📚 A-Level Chemistry: Balancing Equations Using Oxidation Numbers | A-Level化学：氧化数法配平化学方程式
Balancing chemical equations is a core skill in CIE A-Level Chemistry. The oxidation-number method is especially powerful for redox reactions, because it uses the change in oxidation numbers to determine coefficients directly. This article explains the rules and works through examples in acidic, alkaline and disproportionation reactions.
配平化学方程式是 CIE A-Level 化学的核心技能。氧化数法在氧化还原反应中特别有效，因为它直接利用氧化数的变化来确定化学计量系数。本文将讲解规则，并通过酸性介质、碱性介质和歧化反应的例题进行演练。
1. Why Use Oxidation Numbers? | 为什么要用氧化数？
Oxidation numbers are a bookkeeping tool for electrons. Many redox equations cannot be balanced easily by trial and error, especially when several elements change oxidation state. The oxidation-number method links coefficients to electron transfer, making the process systematic.
氧化数是用于“记账”电子的工具。许多氧化还原方程式难以通过试错法直接配平，尤其是当多种元素同时改变氧化态时。氧化数法将化学计量系数与电子转移联系起来，使配平过程更系统化。
In any redox reaction, the total increase in oxidation number equals the total decrease in oxidation number. This conservation statement is the foundation of the method.
在任何氧化还原反应中，氧化数的总升高值一定等于总降低值。这一守恒表述是氧化数法的基础。
2. Rules for Assigning Oxidation Numbers | 氧化数的赋予规则
Before balancing, you must be able to assign oxidation numbers. The following rules are used in CIE examinations.
在配平之前，你必须能够确定氧化数。下面是 CIE 考试中使用的规则。
| Species / Rule | Oxidation Number |
|---|---|
| Free element, e.g. N₂, Cl₂, Cu | 0 |
| Hydrogen in covalent compounds | +1 (except metal hydrides, e.g. NaH, where it is −1) |
| Oxygen | −2 (except peroxides, e.g. H₂O₂, where it is −1; in OF₂ it is +2) |
| Group I metals | +1 |
| Group II metals | +2 |
| Halogens | usually −1, except when combined with oxygen or more electronegative halogens |
| Simple ion | equal to the ionic charge, e.g. Fe³⁺ = +3 |
| Polyatomic ion | sum of oxidation numbers = ionic charge, e.g. SO₄²⁻ equals −2 |
Remember that oxidation numbers are not real charges; they are a conventional method of tracking electrons.
请记住，氧化数并不是真实的电荷，而是一种跟踪电子转移的人为规定方法。
3. Identifying Redox from Oxidation-Number Changes | 通过氧化数变化判断氧化还原
A redox reaction is identified by a change in oxidation number. Consider the reaction between zinc and copper(II) sulfate:
氧化还原反应的特征是氧化数发生变化。以锌与硫酸铜的反应为例：
Zn + CuSO₄ → ZnSO₄ + Cu
Zinc changes from 0 to +2, so it is oxidised. Copper changes from +2 to 0, so it is reduced.
锌的氧化数从 0 变为 +2，因此锌被氧化。铜的氧化数从 +2 变为 0，因此铜被还原。
If no element changes oxidation number, the reaction is not redox. In CIE A-Level questions, you should always annotate oxidation numbers before attempting to balance.
如果没有任何元素改变氧化数，则该反应不是氧化还原反应。在 CIE A-Level 试题中，你应当在配平前先标出氧化数。
4. Step-by-Step Method | 分步配平步骤
The oxidation-number method can be summarised in five steps.
氧化数法可以概括为五个步骤。
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Step 1: Assign oxidation numbers to every atom in the unbalanced equation.
第一步：在不配平的方程式中，标出每个原子的氧化数。
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Step 2: Identify the atoms that change oxidation number and record the change per atom.
第二步：找出氧化数发生变化的原子，并记录每个原子的氧化数变化量。
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Step 3: Multiply the species by whole-number coefficients so that the total increase equals the total decrease.
第三步：在各物质前乘上整数系数，使总升高值等于总降低值。
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Step 4: Balance the remaining atoms by inspection, using H₂O, H⁺ or OH⁻ for acidic or alkaline media.
第四步：通过观察配平其余原子，在酸性或碱性介质中使用 H₂O、H⁺ 或 OH⁻。
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Step 5: Check that atoms and charge are balanced.
第五步：检查原子总数与总电荷是否平衡。
Steps 1 to 3 are the core of the method; steps 4 and 5 are essential for ionic equations.
第 1 至第 3 步是该方法的核心；第 4 和第 5 步对离子方程式不可或缺。
5. Worked Example 1: Copper and Nitric Acid | 例题一：铜与硝酸的反应
Balance the equation for the reaction between copper and concentrated nitric acid.
请配平浓硝酸与铜反应的方程式。
Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O
Step 1: assign oxidation numbers.
第一步：标出氧化数。
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Cu: 0 → +2, so increase = 2 per Cu atom.
Cu：0 → +2，所以每个 Cu 原子升高 2。
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N in HNO₃: +5 → A part stays at +5 in Cu(NO₃)₂; the other part becomes +4 in NO₂. The relevant decrease is 5 → 4, so decrease = 1 per N atom.
HNO₃ 中 N 为 +5；一部分在 Cu(NO₃)₂ 中仍为 +5，另一部分在 NO₂ 中变为 +4。相关的降低是 5 → 4，所以每个 N 原子降低 1。
Step 2: make increase = decrease. Cu × 1 gives +2; NO₂ × 2 gives −2.
第二步：使升高等于降低。Cu × 1 提供 +2；NO₂ × 2 提供 −2。
Cu + HNO₃ → Cu(NO₃)₂ + 2NO₂ + H₂O
Step 3: balance remaining atoms. The nitrate group in Cu(NO₃)₂ supplies 2 N atoms; the 2NO₂ supplies another 2 N atoms, so 4 HNO₃ are needed. This gives 4 H atoms, so 2 H₂O form.
第三步：配平其余原子。Cu(NO₃)₂ 中的硝酸根提供 2 个 N；2NO₂ 再提供 2 个 N，因此需要 4 个 HNO₃。这样共有 4 个 H，所以生成 2 个 H₂O。
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Check: Cu: 1 = 1; H: 4 = 4; N: 4 = 2 + 2; O: 12 = 6 + 4 + 2. The equation is balanced.
检查：Cu：1 = 1；H：4 = 4；N：4 = 2 + 2；O：12 = 6 + 4 + 2。方程式已配平。
6. Worked Example 2: Acidic Medium | 例题二：酸性介质中的离子方程式
Balance the redox reaction between manganate(VII) ions and iron(II) ions in acid.
请配平在酸性介质中高锰酸根离子与亚铁离子之间的氧化还原反应。
MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
Assign oxidation numbers:
标出氧化数：
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Mn: +7 → +2, decrease = 5 per Mn.
Mn：+7 → +2，每个 Mn 降低 5。
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Fe: +2 → +3, increase = 1 per Fe.
Fe：+2 → +3，每个 Fe 升高 1。
To balance 5 electrons and 1 electron, multiply Fe²⁺ by 5:
为了平衡 5 个电子与 1 个电子，Fe²⁺ 应乘以 5：
MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺
Now balance charge. Left: −1 + 10 = +9. Right: +2 + 15 = +17. Add 8H⁺ to the left, and then 4H₂O to the right for oxygen balance.
接着配平电荷。左边：−1 + 10 = +9；右边：+2 + 15 = +17。在左边加入 8 个 H⁺，再由氧原子守恒在右边加入 4 个 H₂O。
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Check: Mn, Fe: 1 and 5 on both sides; O: 4 = 4; H: 8 = 8; charge: +17 = +17.
检查：Mn 与 Fe：两边分别为 1 和 5；O：4 = 4；H：8 = 8；电荷：+17 = +17。
7. Worked Example 3: Alkaline Medium | 例题三：碱性介质中的离子方程式
Balancing in alkaline solution follows the same oxidation-number steps, but charge and oxygen are balanced with OH⁻ and H₂O instead of H⁺.
碱性溶液中的配平使用相同的氧化数步骤，但电荷与氧原子需用 OH⁻ 和 H₂O 来平衡，而不是 H⁺。
MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻
Oxidation-number changes:
氧化数变化：
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Mn: +7 → +4, decrease = 3.
Mn：+7 → +4，降低 3。
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Br: −1 → +5, increase = 6.
Br：−1 → +5，升高 6。
To make increase = decrease, use 2 MnO₄⁻ and 1 Br⁻:
为使升高等于降低，用 2 个 MnO₄⁻ 和 1 个 Br⁻：
2MnO₄⁻ + Br⁻ → 2MnO₂ + BrO₃⁻
Balance oxygen first. Left has 8 O; right has 4 + 3 = 7 O. Add 1 H₂O to the left, which then adds 2 H; add 2 OH⁻ to the right.
先配平氧原子。左边有 8 个 O；右边有 4 + 3 = 7 个 O。在左边加 1 个 H₂O，这样会增加 2 个 H；再在右边加 2 个 OH⁻。
2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻
Check: O: 8 + 1 = 4 + 3 + 2; H: 2 = 2; charge: −2 −1 = −1 −2 = −3. Balanced.
检查：O：8 + 1 = 4 + 3 + 2；H：2 = 2；电荷：−2 −1 = −1 −2 = −3。已配平。
8. Worked Example 4: Disproportionation | 例题四：歧化反应
Disproportionation is a reaction in which the same element is both oxidised and reduced. Chlorine in alkali is a classic CIE example.
歧化反应是指同一元素既被氧化又被还原的反应。氯气在碱中的歧化是 CIE 的经典例子。
Cl₂ + OH⁻ → Cl⁻ + ClO⁻ + H₂O
In Cl₂, each Cl has oxidation number 0. In Cl⁻ it is −1 and in ClO⁻ it is +1.
在 Cl₂ 中，每个 Cl 的氧化数为 0。在 Cl⁻ 中为 −1，在 ClO⁻ 中为 +1。
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One Cl atom is reduced: 0 → −1, decrease = 1.
一个 Cl 原子被还原：0 → −1，降低 1。
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One Cl atom is oxidised: 0 → +1, increase = 1.
另一个 Cl 原子被氧化：0 → +1，升高 1。
Because the increase and decrease are already equal, Cl₂ is used once to give one Cl⁻ and one ClO⁻. Balance charge and atoms with 2 OH⁻:
因为升高与降低已经相等，所以 1 个 Cl₂ 生成 1 个 Cl⁻ 和 1 个 ClO⁻。用 2 个 OH⁻ 来平衡电荷与原子：
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
Check: Cl: 2 = 1 + 1; O: 2 = 1 + 1; H: 2 = 2; charge: −2 = −1 −1. Balanced.
检查：Cl：2 = 1 + 1；O：2 = 1 + 1；H：2 = 2；电荷：−2 = −1 −1。已配平。
9. Checking Your Answer and Common Pitfalls | 检查答案与常见错误
After balancing, always verify both atom balance and charge balance. For ionic equations, a missing H⁺ or OH⁻ often appears as a charge error.
配平后，务必同时检查原子守恒和电荷守恒。对于离子方程式，漏写 H⁺ 或 OH⁻ 通常表现为电荷不平衡。
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Pitfall 1: Forgetting to assign oxidation numbers before changing coefficients. This leads to random trial and error.
常见错误一：没有先标氧化数就改动系数，导致随意试错。
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Pitfall 2: Counting N or Cl atoms
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