📚 Electrolysis for IB and WJEC Chemistry: Key Concepts and Exam Tips | IB WJEC 化学：电解 考点精讲
Electrolysis is a core topic in both IB and WJEC Chemistry, bridging the gap between electricity and chemical change. A clear understanding of how electrical energy drives non-spontaneous redox reactions in molten and aqueous electrolytes will empower you to predict products, perform mole calculations, and interpret industrial processes. This revision guide unpacks every essential concept, from ion migration to Faraday’s laws, with bilingual explanations to reinforce your learning.
电解是 IB 和 WJEC 化学中的核心主题，连接了电与化学变化。清楚理解电能如何在熔融和水溶液电解质中驱动非自发的氧化还原反应，将帮助你预测产物、进行摩尔计算并解释工业过程。本复习指南逐点讲解每个重要概念，从离子迁移到法拉第定律，采用中英双语解释，强化你的学习。
1. What Is Electrolysis? | 什么是电解？
Electrolysis is the process of using direct current (DC) to bring about a non-spontaneous chemical reaction. An external power source forces electrons to flow from the anode to the cathode through the external circuit, while ions move through the electrolyte to complete the circuit.
电解是利用直流电引发非自发化学反应的过程。外部电源迫使电子通过外电路从阳极流向阴极，同时离子在电解质中移动以完成回路。
In electrolysis, electrical energy is converted into chemical energy. Reduction takes place at the cathode (negative electrode), where cations gain electrons. Oxidation occurs at the anode (positive electrode), where anions lose electrons. Remember: RED CAT and AN OX – Reduction at Cathode, Oxidation at Anode – but note that the signs are opposite to those in a galvanic cell.
在电解中，电能转化为化学能。还原反应发生在阴极（负极），阳离子在此获得电子。氧化反应发生在阳极（正极），阴离子在此失去电子。记住：RED CAT 和 AN OX（阴极还原，阳极氧化），但请注意电解池中电极的符号与伏打电池相反。
The electrolyte must be a liquid that contains mobile ions. This can be a molten ionic compound or an aqueous solution. Solid ionic compounds do not conduct electricity because the ions are fixed in a lattice.
电解质必须是含有可自由移动离子的液体。这可以是熔融离子化合物或水溶液。固态离子化合物不导电，因为离子固定在晶格中。
2. Electrolysis of Molten Ionic Compounds | 熔融离子化合物的电解
When a molten salt such as lead(II) bromide (PbBr₂) is electrolysed, only the ions from the compound are present. The metal cation migrates to the cathode and is reduced to the pure metal. The non-metal anion migrates to the anode and is oxidised to the element.
当电解熔融盐如溴化铅 (PbBr₂) 时，仅存在该化合物的离子。金属阳离子移向阴极并被还原为纯金属。非金属阴离子移向阳极并被氧化为单质。
For PbBr₂: at the cathode, Pb²⁺ + 2e⁻ → Pb (silvery liquid lead forms). At the anode, 2Br⁻ → Br₂ + 2e⁻ (brown bromine gas is released). The overall reaction is: PbBr₂(l) → Pb(l) + Br₂(g).
以 PbBr₂ 为例：阴极反应，Pb²⁺ + 2e⁻ → Pb（生成银白色液态铅）。阳极反应，2Br⁻ → Br₂ + 2e⁻（放出红棕色溴蒸气）。总反应为：PbBr₂(l) → Pb(l) + Br₂(g)。
For any molten binary compound, the products are simply the elements from which it is composed. This method is used industrially to extract reactive metals such as sodium, magnesium and aluminium from their ores after conversion to molten salts or oxides.
对于任何熔融二元化合物，产物只是组成它的元素。工业上利用这一方法，在将矿石转化为熔盐或氧化物后，提取钠、镁、铝等活泼金属。
3. Electrolysis of Aqueous Solutions: Competition at Electrodes | 水溶液电解：电极上的竞争
In aqueous electrolysis, water molecules also dissociate slightly to give H⁺ and OH⁻ ions. This introduces competition between the ions from the solute and the ions from water.
在水溶液电解中，水分子也会轻微电离产生 H⁺ 和 OH⁻ 离子。这导致溶质离子与来自水的离子之间产生竞争。
At the cathode, the species with the most positive standard electrode potential (E⁰) is preferentially reduced. Usually, if the metal is less reactive than hydrogen (e.g. Cu²⁺, Ag⁺), the metal is deposited. If the metal is more reactive than hydrogen (e.g. Na⁺, K⁺, Ca²⁺), hydrogen gas is produced from the reduction of water: 2H₂O + 2e⁻ → H₂ + 2OH⁻.
在阴极，标准电极电势 (E⁰) 更正的物种优先被还原。通常，如果金属比氢的活泼性差（如 Cu²⁺、Ag⁺），金属会析出。如果金属比氢更活泼（如 Na⁺、K⁺、Ca²⁺），则水被还原产生氢气：2H₂O + 2e⁻ → H₂ + 2OH⁻。
At the anode, the species with the most negative E⁰ is preferentially oxidised. In a concentrated halide solution (Cl⁻, Br⁻, I⁻), the halide ion is often oxidised to the halogen. In a dilute solution or when a sulfate or nitrate is present, water is oxidised to oxygen: 2H₂O → O₂ + 4H⁺ + 4e⁻. The concentration and electrode material can also affect this.
在阳极，E⁰ 更负的物种优先被氧化。在浓卤化物溶液（Cl⁻、Br⁻、I⁻）中，卤离子常被氧化为卤素。在稀溶液中或存在硫酸根、硝酸根时，水被氧化生成氧气：2H₂O → O₂ + 4H⁺ + 4e⁻。浓度和电极材料也会对此产生影响。
4. Predicting Products Using Standard Electrode Potentials | 利用标准电极电势预测产物
The Standard Electrode Potential (E⁰) table is an indispensable tool for predicting the outcome of electrolysis. The half-equation with the more positive E⁰ value is the one most likely to undergo reduction. However, kinetic factors, overpotential, and concentration can alter the practical outcome, especially for gas evolution.
标准电极电势 (E⁰) 表是预测电解结果的重要工具。具有更正 E⁰ 值的半反应最可能发生还原。但动力学因素、超电势和浓度会影响实际结果，特别是气体产生的情况。
Example: electrolysis of aqueous copper(II) sulfate using inert electrodes. Possible cathode reactions: Cu²⁺ + 2e⁻ → Cu (E⁰ = +0.34 V); 2H₂O + 2e⁻ → H₂ + 2OH⁻ (E⁰ = –0.83 V). Copper is preferentially deposited at the cathode. At the anode: SO₄²⁻ is not oxidised; the oxidation of water occurs: 2H₂O → O₂ + 4H⁺ + 4e⁻ (E⁰ = +1.23 V). So oxygen is produced.
示例：使用惰性电极电解硫酸铜(II)水溶液。可能的阴极反应：Cu²⁺ + 2e⁻ → Cu (E⁰ = +0.34 V)；2H₂O + 2e⁻ → H₂ + 2OH⁻ (E⁰ = –0.83 V)。铜优先在阴极析出。阳极：SO₄²⁻ 不被氧化；发生水的氧化：2H₂O → O₂ + 4H⁺ + 4e⁻ (E⁰ = +1.23 V)。因此产生氧气。
Always remember: the table gives thermodynamic predictions; in exams, describe what you see and note any discrepancy due to overpotential, especially for O₂ and Cl₂ at the anode.
时刻记住：数据表给出的是热力学预测；考试中，要描述观察到的现象，并注意超电势导致的差异，尤其是阳极的氧气和氯气。
5. Electrodes: Inert vs. Active Electrodes | 电极：惰性电极与活性电极
Inert electrodes, such as graphite (carbon) and platinum, do not take part in the chemical reaction. They merely provide a surface for electron transfer. As a result, the products depend solely on the electrolyte.
惰性电极如石墨（碳）和铂不参与化学反应，仅提供电子转移的表面。因此，产物仅取决于电解质。
Active electrodes, such as copper electrodes in copper(II) sulfate electrolysis, participate in the reaction. At the anode, copper metal is oxidised: Cu(s) → Cu²⁺ + 2e⁻. At the cathode, Cu²⁺ ions are reduced and deposited: Cu²⁺ + 2e⁻ → Cu(s). This is the principle of electroplating and electrorefining.
活性电极如电解硫酸铜时的铜电极会参与反应。阳极的铜被氧化：Cu(s) → Cu²⁺ + 2e⁻。阴极的 Cu²⁺ 被还原并沉积：Cu²⁺ + 2e⁻ → Cu(s)。这就是电镀和电解精炼的原理。
With active anodes, the anode loses mass while the cathode gains mass. The concentration of the electrolyte remains constant because the number of ions entering the solution equals the number leaving it.
使用活性阳极时，阳极质量减少，阴极质量增加。电解质的浓度保持不变，因为进入溶液的离子数与离开的离子数相等。
6. Electroplating and Electrorefining | 电镀与电解精炼
Electroplating is used to coat a thin layer of metal onto a surface for corrosion resistance, decoration, or improved conductivity. The object to be plated is made the cathode, and the plating metal is made the anode. The electrolyte is a solution containing ions of the plating metal.
电镀用于在物体表面覆盖一薄层金属，以达到防腐蚀、装饰或提高导电性的目的。被镀物体作为阴极，镀层金属作为阳极。电解质为含有该金属离子的溶液。
For silver plating, the anode is silver, the cathode is the object (e.g. a spoon), and the electrolyte is a silver nitrate solution. Silver dissolves at the anode and deposits on the cathode, giving a uniform layer.
对于镀银，阳极为银，阴极为被镀物体（如勺子），电解质为硝酸银溶液。银在阳极溶解并在阴极沉积，形成均匀涂层。
Electrorefining purifies impure metals such as copper. The impure copper is the anode, which dissolves; pure copper is deposited at the cathode. Insoluble impurities fall to the bottom as anode sludge, while more reactive metals remain in solution.
电解精炼用于提纯不纯金属，如铜。不纯的铜作为阳极并溶解；纯铜在阴极沉积。不溶性杂质沉到底部形成阳极泥，较活泼的金属留在溶液中。
7. Quantitative Electrolysis and Faraday’s Laws | 定量电解与法拉第定律
Faraday’s laws link the amount of substance produced at an electrode to the quantity of electric charge passed. The first law states: the mass (m) of a substance liberated is directly proportional to the charge (Q): m ∝ Q. The second law states: for the same charge, the masses of different substances liberated are proportional to their equivalent weights.
法拉第定律将电极上产生的物质量与通过的电量联系起来。第一定律：析出物质的质量 (m) 与电荷量 (Q) 成正比：m ∝ Q。第二定律：对于相同的电荷量，不同物质析出的质量与其当量成正比。
The charge is calculated by Q = I × t, where I is current in amperes (A) and t is time in seconds (s). The unit of charge is the coulomb (C). The Faraday constant (F) is the charge of one mole of electrons: F = 96 500 C mol⁻¹ (often approximated to 96 500 C mol⁻¹; IB and WJEC may accept 96 500 or 9.65 × 10⁴).
电荷量计算为 Q = I × t，其中 I 为电流（安培，A），t 为时间（秒，s）。电荷单位为库仑 (C)。法拉第常数 (F) 是一摩尔电子所带的电荷：F = 96 500 C mol⁻¹（常近似为 96 500；IB 和 WJEC 可能接受 96 500 或 9.65 × 10⁴）。
The mole of electrons (n(e⁻)) transferred is calculated by: n(e⁻) = Q / F. To find the mass of metal deposited, relate the electrons to the stoichiometry of the half-reaction. For example, for Ag⁺ + e⁻ → Ag, 1 mol e⁻ deposits 1 mol Ag (107.87 g). For Cu²⁺ + 2e⁻ → Cu, 2 mol e⁻ deposit 1 mol Cu (63.55 g).
转移的电子的物质的量 n(e⁻) 计算公式：n(e⁻) = Q / F。要计算析出金属的质量，需将电子与半反应的化学计量比关联。例如，对于 Ag⁺ + e⁻ → Ag，1 mol e⁻ 沉积 1 mol Ag (107.87 g)。对于 Cu²⁺ + 2e⁻ → Cu，2 mol e⁻ 沉积 1 mol Cu (63.55 g)。
Exam tip: always write the balanced half-equation, calculate n(e⁻), then use the mole ratio to find moles of the substance, and finally convert to mass or volume (for gases, use 24 000 cm³ mol⁻¹ at RTP).
考试技巧：始终写出配平的半反应式，计算 n(e⁻)，再根据摩尔比求出物质的量，最后转换为质量或体积（对于气体，在室温常压下使用 24 000 cm³ mol⁻¹）。
8. Calculating Charge, Mass and Gas Volume | 电荷、质量与气体体积计算
Worked example: A current of 0.50 A is passed through molten Al₂O₃ for 1930 seconds. Calculate the mass of aluminium produced and the volume of oxygen gas released at RTP. Half‑equations: Al³⁺ + 3e⁻ → Al; 2O²⁻ → O₂ + 4e⁻.
计算示例：0.50 A 的电流通过熔融 Al₂O₃ 达 1930 秒。计算生成的铝的质量和在室温常压下释放的氧气体积。半反应式：Al³⁺ + 3e⁻ → Al；2O²⁻ → O₂ + 4e⁻。
Q = 0.50 A × 1930 s = 965 C. n(e⁻) = 965 / 96 500 = 0.0100 mol. For Al: mol Al = 0.0100 / 3 = 0.00333 mol; mass Al = 0.00333 × 27.0 = 0.090 g. For O₂: mol O₂ = 0.0100 / 4 = 0.00250 mol; volume O₂ = 0.00250 × 24 000 = 60 cm³.
Q = 0.50 × 1930 = 965 C。n(e⁻) = 965 / 96 500 = 0.0100 mol。对于 Al：Al 的物质的量 = 0.0100 / 3 = 0.00333 mol；Al 的质量 = 0.00333 × 27.0 = 0.090 g。对于 O₂：O₂ 的物质的量 = 0.0100 / 4 = 0.00250 mol；O₂ 体积 = 0.00250 × 24 000 = 60 cm³。
When dealing with mixtures or multiple cells in series, remember that the same charge passes through each cell. All calculations begin with Q = I × t, followed by the mole‑ratio method. Use Avogadro’s law for gas volumes at the same temperature and pressure.
处理混合物或串联电解池时，记住通过每个电解池的电荷量相同。所有计算都从 Q = I × t 开始，接着运用摩尔比方法。对于相同温压下的气体体积，可运用阿伏加德罗定律。
9. Electrolysis of Water and Important Aqueous Systems | 水的电解及重要水溶液体系
Electrolysis of water acidified with sulfuric acid (or using sodium sulfate) yields hydrogen at the cathode and oxygen at the anode in a 2:1 volume ratio. This confirms the formula of water as H₂O. The half‑reactions are: 4H⁺ + 4e⁻ → 2H₂ (or 2H₂O + 2e⁻ → H₂ + 2OH⁻) and 2H₂O → O₂ + 4H⁺ + 4e⁻.
用硫酸酸化（或使用硫酸钠）的水进行电解，阴极产生氢气，阳极产生氧气，体积比为 2:1。这证实了水的化学式为 H₂O。半反应为：4H⁺ + 4e⁻ → 2H₂（或 2H₂O + 2e⁻ → H₂ + 2OH⁻）和 2H₂O → O₂ + 4H⁺ + 4e⁻。
Other common systems: electrolysis of concentrated aqueous sodium chloride (brine) produces chlorine at the anode, hydrogen at the cathode, and sodium hydroxide remains in solution. This is the chlor-alkali process. For dilute NaCl, oxygen may be produced instead of chlorine.
其他常见体系：电解浓氯化钠水溶液（盐水）在阳极产生氯气、阴极产生氢气，氢氧化钠留在溶液中。这是氯碱法。对于稀 NaCl，可能产生氧气而非氯气。
Electrolysis of aqueous silver nitrate with silver electrodes is used for quantitative analysis (coulometry). With inert electrodes, silver deposits at the cathode and oxygen is released at the anode.
使用银电极电解硝酸银水溶液用于定量分析（库仑法）。使用惰性电极时，银在阴极析出，氧气在阳极释放。
10. Factors Affecting the Electrolysis Products | 影响电解产物的因素
Beyond standard electrode potentials, three main factors dictate the actual products: the nature of the electrode (inert or active), the concentration of ions in solution, and overpotential.
除标准电极电势外，三个主要因素决定实际产物：电极的性质（惰性或活性）、溶液中离子的浓度以及超电势。
Concentrated solutions of halides favour halogen discharge. For example, concentrated NaCl gives Cl₂, but very dilute NaCl gives O₂. For aqueous solutions containing multiple cations, the one with the least negative (or most positive) E⁰ is reduced first, provided concentrations are roughly 1 mol dm⁻³.
浓卤化物溶液有利于卤素放电。例如，浓 NaCl 产生 Cl₂，但很稀的 NaCl 产生 O₂。对于含多种阳离子的水溶液，如果浓度约为 1 mol dm⁻³，E⁰ 最不显负（或最正）的离子先被还原。
Overpotential is the extra voltage beyond the thermodynamic requirement needed to drive a reaction at a practical rate. Oxygen evolution typically has a high overpotential, meaning chlorine is often produced at lower voltages from brine despite the E⁰ values favouring oxygen.
超电势是指在实际反应速率下驱动反应所需超过热力学电压的额外电压。氧气的析出通常有较高的超电势，这意味着在盐水中，尽管 E⁰ 值倾向于析出氧气，但在较低电压下常常产生氯气。
11. Industrial Applications of Electrolysis | 电解的工业应用
Aluminium extraction by the Hall-Héroult process uses electrolysis of purified Al₂O₃ dissolved in molten cryolite (Na₃AlF₆) at about 950°C. This reduces the melting point and improves conductivity. Carbon anodes are consumed by reaction with the oxygen produced, generating CO₂.
霍尔-埃鲁法炼铝是在约 950°C 下电解溶于熔融冰晶石 (Na₃AlF₆) 的纯 Al₂O₃。这降低了熔点并提高了导电率。碳阳极与产生的氧气反应而被消耗，生成 CO₂。
The chlor-alkali process produces chlorine, hydrogen, and sodium hydroxide, all of which are essential industrial chemicals. Membrane cells are now used to prevent mixing of products and to produce high-purity NaOH.
氯碱法生产氯气、氢气和氢氧化钠，它们都是重要的工业化学品。目前使用膜电解槽以防止产品混合，并生产高纯度的 NaOH。
Electrorefining of copper yields 99.99% pure copper, crucial for electrical wiring. Electroplating is widely applied in the automotive, electronics, and jewellery industries.
铜的电解精炼可产生 99.99% 的纯铜，对电线制造至关重要。电镀在汽车、电子和珠宝行业中被广泛应用。
12. Common Exam Pitfalls and Tips | 常见考试误区与技巧
Pitfall: confusing the sign of electrodes in electrolytic and galvanic cells. Remember: electrolysis cathode = negative (attracts cations), anode = positive. In a galvanic cell, the signs are opposite: cathode = positive, anode = negative.
误区：混淆电解池与伏打电池中电极的符号。记住：电解池中阴极为负极（吸引阳离子），阳极为正极。伏打电池中符号相反：阴极为正极，阳极为负极。
Pitfall: forgetting to convert time to seconds when using Q = I × t. Always double-check units.
误区：使用 Q = I × t 时忘记将时间换算成秒。务必仔细检查单位。
Pitfall: not using the mole ratio correctly. Write the half-equation and label the number of electrons. Cross‑multiply carefully. Avoid rounding errors during intermediate steps.
误区：摩尔比使用不正确。写出半反应式并标明电子数。交叉相乘要仔细。避免在中间步骤产生舍入误差。
Pitfall: assuming that water is always reduced/oxidized. Always list the possible competing species and compare their E⁰ values (or remember the activity series and the preferential discharge series).
误区：想当然地认为水总是被还原/氧化。始终列出所有可能的竞争物种，并比较它们的 E⁰ 值（或记住金属活动性顺序和优先放电顺序）。
Tip: if a gas is produced at an electrode, you may be asked to test it. Hydrogen – ‘pop’ with a lighted splint; oxygen – relights a glowing splint; chlorine – bleaches damp litmus paper.
技巧：如果电极上产生气体，可能要求你进行检验。氢气——燃着的木条靠近有爆鸣声；氧气——使带火星的木条复燃；氯气——使湿润的石蕊试纸褪色。
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