📚 Electrolysis for IGCSE OCR Chemistry | IGCSE OCR 化学：电解 考点精讲
Electrolysis is a core topic in IGCSE OCR Chemistry that explains how electricity drives chemical changes. It covers essential concepts such as electrolytes, electrodes, and the reactions occurring in both molten and aqueous systems, alongside key industrial applications like aluminium extraction and electroplating. Mastering this topic means understanding ion movement, discharge rules, and half-equation writing. This article provides a structured revision guide to help you tackle every electrolysis question with confidence.
电解是 IGCSE OCR 化学的核心主题，它解释了电能如何驱动化学变化。内容涵盖电解质、电极、熔融及水溶液体系中的反应，以及铝的提取和电镀等关键工业应用。掌握这部分知识需要理解离子的移动、放电顺序和半方程式的书写。本文提供结构化复习指南，帮助你自信应对每一道电解考题。
1. What is Electrolysis? | 什么是电解？
Electrolysis is the process of using a direct electric current to drive a non-spontaneous chemical reaction. An ionic compound must be molten or dissolved in water for electrolysis to occur, because the ions need to be free to move and carry charge.
电解是利用直流电驱动非自发化学反应的过程。离子化合物必须处于熔融状态或溶于水中才能发生电解，因为离子需要自由移动并携带电荷。
The substance that is broken down is called the electrolyte. The positive electrode is the anode, and the negative electrode is the cathode. Cations (positive ions) migrate towards the cathode, where they gain electrons (reduction). Anions (negative ions) migrate towards the anode, where they lose electrons (oxidation).
被分解的物质称为电解质。正极是阳极，负极是阴极。阳离子（正离子）向阴极移动，在那里获得电子（还原反应）。阴离子（负离子）向阳极移动，在那里失去电子（氧化反应）。
A common mnemonic is ‘CATIONS to CAThode, ANIONS to ANode’ or ‘RED CAT, AN OX’: Reduction at Cathode, Oxidation at Anode.
常见助记口诀是“阳离子去阴极，阴离子去阳极”或“RED CAT, AN OX”：阴极还原，阳极氧化。
2. Electrolytic Cell Components | 电解池的组成
A typical electrolytic cell consists of a power source (direct current), two electrodes (usually graphite or platinum when inertness is required), and an electrolyte. The electrodes are connected to the power supply, and the circuit is completed by the movement of ions in the electrolyte.
典型的电解池由直流电源、两个电极（需要惰性时通常为石墨或铂）和电解质组成。电极连接到电源，电解质中离子的移动使电路形成闭合回路。
Inert electrodes do not react with the electrolyte or products. Active electrodes, such as copper in copper purification, can take part in the electrode reactions. The electrode material can influence the products obtained at the anode.
惰性电极不与电解质或产物发生反应。活性电极，比如铜精炼中的铜电极，会参与电极反应。电极材料会影响阳极得到的产物。
Diagrams are frequently tested: you must be able to label the anode, cathode, electrolyte, and direction of ion flow (cations to cathode, anions to anode), as well as the direction of electron flow in the external circuit (from anode to cathode).
图示是常考内容：你需要能够标注阳极、阴极、电解质、离子流动方向（阳离子去阴极，阴离子去阳极）以及外电路中电子的流动方向（从阳极到阴极）。
3. Electrolysis of Molten Lead(II) Bromide | 熔融溴化铅的电解
The electrolysis of molten lead(II) bromide (PbBr₂) is a classic example used to demonstrate the decomposition of an ionic compound. The electrolyte is heated until it melts, enabling Pb²⁺ and Br⁻ ions to move freely.
熔融溴化铅（PbBr₂）的电解是用于演示离子化合物分解的经典示例。电解质被加热直至熔化，使 Pb²⁺ 和 Br⁻ 离子能够自由移动。
At the cathode, Pb²⁺ ions gain two electrons: Pb²⁺ + 2e⁻ → Pb (liquid lead, reduced). Silvery droplets of molten lead form at the bottom. At the anode, Br⁻ ions lose electrons: 2Br⁻ → Br₂ + 2e⁻. Brown bromine gas is observed at the anode.
在阴极，Pb²⁺ 离子得到两个电子：Pb²⁺ + 2e⁻ → Pb（液态铅，被还原）。银白色熔融铅滴在底部生成。在阳极，Br⁻ 离子失去电子：2Br⁻ → Br₂ + 2e⁻。在阳极可观察到红棕色的溴气。
This reaction confirms that molten ionic compounds conduct electricity and undergo decomposition. It also highlights the production of a metal at the cathode and a non-metal at the anode.
该反应证实熔融离子化合物可以导电并发生分解。同时也突出了在阴极生成金属、在阳极生成非金属的特征。
4. Electrolysis of Aqueous Solutions: General Rules | 水溶液电解的一般规则
When an ionic compound is dissolved in water, the situation becomes more complex because water itself can be electrolysed, producing H⁺ and OH⁻ ions. The products at the electrodes depend on the relative ease of discharge of the ions present.
当离子化合物溶于水时，情况变得更复杂，因为水本身也可被电解，产生 H⁺ 和 OH⁻ 离子。电极产物取决于所存在离子的相对放电难易程度。
At the cathode, if the metal is more reactive than hydrogen (e.g., Na⁺, K⁺, Ca²⁺, Mg²⁺, Al³⁺), hydrogen gas (H₂) is produced from the discharge of H⁺ ions from water: 2H⁺ + 2e⁻ → H₂. If the metal is less reactive than hydrogen (e.g., Cu²⁺, Ag⁺), the metal itself is deposited.
在阴极，如果金属比氢更活泼（如 Na⁺, K⁺, Ca²⁺, Mg²⁺, Al³⁺），则会从水中 H⁺ 离子放电产生氢气（H₂）：2H⁺ + 2e⁻ → H₂。如果金属不如氢活泼（如 Cu²⁺, Ag⁺），则金属本身会被沉积出来。
At the anode, if the anion is a halide (Cl⁻, Br⁻, I⁻), the halogen is produced (e.g., 2Cl⁻ → Cl₂ + 2e⁻). If the anion is a sulfate (SO₄²⁻) or nitrate (NO₃⁻), oxygen gas (O₂) is produced from the discharge of OH⁻ ions from water: 4OH⁻ → O₂ + 2H₂O + 4e⁻.
在阳极，如果阴离子是卤素离子（Cl⁻, Br⁻, I⁻），则生成对应的卤素单质（如 2Cl⁻ → Cl₂ + 2e⁻）。如果阴离子是硫酸根（SO₄²⁻）或硝酸根（NO₃⁻），则从水中 OH⁻ 离子放电产生氧气（O₂）：4OH⁻ → O₂ + 2H₂O + 4e⁻。
5. Preferential Discharge Series | 离子优先放电顺序
The preferential discharge series helps predict which ion will be discharged at each electrode when multiple ions are present. Memorising this series is crucial for IGCSE OCR Chemistry.
优先放电顺序有助于预测存在多种离子时每个电极上哪种离子会放电。熟记此顺序对 IGCSE OCR 化学至关重要。
Cathode (reduction) ease of discharge: Ag⁺ > Cu²⁺ > H⁺ > Pb²⁺ > Fe²⁺ > Zn²⁺ > Al³⁺ > Mg²⁺ > Ca²⁺ > Na⁺ > K⁺. In aqueous solutions, H⁺ competes with metal cations. Metals above hydrogen will not plate out; hydrogen gas is evolved instead.
阴极（还原）放电容易程度： Ag⁺ > Cu²⁺ > H⁺ > Pb²⁺ > Fe²⁺ > Zn²⁺ > Al³⁺ > Mg²⁺ > Ca²⁺ > Na⁺ > K⁺。在水溶液中，H⁺ 与金属阳离子竞争。活泼性排在氢之前的金属不会析出；而是生成氢气。
Anode (oxidation) ease of discharge: I⁻ > Br⁻ > Cl⁻ > OH⁻ > NO₃⁻ > SO₄²⁻ > F⁻. Halide ions are discharged in preference to hydroxide ions. If no halide is present, oxygen is evolved from OH⁻.
阳极（氧化）放电容易程度： I⁻ > Br⁻ > Cl⁻ > OH⁻ > NO₃⁻ > SO₄²⁻ > F⁻。卤素离子优先于氢氧根离子放电。如果不存在卤素离子，则从 OH⁻ 生成氧气。
Concentration can affect discharge order: a concentrated chloride solution can discharge Cl⁻ even though OH⁻ is theoretically easier to discharge according to the series, due to the abundance of chloride ions. Be mindful of exam contexts mentioning ‘concentrated’ or ‘dilute’.
浓度会影响放电顺序：浓氯化物溶液可能使 Cl⁻ 放电，即便从系列上看 OH⁻ 理论放电更容易，这是因为氯离子浓度极高。注意考题中是否提及“浓”或“稀”。
6. Electrolysis of Specific Aqueous Solutions | 特定水溶液的电解
Let us apply the rules to two frequently examined solutions: copper(II) sulfate (CuSO₄) with inert electrodes, and concentrated sodium chloride (brine).
让我们将规则应用于两个常考溶液：硫酸铜（CuSO₄）使用惰性电极，以及浓氯化钠溶液（盐水）。
Copper(II) sulfate with graphite electrodes: Ions present: Cu²⁺, SO₄²⁻, H⁺, OH⁻. At the cathode, Cu²⁺ is less reactive than hydrogen, so Cu is deposited: Cu²⁺ + 2e⁻ → Cu (pink-brown metal). At the anode, OH⁻ is discharged in preference to SO₄²⁻, giving O₂: 4OH⁻ → O₂ + 2H₂O + 4e⁻. The blue colour fades as Cu²⁺ ions are removed, and the solution eventually becomes sulfuric acid.
硫酸铜溶液使用石墨电极： 存在的离子：Cu²⁺, SO₄²⁻, H⁺, OH⁻。在阴极，Cu²⁺ 不如氢活泼，因此铜被沉积：Cu²⁺ + 2e⁻ → Cu（红棕色金属）。在阳极，OH⁻ 优先于 SO₄²⁻ 放电，生成 O₂：4OH⁻ → O₂ + 2H₂O + 4e⁻。随着 Cu²⁺ 被移除，蓝色逐渐褪去，溶液最终变为硫酸。
Concentrated sodium chloride (brine) with inert electrodes: Ions: Na⁺, Cl⁻, H⁺, OH⁻. Cathode: H⁺ discharged (Na⁺ too reactive), 2H⁺ + 2e⁻ → H₂. Anode: Cl⁻ discharged because it is concentrated, 2Cl⁻ → Cl₂ + 2e⁻. Left in solution: Na⁺ and OH⁻, forming sodium hydroxide (NaOH). This is the chlor-alkali industry.
浓氯化钠溶液（盐水）使用惰性电极： 离子：Na⁺, Cl⁻, H⁺, OH⁻。阴极：H⁺ 放电（Na⁺ 太活泼），2H⁺ + 2e⁻ → H₂。阳极：因浓度高，Cl⁻ 放电，2Cl⁻ → Cl₂ + 2e⁻。溶液中剩下 Na⁺ 和 OH⁻，形成氢氧化钠（NaOH）。这就是氯碱工业。
7. Writing Half Equations | 书写半方程式
Half equations show the electron transfer at each electrode. They must balance atoms and charge. Practice is essential.
半方程式展示每个电极上的电子转移过程，必须配平原子和电荷。练习至关重要。
For the cathode, write the cation plus the correct number of electrons to form the neutral atom or molecule. Example: Al³⁺ + 3e⁻ → Al. For hydrogen: 2H⁺ + 2e⁻ → H₂.
对于阴极，写出阳离子加上正确数量的电子生成中性原子或分子。例如：Al³⁺ + 3e⁻ → Al。对于氢：2H⁺ + 2e⁻ → H₂。
For the anode, write the anion on the left losing electrons to form the non-metal. For halogens: 2X⁻ → X₂ + 2e⁻. For oxygen from OH⁻: 4OH⁻ → O₂ + 2H₂O + 4e⁻. Always use state symbols (s, l, g, aq) where requested in the exam.
对于阳极，写出阴离子在左侧失去电子生成非金属单质。卤素：2X⁻ → X₂ + 2e⁻。对于从 OH⁻ 生成的氧气：4OH⁻ → O₂ + 2H₂O + 4e⁻。若考题要求，务必标注状态符号（s, l, g, aq）。
A common mistake is attempting to include water molecules directly in the half equation for metal deposition—only water’s H⁺ or OH⁻ ions participate. Balance oxygen and hydrogen using H₂O and H⁺/OH⁻ as appropriate, but follow the IGCSE convention of using H⁺ in acidic conditions and OH⁻ in alkaline conditions. For neutral solutions producing oxygen, IGCSE generally accepts 4OH⁻ → O₂ + 2H₂O + 4e⁻.
常见错误是直接将水分子写入金属沉积的半方程式中——只有水中的 H⁺ 或 OH⁻ 参与反应。使用 H₂O 和 H⁺/OH⁻ 适当配平氧和氢，但在 IGCSE 中，酸性条件用 H⁺，碱性条件用 OH⁻。对于中性溶液产生氧气，IGCSE 通常接受 4OH⁻ → O₂ + 2H₂O + 4e⁻。
8. Extraction of Aluminium | 铝的提取
Aluminium is extracted from its ore bauxite (Al₂O₃) by electrolysis. The process is called the Hall-Héroult process. Because aluminium oxide has a very high melting point, it is dissolved in molten cryolite (Na₃AlF₆) to lower the temperature and reduce energy costs.
铝是通过电解从其矿石铝土矿（Al₂O₃）中提取出来的。该过程称为霍尔-埃鲁法。由于氧化铝熔点极高，将其溶于熔融冰晶石（Na₃AlF₆）中以降低温度并减少能耗。
The electrolytic cell has a steel cathode and graphite anodes. Molten aluminium forms at the cathode: Al³⁺ + 3e⁻ → Al. The aluminium sinks to the bottom and is tapped off. At the anode, oxygen is produced: 2O²⁻ → O₂ + 4e⁻. The oxygen reacts with the graphite anodes to form CO₂, so the anodes need periodic replacement.
电解池使用钢制阴极和石墨阳极。熔融铝在阴极生成：Al³⁺ + 3e⁻ → Al。铝沉入底部并被抽出。在阳极生成氧气：2O²⁻ → O₂ + 4e⁻。氧气与石墨阳极反应生成 CO₂，因此阳极需要定期更换。
Key points for exams: cryolite reduces the melting point from over 2000°C to about 950°C; the process requires a huge amount of electricity, so aluminium smelters are often located near hydroelectric power sources. Aluminium is expensive because of the high energy demand.
考试要点：冰晶石将熔点从超过 2000°C 降低到约 950°C；该过程需要大量电能，因此铝冶炼厂通常建在水电站附近。铝昂贵的原因在于高能耗。
9. Electroplating | 电镀
Electroplating uses electrolysis to coat a metal object with a thin layer of another metal. The object to be plated is made the cathode, the plating metal is the anode, and the electrolyte contains ions of the plating metal.
电镀利用电解在金属物体表面镀上一层薄薄的另一种金属。待镀物件作为阴极，镀层金属作为阳极，电解质含有镀层金属的离子。
For example, to electroplate a steel fork with silver, the fork is the cathode, a silver bar is the anode, and the electrolyte is silver nitrate solution. At the cathode: Ag⁺ + e⁻ → Ag (silver deposited). At the anode: Ag → Ag⁺ + e⁻ (silver dissolves, replenishing the electrolyte). The concentration of Ag⁺ remains constant.
例如，给钢叉镀银，钢叉为阴极，银棒为阳极，电解质为硝酸银溶液。阴极：Ag⁺ + e⁻ → Ag（银沉积）。阳极：Ag → Ag⁺ + e⁻（银溶解，补充电解质）。Ag⁺ 浓度保持恒定。
Electroplating is used for decoration (jewellery, car parts) and protection against corrosion (zinc plating on iron, known as galvanising). It requires careful control of current and time to achieve an even coat.
电镀用于装饰（珠宝、汽车部件）和防腐蚀（铁上镀锌，即镀锌）。需要精确控制电流和时间以获得均匀镀层。
10. Purification of Copper | 铜的精炼
Copper purification is another application of electrolysis. Impure copper is used as the anode, a thin sheet of pure copper as the cathode, and copper(II) sulfate solution as the electrolyte.
铜的精炼是电解的又一应用。不纯的铜作为阳极，纯铜薄片作为阴极，硫酸铜溶液作为电解质。
At the anode, copper atoms lose electrons: Cu → Cu²⁺ + 2e⁻. Impure copper dissolves, and impurities such as silver and gold fall to the bottom as ‘anode sludge’, while more reactive metals like zinc and iron also dissolve but are not deposited at the cathode because they are more reactive than copper.
在阳极，铜原子失去电子：Cu → Cu²⁺ + 2e⁻。不纯铜溶解，银和金等杂质落到底部成为“阳极泥”，而锌、铁等更活泼的金属也溶解但不会在阴极析出，因为它们比铜更活泼。
At the cathode, Cu²⁺ ions gain electrons: Cu²⁺ + 2e⁻ → Cu. Pure copper is deposited, increasing the thickness of the cathode. The concentration of Cu²⁺ remains roughly unchanged. This process yields copper of very high purity ( > 99.9%).
在阴极，Cu²⁺ 离子得到电子：Cu²⁺ + 2e⁻ → Cu。纯铜被沉积，阴极增厚。Cu²⁺ 的浓度基本保持不变。此过程可得到纯度极高的铜（>99.9%）。
11. Factors Affecting Electrolysis and Quantitative Aspects | 影响电解的因素与定量关系
The amount of substance produced during electrolysis depends on the current and the time for which it flows. This relationship is expressed by Faraday’s laws, but for IGCSE OCR, qualitative understanding is sufficient: increasing current or time increases the mass of product at an electrode.
电解过程中生成物的量取决于电流强度和时间。这一关系由法拉第定律表达，但对于 IGCSE OCR 而言，定性理解即可：增大电流或延长通电时间会增加电极上生成物的质量。
The charge (Q) is calculated as current (I) multiplied by time (t): Q = I × t, where Q is in coulombs, I in amperes, and t in seconds. A higher charge transfers more electrons, leading to more product. Calculations involving Faraday constant may appear in extended papers.
电荷量（Q）等于电流（I）乘以时间（t）：Q = I × t，其中 Q 的单位是库仑，I 是安培，t 是秒。较高的电荷量传递更多电子，从而生成更多产物。涉及法拉第常数的计算可能出现在拓展试卷中。
Temperature and concentration can also influence the rate of electrolysis. Higher temperature increases ion mobility, while higher concentration provides more ions, often accelerating the process. However, these factors do not change the products, only the rate.
温度和浓度也会影响电解速率。较高温度增加离子迁移率，较高浓度提供更多离子，常会加速过程。但这些因素只改变速率，不改变产物。
12. Common Exam Questions and Tips | 常见考题与备考技巧
IGCSE OCR exam questions on electrolysis typically ask you to predict products at electrodes, write half equations, label apparatus, or explain observations using preferential discharge. You may also be asked to compare electrolytic cells with chemical cells (batteries).
IGCSE OCR 电解考题通常要求预测电极产物，书写半方程式，标注装置，或用优先放电解释观察结果。还可能会要求比较电解池与化学电池（原电池）。
Always identify the ions present in the electrolyte first, then apply the discharge series. For aqueous solutions, remember to include H⁺ and OH⁻ from water. Use the series to decide which cation and anion are discharged. If the exam mentions ‘inert electrodes’, assume no electrode participation; if copper or silver electrodes are specified, consider the anode dissolving.
首先确认电解质中存在的离子，然后应用放电顺序。对于水溶液，别忘了计入水中的 H⁺ 和 OH⁻。利用顺序决定哪种阳离子和阴离子放电。如果考题提到“惰性电极”，假设电极不参与反应；若指定了铜或银电极，则考虑阳极溶解。
Diagrams must show the power supply correctly oriented; label the anode as the positive electrode connected to the positive terminal. Arrows for ion movement must point correctly. In half equations, ensure charges and atoms balance and use the correct number of electrons. Finally, practice past paper questions, especially on aluminium extraction and brine electrolysis.
图示必须正确标出电源方向；阳极标注为正极，并连接到电源正极。离子移动箭头方向要正确。在半方程式中，确保电荷与原子配平并使用正确的电子数。最后，多做历年真题，尤其关注铝提取和盐水电解。
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