Electrolysis in A-Level Edexcel Chemistry | A-Level Edexcel 化学：电解考点精讲

📚 Electrolysis in A-Level Edexcel Chemistry | A-Level Edexcel 化学：电解考点精讲

Electrolysis is a core topic in the Edexcel A-Level Chemistry specification, linking fundamental concepts of redox chemistry, energetic feasibility, and quantitative calculations. This article provides a thorough breakdown of the key principles, common electrolyte systems, Faraday’s laws, and industrial applications — all tailored to the demands of Edexcel examination questions.

电解是 Edexcel A-Level 化学大纲中的核心主题，连接了氧化还原化学、能量可行性和定量计算等基础概念。本文彻底剖析关键原理、常见电解质体系、法拉第定律和工业应用，全部紧扣 Edexcel 考试要求。

Published by TutorHao | Chemistry Revision Series | aleveler.com

1. Fundamentals of Electrolysis | 电解的基本原理

Electrolysis is the process of driving a non-spontaneous chemical reaction using direct current (DC) electricity. An electrolytic cell consists of two electrodes dipped in an electrolyte — a liquid containing mobile ions. The positive electrode (anode) attracts anions, where oxidation occurs; the negative electrode (cathode) attracts cations, where reduction occurs. Remember: Red Cat, An Ox — Reduction at Cathode, Oxidation at Anode.

电解是利用直流电驱动非自发化学反应的过程。电解池由两根浸入电解液（含可移动离子的液体）的电极组成。正极（阳极）吸引阴离子，发生氧化反应；负极（阴极）吸引阳离子，发生还原反应。记住口诀：Red Cat, An Ox —— 阴极还原，阳极氧化。

The external power source pushes electrons onto the cathode and removes them from the anode. In the external circuit, current flows from anode to cathode (conventional), but electrons flow from cathode to anode via the wires. The electrolyte must contain freely moving ions, which can be produced by melting an ionic compound or dissolving it in water.

外部电源将电子推向阴极并从阳极抽走电子。在外电路中，电流从阳极流向阴极（常规方向），但电子通过导线从阴极流向阳极。电解质必须含有可自由移动的离子，可通过熔化离子化合物或将其溶于水产生。

Ion migration completes the circuit: cations move toward the cathode, anions toward the anode. The chemical changes at the electrodes are always redox processes, and the overall reaction is the sum of the two half‑equations.

离子迁移使电路完整：阳离子向阴极移动，阴离子向阳极移动。电极上的化学变化始终是氧化还原过程，总反应是两个半反应方程式之和。

2. Electrolysis of Molten Compounds | 熔融化合物的电解

When an ionic compound is heated to its melting point, the ions become mobile. In the electrolysis of molten lead(II) bromide (PbBr₂), the cathode product is lead metal from Pb²⁺ reduction, and the anode product is bromine gas from Br⁻ oxidation. No other species compete for discharge.

当离子化合物加热至熔点，离子变得可移动。在熔融溴化铅（PbBr₂）的电解中，阴极产物是 Pb²⁺ 还原得到的铅金属，阳极产物是 Br⁻ 氧化得到的溴气。没有其他物质竞争放电。

Half‑equations:
Cathode: Pb²⁺(l) + 2e⁻ → Pb(l)
Anode: 2Br⁻(l) → Br₂(g) + 2e⁻

半反应方程式：
阴极：Pb²⁺(l) + 2e⁻ → Pb(l)
阳极：2Br⁻(l) → Br₂(g) + 2e⁻

This simple system illustrates the principle: the cation is reduced at the cathode, the anion oxidised at the anode. For aluminium oxide (Al₂O₃) in the Hall–Héroult process, cryolite is added to lower the melting point, and the cathode produces molten aluminium while the graphite anode forms CO₂.

这个简单系统说明了原理：阳离子在阴极被还原，阴离子在阳极被氧化。对于氧化铝（Al₂O₃）在霍尔–埃鲁法中，加入冰晶石降低熔点，阴极产生熔融铝，石墨阳极生成 CO₂。

3. Discharge Series in Aqueous Solutions | 水溶液中的放电顺序

In aqueous electrolysis, water molecules can also be oxidised or reduced, leading to competing reactions. The species that actually discharges depends on its position in the electrochemical series, the concentration, and the electrode material.

在水溶液电解中，水分子也可被氧化或还原，导致竞争反应。实际放电的物质取决于其在电化学序中的位置、浓度和电极材料。

For cations: if a metal is more reactive than hydrogen (e.g. Na⁺, Mg²⁺, Al³⁺), hydrogen gas is produced at the cathode from the reduction of water. If the metal is less reactive (e.g. Cu²⁺, Ag⁺), the metal ions are reduced to the metal.

对于阳离子：如果金属比氢活泼（如 Na⁺、Mg²⁺、Al³⁺），则阴极产生氢气（来自水的还原）。如果金属不如氢活泼（如 Cu²⁺、Ag⁺），则金属离子被还原为金属单质。

For anions: sulfate and nitrate ions (SO₄²⁻, NO₃⁻) are not easily oxidised; instead, water is oxidised to form oxygen gas at the anode. Halide ions (Cl⁻, Br⁻, I⁻) are discharged if their concentration is sufficiently high, producing the halogen.

对于阴离子：硫酸根和硝酸根离子（SO₄²⁻、NO₃⁻）不易被氧化；取而代之的是水被氧化再阳极生成氧气。卤素离子（Cl⁻、Br⁻、I⁻）若浓度足够高则放电，生成卤素单质。

In concentrated NaCl(aq), chloride ions are oxidised to Cl₂ at the anode, but in dilute solution, oxygen is the main anode product. Hydroxide ions (OH⁻) can also be discharged: 4OH⁻ → O₂ + 2H₂O + 4e⁻.

在浓 NaCl 溶液中，氯离子在阳极被氧化为 Cl₂；但在稀溶液中，氧气是主要阳极产物。氢氧根离子也可放电：4OH⁻ → O₂ + 2H₂O + 4e⁻。

4. Electrolysis of Aqueous Copper(II) Sulfate | 硫酸铜水溶液的电解

With inert electrodes (e.g. platinum or graphite), the Cu²⁺ ions are reduced at the cathode: Cu²⁺(aq) + 2e⁻ → Cu(s). At the anode, water is oxidised: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻. The solution gradually becomes acidic due to the formation of sulfuric acid.

使用惰性电极（如铂或石墨），Cu²⁺ 离子在阴极被还原：Cu²⁺(aq) + 2e⁻ → Cu(s)。在阳极，水被氧化：2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻。由于硫酸生成，溶液逐渐变为酸性。

With active copper electrodes, the cathode still plates copper, but the anode itself dissolves: Cu(s) → Cu²⁺(aq) + 2e⁻. The concentration of Cu²⁺ ions in the solution remains constant. This is used in copper electrorefining.

使用活性铜电极，阴极仍然镀铜，但阳极本身溶解：Cu(s) → Cu²⁺(aq) + 2e⁻。溶液中 Cu²⁺ 浓度保持不变。这用于铜的电解精炼。

Colour changes: the blue colour from Cu²⁺ ions fades with inert electrodes as copper ions are removed, but remains unchanged with copper electrodes. The oxygen produced at the inert anode can be tested with a glowing splint.

颜色变化：使用惰性电极时，随着铜离子被消耗，Cu²⁺ 产生的蓝色会褪去；但使用铜电极时颜色不变。惰性阳极生成的氧气可用带火星的木条检验。

5. Electrolysis of Aqueous Sodium Chloride | 氯化钠水溶液的电解

The discharge of ions depends on concentration. In dilute NaCl(aq), hydrogen and oxygen are the main products:

Cathode: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)
Anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻

Overall, water is effectively electrolysed. The solution near the cathode becomes alkaline due to OH⁻.

离子的放电取决于浓度。在稀 NaCl 溶液中，主要产物是氢和氧：
阴极：2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)
阳极：2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻
总体而言，实际上是水被电解。阴极附近溶液因 OH⁻ 而变碱性。

In concentrated NaCl(aq), chloride ions compete successfully at the anode: 2Cl⁻(aq) → Cl₂(g) + 2e⁻. The cathode still produces hydrogen gas and hydroxide ions. Thus the overall products are H₂, Cl₂, and NaOH — the basis of the chlor‑alkali industry.

在浓 NaCl 溶液中，氯离子在阳极成功竞争放电：2Cl⁻(aq) → Cl₂(g) + 2e⁻。阴极仍然产生氢气和氢氧根离子。因此总产物是 H₂、Cl₂ 和 NaOH——氯碱工业的基础。

The membrane or diaphragm cell keeps the products separate to prevent the reaction between Cl₂ and NaOH which would form unwanted sodium chlorate(I). The net reaction is: 2NaCl + 2H₂O → H₂ + Cl₂ + 2NaOH.

膜或隔膜电解槽使产物分开，防止 Cl₂ 与 NaOH 反应生成不需要的次氯酸钠。净反应为：2NaCl + 2H₂O → H₂ + Cl₂ + 2NaOH。

6. Faraday’s Laws of Electrolysis | 法拉第电解定律

Faraday’s First Law states that the mass (m) of a substance produced or consumed at an electrode is directly proportional to the quantity of electricity (Q) that passes through the cell: m ∝ Q.

法拉第第一定律指出，在电极上产生或消耗的物质的质量 (m) 与通过电解池的电量 (Q) 成正比：m ∝ Q。

Faraday’s Second Law states that the masses of different substances liberated by the same quantity of electricity are proportional to their equivalent weights (molar mass divided by the number of electrons per ion).

法拉第第二定律指出，相同电量释放的不同物质的质量与其当量（摩尔质量除以每离子的电子数）成正比。

Quantitatively, Q = I × t, where I is current in amperes (A) and t is time in seconds (s). The amount of electrons in moles, n(e⁻) = Q / F, where F is the Faraday constant ≈ 96 500 C mol⁻¹.

定量上，Q = I × t，其中 I 是电流（安培），t 是时间（秒）。电子的物质的量 n(e⁻) = Q / F，F 为法拉第常数 ≈ 96 500 C mol⁻¹。

The mass of a product can be found using stoichiometry. For instance, to deposit 1 mole of Cu from Cu²⁺ requires 2 moles of electrons; thus m = (I × t × M) / (z × F), where z is the number of electrons per ion and M is molar mass.

产物质量可通过化学计量求得。例如，从 Cu²⁺ 沉积 1 摩尔 Cu 需要 2 摩尔电子；因此 m = (I × t × M) / (z × F)，其中 z 是每离子的电子数，M 是摩尔质量。

7. Calculations Involving Quantity of Electricity | 涉及电量的计算

A typical exam question might ask: “Calculate the mass of silver deposited when a current of 0.50 A flows for 30 minutes through AgNO₃ solution.” First, convert time to seconds: t = 30 × 60 = 1800 s. Q = 0.50 A × 1800 s = 900 C.

一道典型的考题可能问：“当 0.50 A 电流通过 AgNO₃ 溶液 30 分钟，计算沉积银的质量。”首先，时间换算为秒：t = 30 × 60 = 1800 s。Q = 0.50 A × 1800 s = 900 C。

n(e⁻) = 900 C / 96 500 C mol⁻¹ ≈ 9.33 × 10⁻³ mol. Since Ag⁺ + e⁻ → Ag, 1 mol e⁻ yields 1 mol Ag. n(Ag) = 9.33 × 10⁻³ mol. m(Ag) = 9.33 × 10⁻³ mol × 107.9 g mol⁻¹ ≈ 1.01 g.

n(e⁻) = 900 C / 96 500 C mol⁻¹ ≈ 9.33 × 10⁻³ mol。因为 Ag⁺ + e⁻ → Ag，1 mol 电子得到 1 mol Ag。n(Ag) = 9.33 × 10⁻³ mol。m(Ag) = 9.33 × 10⁻³ mol × 107.9 g mol⁻¹ ≈ 1.01 g。

For gases, use the volume equivalents. Under standard conditions, 1 mol of gas occupies 24.0 dm³ (Edexcel data book). So if hydrogen is produced: n(H₂) = n(e⁻)/2, volume = n(H₂) × 24.0 dm³.

对于气体，使用体积当量。标准状况下，1 mol 气体占 24.0 dm³（Edexcel 数据手册）。因此若生成氢气：n(H₂) = n(e⁻)/2，体积 = n(H₂) × 24.0 dm³。

Watch out for multiple steps: e.g. Al³⁺ + 3e⁻ → Al, so z = 3. Always check the half‑equation to determine the electron‑to‑product ratio.

注意多步过程：例如 Al³⁺ + 3e⁻ → Al，因此 z = 3。始终检查半反应方程式以确定电子与产物的比例。

8. Industrial Application: Extraction of Aluminium | 工业应用：铝的提取

Aluminium is extracted by the electrolysis of purified Al₂O₃ dissolved in molten cryolite (Na₃AlF₆) at about 950 °C. Cryolite reduces the melting point from over 2000 °C to about 950 °C and increases conductivity, drastically lowering energy costs.

铝通过电解溶解在熔融冰晶石（Na₃AlF₆）中的精炼 Al₂O₃ 提取，温度约 950 °C。冰晶石将熔点从超过 2000 °C 降至约 950 °C 并提高电导率，大幅降低能耗。

The cell uses graphite anodes and a graphite‑lined steel cathode. At the cathode: Al³⁺ + 3e⁻ → Al(l). At the anode: 2O²⁻ → O₂ + 4e⁻, but the oxygen reacts with the carbon anode: C(s) + O₂(g) → CO₂(g). The anodes are thus consumed and must be replaced periodically.

电解槽使用石墨阳极和石墨衬里的钢阴极。阴极：Al³⁺ + 3e⁻ → Al(l)。阳极：2O²⁻ → O₂ + 4e⁻，但氧气与碳阳极反应：C(s) + O₂(g) → CO₂(g)。因此阳极不断消耗，需定期更换。

Edexcel often asks why cryolite is used, why anodes need replacing, and to write the overall equation: 2Al₂O₃ + 3C → 4Al + 3CO₂. Environmental concerns include CO₂ emissions and the disposal of spent pot linings.

Edexcel 常问为何使用冰晶石、为何阳极需更换，并写出总方程式：2Al₂O₃ + 3C → 4Al + 3CO₂。环境问题包括 CO₂ 排放和废槽衬的处置。

9. Industrial Application: Chlor‑Alkali Industry | 工业应用：氯碱工业

The chlor‑alkali process electrolyses concentrated brine (NaCl solution) using a membrane cell or diaphragm cell. The membrane is a selective ion‑exchange polymer that allows Na⁺ ions to pass but prevents the migration of Cl⁻ and OH⁻, keeping products separate.

氯碱工业使用膜电解槽或隔膜电解槽电解浓盐水（NaCl 溶液）。膜是一种选择性离子交换聚合物，允许 Na⁺ 通过但阻止 Cl⁻ 和 OH⁻ 迁移，使产物保持分离。

Cathode (often steel): 2H₂O + 2e⁻ → H₂ + 2OH⁻. Sodium ions pass through the membrane to combine with hydroxide ions, forming sodium hydroxide solution. Anode (titanium coated with RuO₂): 2Cl⁻ → Cl₂ + 2e⁻.

阴极（通常为钢）：2H₂O + 2e⁻ → H₂ + 2OH⁻。钠离子通过膜与氢氧根离子结合，形成氢氧化钠溶液。阳极（钛涂覆 RuO₂）：2Cl⁻ → Cl₂ + 2e⁻。

Products and their uses: chlorine for PVC, disinfectants, bleach; hydrogen for ammonia (Haber process), hydrogenation, fuel cells; sodium hydroxide for soap, paper, detergents, alumina extraction.

产物及其用途：氯用于 PVC、消毒剂、漂白剂；氢用于氨（哈伯法）、加氢、燃料电池；氢氧化钠用于肥皂、造纸、洗涤剂、氧化铝提取。

Edexcel expects students to compare the membrane cell with the older mercury cell (cathode: Na⁺ + e⁻ → Na, which dissolves in mercury; then reacts with water). The membrane cell is more environmentally friendly as it avoids toxic mercury.

Edexcel 要求学生比较膜电解槽与旧式汞电解槽（阴极：Na⁺ + e⁻ → Na，溶于汞；然后与水反应）。膜电解槽更环保，避免使用有毒汞。

10. Electrorefining and Electroplating | 电解精炼与电镀

Electrorefining of copper uses an impure copper anode and a pure copper cathode in CuSO₄/H₂SO₄ electrolyte. The anode dissolves: Cu → Cu²⁺ + 2e⁻; pure copper is deposited on the cathode: Cu²⁺ + 2e⁻ → Cu. Impurities like gold and silver fall as anode sludge.

铜的电解精炼使用不纯铜阳极和纯铜阴极，在 CuSO₄/H₂SO₄ 电解液中进行。阳极溶解：Cu → Cu²⁺ + 2e⁻；纯铜在阴极沉积：Cu²⁺ + 2e⁻ → Cu。金和银等杂质以阳极泥形式落下。

Electroplating uses electrolysis to coat a conductive object with a thin layer of metal (e.g. silver, nickel). The object to be plated is made the cathode, the anode is the plating metal, and the electrolyte contains ions of that metal. Conditions must be carefully controlled for an even, adherent layer.

电镀是利用电解在导电物体上覆盖一薄层金属（如银、镍）。待镀物体作阴极，阳极是镀层金属，电解质含该金属的离子。必须严格控制条件以获得均匀、附着的镀层。

For example, silver plating: anode = Ag(s), cathode = spoon, electrolyte = AgNO₃(aq). Cathode: Ag⁺ + e⁻ → Ag(s); anode: Ag(s) → Ag⁺ + e⁻. The solution concentration remains constant.

例如镀银：阳极 = Ag(s)，阴极 = 勺子，电解液 = AgNO₃(aq)。阴极：Ag⁺ + e⁻ → Ag(s)；阳极：Ag(s) → Ag⁺ + e⁻。溶液浓度保持不变。

11. Writing Half‑Equations and Overall Reactions | 书写半反应方程式和总反应

Mastering half‑equations is essential. Always balance atoms first, then add electrons to balance charge. For the oxidation of water to oxygen in acidic conditions: 2H₂O → O₂ + 4H⁺ + 4e⁻. In alkaline conditions: 4OH⁻ → O₂ + 2H₂O + 4e⁻.

掌握半反应方程式至关重要。总是先配平原子，然后加电子配平电荷。酸性条件下水氧化为氧：2H₂O → O₂ + 4H⁺ + 4e⁻。碱性条件下：4OH⁻ → O₂ + 2H₂O + 4e⁻。

When combining half‑equations, ensure the number of electrons matches. For the electrolysis of water: cathode 2H₂O + 2e⁻ → H₂ + 2OH⁻ (×2) and anode 4OH⁻ → O₂ + 2H₂O + 4e⁻. Cancel electrons and combine: 4H₂O + 4e⁻ → 2H₂ + 4OH⁻ + O₂ + 2H₂O + 4e⁻ − 4OH⁻ simplifies to 2H₂O → 2H₂ + O₂.

合并半反应方程式时，确保电子数相等。电解水：阴极 2H₂O + 2e⁻ → H₂ + 2OH⁻（×2），阳极 4OH⁻ → O₂ + 2H₂O + 4e⁻。消去电子并合并：得 2H₂O → 2H₂ + O₂。

For unfamiliar electrolytes, apply the discharge rules: determine which cation and anion (or water) are discharged, write the appropriate half‑equations, and then balance. Edexcel may ask for half‑equations in context (e.g. Al₂O₃ in molten cryolite).

对于不熟悉的电解质，应用放电规则：确定哪种阳离子和阴离子（或水）放电，写出适当的半反应方程式，然后配平。Edexcel 可能在情境中要求写半反应方程式（如 Al₂O₃ 在冰晶石中）。

12. Common Pitfalls and Exam Tips | 常见错误与考试提示

Avoid confusing the direction of electron flow with conventional current. Electrons travel from cathode (negative) to anode (positive) in the external circuit, while current flows the opposite way. In the electrolyte, ions carry the current.

避免混淆电子流动方向与常规电流方向。电子在外电路中从阴极（负极）流向阳极（正极），而电流方向相反。在电解液中，离子携带电流。

Do not forget to convert time to seconds (Q = I × t, t in seconds) — a common mistake that leads to answers wrong by a factor of 60. Always check units.

不要忘记将时间转换为秒（Q = I × t，t 以秒计）——这是常见错误，导致答案差 60 倍。始终检查单位。

When predicting products in aqueous solutions, students often forget to consider the concentration of halides or the electrode material. Scrutinise the conditions given: “concentrated NaCl” gives Cl₂, “dilute NaCl” gives O₂; “inert electrodes” versus “copper electrodes” completely change the anode reaction.

预测水溶液产物时，学生常忘记考虑卤化物浓度或电极材料。仔细审视给定条件：“浓 NaCl”产生 Cl₂，“稀 NaCl”产生 O₂；“惰性电极”与“铜电极”会完全改变阳极反应。

For Faraday law calculations, always write the balanced half‑equation first to find z (electrons per ion). For example, Al³⁺ requires 3e⁻, but Fe²⁺ → Fe only needs 2e⁻. Mixing up z will give a wrong mass ratio.

进行法拉第定律计算时，务必先写出配平的半反应方程式以确定 z（每离子电子数）。例如 Al³⁺ 需要 3e⁻，而 Fe²⁺ → Fe 只需 2e⁻。混淆 z 将导致错误的质量比。

Be precise with terminology: oxidation occurs at the anode, reduction at the cathode, regardless of the cell type (electrolytic or galvanic). Edexcel mark schemes penalise vague language like “the anode is positive” without linking to the type of cell.

术语要精确：氧化发生在阳极，还原发生在阴极，不论电池类型（电解池还是原电池）。Edexcel 评分标准会惩罚模糊表述，如只说“阳极是正极”而未关联电池类型。

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Electrolysis in A-Level Edexcel Chemistry | A-Level Edexcel 化学：电解 考点精讲