Oxidation, Reduction and Redox Equations | 氧化、还原与氧化还原方程式

📚 Oxidation, Reduction and Redox Equations | 氧化、还原与氧化还原方程式

Oxidation and reduction are fundamental concepts in chemistry. Understanding redox reactions allows you to predict reaction behaviour, balance complex equations and explain the chemistry of electrochemical cells.

氧化与还原是化学中的基础概念。理解氧化还原反应有助于预测反应行为、配平复杂方程式，并解释电化学电池中的化学原理。

1. Oxidation and Reduction | 氧化与还原的定义

Oxidation is the loss of electrons from a species. Reduction is the gain of electrons by a species. These two processes always occur together in a redox reaction.

氧化是指物质失去电子的过程；还原是指物质获得电子的过程。这两个过程在氧化还原反应中总是同时发生。

The mnemonic OIL RIG is useful: Oxidation Is Loss, Reduction Is Gain (of electrons).

助记词 OIL RIG 非常实用：OIL 表示氧化是失去电子，RIG 表示还原是得到电子。

For example, in the reaction between sodium and chlorine:

例如，在钠与氯气的反应中：

2Na + Cl₂ → 2Na⁺Cl⁻

Na loses one electron to form Na⁺; this is oxidation.
Cl gains one electron to form Cl⁻; this is reduction.

Na 失去一个电子形成 Na⁺，这是氧化。
Cl 获得一个电子形成 Cl⁻，这是还原。

2. Oxidation Numbers | 氧化数

Oxidation number (or oxidation state) is a bookkeeping device used to track electron transfer in reactions. It is the charge an atom would have if all bonds were treated as fully ionic.

氧化数（又称氧化态）是一种用于追踪反应中电子转移的记账工具。它表示如果所有键都被视为完全离子键时，原子所带有的电荷。

The key rules for assigning oxidation numbers are:

确定氧化数的关键规则如下：

Rule	Example
Element in its free state has oxidation number 0.	O₂, Na, Cl₂ all have oxidation number 0.
Monatomic ion: oxidation number = charge.	Fe³⁺ has +3; Cl⁻ has −1.
Oxygen is usually −2 (except in peroxides and OF₂).	H₂O: O = −2; H₂O₂: O = −1.
Hydrogen is usually +1 (except in metal hydrides).	HCl: H = +1; NaH: H = −1.
The sum of oxidation numbers in a neutral compound is 0.	CO₂: C + 2(−2) = 0 → C = +4.
The sum of oxidation numbers in a polyatomic ion equals the charge.	SO₄²⁻: S + 4(−2) = −2 → S = +6.

3. Using Oxidation Numbers to Define Redox | 用氧化数定义氧化还原

An increase in oxidation number corresponds to oxidation, while a decrease corresponds to reduction.

氧化数升高对应氧化，氧化数降低对应还原。

Consider the reaction:

考虑以下反应：

2Mg + O₂ → 2MgO

Mg changes from 0 to +2 (oxidised). O changes from 0 to −2 (reduced).

Mg 的氧化数由 0 变为 +2（被氧化）；O 的氧化数由 0 变为 −2（被还原）。

Oxidation numbers also identify oxidising and reducing agents. The oxidising agent is the species that is reduced; the reducing agent is the species that is oxidised.

氧化数还有助于识别氧化剂和还原剂。被还原的物质是氧化剂，被氧化的物质是还原剂。

4. Oxidising and Reducing Agents | 氧化剂与还原剂

An oxidising agent (oxidant) accepts electrons and is itself reduced. A reducing agent (reductant) donates electrons and is itself oxidised.

氧化剂接受电子，自身被还原；还原剂给出电子，自身被氧化。

Common oxidising agents include KMnO₄, K₂Cr₂O₇ and H₂O₂. Common reducing agents include Fe²⁺, I⁻ and SO₂.

常见的氧化剂包括 KMnO₄、K₂Cr₂O₇ 和 H₂O₂。常见的还原剂包括 Fe²⁺、I⁻ 和 SO₂。

In the reaction between Fe²⁺ and MnO₄⁻ in acid:

在酸性条件下 Fe²⁺ 与 MnO₄⁻ 的反应中：

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

MnO₄⁻ is the oxidising agent (Mn decreases from +7 to +2).
Fe²⁺ is the reducing agent (Fe increases from +2 to +3).

MnO₄⁻ 是氧化剂（Mn 的氧化数从 +7 降至 +2）。
Fe²⁺ 是还原剂（Fe 的氧化数从 +2 升至 +3）。

5. Half Equations | 半方程式

Redox reactions can be split into two half equations: one representing oxidation and one representing reduction. Each half equation shows electrons explicitly.

氧化还原反应可以拆分为两个半方程式：一个表示氧化，一个表示还原。每个半方程式都明确写出电子。

For the oxidation of iron(II):

对于铁(II)的氧化：

Fe²⁺ → Fe³⁺ + e⁻

For the reduction of manganate(VII) in acid:

对于酸性条件下高锰酸根(VII)的还原：

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Notice that atoms and charges must balance. In acid solution, use H⁺ and H₂O to balance oxygen and hydrogen. In alkaline solution, use OH⁻ and H₂O.

注意原子和电荷必须平衡。在酸性溶液中，使用 H⁺ 和 H₂O 来平衡氧和氢；在碱性溶液中，使用 OH⁻ 和 H₂O。

6. Balancing Redox Equations Using Half Equations | 用半方程式配平氧化还原方程式

To combine half equations into a full redox equation, follow these steps:

将半方程式合并为完整氧化还原方程式的步骤如下：

Identify the oxidation and reduction half equations.
Balance atoms other than O and H.
Balance O with H₂O and H with H⁺ (acidic) or OH⁻ (alkaline).
Balance charge by adding electrons to the more positive side.
Multiply each half equation so that electrons cancel.
Add the half equations and simplify.

确定氧化和还原的半方程式。
配平除 O 和 H 以外的原子。
用 H₂O 配平 O，用 H⁺（酸性）或 OH⁻（碱性）配平 H。
在电荷较多的一侧添加电子以配平电荷。
将每个半方程式乘以适当系数，使电子数目相等。
将两个半方程式相加并化简。

7. Worked Example: Acidified KMnO₄ with Fe²⁺ | 示例：酸化高锰酸钾与 Fe²⁺ 的反应

Oxidation half equation:

氧化半方程式：

Fe²⁺ → Fe³⁺ + e⁻

Reduction half equation:

还原半方程式：

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Multiply the oxidation half equation by 5:

将氧化半方程式乘以 5：

5Fe²⁺ → 5Fe³⁺ + 5e⁻

Add both half equations and cancel electrons:

合并两个半方程式并消去电子：

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

Check that atoms and charges balance: left charge = −1 + 8 + 10 = +17; right charge = 2 + 0 + 15 = +17. Correct.

检查原子和电荷是否平衡：左侧电荷 = −1 + 8 + 10 = +17；右侧电荷 = 2 + 0 + 15 = +17。正确。

8. Balancing in Alkaline Conditions | 碱性条件下的配平

In alkaline solution, OH⁻ and H₂O are used instead of H⁺. For example, the reduction of MnO₄⁻ to MnO₂ in alkaline medium:

在碱性溶液中，使用 OH⁻ 和 H₂O 代替 H⁺。例如，碱性介质中 MnO₄⁻ 被还原为 MnO₂：

MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

To derive this, first balance in acidic form, then add enough OH⁻ to neutralise H⁺ on both sides.

推导方法：先用酸性形式配平，然后两侧加入足够 OH⁻ 中和 H⁺。

Practice converting between acidic and alkaline half equations is essential for AQA exams.

在 AQA 考试中，练习酸性与碱性半方程式之间的相互转化至关重要。

9. Redox Equations in Terms of Oxidation Numbers | 用氧化数配平氧化还原方程式

An alternative method for balancing redox equations is based on changes in oxidation number. The total increase in oxidation number must equal the total decrease.

另一种配平氧化还原方程式的方法基于氧化数的变化。氧化数的总升高量必须等于总降低量。

For example, in the reaction of iodine with sodium thiosulphate:

例如，碘与硫代硫酸钠的反应：

2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻

Sulfur changes from +2 in S₂O₃²⁻ to +2.5 on average in S₄O₆²⁻ (increase of 0.5 per S, total increase of 1 for two S atoms). Iodine changes from 0 to −1 (decrease of 1 per I, total decrease of 2 for I₂). The ratio is therefore 2 S₂O₃²⁻ : 1 I₂.

S₂O₃²⁻ 中硫的氧化数为 +2，S₄O₆²⁻ 中硫的平均氧化数为 +2.5（每个 S 升高 0.5，两个 S 总升高 1）。碘从 0 变为 −1（每个 I 降低 1，I₂ 总降低 2）。因此 S₂O₃²⁻ 与 I₂ 的配比是 2:1。

10. Disproportionation | 歧化反应

Disproportionation is a redox reaction in which the same species is both oxidised and reduced. Chlorine with water is a classic example:

歧化反应是指同一物质在同一反应中既被氧化又被还原。氯气与水的反应是经典例子：

Cl₂ + H₂O ⇌ HCl + HClO

The oxidation number of chlorine changes from 0 in Cl₂ to −1 in HCl (reduction) and +1 in HClO (oxidation).

氯的氧化数从 Cl₂ 中的 0 变为 HCl 中的 −1（还原）和 HClO 中的 +1（氧化）。

Another example is the reaction of chlorine with cold dilute sodium hydroxide:

另一个例子是氯气与冷稀氢氧化钠的反应：

Cl₂ + 2NaOH → NaCl + NaClO + H₂O

Recognising disproportionation helps identify products and balance equations.

识别歧化反应有助于推断产物并配平方程式。

11. Common Redox Titration Calculations | 常见氧化还原滴定计算

Redox titrations are a popular AQA practical topic. For example, the titration of iodine with sodium thiosulphate:

氧化还原滴定是 AQA 常见的实验考点。例如，碘与硫代硫酸钠的滴定：

2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻

The stoichiometric ratio is 2:1, meaning 2 moles of thiosulphate react with 1 mole of iodine. Use this ratio to calculate concentrations.

化学计量比为 2:1，即 2 mol 硫代硫酸根与 1 mol 碘反应。利用该比例可计算浓度。

For example, if 25.0 cm³ of iodine solution requires 20.0 cm³ of 0.100 mol dm⁻³ Na₂S₂O₃, the iodine concentration is:

例如，若 25.0 cm³ 碘溶液需要 20.0 cm³ 0.100 mol dm⁻³ Na₂S₂O₃，则碘的浓度为：

n(S₂O₃²⁻) = 0.100 × 20.0/1000 = 0.00200 mol

n(I₂) = 0.00200 / 2 = 0.00100 mol

[I₂] = 0.00100 / 0.0250 = 0.0400 mol dm⁻³

Always state your half equations before performing titration calculations.

进行滴定计算前，务必先写出对应的半方程式。

12. Summary and Exam Tips | 总结与考试提示

Redox chemistry requires confidence in assigning oxidation numbers, writing half equations and combining them correctly.

氧化还原化学需要熟练掌握氧化数的确定、半方程式的书写以及正确合并它们。

Always check that atoms and charges balance in half equations.
Learn common oxidation states: MnO₄⁻ → Mn²⁺ (+7 to +2), Cr₂O₇²⁻ → Cr³⁺ (+6 to +3).
Use OIL RIG to avoid confusing oxidation and reduction.
In disproportionation, the same element appears in both oxidised and reduced products.

始终检查半方程式中原子和电荷是否平衡。
熟记常见氧化态变化：MnO₄⁻ → Mn²⁺（+7 到 +2），Cr₂O₇²⁻ → Cr³⁺（+6 到 +3）。
使用 OIL RIG 避免混淆氧化与还原。
在歧化反应中，同一元素同时出现在氧化产物和还原产物中。

Regular practice with redox equations will help you answer both calculation and practical-based questions with confidence.

定期练习氧化还原方程式，将帮助你自信应对计算题和实验类题目。

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Oxidation, Reduction and Redox Equations | 氧化、还原与氧化还原方程式