Redox Reactions Revisited | 氧化还原反应全面回顾

📚 Redox Reactions Revisited | 氧化还原反应全面回顾

Redox reactions are the central thread connecting energy, synthesis and analysis in chemistry. Revisiting them at A-Level means moving far beyond 'OIL RIG' to a quantitative, electrode-potential-driven understanding that includes half-equations, electrochemical cells, the Nernst equation, electrolysis and redox titrations. This article consolidates every major concept, always linking back to applications and Cambridge exam requirements.

氧化还原反应是连接化学中能量、合成与分析的核心线索。在A-Level阶段重新探讨氧化还原，意味着要从简单的 'OIL RIG' 记忆口号跃升到定量的、由电极电势推动的理解，涵盖半反应、电化学电池、能斯特方程、电解和氧化还原滴定。本文将巩固每一个重要概念，并始终紧扣应用与剑桥考试的要求。

---

1. Defining Redox & Oxidation Numbers | 氧化还原定义与氧化数

The modern definition of reduction and oxidation centres on electron transfer: oxidation is the loss of electrons, reduction is the gain of electrons. Since electron transfer is not always obvious, we use oxidation numbers (states) as a book-keeping tool. Oxidation is an increase in oxidation number; reduction is a decrease.

还原和氧化的现代定义以电子转移为核心：氧化是失去电子，还原是得到电子。由于电子转移并非总是显而易见，我们使用氧化数（氧化态）作为记录工具。氧化意味着氧化数升高，还原意味着氧化数降低。

Rule 1: For an element in its free (uncombined) state, the oxidation number is 0 (e.g., O₂, Na, P₄).

规则1：单质（未化合）状态下的元素，氧化数为0（如 O₂、Na、P₄）。

le monatomic ion, the oxidation number equals the charge on the ion (e.g., Na⁺ = +1, Cl⁻ = –1).

规则2：简单单原子离子的氧化数等于离子电荷（如 Na⁺ = +1，Cl⁻ = –1）。

Rule 3: Oxygen normally has oxidation number –2, except in peroxides (–1, as in H₂O₂) and in OF₂ (+2).

规则3：氧通常为 –2，过氧化物中为 –1（如 H₂O₂），在 OF₂ 中为 +2。

Rule 4: Hydrogen is +1 when bonded to non-metals, but –1 in metal hydrides (e.g., NaH, CaH₂).

规则4：氢与非金属结合时为 +1，但在金属氢化物中为 –1（如 NaH、CaH₂）。

Rule 5: The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion it equals the ion's charge.

规则5：中性化合物中氧化数总和为0；多原子离子中等于离子电荷。

---

2. Identifying Oxidation Numbers in Compounds | 化合物的氧化数确定

Applying these rules allows you to deduce the oxidation number of any element in a compound. For example, in KMnO₄: K is +1, O₄ gives –8 total, so Mn must be +7 to give a net charge of 0. In K₂Cr₂O₇: each K +1, O₇ –14, total for two Cr atoms +12, so Cr is +6.

运用这些规则可以推断出化合物中任何元素的氧化数。例如，KMnO₄ 中：K 为 +1，4个O共 –8，因此 Mn 必须为 +7 才能总和为0。在 K₂Cr₂O₇ 中：每个 K +1，7个O –14，两个 Cr 总和 +12，所以 Cr 为 +6。

Transition metals often exhibit variable oxidation states; for instance, iron in Fe²⁺ (+2) and Fe³⁺ (+3), or copper in Cu⁺ (+1) and Cu²⁺ (+2). Recognising these is essential for writing half-equations and for naming compounds using the Stock notation, e.g., iron(II) sulfate and iron(III) chloride.

过渡金属常表现出可变的氧化态；例如铁有 Fe²⁺ (+2) 和 Fe³⁺ (+3)，铜有 Cu⁺ (+1) 和 Cu²⁺ (+2)。识别这些对于书写半反应以及使用斯托克命名法（如硫酸铁(II)和氯化铁(III)）至关重要。

---

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

An oxidising agent (oxidant) causes oxidation and is itself reduced; it gains electrons. Common oxidising agents include acidified MnO₄⁻ (permanganate, reduced to Mn²⁺), acidified Cr₂O₇²⁻ (dichromate, reduced to Cr³⁺), halogens (Cl₂, Br₂), and hydrogen peroxide H₂O₂ (reduced to H₂O or OH⁻).

氧化剂（氧化试剂）使其他物质氧化，自身被还原，接受电子。常见的氧化剂包括酸化 MnO₄⁻（高锰酸根，被还原为 Mn²⁺）、酸化 Cr₂O₇²⁻（重铬酸根，被还原为 Cr³⁺）、卤素（Cl₂、Br₂）以及过氧化氢 H₂O₂（被还原为 H₂O 或 OH⁻）。

sed to Zn²⁺, Fe²⁺), iron(II) salts (Fe²⁺ → Fe³⁺), iodide ions (2I⁻ → I₂), and thiosulfate ions S₂O₃²⁻ (oxidised to S₄O₆²⁻ in iodine titrations).

还原剂（还原试剂）使其他物质还原，自身被氧化，失去电子。典型的还原剂包括金属如 Zn 和 Fe（被氧化为 Zn²⁺、Fe²⁺）、铁(II)盐（Fe²⁺ → Fe³⁺）、碘离子（2I⁻ → I₂）和硫代硫酸根离子 S₂O₃²⁻（在碘滴定中被氧化为 S₄O₆²⁻）。

---

4. Writing Half-Equations and Balancing Redox Reactions | 书写半反应及配平氧化还原反应

Half-equations show clearly the electron loss or gain. In acid solution, the ion-electron method is used:

半反应清晰地显示电子的失去或得到。在酸性溶液中采用离子-电子法：

Step 1: Write the unbalanced half-equation for the reduction, e.g., MnO₄⁻ → Mn²⁺.

步骤1：写出还原的未配平半反应，例如 MnO₄⁻ → Mn²⁺。

Step 2: Balance atoms other than O and H (already balanced for Mn).

步骤2：配平除 O 和 H 以外的原子（Mn 已配平）。

Step 3: Balance O by adding H₂O molecules: MnO₄⁻ → Mn²⁺ + 4H₂O.

步骤3：通过添加 H₂O 分子配平 O：MnO₄⁻ → Mn²⁺ + 4H₂O。

Step 4: Balance H by adding H⁺ ions: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O.

步骤4：通过添加 H⁺ 配平 H：MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O。

arge by adding electrons: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (total charge +2 on both sides).

步骤5：添加电子配平电荷：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O（两侧总电荷均为 +2）。

s, then add to give the overall ionic equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

氧化的半反应为 Fe²⁺ → Fe³⁺ + e⁻。将铁半反应乘以5使电子数相等，然后相加得到总离子方程式：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。

---

广告 / Sponsored

5. Electrochemical Cells: Galvanic (Voltaic) Cells | 电化学电池：原电池

A galvanic cell converts chemical energy into electrical energy through spontaneous redox reactions. It consists of two half-cells connected by a salt bridge and an external wire. The electrode where oxidation occurs is the anode (negative), and reduction occurs at the cathode (positive). Electrons flow through the external circuit from anode to cathode.

原电池通过自发的氧化还原反应将化学能转化为电能。它由两个半电池组成，通过盐桥和外导线连接。发生氧化的电极是阳极（负极），还原发生在阴极（正极）。电子通过外电路从阳极流向阴极。

The well-known Daniell cell uses Zn|Zn²⁺(aq) and Cu| Zn²⁺ + 2e⁻; copper ions are reduced: Cu²⁺ + 2e⁻ → Cu. The cell diagram is written as Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)| e measured is the cell emf, a crucial quantity predicted by standard electrode potentials.

著名的丹尼尔电池使用 Zn|Zn²⁺(aq) 和 Cu|Cu²⁺(aq) 半电池。锌被氧化：Zn → Zn²⁺ + 2e⁻；铜离子被还原：Cu²⁺ + 2e⁻ → Cu。电池表示为 Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s)，阳极写在左侧。测得的电位差就是电池电动势，它是标准电极电势预测的关键物理量。

---

6. Standard Electrode Potentials and the Standard Hydrogen Electrode | 标准电极电势与标准氢电极

The standard electrode potential, E°, measures the tendency of a half-cell to be reduced (gain electrons) compared to the standard hydrogen electrode (SHE). Standard conditions are 298 K, all ion concentrations 1 mol dm⁻³, gas pressure 100 kPa, and solids/liquids in their standard states. The SHE is assigned an arbitrary E° of 0.00 V: 2H⁺(aq) + 2e⁻ ⇌ H₂(g).

标准电极电势 E° 衡量半电池相对于标准氢电极（SHE）发生还原（得电子）趋势的大小。标准条件为 298 K、所有离子浓度 1 mol dm⁻³、气体压力 100 kPa、以及固体/液体处于标准状态。SHE 被人为指定为 E° = 0.00 V：2H⁺(aq) + 2e⁻ ⇌ H₂(g)。

To measure E° for a Zn²⁺/Zn half-cell, it is connected to the SHE. The zinc electrode tends to release electrons more readily, so a potential difference is registered. By convention, E° values are listed as reduction potentials. A more positive E° means a stronger tendency to undergo reduction; a more negative E° means a greater tendency to be oxidised.

要测量 Zn²⁺/Zn 半电池的 E°，将其与 SHE 连接。锌电极更容易释放电子，因此会产生电位差。按照惯例，E° 值均以还原电势形式列出。E° 值越正，还原趋势越强；E° 值越负，被氧化的趋势越大。

---

7. The Electrochemical Series and Predicting Feasibility | 电化学系列与反应可行性预测

half-reactions by their E° values. At the bottom, strong reducing agents (e.g., Li⁺/Li at – tion is feasible, calculate the standard cell potential: E°cell = E°(cathode) - E°(anode), both taken as reduction potentials. If E°cell > 0, the reaction is thermodynamically spontaneous.

电化学系列按 E° 值排列半反应。底部是强还原剂（如 Li⁺/Li，–3.04 V），容易失去电子；顶部是强氧化剂（如 F₂/F⁻，+2.87 V），容易得到电子。预测氧化还原反应是否可行时，需计算标准电池电势：E°cell = E°(阴极) – E°(阳极)，两者均为还原电势。若 E°cell > 0，则反应在热力学上是自发的。

For example, will zinc displace copper from Cu²⁺? Anode (oxidation): Zn → Zn²⁺ + 2e⁻, E°anode = –0.76 V. Cathode (reduction): Cu²⁺ + 2e⁻ → Cu, E°cathode = +0.34 V. E°cell = +0.34 – (– E°cell does not guarantee a fast reaction; kinetic barriers may prevent it from occurring.

例如，锌能否从 Cu²⁺ 中置换出铜？阳极（氧化）：Zn → Zn²⁺ + 2e⁻，E°阳极 = –0.76 V。阴极（还原）：Cu²⁺ + 2e⁻ → Cu，E°阴极 = +0.34 V。E°cell = +0.34 – (–0.76) = +1.10 V > 0，因此反应可行。但正的 E°cell 并不保证反应速率快，动力学障碍可能阻止其发生。

---

8. The Nernst Equation: Effect of Concentration on Electrode Potential | 能斯特方程：浓度对电极电势的影响

When concentrations differ from 1 mol dm⁻³, the electrode potential shifts according to the Nernst equation. For a general half-reaction aOx + ne⁻ ⇌ bRed, the potential E is given by:

当浓度偏离 1 mol dm⁻³ 时，电极电势根据能斯特方程发生变化。对于一般半反应 aOx + ne⁻ ⇌ bRed，电势 E 为：

E = E° + (RT/nF) ln([Ox]^a / [Red]^b)

At 298 K, replacing RT/F with the appropriate constants gives a commonly used form in volts:

在 298 K 下，将常数代入后得到常用的伏特形式：

E = E° + (0.0592 V / n) log₁₀([Ox]^a / [Red]^b)

As an example, consider the Cu²⁺/Cu half-cell. If [Cu²⁺] is increased, log₁₀([Cu²⁺]/1) becomes positive, making E more positive than E°, which favours reduction. Decreasing [Cu²⁺] makes the half-cell a stronger reducing agent. The Nernst equation explains why a cell's voltage drops during discharge and underpins the calculation of equilibrium constants from E° values.

以 Cu²⁺/Cu 半电池为例，如果 [Cu²⁺] 增大，log₁₀([Cu²⁺]/1) 为正，E 将比 E° 更正，有利于还原。降低 [Cu²⁺] 会使该半电池成为更强的还原剂。能斯特方程解释了为什么电池在放电过程中电压会下降，并为从 E° 计算平衡常数奠定了基础。

---

9. Electrolysis and Faraday's Laws | 电解与法拉第定律

Electrolysis drives a non-spontaneous redox reaction using an external power source. The cathode is now the negative electrode (reduction occurs), and the anode is positive (oxidation occurs). In aqueous solutions, the products depend on the relative electrode potentials of possible reactants, including water. For concentrated NaCl(aq), chlorine gas

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

更多咨询请联系16621398022（同微信）