Electrochemistry for IGCSE CIE Chemistry | IGCSE CIE 化学：电化学考点精讲

📚 Electrochemistry for IGCSE CIE Chemistry | IGCSE CIE 化学：电化学考点精讲

Electrochemistry is the study of chemical reactions that involve the transfer of electrons. It covers two main areas: electrolysis, where electrical energy drives non-spontaneous chemical changes, and electrochemical cells (including simple cells and fuel cells), where chemical energy is converted into electrical energy. For IGCSE CIE Chemistry, you must be able to predict products of electrolysis for molten and aqueous electrolytes, explain industrial processes like electroplating and aluminium extraction, and understand the hydrogen-oxygen fuel cell.

电化学是研究涉及电子转移的化学反应的一门学科。它主要包括两个领域：电解，即电能驱动非自发的化学变化；以及电化学电池（包括简易电池和燃料电池），将化学能转化为电能。在 IGCSE CIE 化学中，你必须能够预测熔融态和水溶液电解质的电解产物，解释电镀与铝的提取等工业过程，并理解氢氧燃料电池。

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1. What is Electrolysis? | 什么是电解？

Electrolysis is the decomposition of an ionic compound, either molten or in aqueous solution, by the passage of an electric current. A direct current (d.c.) power supply forces electrons into one electrode and removes them from the other. This causes non-spontaneous redox reactions to occur at the electrodes.

电解是通过电流使离子化合物（熔融态或水溶液）发生分解的过程。直流电源将电子强行推入一个电极，并从另一电极移除电子，从而使非自发的氧化还原反应在电极上发生。

The overall process converts electrical energy into chemical energy. Ions must be free to move; this is why the electrolyte is either molten or in solution — the ions are mobile and can carry the current.

整个电解过程将电能转化为化学能。离子必须能够自由移动，因此电解质要么是熔融态，要么是溶液状态——离子可以流动并能传导电流。

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2. Key Terminology: Electrodes, Electrolyte and Ions | 关键术语：电极、电解质与离子

The electrolyte is the molten or aqueous ionic compound that conducts electricity and is decomposed. The cathode is the negative electrode, connected to the negative terminal of the power supply. Cations (positive ions) migrate to the cathode and gain electrons (reduction). The anode is the positive electrode, connected to the positive terminal. Anions (negative ions) migrate to the anode and lose electrons (oxidation).

电解质是能导电并发生分解的熔融态或水溶液离子化合物。阴极是与电源负极相连的负极，阳离子（正离子）移向阴极并获得电子（还原）。阳极是与电源正极相连的正极，阴离子（负离子）移向阳极并失去电子（氧化）。

Remember: Red Cat (Reduction at the Cathode) and An Ox (Anode, Oxidation). In electrolysis, the cathode supplies electrons, so reduction occurs; the anode removes electrons, so oxidation occurs.

记住：Red Cat（阴极发生还原）和 An Ox（阳极发生氧化）。在电解中，阴极提供电子，发生还原；阳极提取电子，发生氧化。

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3. Electrolysis of Molten Lead(II) Bromide | 熔融溴化铅的电解

Lead(II) bromide, PbBr₂, is an ionic solid that does not conduct electricity because the ions are held tightly in a lattice. When heated until molten, the ions become mobile. During electrolysis using inert graphite electrodes:

溴化铅 PbBr₂ 是一种离子固体，因离子被紧紧束缚在晶格中而不能导电。当加热至熔融状态时，离子变得可以自由移动。使用惰性石墨电极进行电解时：

• Cathode (reduction): Pb²⁺ + 2e⁻ → Pb (molten lead collects at the bottom)

  阴极（还原）：Pb²⁺ + 2e⁻ → Pb（熔融铅聚集在底部）

• Anode (oxidation): 2Br⁻ → Br₂ + 2e⁻ (brown bromine gas is released)

  阳极（氧化）：2Br⁻ → Br₂ + 2e⁻（释放红棕色的溴气）

The overall reaction is: PbBr₂(l) → Pb(l) + Br₂(g). This is a decomposition reaction driven by electricity.

总反应为：PbBr₂(l) → Pb(l) + Br₂(g)。这是一个由电能驱动的分解反应。

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4. Electrolysis of Molten Sodium Chloride | 熔融氯化钠的电解

Molten sodium chloride (NaCl) is used to extract reactive sodium metal and chlorine gas. The electrolyte contains Na⁺ and Cl⁻ ions only. Products:

电解熔融氯化钠 (NaCl) 用于提取活泼金属钠和氯气。电解质中仅含有 Na⁺ 和 Cl⁻ 离子。产物：

• Cathode: Na⁺ + e⁻ → Na (sodium metal, collected under an inert atmosphere as it is highly reactive)

  阴极：Na⁺ + e⁻ → Na（钠金属，因性质极为活泼需在惰性气氛下收集）

• Anode: 2Cl⁻ → Cl₂ + 2e⁻ (chlorine gas)

  阳极：2Cl⁻ → Cl₂ + 2e⁻（氯气）

No other ions are present, so the prediction is straightforward. The melting point of NaCl is high, so calcium chloride is often added to lower it and save energy.

因为没有其他离子存在，产物的预测十分直接。NaCl 的熔点很高，因此常加入氯化钙以降低熔点、节约能源。

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5. Electrolysis of Aqueous Solutions: General Rules | 水溶液的电解：一般规则

When an ionic compound is dissolved in water, the solution contains not only the ions from the compound but also H⁺ and OH⁻ ions from the self-ionisation of water. At each electrode, more than one ion may be attracted, so we must use the discharge series to determine which ion is preferentially discharged.

当离子化合物溶解于水中时，溶液中不仅存在来自溶质的离子，还存在由水的自解离产生的 H⁺ 和 OH⁻ 离子。在每个电极处，可能有一种以上离子被吸引，因此我们需要用放电顺序来判断哪种离子优先放电。

The ion more easily discharged (lower in the series) will react. The concentration of the solution can also affect the discharge order for anions — a concentrated chloride solution, for example, can produce chlorine instead of oxygen.

越容易放电的离子（在顺序中位置越低）越优先反应。溶液的浓度也会影响阴离子的放电顺序——例如，浓的氯化物溶液可能产生氯气而非氧气。

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6. Discharge Series for Cations and Anions | 阳离子与阴离子的放电顺序

The ease of discharge depends on the reactivity of the element and the position in the electrochemical series. For CIE IGCSE, the key rules are:

放电的容易程度取决于元素的活泼性及其在电化学系列中的位置。CIE IGCSE 的关键规则如下：

Cations	Ease of discharge
K⁺, Na⁺, Ca²⁺, Mg²⁺, Al³⁺	Never discharged in aqueous solution; H⁺ from water is reduced instead
Zn²⁺, Fe²⁺, Pb²⁺	Can be discharged if no more reactive cations present
Cu²⁺, Ag⁺	Discharged easily

Anions	Ease of discharge
SO₄²⁻, NO₃⁻	Very difficult to discharge; OH⁻ oxidised instead → O₂ + 2H₂O + 4e⁻
Cl⁻, Br⁻, I⁻	Can be discharged if concentrated; otherwise OH⁻ oxidised
OH⁻	Easily oxidised to oxygen: 4OH⁻ → O₂ + 2H₂O + 4e⁻

Cathode (reduction) product in aqueous solution: if the metal is more reactive than manganese, hydrogen gas is produced. Anode (oxidation) product: if halide is present and concentrated, halogen is produced; otherwise oxygen from OH⁻.

水溶液中阴极（还原）产物：若金属活泼性比锰更强，则产生氢气。阳极（氧化）产物：若存在卤离子且浓度较大，则产生卤素；否则由 OH⁻ 氧化产生氧气。

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7. Electrolysis of Concentrated Sodium Chloride Solution | 浓氯化钠溶液的电解

Concentrated brine (NaCl(aq)) contains Na⁺, Cl⁻, H⁺ and OH⁻ ions. According to the discharge series:

浓盐水 (NaCl(aq)) 含有 Na⁺、Cl⁻、H⁺ 和 OH⁻ 离子。根据放电顺序：

• Cathode: H⁺ is discharged because Na⁺ is too reactive. 2H⁺ + 2e⁻ → H₂ (hydrogen gas bubbles off)

  阴极：H⁺ 放电，因为 Na⁺ 太活泼。2H⁺ + 2e⁻ → H₂（氢气气泡逸出）

• Anode: Cl⁻ is discharged because the solution is concentrated. 2Cl⁻ → Cl₂ + 2e⁻ (chlorine gas)

  阳极：Cl⁻ 放电，因为溶液浓度高。2Cl⁻ → Cl₂ + 2e⁻（氯气）

The remaining solution becomes sodium hydroxide (NaOH), an important industrial product. This process is used in the chlor-alkali industry.

残留的溶液变成氢氧化钠 (NaOH)，这是一种重要的工业产品。该工艺被用于氯碱工业。

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8. Electrolysis of Copper(II) Sulfate Solution | 硫酸铜溶液的电解

Copper(II) sulfate solution can be electrolysed using two types of electrodes: inert (e.g., graphite or platinum) and reactive copper electrodes. The products differ significantly.

电解硫酸铜溶液可以分别采用惰性电极（如石墨或铂）和活泼性铜电极。产物有很大差异。

• With inert electrodes (graphite): Cathode: Cu²⁺ + 2e⁻ → Cu (pink-brown copper metal deposits). Anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻ (oxygen gas, as SO₄²⁻ is not discharged). The blue colour fades as Cu²⁺ ions are removed.

  使用惰性电极（石墨）时： 阴极：Cu²⁺ + 2e⁻ → Cu（粉棕色铜金属析出）。阳极：4OH⁻ → O₂ + 2H₂O + 4e⁻（产生氧气，因为 SO₄²⁻ 不被放电）。随着 Cu²⁺ 被移除，蓝色变浅。

• With copper electrodes: Cathode: Cu²⁺ + 2e⁻ → Cu (copper is deposited, mass increases). Anode: Cu → Cu²⁺ + 2e⁻ (copper dissolves, mass decreases). The concentration of Cu²⁺ remains constant, and the colour of the solution does not fade. This is the basis of copper electrorefining and electroplating.

  使用铜电极时： 阴极：Cu²⁺ + 2e⁻ → Cu（铜被沉积，质量增加）。阳极：Cu → Cu²⁺ + 2e⁻（铜溶解，质量减少）。溶液中 Cu²⁺ 的浓度保持恒定，颜色不变。这是铜电解精炼和电镀的基础。

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9. Electroplating and Electrorefining | 电镀与电解精炼

Electroplating uses electrolysis to coat a metal object with a thin layer of another metal, typically for corrosion resistance or decoration. The object to be plated is made the cathode, the plating metal is the anode, and the electrolyte contains ions of the plating metal. For silver plating: anode is silver, cathode is the object, electrolyte is silver nitrate solution.

电镀利用电解在金属物体表面覆盖一层薄的其他金属，通常为了抗腐蚀或装饰。被镀物件作为阴极，镀层金属作为阳极，电解质含有该镀层金属的离子。以镀银为例：阳极为银，阴极为物件，电解质为硝酸银溶液。

Electrorefining of copper: Impure copper is made the anode, pure copper is the cathode, and the electrolyte is copper(II) sulfate with some sulfuric acid. During electrolysis, copper dissolves from the anode and pure copper deposits on the cathode. Impurities like gold and silver fall as ‘anode slime’ and are collected.

铜的电解精炼： 将不纯的铜作阳极，纯铜作阴极，电解质为含少量硫酸的硫酸铜溶液。电解过程中，铜从阳极溶解，纯铜沉积在阴极。金、银等杂质以“阳极泥”形式掉落并被收集。

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10. Extraction of Aluminium from Bauxite | 从铝土矿中提取铝

Aluminium is extracted by electrolysis of purified aluminium oxide (Al₂O₃) dissolved in molten cryolite (Na₃AlF₆). The cryolite lowers the melting point from over 2000 °C to about 950 °C, saving energy and reducing costs. The process uses carbon anodes and a carbon-lined steel cathode.

铝是通过电解溶解在熔融冰晶石 (Na₃AlF₆) 中的纯净氧化铝 (Al₂O₃) 提取的。冰晶石将熔点从 2000 °C 以上降低到约 950 °C，从而节约能源并降低成本。该工艺使用碳阳极和碳衬里的钢阴极。

• Cathode (reduction): Al³⁺ + 3e⁻ → Al (molten aluminium collects at the bottom and is tapped off)

  阴极（还原）：Al³⁺ + 3e⁻ → Al（熔融铝聚集在底部并被出料）

• Anode (oxidation): 2O²⁻ → O₂ + 4e⁻; the oxygen produced reacts with the carbon anodes: C + O₂ → CO₂, so the anodes are gradually consumed and need replacement.

  阳极（氧化）：2O²⁻ → O₂ + 4e⁻；产生的氧气与碳阳极反应：C + O₂ → CO₂，因此阳极逐渐消耗，需要更换。

Overall reaction: 2Al₂O₃(l) → 4Al(l) + 3O₂(g). This process is known as the Hall-Héroult process.

总反应：2Al₂O₃(l) → 4Al(l) + 3O₂(g)。此过程称为霍尔-埃鲁法。

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11. Simple Cells and the Hydrogen Fuel Cell | 简易电池与氢燃料电池

In a simple cell, two different metals are dipped into an electrolyte. The more reactive metal acts as the negative electrode (loses electrons) and the less reactive metal is the positive electrode. Electrons flow through the external circuit. For example, a zinc-copper cell in dilute sulfuric acid produces about 1.0 V; zinc dissolves while hydrogen gas forms on the copper electrode.

在简易电池中，将两种不同金属浸入电解质中。较活泼的金属充当负极（失去电子），较不活泼的金属为正极。电子通过外电路流动。例如，锌-铜电池在稀硫酸中可产生约 1.0 V 电压；锌溶解，而氢气在铜电极上产生。

The hydrogen-oxygen fuel cell converts chemical energy from the reaction of hydrogen and oxygen directly into electrical energy, with water as the only product. In an alkaline electrolyte:

氢氧燃料电池将氢和氧反应的化学能直接转化为电能，唯一的产物是水。在碱性电解质中：

• Anode (negative, oxidation): 2H₂ + 4OH⁻ → 4H₂O + 4e⁻

  阳极（负极，氧化）：2H₂ + 4OH⁻ → 4H₂O + 4e⁻

• Cathode (positive, reduction): O₂ + 2H₂O + 4e⁻ → 4OH⁻

  阴极（正极，还原）：O₂ + 2H₂O + 4e⁻ → 4OH⁻

Overall: 2H₂ + O₂ → 2H₂O. Fuel cells do not need recharging and provide continuous electricity as long as fuel and oxygen are supplied. They are efficient and produce no pollutants (just water), making them suitable for vehicles and spacecraft.

总反应：2H₂ + O₂ → 2H₂O。燃料电池不需要充电，只要持续供应燃料和氧气就能连续发电。它们效率高，不产生污染物（仅水），因此适用于车辆和航天器。

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12. Quantitative Electrolysis (Extended Only) | 定量电解（仅限扩展）

The amount of substance produced at an electrode is directly proportional to the quantity of electric charge passed. The charge (Q, in coulombs) is given by Q = I × t, where I is current in amperes and t is time in seconds. The constant 1 faraday (F) = 96 500 C mol⁻¹ of electrons.

在电极上生成的物质质量与通过的电量成正比。电量（Q，单位为库仑）由 Q = I × t 给出，其中 I 为电流（安培），t 为时间（秒）。1 法拉第常数 (F) = 96 500 C mol⁻¹ 电子。

Steps to calculate the mass of a product:

1. Q = I × t ; 2. moles of electrons = Q / F ; 3. use half-equation to find moles of substance ; 4. mass = moles × Mᵣ

计算产物质量的步骤：

1. Q = I × t ; 2. 电子的物质的量 = Q / F ; 3. 利用半反应求物质物质的量 ; 4. 质量 = 物质的量 × Mᵣ

Example: Calculate the mass of copper deposited when a current of 2.0 A is passed through CuSO₄(aq) for 1 hour. Q = 2.0 × 3600 = 7200 C. Moles of e⁻ = 7200 / 96 500 ≈ 0.0746 mol. Cu²⁺ + 2e⁻ → Cu, so 2 mol e⁻ give 1 mol Cu. Moles of Cu = 0.0746 / 2 = 0.0373 mol. Mass of Cu = 0.0373 × 63.5 ≈ 2.37 g.

例题：将 2.0 A 电流通入 CuSO₄(aq) 1 小时，求析出铜的质量。Q = 2.0 × 3600 = 7200 C。电子物质的量 = 7200 / 96 500 ≈ 0.0746 mol。Cu²⁺ + 2e⁻ → Cu，故 2 mol e⁻ 产生 1 mol Cu。Cu 的物质的量 = 0.0746 / 2 = 0.0373 mol。Cu 的质量 = 0.0373 × 63.5 ≈ 2.37 g。

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