📚 A-Level CIE Chemistry: Electrolysis Key Concepts | A-Level CIE 化学：电解 考点精讲
Electrolysis is a core topic in CIE A-Level Chemistry (9701), combining redox theory, energetics and quantitative calculations. It appears frequently in structured and multiple-choice questions, requiring a precise understanding of electrode reactions, ion discharge priorities and Faraday's laws. Mastery of practical applications such as aluminium extraction and the chlor-alkali industry is also essential. This revision guide distills every key point you need for the exam, with clear English‑Chinese parallel explanations.
电解是 CIE A-Level 化学（9701）的核心章节，融合了氧化还原理论、能量学和定量计算。它在简答题和选择题中反复出现，要求精确掌握电极反应、离子放电顺序和法拉第定律。掌握铝的冶炼、氯碱工业等实际应用也必不可少。这份复习指南以英中对照的形式提炼了考试必备的每一个关键点。
1. What is Electrolysis? | 什么是电解？
Electrolysis is the process of using a direct electric current to bring about a non‑spontaneous chemical reaction. An electrolytic cell consists of two electrodes connected to a power supply and immersed in an electrolyte.
电解是利用直流电引发非自发化学反应的过程。电解池由连接电源并浸入电解质中的两个电极组成。
Unlike a galvanic cell, where chemical energy is converted into electrical energy, an electrolytic cell converts electrical energy into chemical energy. The anode is the positive electrode (oxidation occurs), and the cathode is the negative electrode (reduction occurs).
与原电池将化学能转化为电能不同，电解池将电能转化为化学能。阳极是正极（发生氧化反应），阴极是负极（发生还原反应）。
For electrolysis to occur, the electrolyte must contain mobile ions – either as a molten ionic compound or as an aqueous solution of an ionic compound. The external circuit supplies electrons to the cathode and removes them from the anode.
电解的发生要求电解质含有可自由移动的离子——呈熔融态离子化合物或其水溶液。外电路向阴极提供电子并从阳极移走电子。
Electrons flow through the external circuit from the anode to the cathode, while cations migrate to the cathode and anions migrate to the anode inside the cell.
电子在外电路中从阳极流向阴极，而在电解池内部阳离子移向阴极，阴离子移向阳极。
2. Components and Conduction | 电解池的组成与导电方式
An electrolytic cell must have: (i) a d.c. power source, (ii) two electrodes (inert or active), (iii) an electrolyte containing free ions, and (iv) connecting wires. The electrodes are typically graphite (inert) or metals such as copper, platinum or steel.
电解池必须包含：(i) 直流电源，(ii) 两个电极（惰性或活性），(iii) 含自由离子的电解质，(iv) 导线。电极通常为石墨（惰性）或铜、铂、钢等金属。
Conduction in the electrolyte occurs by the movement of ions, not electrons. Cations move to the cathode to gain electrons (reduction), while anions move to the anode to lose electrons (oxidation). This ionic conduction completes the circuit internally.
电解质中的导电靠离子移动而非电子。阳离子移向阴极获得电子（还原），阴离子移向阳极失去电子（氧化）。这种离子导电使内部电路完整。
In the external circuit, electrons are pushed from the anode to the cathode by the power supply. The anode is positive because it is connected to the positive terminal of the battery, which attracts anions. The cathode is negative, attracting cations.
在外电路中，电源推动电子从阳极流向阴极。阳极为正极，因为它连接电池正极，吸引阴离子；阴极为负极，吸引阳离子。
3. Electrode Reactions and Discharge Series | 电极反应与放电顺序
At each electrode, the ion that is discharged is the one that is most readily reduced (at the cathode) or most readily oxidised (at the anode), depending on the nature of the electrolyte, the concentration and the electrode material.
在每个电极上，优先放电的离子是最容易还原的（阴极）或最容易氧化的（阳极），这取决于电解质的种类、浓度和电极材料。
The cathode discharge series for aqueous solutions:
K⁺ < Na⁺ < Ca²⁺ < Mg²⁺ < Al³⁺ < (H⁺ from water) < Zn²⁺ < Fe²⁺ < Sn²⁺ < Pb²⁺ < H⁺ (from acid) < Cu²⁺ < Ag⁺ < Au³⁺
水溶液中阳离子放电顺序：
K⁺ < Na⁺ < Ca²⁺ < Mg²⁺ < Al³⁺ < (水中 H⁺) < Zn²⁺ < Fe²⁺ < Sn²⁺ < Pb²⁺ < (酸中 H⁺) < Cu²⁺ < Ag⁺ < Au³⁺
H₂ + 2OH⁻. In a concentrated solution of salts of less reactive metals, the metal ion itself may be discharged.
对于极活泼金属的阳离子（K 至 Al），实际放电的是水分子，产生 H₂ 和 OH⁻：2H₂O + 2e⁻ → H₂ + 2OH⁻。在较不活泼金属盐的浓溶液中，金属离子自身可能放电。
The anode discharge series for common anions:
SO₄²⁻ ~ NO₃⁻ ~ F⁻ < OH⁻ < Cl⁻ < Br⁻ < I⁻
2H₂O → O₂ + 4H⁺ + 4e⁻, unless a halide ion is present in sufficiently high concentration.
常见阴离子放电顺序：
SO₄²⁻ ~ NO₃⁻ ~ F⁻ < OH⁻ < Cl⁻ < Br⁻ < I⁻
硫酸根、硝酸根和氟离子在水溶液中不易放电，而是水分子被氧化：2H₂O → O₂ + 4H⁺ + 4e⁻，除非卤离子浓度足够高。
| Cathode (reduction) | Observation |
|---|---|
| Cu²⁺ + 2e⁻ → Cu | Pink-brown copper deposit |
| 2H⁺ + 2e⁻ → H₂ | Colourless gas, 'pop' with lighted splint |
| 2H₂O + 2e⁻ → H₂ + 2OH⁻ | Gas at cathode, solution becomes alkaline |
| Anode (oxidation) | Observation |
|---|---|
| 2Cl⁻ → Cl₂ + 2e⁻ | Greenish-yellow gas, bleaches damp litmus |
| 4OH⁻ → O₂ + 2H₂O + 4e⁻ | Colourless gas, relights glowing splint |
| 2H₂O → O₂ + 4H⁺ + 4e⁻ | Gas at anode, solution becomes acidic |
4. Molten Electrolytes vs Aqueous Solutions | 熔融电解质与溶液电解的比较
When a molten ionic compound is electrolysed, there is no water present, so the only ions present are those of the compound. The metal cation is reduced at the cathode and the non‑metal anion is oxidised at the anode.
电解熔融离子化合物时，不存在水，因此体系中只有该化合物的离子。金属阳离子在阴极被还原，非金属阴离子在阳极被氧化。
ields lead metal at the cathode and bromine gas at the anode. Electrode reactions: Cathode – Pb²⁺ + 2e⁻ → Pb(l); Anode – 2Br⁻ → Br₂(g) + 2e⁻.
例如，熔融溴化铅 PbBr₂ 在阴极产生金属铅，阳极产生溴气。电极反应：阴极 – Pb²⁺ + 2e⁻ → Pb(l)；阳极 – 2Br⁻ → Br₂(g) + 2e⁻。
In aqueous solutions, water molecules also provide H⁺ and OH⁻ ions through self‑ionisation, which compete for discharge. The product at each electrode depends on the relative positions of the ions in the discharge series and their concentrations.
在水溶液中，水分子通过自电离提供 H⁺ 和 OH⁻ 参与竞争放电。每个电极的产物取决于离子在放电顺序中的相对位置及其浓度。
Consequently, electrolysis of aqueous copper(II) sulfate with inert electrodes gives copper at the cathode (Cu²⁺ discharged) and oxygen at the anode (OH⁻ discharged, since SO₄²⁻ does not discharge).
因此，用惰性电极电解硫酸铜水溶液时，阴极析出铜（Cu²⁺ 放电），阳极析出氧气（OH⁻ 放电，SO₄²⁻ 不放电）。
5. Effect of Concentration | 浓度对产物的影响
us sodium chloride, Cl⁻ is discharged at the anode forming chlorine gas: 2Cl⁻ → Cl₂ + 2e⁻.
对于阴离子，高浓度的卤离子会优先于氢氧根放电。在浓氯化钠水溶液中，阳极 Cl⁻ 放电生成氯气：2Cl⁻ → Cl₂ + 2e⁻。
O₂ + 2H₂O + 4e⁻. The cathode always gives hydrogen in both cases: 2H₂O + 2e⁻ → H₂ + 2OH⁻.
而在稀氯化钠溶液中，阳极放电的是 OH⁻，产生氧气：4OH⁻ → O₂ + 2H₂O + 4e⁻。两种情况下阴极都产生氢气：2H₂O + 2e⁻ → H₂ + 2OH⁻。
For cations, changing the concentration (e.g., Cu²⁺ concentration) does not usually alter the discharge product when inert electrodes are used, because Cu²⁺ is much easier to reduce than H⁺ from water.
对阳离子而言，改变浓度（如 Cu²⁺ 浓度）通常不会改变惰性电极下的放电产物，因为 Cu²⁺ 比水中 H⁺ 容易还原得多。
However, in the chlor‑alkali cell, a concentrated brine is essential to obtain chlorine at the anode; otherwise, oxygen would form, contaminating the product.
但在氯碱工业中，必须使用浓盐水才能使阳极得到氯气；否则会产生氧气，污染产物。
6. Inert and Active Electrodes | 惰性电极与活性电极
Inert electrodes (graphite, carbon or platinum) do not participate in the reactions. They simply provide a surface for electron transfer. Products depend solely on the electrolyte.
惰性电极（石墨、碳或铂）本身不参与反应，仅提供电子转移的表面。产物完全取决于电解质。
he impure copper anode dissolves: Cu(s) → Cu²⁺ + 2e⁻. Pure copper deposits on the cathode.
活性电极，如铜精炼中的铜阳极，自身可能被氧化。在铜的电精炼中，不纯的铜阳极发生溶解：Cu(s) → Cu²⁺ + 2e⁻。纯铜在阴极析出。
When an active anode is used in aqueous copper(II) sulfate (e.g., two copper electrodes), the anode loses mass as copper goes into solution, while the cathode gains the same amount of pure copper. The concentration of Cu²⁺ remains constant.
若在硫酸铜水溶液中使用铜作电极，阳极不断溶解，质量减少，阴极易析出等质量的纯铜。Cu²⁺ 的浓度保持不变。
In the Hall‑Hé uced: C(s) + O₂(g) → CO₂(g), so the carbon anode is active and must be replaced regularly.
在霍尔-埃鲁铝冶炼法中，碳阳极与生成的氧气反应被消耗：C(s) + O₂(g) → CO₂(g)，因此碳阳极是活性的，需定期更换。
7. Faraday's Laws and Quantitative Electrolysis | 法拉第定律与定量电解
Faraday's First Law: The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electric charge passed.
法拉第第一定律：电极上析出或释放的物质质量与通过的电量成正比。
Q = I × t
Q = I × t
where Q is the charge in coulombs (C), I is the current in amperes (A) and t is the time in seconds (s).
其中 Q 为电量（库仑 C），I 为电流（安培 A），t 为时间（秒 s）。
Faraday's Second Law: When the same quantity of electricity is passed through different electrolytes, the masses of substances produced are proportional to their equivalent masses (M/n, where n is the number of electrons in the half‑equation).
法拉第第二定律：相同的电量通过不同电解质时，各电极产物的质量与它们的当量质量（M/n，n 为半反应电子数）成正比。
n(e⁻) = Q / F
n(e⁻) = Q / F
here F = 96 500 C mol⁻¹ (Faraday constant). Then the mass deposited m = (Q × M) / (n × F).
其中 F = 96 500 C mol⁻¹（法拉第常数）。则析出质量 m = (Q × M) / (n × F)。
(I × t × 24) / (n × F) dm³, provided the gas is formed at a rate of n(e⁻) per mol of gas.
对于常温常压下的气体，体积 V = [n(e⁻) × 24] / n = (I × t × 24) / (n × F) dm³，条件是每摩尔气体对应 n(e⁻) 的电子转移。
Exam questions often combine electrolysis with acid‑base back titrations or ask for the thickness of a plating. Always convert time to seconds and use the stoichiometric ratio from the half‑equation.
考题常将电解与酸碱返滴定结合，或要求计算镀层厚度。务必将时间换算成秒，并根据半方程中的计量比进行计算。
8. Electroplating and Refining Copper | 电镀与铜的精炼
In electroplating, an object is made the cathode and the plating metal the anode, both in a solution of the metal's ions. During plating, the anode dissolves, replenishing the metal ions, and the object gets coated evenly.
电镀时，被镀物件作阴极，镀层金属作阳极，同置于该金属离子的溶液中。电镀过程中阳极溶解，补充金属离子，物件表面得到均匀镀层。
In copper refining, the impure copper anode contains impurities such as Zn, Fe, Ag and Au. More reactive metals (Zn, Fe) also dissolve, but less reactive ones (Ag, Au) fall off as anode mud.
铜精炼时，不纯铜阳极含有 Zn、Fe、Ag、Au 等杂质。较活泼金属（Zn、Fe）也溶解，而较不活泼的（Ag、Au）脱落成为阳极泥。
At the cathode, only Cu²⁺ is discharged because it lies below H⁺ in the series and the voltage is controlled. The net result is transfer of pure copper from anode to cathode.
阴极上只有 Cu²⁺ 放电，因为它位于放电顺序中 H⁺ 之下且电压受到控制。最终结果是纯铜从阳极转移到阴极。
The electrolyte, typically aqueous CuSO₄ acidified with H₂SO₄, remains unchanged in Cu²⁺ concentration because the rate of dissolution of Cu at the anode equals the rate of deposition at the cathode.
电解质通常为酸化 CuSO₄ 溶液，Cu²⁺ 浓度保持不变，因为阳极铜溶解速率与阴极析出速率相等。
9. Extraction of Aluminium (Hall‑Héroult Process) | 铝的冶炼（霍尔-埃鲁法）
Aluminium is extracted by electrolysis of purified Al₂O₃ dissolved in molten cryolite, Na₃AlF₆, at about 950 °C. Cryolite lowers the melting point and improves conductivity, making the process economically viable.
铝通过电解溶于熔融冰晶石（Na₃AlF₆）中的纯氧化铝 Al₂O₃ 进行冶炼，温度约 950 °C。冰晶石降低熔点并提高导电性，使该法具备经济可行性。
l). Liquid aluminium is denser than the electrolyte and collects at the bottom of the cell.
阴极反应：Al³⁺ + 3e⁻ → Al(l)。液态铝密度大于电解质，聚集于电解槽底部。
, forming CO₂: C(s) + O₂(g) → CO₂(g). The overall cell reaction: 2Al₂O₃ + 3C → 4Al + 3CO₂.
阳极反应：2O²⁻ → O₂ + 4e⁻。氧气立即与碳阳极反应生成 CO₂：C(s) + O₂(g) → CO₂(g)。总反应：2Al₂O₃ + 3C → 4Al + 3CO₂。
Because the carbon anodes are consumed, they must be continually replaced. The cell operates continuously, with fresh alumina added at the top.
由于碳阳极被消耗，必须定期更换。电解槽连续运行，从顶部添加新鲜氧化铝。
A major exam focus is explaining why aluminium cannot be extracted by carbon reduction (it is more reactive than carbon) and the role of cryolite (lowering the operating temperature and energy costs).
考点常涉及为什么铝不能通过碳还原法
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