另一个陷阱是假设氧的氧化数总是 −2，而忽略过氧化物和超氧化物中的特殊情况。在确定数值前务必检查键合环境。还要记住，金属氢化物中氢的氧化数为 −1 而不是 +1。
In examinations, always show your working when calculating oxidation numbers. State the rule you are using, write the equation, and check that the sum is consistent with the overall charge. For redox questions, clearly identify which species is oxidised and which is reduced.
在考试中，计算氧化数一定要写出计算过程。说明所用规则，写出方程，并检查总和是否与整体电荷一致。对于氧化还原问题，要明确指出哪种物质被氧化、哪种物质被还原。
11. Practice Questions | 练习与自测
Question 1: Determine the oxidation number of nitrogen in NH₄⁺, NO₂⁻, and HNO₃. Question 2: Which species is the oxidising agent in the reaction 2Br⁻ + Cl₂ → Br₂ + 2Cl⁻? Question 3: Show that the decomposition of thiosulfate in acid, S₂O₃²⁻ → S + SO₃²⁻, is a disproportionation reaction.
练习一：计算 NH₄⁺、NO₂⁻ 和 HNO₃ 中氮的氧化数。练习二：在反应 2Br⁻ + Cl₂ → Br₂ + 2Cl⁻ 中，哪种物质是氧化剂？练习三：证明硫代硫酸根在酸中分解的反应 S₂O₃²⁻ → S + SO₃²⁻ 属于歧化反应。
Answers: In NH₄⁺, nitrogen is −3. In NO₂⁻, nitrogen is +3. In HNO₃, nitrogen is +5. Chlorine is reduced from 0 to −1, so Cl₂ is the oxidising agent; bromide is oxidised from −1 to 0. In S₂O₃²⁻, the average oxidation number of sulfur is +2, but in elemental S it is 0 and in SO₃²⁻ it is +4, so sulfur is both reduced and oxidised.
参考答案：NH₄⁺ 中氮为 −3；NO₂⁻ 中氮为 +3；HNO₃ 中氮为 +5。氯从 0 被还原到 −1，因此 Cl₂ 是氧化剂；溴从 −1 被氧化到 0。S₂O₃²⁻ 中硫的平均氧化数为 +2，但单质 S 中为 0，SO₃²⁻ 中为 +4，因此硫既被还原又被氧化，属于歧化反应。
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Cancel reply