📚 Electrolysis of Aqueous Solutions | 水溶液的电解
In Edexcel IGCSE Chemistry, topic 5.6.4 covers the electrolysis of aqueous solutions, a fascinating process where water molecules themselves can be split and compete with dissolved ions to produce unexpected products. Unlike molten electrolysis, where only one type of cation and anion are present, aqueous systems introduce H⁺ and OH⁻ ions from water, making predictions more challenging and exciting. This article unpacks the key concepts, the electrochemical series, and worked examples to help you master the topic.
在 Edexcel IGCSE 化学课程中，专题 5.6.4 讲授水溶液的电解，这是一个引人入胜的过程，水分子本身可以分解并与溶解的离子竞争，生成意想不到的产物。与熔融电解中只存在一种阳离子和一种阴离子不同，水溶液体系由于来自水的 H⁺ 和 OH⁻ 离子的存在，使产物的预测更具挑战性和趣味性。本文梳理关键概念、电化学序，并通过实例讲解，帮助你掌握这一主题。
1. What Is Electrolysis of Aqueous Solutions? | 什么是水溶液的电解？
When an ionic compound dissolves in water, it dissociates into free-moving cations and anions. Applying a direct electric current via inert electrodes forces these ions to move: cations migrate to the cathode (negative electrode) and anions to the anode (positive electrode). However, water itself undergoes slight ionisation, generating low concentrations of H⁺ and OH⁻ ions. During electrolysis, these water-derived ions can be discharged instead of the solute ions, depending on their positions in the electrochemical series.
当离子化合物溶于水时，它会解离成自由移动的阳离子和阴离子。通过惰性电极施加直流电会迫使这些离子移动：阳离子向阴极（负极）迁移，阴离子向阳极（正极）迁移。然而，水本身会发生微弱的电离，产生低浓度的 H⁺ 和 OH⁻ 离子。在电解过程中，这些来自水的离子会根据它们在电化序中的位置，被优先放电而非溶质离子。
2. Electrolysis Basics: Electrodes, Ions and Current | 电解基础：电极、离子与电流
An electrolytic cell consists of two inert electrodes (often graphite or platinum) connected to a DC power supply, dipped into an electrolyte – the aqueous solution. The external circuit pushes electrons from the positive anode to the negative cathode. At the electrodes, reduction occurs at the cathode (gain of electrons) and oxidation at the anode (loss of electrons). For aqueous solutions, the electrolyte contains four types of ions: cations from the solute, anions from the solute, and H⁺ and OH⁻ from water.
电解池由两个惰性电极（通常为石墨或铂）与直流电源相连，并浸入电解质（即水溶液）中。外电路将电子从正极阳极推向负极阴极。在电极上，阴极发生还原反应（得电子），阳极发生氧化反应（失电子）。对于水溶液，电解质中含有四种离子：来自溶质的阳离子和阴离子，以及来自水的 H⁺ 和 OH⁻ 离子。
3. Dissociation of Ionic Compounds in Water | 离子化合物在水中的解离
When an ionic solid like copper(II) sulfate (CuSO₄) or sodium chloride (NaCl) dissolves in water, the ions separate and become surrounded by water molecules. For aqueous NaCl, the dissociation gives:
NaCl(s) → Na⁺(aq) + Cl⁻(aq)
Meanwhile, water itself undergoes self-ionisation:
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
Although the concentration of H⁺ and OH⁻ ions from water is very low (about 10⁻⁷ mol dm⁻³ in pure water), they play a decisive role when the solute ions are harder to discharge.
当像硫酸铜 (CuSO₄) 或氯化钠 (NaCl) 这样的离子固体溶于水时，离子分离并被水分子包围。对于氯化钠水溶液，解离方程式为：NaCl(s) → Na⁺(aq) + Cl⁻(aq)。同时，水本身发生自电离：H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)。尽管来自水的 H⁺ 和 OH⁻ 离子浓度极低（纯水中约为 10⁻⁷ mol dm⁻³），但当溶质离子较难放电时，它们会起到决定性作用。
4. Competing Ions: The Electrochemical Series | 竞争离子：电化学序
The key to predicting products is the reactivity or electrochemical series for cations and a similar discharge series for anions. The easier an ion gains or loses electrons, the more likely it is to be discharged. For cations at the cathode, the series from hardest to easiest to discharge (i.e. most reactive metal ions resist discharge) is:
预测产物的关键在于阳离子的反应性（或电化序）以及阴离子类似的放电顺序。离子越容易得到或失去电子，就越有可能优先放电。对于阴极处的阳离子，从最难放电到最易放电（即最活泼的金属离子最不易被还原）的顺序为：
| Cation | Difficulty to discharge |
|---|---|
| K⁺, Na⁺, Ca²⁺, Mg²⁺, Al³⁺ | Very hard – discharged only in molten, never from aqueous solution |
| Zn²⁺, Fe²⁺, Pb²⁺ | Harder than H⁺ |
| H⁺ (from water or acid) | Medium ease |
| Cu²⁺, Ag⁺ | Easily discharged (below hydrogen) |
For anions at the anode, the discharge order (easiest oxidised first) is typically:
对于阳极处的阴离子，放电顺序（最易被氧化的优先）通常为：
| Anion | Ease of discharge |
|---|---|
| SO₄²⁻, NO₃⁻ | Very difficult – never discharged from aqueous solution |
| OH⁻ (from water) | Moderate ease |
| Cl⁻, Br⁻, I⁻ | Easier than OH⁻ (when concentrated; see concentration effects) |
Remember: sulfate and nitrate ions are not discharged in aqueous solutions; instead, oxygen is produced from the oxidation of OH⁻. Halide ions can compete with OH⁻ depending on concentration.
请记住：硫酸根和硝酸根离子在水溶液中不会被放电；而是通过 OH⁻ 的氧化生成氧气。卤素离子能否与 OH⁻ 竞争取决于浓度。
5. Rules for Predicting Products at Each Electrode | 各电极产物预测规则
To determine what is formed, follow these steps:
要确定产物，可遵循以下步骤：
Step 1: Identify all ions present – cations from solute and H⁺; anions from solute and OH⁻.
步骤1：识别所有存在的离子——来自溶质的阳离子和 H⁺；来自溶质的阴离子和 OH⁻。
Step 2: At the cathode, the ion that is lower in the reactivity series (or higher in the electrochemical series of discharge) will gain electrons. If the metal is more reactive than hydrogen (e.g. Na⁺, K⁺, Mg²⁺), H⁺ from water is discharged to form hydrogen gas. If the metal is less reactive than hydrogen (e.g. Cu²⁺, Ag⁺), the metal cation is discharged and deposited.
步骤2：在阴极，位于反应性顺序较低（或放电电化序较高）的离子将获得电子。如果金属比氢更活泼（例如 Na⁺、K⁺、Mg²⁺），则水中的 H⁺ 放电生成氢气。如果金属不如氢活泼（如 Cu²⁺、Ag⁺），则金属阳离子放电并沉积。
Step 3: At the anode, if the anion is a halide (Cl⁻, Br⁻, I⁻) and the solution is reasonably concentrated, the halide ion is discharged to give the halogen. If the solution is dilute or the anion is sulfate/nitrate, then OH⁻ from water is discharged, forming oxygen gas and water.
步骤3：在阳极，如果阴离子为卤素离子（Cl⁻、Br⁻、I⁻）且溶液浓度足够高，卤素离子放电生成卤素单质。如果溶液较稀或阴离子为硫酸根/硝酸根，则来自水的 OH⁻ 放电，生成氧气和水。
6. Case Study 1: Electrolysis of Copper(II) Sulfate Solution | 案例研究1：硫酸铜溶液的电解
Electrolyte: aqueous CuSO₄ (using graphite electrodes).
电解质：硫酸铜水溶液（使用石墨电极）。
Ions present: Cu²⁺, SO₄²⁻, H⁺, OH⁻. At the cathode, Cu²⁺ is less reactive than H⁺, so Cu²⁺ ions are discharged and copper metal plates onto the electrode:
存在的离子：Cu²⁺、SO₄²⁻、H⁺、OH⁻。在阴极，Cu²⁺ 不如 H⁺ 活泼，因此 Cu²⁺ 放电，铜金属镀在电极上：
Cu²⁺(aq) + 2e⁻ → Cu(s)
At the anode, sulfate ions are not discharged; instead, OH⁻ ions give up electrons, producing oxygen gas:
在阳极，硫酸根离子不被放电；而是 OH⁻ 离子失去电子，生成氧气：
4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
The solution gradually changes from blue to paler blue as the concentration of Cu²⁺ decreases; the cathode gains a pink-brown coating of copper.
随着 Cu²⁺ 浓度降低，溶液由蓝色逐渐变为浅蓝色；阴极出现粉棕色的铜镀层。
7. Case Study 2: Electrolysis of Sodium Chloride Solution (Brine) | 案例研究2：氯化钠溶液（盐水）的电解
When concentrated NaCl solution (brine) is electrolysed, the products differ dramatically from those in molten NaCl.
当电解浓氯化钠溶液（盐水）时，产物与熔融 NaCl 的电解产物截然不同。
Ions present: Na⁺, Cl⁻, H⁺, OH⁻. At the cathode: Na⁺ is highly reactive, so H⁺ from water is discharged instead, giving hydrogen gas:
存在的离子：Na⁺、Cl⁻、H⁺、OH⁻。在阴极：Na⁺ 极为活泼，因此由水中的 H⁺ 放电，生成氢气：
2H⁺(aq) + 2e⁻ → H₂(g)
At the anode: concentrated chloride ions are discharged in preference to OH⁻, producing chlorine gas:
在阳极：浓溶液中氯离子优先于 OH⁻ 放电，产生氯气：
2Cl⁻(aq) → Cl₂(g) + 2e⁻
The remaining solution becomes sodium hydroxide (NaOH), which is an important industrial outcome (chlor-alkali process). If a dilute NaCl solution is used instead, the anode product can shift to oxygen, because low Cl⁻ concentration makes OH⁻ discharge more favourable.
剩余的溶液变成氢氧化钠 (NaOH)，这是重要的工业成果（氯碱工艺）。如果使用稀 NaCl 溶液，阳极产物可能转为氧气，因为低浓度 Cl⁻ 使 OH⁻ 放电更有利。
8. Electrolysis of Dilute Sulfuric Acid | 稀硫酸的电解
Dilute sulfuric acid (H₂SO₄) provides an instructive example where only water-derived ions are discharged. Ions present: H⁺, SO₄²⁻ and OH⁻ from water. At the cathode, H⁺ ions (which are plentiful from the acid) gain electrons to form hydrogen gas:
稀硫酸 (H₂SO₄) 提供了一个有启发性的例子，其中只有来自水的离子被放电。存在的离子：H⁺、SO₄²⁻ 和来自水的 OH⁻。在阴极，H⁺ 离子（大量来自酸）得到电子生成氢气：
2H⁺(aq) + 2e⁻ → H₂(g)
At the anode, sulfate ions are not discharged; OH⁻ is oxidised, producing oxygen:
在阳极，硫酸根不被放电；OH⁻ 被氧化生成氧气：
4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
g). The volume of hydrogen collected is twice that of oxygen, and the acid concentration effectively increases slightly as water is consumed.
总反应是水的分解：2H₂O(l) → 2H₂(g) + O₂(g)。收集到的氢气体积是氧气的两倍，且由于水被消耗，酸的浓度实际上略微升高。
9. The Effect of Concentration on Products | 浓度对产物的影响
Concentration can overturn the normal discharge order. For anions, a high concentration of chloride ions (e.g., in brine) means Cl⁻ is discharged at the anode even though OH⁻ lies above it in the series. The situation is similar for bromides and iodides. For very dilute halide solutions, however, OH⁻ discharge is favoured, giving oxygen. In the laboratory, students must note the concentration given in the question to predict the correct anode product. At the cathode, concentration rarely changes the cation discharge order because H⁺ from water is always present, but a very high concentration of an acid can only increase the rate of hydrogen evolution.
浓度可以颠覆正常的放电顺序。对于阴离子，高浓度的氯离子（如在盐水中）意味着 Cl⁻ 在阳极被放电，即使 OH⁻ 在电化序中更易放电。对于溴化物和碘化物情况类似。然而，对于极稀的卤化物溶液，OH⁻ 放电则更为有利，产生氧气。在实验室中，学生必须注意题目中给出的浓度，以便正确预测阳极产物。在阴极，浓度通常不会改变阳离子的放电顺序，因为来自水的 H⁺ 始终存在，但极高浓度的酸只会加速氢气的生成。
10. Industrial Applications and Exam Tips | 工业应用与考试技巧
he electrochemical series to justify your cathode and anode choices, and state any observations like gas bubbling, colour changes or electrode coatings. Practice half-equations using the correct charges (e.g. 2Cl⁻ → Cl₂ + 2e⁻) and remember that sulfate and nitrate ions are never discharged in aqueous solutions.
理解水溶液电解对于氯碱工业（从盐水中生产氯气、氢气和氢氧化钠）、电镀乃至铜的精炼等工业过程至关重要。在回答考试题目时，务必列出所有四种离子，运用电化序来论证你对阴极和阳极产物的选择，并描述观察到的现象，如气泡冒出、颜色变化或电极镀层。练习使用正确电荷的半反应式（例如 2Cl⁻ → Cl₂ + 2e⁻），并牢记硫酸根和硝酸根离子在水溶液中绝不会被放电。
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