Balancing Chemical Equations Using Oxidation Numbers | 用氧化数法配平化学方程式

📚 Balancing Chemical Equations Using Oxidation Numbers | 用氧化数法配平化学方程式

At Cambridge A-Level, some redox equations are too complex to balance by simple trial and error. The oxidation-number method gives a reliable route: assign oxidation numbers, identify which atoms are oxidised and reduced, equalise the total electrons lost and gained, and then add H₂O, H⁺ or OH⁻ to complete the balancing.

在剑桥 A-Level 化学中，有些氧化还原方程式过于复杂，难以用试凑法配平。氧化数法提供了一条可靠路径：先标出氧化数，判断哪些原子被氧化、哪些被还原，使电子得失总数相等，然后补充 H₂O、H⁺ 或 OH⁻ 完成配平。

1. What Are Oxidation Numbers? | 什么是氧化数？

An oxidation number is the charge an atom would have if all shared electrons in a covalent bond were assigned to the more electronegative atom. It is a bookkeeping tool used to track electron transfer, not necessarily a real ionic charge.

氧化数是假设共价键中的共用电子全部归属于电负性较大的原子时，该原子所带的电荷。它是一种用于追踪电子转移的记账工具，不一定是真实离子电荷。

Oxidation-number rule	氧化数规则
Free elements have oxidation number 0, e.g. O₂, Fe, Cl₂.	游离态单质的氧化数为 0，如 O₂、Fe、Cl₂。
A simple ion has an oxidation number equal to its charge, e.g. Fe³⁺ = +3, Cl⁻ = -1.	简单离子的氧化数等于其电荷，如 Fe³⁺ 为 +3，Cl⁻ 为 -1。
Hydrogen is usually +1, but in metal hydrides such as NaH it is -1.	氢通常为 +1，但在 NaH 等金属氢化物中为 -1。
Oxygen is usually -2, but in peroxides it is -1 and in OF₂ it is +2.	氧通常为 -2，但在过氧化物中为 -1，在 OF₂ 中为 +2。
Fluorine is always -1 in compounds.	氟在化合物中总是 -1。
The sum of oxidation numbers equals the overall charge; it is zero for a neutral molecule.	氧化数总和等于整体电荷；中性分子为零。

2. Oxidation and Reduction in Terms of Oxidation Numbers | 用氧化数判断氧化与还原

Oxidation is an increase in oxidation number, and reduction is a decrease in oxidation number. If the oxidation number becomes more positive, the atom has lost electrons; if it becomes more negative, the atom has gained electrons.

氧化是氧化数升高，还原是氧化数降低。若氧化数变得更正，原子失去电子；若变得更负，原子得到电子。

For example, Fe²⁺ changes from +2 to +3 when it is oxidised to Fe³⁺. In the same reaction, an oxidising agent accepts electrons and is itself reduced, while a reducing agent donates electrons and is itself oxidised.

例如，Fe²⁺ 从 +2 变为 +3 时被氧化为 Fe³⁺。在同一反应中，氧化剂接受电子而本身被还原，还原剂提供电子而本身被氧化。

3. The Core Principle: Electron Balance | 核心原则：电子守恒

The oxidation-number method is based on one key idea: the total increase in oxidation number must equal the total decrease in oxidation number. This reflects conservation of electrons, because electrons lost by the reducing agent must be gained by the oxidising agent.

氧化数法基于一个核心思想：氧化数的总升高值必须等于总降低值。这体现了电子守恒，因为还原剂失去的电子必须由氧化剂获得。

Once the electron changes are equalised, the remaining atoms, hydrogen, oxygen and charge can be balanced using H₂O and H⁺ in acidic solution, or H₂O and OH⁻ in basic solution.

一旦电子变化相等，剩余的原子、氢、氧和电荷就可以在酸性溶液中使用 H₂O 和 H⁺，或在碱性溶液中使用 H₂O 和 OH⁻ 来配平。

4. Step-by-Step Method for Acidic or Neutral Solutions | 酸性或中性溶液中的分步配平法

Step 1: Assign oxidation numbers to all atoms in the skeleton equation.

第1步：给骨架方程式中的所有原子标出氧化数。

Step 2: Identify which atoms are oxidised and which are reduced by comparing oxidation numbers on both sides.

第2步：通过比较两侧氧化数，确定哪些原子被氧化、哪些被还原。

Step 3: Calculate the change in oxidation number per atom.

第3步：计算每个原子的氧化数变化值。

Step 4: Multiply the change per atom by the number of atoms of that element in one formula unit to obtain the total electron change per species.

第4步：将每个原子的变化值乘以一个化学式中该元素的原子数，得到每物种的总电子变化。

Step 5: Choose coefficients for the oxidised and reduced species so that the total increase equals the total decrease, using the lowest common multiple.

第5步：选择被氧化和被还原物种的系数，使总升高值等于总降低值，采用最小公倍数。

Step 6: Insert these coefficients and balance any other atoms except O and H.

第6步：代入这些系数，配平除 O 和 H 之外的其他原子。

Step 7: Balance oxygen by adding H₂O, then balance hydrogen by adding H⁺ in acidic or neutral solution.

第7步：在酸性或中性溶液中，加入 H₂O 配平氧，再加入 H⁺ 配平氢。

Step 8: Check that both atoms and total charge are balanced on both sides.

第8步：检查两侧原子数和总电荷是否都配平。

5. Worked Example 1: Iron(II) with Dichromate(VI) in Acid | 示例 1：酸性溶液中铁(II)与重铬酸根反应

Balance the reaction between iron(II) ions and dichromate(VI) ions in acidic solution. The skeleton equation is:

配平酸性溶液中铁(II)离子与重铬酸根离子的反应。骨架方程式为：

Cr₂O₇²⁻ + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O

Chromium in Cr₂O₇²⁻ has oxidation number +6, and it is reduced to Cr³⁺ with oxidation number +3. Each Cr gains 3 electrons, so one Cr₂O₇²⁻ ion gains 6 electrons in total.

Cr₂O₇²⁻ 中铬的氧化数为 +6，被还原为氧化数 +3 的 Cr³⁺。每个 Cr 得到 3 个电子，因此一个 Cr₂O₇²⁻ 离子总共得到 6 个电子。

Iron in Fe²⁺ has oxidation number +2, and it is oxidised to Fe³⁺ with oxidation number +3. Each Fe²⁺ loses 1 electron, so 6 Fe²⁺ ions are needed to balance the 6 electrons gained by one Cr₂O₇²⁻.

Fe²⁺ 中铁的氧化数为 +2，被氧化为氧化数 +3 的 Fe³⁺。每个 Fe²⁺ 失去 1 个电子，因此需要 6 个 Fe²⁺ 来平衡一个 Cr₂O₇²⁻ 得到的 6 个电子。

Inserting the redox coefficients gives Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺. There are 7 oxygen atoms on the left, so add 7H₂O on the right, then add 14H⁺ on the left to balance hydrogen.

代入氧化还原系数得到 Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺。左边有 7 个氧原子，因此在右边加 7H₂O，然后在左边加 14H⁺ 来配平氢。

Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺

Charge check: left side = -2 + 14 + 12 = +24; right side = 2(+3) + 6(+3) = +24. Atoms and charge are balanced.

电荷检查：左边 = -2 + 14 + 12 = +24；右边 =

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）