📚 Electrolysis and Electrolytic Cells | 电解原理与电解池
Electrolysis is the decomposition of a compound into its elements by passing a direct electric current through it. The process takes place in an electrolytic cell, where electrical energy is converted into chemical energy. This article covers the fundamental principles, competitive reactions, Faraday’s laws, and common applications required for the CIE A-Level Chemistry syllabus.
电解是指通过直流电流使化合物分解为其组成元素的过程。该过程在电解池中发生,将电能转化为化学能。本文围绕 CIE A-Level 化学大纲,系统讲解电解的基本原理、离子放电竞争、法拉第定律及常见应用。
1. Structure of an Electrolytic Cell | 电解池的结构
An electrolytic cell consists of two electrodes (anode and cathode), an electrolyte, and a DC power supply. The cathode is connected to the negative terminal of the battery, and the anode is connected to the positive terminal. Electrons flow from the battery into the cathode, then through the external circuit from the anode back to the battery.
电解池包含两个电极(阳极和阴极)、电解质和直流电源。阴极连接电源负极,阳极连接电源正极。电子从电源负极流入阴极,在阳极经外电路返回电源正极。
- Cathode (negative electrode): supplies electrons to cations, causing reduction.
- 阴极(负极):向阳离子提供电子,使其发生还原反应。
- Anode (positive electrode): accepts electrons from anions or from the electrode itself, causing oxidation.
- 阳极(正极):从阴离子或电极本身接受电子,使其发生氧化反应。
- Electrolyte: a molten ionic compound or an aqueous solution containing mobile ions.
- 电解质:熔融离子化合物或含有可自由移动离子的水溶液。
Oxidation occurs at the anode; Reduction occurs at the cathode.
阳极发生氧化;阴极发生还原。
A common mnemonic is “An Ox, Red Cat” — anode is oxidation, cathode is reduction.
常用的助记短语是 “An Ox, Red Cat”——阳极氧化,阴极还原。
2. Electrode Types: Inert vs. Reactive | 电极类型:惰性电极与活性电极
Electrodes can be inert or reactive. Inert electrodes, such as graphite and platinum, do not participate in the chemical reaction; they only transfer electrons. Reactive electrodes, such as copper or zinc, may themselves undergo oxidation, especially during electrorefining or electroplating.
电极分为惰性电极和活性电极。惰性电极如石墨和铂,不参与化学反应,仅传递电子。活性电极如铜或锌,可能自身发生氧化,常见于电解精炼或电镀中。
| Inert Electrode | 惰性电极 | Reactive Electrode | 活性电极 |
| Carbon (graphite), platinum | Copper, zinc, nickel, silver |
| Does not react; only electrocatalyses | May dissolve or deposit during electrolysis |
3. Electrolysis of Molten Ionic Compounds | 熔融离子化合物的电解
In a molten salt, ions are free to move. For example, molten sodium chloride (NaCl) is electrolysed using graphite electrodes. At the cathode, sodium ions are reduced to sodium metal; at the anode, chloride ions are oxidised to chlorine gas.
熔融盐中离子可以自由移动。例如,使用石墨电极电解熔融氯化钠:阴极上钠离子被还原为金属钠;阳极上氯离子被氧化为氯气。
Cathode: 2Na⁺ + 2e⁻ → 2Na
阴极:2Na⁺ + 2e⁻ → 2Na
Anode: 2Cl⁻ → Cl₂ + 2e⁻
阳极:2Cl⁻ → Cl₂ + 2e⁻
Overall reaction: 2NaCl(l) → 2Na(s) + Cl₂(g). This method is used to extract highly reactive metals like sodium, magnesium, and aluminium from their molten ores.
总反应:2NaCl(l) → 2Na(s) + Cl₂(g)。该方法用于从熔融矿石中提取高活性金属,如钠、镁和铝。
4. Electrolysis of Aqueous Solutions: Competition | 水溶液电解:离子放电竞争
In aqueous solution, water can also be reduced or oxidised. Therefore, more than one cation and more than one anion are present, leading to competition for discharge.
在水溶液中,水本身也可能被还原或氧化,因此存在多种阳离子和多种阴离子,发生放电竞争。
- At the cathode (reduction): the cation with the least negative (or most positive) standard electrode potential is discharged. The order usually follows the electrochemical series: Ag⁺ > Cu²⁺ > H⁺ > Fe²⁺ > Zn²⁺ > Na⁺ > K⁺.
- 在阴极(还原):标准电极电势最高(或最不负)的阳离子优先放电。常见顺序:Ag⁺ > Cu²⁺ > H⁺ > Fe²⁺ > Zn²⁺ > Na⁺ > K⁺。
- At the anode (oxidation): if the anion is a halide (Cl⁻, Br⁻, I⁻), it is usually oxidised to the halogen. Otherwise, water is oxidised to oxygen gas.
- 在阳极(氧化):如果阴离子是卤离子(Cl⁻、Br⁻、I⁻),通常被氧化为卤素;否则水被氧化为氧气。
Anode (water oxidation): 2H₂O → O₂ + 4H⁺ + 4e⁻
阳极(水的氧化):2H₂O → O₂ + 4H⁺ + 4e⁻
5. Electrolysis of Water | 电解水
Pure water is a very weak electrolyte. To electrolyse water, a small amount of dilute sulfuric acid or sodium sulfate is added to increase conductivity. At the cathode hydrogen gas is produced, and at the anode oxygen gas is produced.
纯水是极弱的电解质。为增强导电性,常加入少量稀硫酸或硫酸钠。电解时阴极产生氢气,阳极产生氧气。
Cathode: 4H⁺ + 4e⁻ → 2H₂
阴极:4H⁺ + 4e⁻ → 2H₂
Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻
阳极:2H₂O → O₂ + 4H⁺ + 4e⁻
The volume of hydrogen collected is twice the volume of oxygen, because the mole ratio is 2:1.
收集到的氢气体积是氧气的两倍,因为它们的物质的量之比为 2:1。
6. Electrolysis of Concentrated Sodium Chloride Solution | 浓氯化钠溶液的电解(氯碱工业)
When concentrated aqueous NaCl is electrolysed with inert electrodes, the possible cathode reactions are reduction of Na⁺ or water; the possible anode reactions are oxidation of Cl⁻ or water. In practice, H₂ is produced at the cathode and Cl₂ at the anode, leaving NaOH in solution.
使用惰性电极电解浓 NaCl 溶液时,阴极可能的反应是 Na⁺ 或水的还原;阳极可能的反应是 Cl⁻ 或水的氧化。实际上阴极产生 H₂,阳极产生 Cl₂,溶液中剩余 NaOH。
Cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻
阴极:2H₂O + 2e⁻ → H₂ + 2OH⁻
Anode: 2Cl⁻ → Cl₂ + 2e⁻
阳极:2Cl⁻ → Cl₂ + 2e⁻
Overall: 2NaCl(aq) + 2H₂O(l) → Cl₂(g) + H₂(g) + 2NaOH(aq). This is the basis of the chlor-alkali industry.
总反应:2NaCl(aq) + 2H₂O(l) → Cl₂(g) + H₂(g) + 2NaOH(aq)。这是氯碱工业的基础。
If the solution is dilute, oxygen is produced at the anode instead of chlorine, because water molecules outnumber chloride ions and, under dilute conditions, OH⁻ is preferentially oxidised.
如果溶液为稀溶液,阳极将产生氧气而非氯气,因为稀溶液中水分子远多于氯离子,OH⁻ 优先被氧化。
7. Electrorefining of Copper | 铜的电解精炼
Copper obtained from smelting is impure (about 99%). Electrorefining removes impurities and produces pure copper (99.99%).
冶炼得到的粗铜纯度约为 99%。通过电解精炼可除去杂质并获得纯铜(99.99%)。
- Anode: impure copper; copper atoms lose electrons and dissolve as Cu²⁺ into solution.
- 阳极为粗铜:铜原子失去电子,以 Cu²⁺ 形式进入溶液。
- Cathode: pure copper; Cu²⁺ ions gain electrons and deposit as pure copper metal.
- 阴极为纯铜:Cu²⁺ 获得电子,以纯铜金属形式沉积。
- Electrolyte: CuSO₄ solution acidified with H₂SO₄.
- 电解质:加入 H₂SO₄ 酸化的 CuSO₄ 溶液。
Anode: Cu → Cu²⁺ + 2e⁻
阳极:Cu → Cu²⁺ + 2e⁻
Cathode: Cu²⁺ + 2e⁻ → Cu
阴极:Cu²⁺ + 2e⁻ → Cu
Solution concentration remains roughly constant. Less reactive metals (Ag, Au) fall as anode mud; more reactive metals (Fe, Zn) stay in solution.
溶液浓度基本保持不变。较不活泼的金属(Ag、Au)以阳极泥形式沉降;较活泼的金属(Fe、Zn)留在溶液中。
8. Faraday’s Laws of Electrolysis | 法拉第电解定律
The amount of substance produced during electrolysis is directly proportional to the quantity of electric charge passed through the cell.
电解过程中生成物的物质的量与通过电解池的电量成正比。
Q = I × t
Q = I × t
where Q is charge in coulombs (C), I is current in amperes (A), and t is time in seconds (s). One mole of electrons carries a charge equal to the Faraday constant, F ≈ 96500 C mol⁻¹.
其中 Q 为电量(库仑 C),I 为电流(安培 A),t 为时间(秒 s)。一摩尔电子所带的电荷量等于法拉第常数 F ≈ 96500 C mol⁻¹。
n(e⁻) = It / F
n(e⁻) = It / F
The mass of a substance deposited or liberated is calculated from the half-equation stoichiometry. For example, to deposit 1 mol of Cu requires 2 mol of electrons, so m = (It / F) × (M / z), where z is the number of electrons transferred per ion.
沉积或释放物质的量可根据半反应计量关系计算。例如,沉积 1 mol Cu 需要 2 mol 电子,因此 m = (It / F) × (M / z),其中 z 是每个离子转移的电子数。
9. Quantitative Example | 定量计算示例
Calculate the mass of copper deposited at the cathode when a current of 2.00 A flows through CuSO₄ solution for 30.0 minutes. (Ar(Cu) = 63.5, F = 96500 C mol⁻¹)
计算当 2.00 A 电流通过 CuSO₄ 溶液 30.0 分钟时,阴极沉积铜的质量。(Cu 的相对原子质量为 63.5,F = 96500 C mol⁻¹)
Q = I × t = 2.00 × 30.0 × 60 = 3600 C
Q = I × t = 2.00 × 30.0 × 60 = 3600 C
n(e⁻) = Q / F = 3600 / 96500 = 0.0373 mol
n(e⁻) = Q / F = 3600 / 96500 = 0.0373 mol
Half-reaction: Cu²⁺ + 2e⁻ → Cu; n(Cu) = n(e⁻) / 2 = 0.01865 mol
半反应:Cu²⁺ + 2e⁻ → Cu;n(Cu) = n(e⁻) / 2 = 0.01865 mol
m(Cu) = 0.01865 × 63.5 = 1.18 g
m(Cu) = 0.01865 × 63.5 = 1.18 g
Be careful to convert minutes to seconds and to use the correct stoichiometric factor.
注意将分钟转换为秒,并使用正确的化学计量系数。
10. Common Mistakes and Exam Tips | 常见错误与考试提示
Candidates often confuse anode/cathode names or write incorrect half-equations. Here are high-yield reminders:
考生经常混淆阳极/阴极名称或写错半反应式。以下为高频考点提醒:
- Always state the electrode: anode is oxidation, cathode is reduction.
- 务必说明电极:阳极为氧化,阴极为还原。
- In aqueous electrolysis, check which species is preferentially discharged using the electrochemical series and concentration effects.
- 在水溶液电解中,需结合电化学顺序和浓度效应判断优先放电的物种。
- Write half-equations separately, balancing atoms and charge with electrons.
- 分别写出半反应式,利用电子配平原子和电荷。
- For Faraday calculations, always calculate n(e⁻) = It/F first, then apply stoichiometry.
- 法拉第计算中,先求 n(e⁻) = It/F,再进行化学计量换算。
- Remember that reactive electrodes can dissolve: e.g., copper anode in electrorefining.
- 记住活性电极可溶解:例如电解精炼中的铜阳极。
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