📚 Quantitative Electrolysis | 定量电解
Quantitative electrolysis links electric current, time, and the charge on one mole of electrons to the mass, volume, or concentration of products formed at electrodes. By using Faraday's laws and balanced half-equations, you can predict exactly how much metal is deposited or how much gas is evolved.
定量电解将电流、时间和每摩尔电子的电荷与电极上产物的质量、体积或浓度联系起来。利用法拉第定律和配平的半反应，你可以准确预测沉积多少金属或放出多少气体。
The topic sits at the junction of two big ideas in A-Level chemistry: electrode reactions (redox and half-equations) and the mole concept (amount of substance and stoichiometry). Once you can convert charge into moles of electrons, every quantitative electrolysis question becomes a short chain of simple steps.
这个主题处于 A-Level 化学两大核心思想的交汇处：电极反应（氧化还原与半反应）和物质的量（摩尔与化学计量）。一旦你学会把电荷转换成电子物质的量，所有定量电解题目都会变成一串简短的步骤。
1. Electrolysis and Charge | 电解与电荷
Electrolysis is the decomposition of a molten or dissolved ionic compound by a direct electric current. Oxidation occurs at the anode and reduction occurs at the cathode, because anions migrate to the positive anode and cations migrate to the negative cathode.
电解是直流电使熔融或溶解的离子化合物分解的过程。阳极发生氧化，阴极发生还原，因为阴离子移向正极的阳极，阳离子移向负极的阴极。
The quantity of product formed depends on the total charge passed through the cell, not on the size of the electrodes or the volume of the electrolyte. The charge Q is measured in coulombs (C).
生成的产物量取决于通过电池的总电荷，而与电极的大小或电解液的体积无关。电荷 Q 的单位是库仑（C）。
elow. Note that 1 A = 1 C s⁻¹, so current multiplied by time automatically gives coulombs.
如果稳定电流 I 以安培（A）为单位，时间 t 以秒（s）为单位，则总电荷由下式给出。注意 1 A = 1 C s⁻¹，因此电流乘以时间直接得到库仑数。
Q = I × t
mmon conversions include 1 hour = 3600 s, 1 minute = 60 s, and 1 milliampere = 1 × 10⁻³ A.
使用该式前一定要把时间换算成秒。常见换算是 1 小时 = 3600 s，1 分钟 = 60 s，1 毫安 = 1 × 10⁻³ A。
- Charge: Q = I × t, unit C.
- Current: I = Q ÷ t, unit A.
- Time: t = Q ÷ I, unit s.
- 物质的量关系：Q = I × t，单位 C。
- 电流：I = Q ÷ t，单位 A。
- 时间：t = Q ÷ I，单位 s。
2. Faraday's First Law | 法拉第第一定律
Faraday's first law states that the mass of a substance produced at an electrode is directly proportional to the total electric charge passed through the electrolyte. Doubling the charge doubles the mass deposited.
法拉第第一定律指出，电极上生成物质的质量与通过电解质的总电荷成正比。电荷加倍，沉积的质量也加倍。
This is because each ion requires a fixed number of electrons to become a neutral atom or molecule. More charge means more electrons, and therefore more product is formed at the electrode surface.
这是因为每个离子变成中性原子或分子需要固定数量的电子。电荷越多，电子越多，因此在电极表面生成的产物也越多。
For example, if a copper(II) sulfate solution is electrolysed with inert electrodes, doubling the current with the same time, or doubling the time with the same current, doubles the mass of copper deposited at the cathode.
例如，用惰性电极电解硫酸铜(II)溶液时，在时间不变的情况下电流加倍，或电流不变的情况下时间加倍，都会使阴极析出的铜质量加倍。
Strictly speaking, the law assumes that a single reduction or oxidation reaction occurs with 100 percent current efficiency. In real cells some charge may be consumed by side reactions, such as reduction of dissolved oxygen or reduction of water instead of the metal ion.
严格来说，该定律假设只发生单一的还原或氧化反应，且电流效率为 100%。在真实的电池中，部分电荷可能被副反应消耗，例如溶解氧的还原或水被还原而不是金属离子被还原。
3. The Mole of Electrons and the Faraday Constant | 电子摩尔与法拉第常数
ol⁻¹ (more precisely 96485 C mol⁻¹). It is the product of the Avogadro constant and the elementary charge, as shown below.
要进行计算，我们需要知道 1 摩尔电子所带的电荷。这就是法拉第常数 F = 96500 C mol⁻¹（更精确为 96485 C mol⁻¹）。它是阿伏加德罗常数与元电荷的乘积，如下式所示。
F = NA × e = 6.02 × 10²³ mol⁻¹ × 1.60 × 10⁻¹⁹ C = 96500 C mol⁻¹
The number of moles of electrons transferred is therefore the total charge divided by the Faraday constant. This single relationship is the bridge between electricity and chemistry.
因此，转移的电子的物质的量等于总电荷除以法拉第常数。这一个关系式就是电学与化学之间的桥梁。
n(e⁻) = Q ÷ F = (I × t) ÷ 96500
by the Avogadro constant to get moles of electrons, then multiply by F to get the charge.
如果题目直接给出电子数，例如 1.20 × 10²² 个电子，先除以阿伏加德罗常数得到电子的物质的量，再乘以 F 得到电荷。
4. Half-Equations and Electron Stoichiometry | 半反应与电子计量
Every half-equation tells you how many moles of electrons are needed to form one mole of product. That number, often called z or n(e⁻) per mole, is the conversion factor between moles of electrons and moles of substance.
每个半反应都告诉你生成 1 摩尔产物需要多少摩尔电子。这个数字（常写作 z，或每摩尔产物所需电子的物质的量）就是电子物质的量与物质物质的量之间的换算系数。
n(substance) = n(e⁻) ÷ z
The table below lists the half-equations most frequently used in A-Level quantitative electrolysis questions, together with the value of z for each product.
下表列出了 A-Level 定量电解题目中最常用的半反应，以及每种产物对应的 z 值。
| Half-equation | Electrode | z |
|---|---|---|
| Ag⁺ + e⁻ → Ag | cathode | 1 |
| Cu²⁺ + 2e⁻ → Cu | cathode | 2 |
| Pb²⁺ + 2e⁻ → Pb | cathode | 2 |
| Al³⁺ + 3e⁻ → Al | cathode | 3 |
| 2H⁺ + 2e⁻ → H₂ | cathode | 2 |
| 2Cl⁻ → Cl₂ + 2e⁻ | anode | 2 |
| 4OH⁻ → O₂ + 2H₂O + 4e⁻ | anode | 4 |
| 2H₂O → O₂ + 4H⁺ + 4e⁻ | anode | 4 |
O₂, so z = 4. Always balance the half-equation first and then read off the electron count.
注意电子数是相对于每摩尔产物而言，而不是每摩尔离子。对于氧气，4 摩尔电子生成 1 摩尔 O₂，所以 z = 4。一定要先配平半反应，再读出电子数。
5. Faraday's Second Law | 法拉第第二定律
Faraday's second law states that when the same quantity of charge is passed through different electrolytes, the amounts of the substances liberated at the electrodes are in the ratio of their molar masses divided by the number of electrons in the half-equation.
法拉第第二定律指出，当相同的电荷通过不同的电解质时，电极上析出的各物质的量之比，等于它们的摩尔质量除以半反应中电子数的比。
n(substance) = Q ÷ (z × F), and m = (Q × M) ÷ (z × F)
of electrons. Using 1 mole of electrons (1 F = 96500 C) gives the following simple values.
M ÷ z 有时被称为该物质的当量质量，即 1 摩尔电子所沉积的质量。使用 1 摩尔电子（1 F = 96500 C）可得到以下简单数值。
- 1 F deposits 1 mol of Ag = 108 g of silver.
- 1 F deposits 0.5 mol of Cu = 0.5 × 63.5 = 31.75 g of copper.
- 1 F deposits 0.5 mol of H₂ = 1.0 g of hydrogen gas.
- 1 F deposits 1 mol of 银 = 108 g 银。
- 1 F 沉积 0.5 mol 铜 = 0.5 × 63.5 = 31.75 g 铜。
- 1 F 沉积 0.5 mol 氢气 = 1.0 g 氢气。
This law explains why silver electroplating is fast and economical: only one mole of electrons is needed per mole of silver, whereas aluminium requires three moles of electrons per mole of metal. That is one reason aluminium extraction is so energy intensive.
这一定律解释了为什么镀银既快又经济：每摩尔银只需 1 摩尔电子，而每摩尔铝需要 3 摩尔电子。这也是铝的电解提取如此耗能的原因之一。
6. Deciding the Electrode Reactions in Aqueous Solution | 水溶液中电极反应的判断
Faraday's laws only work after you have identified the actual electrode reactions. In aqueous solutions, water and dissolved ions compete, so you must use electrode potentials and concentration effects to decide which species is discharged.
法拉第定律只有在你确定了实际电极反应之后才适用。在水溶液中，水分子与溶解的离子会竞争放电，因此必须利用电极电势和浓度效应判断哪种微粒放电。
At the cathode, the species with the more positive (less negative) reduction potential is usually reduced first. At the anode, the species with the more negative (less positive) oxidation potential is usually oxidised first, unless a reactive metal anode dissolves instead.
在阴极，还原电势较正（较不负）的微粒通常先被还原。在阳极，氧化电势较负（较不正）的微粒通常先被氧化，除非可反应金属阳极自身溶解。
- ): 2H⁺ + 2e⁻ → H₂ at the cathode, 2Cl⁻ → Cl₂ + 2e⁻ at the anode; NaOH remains in solution.
- Dilute H₂SO₄(aq): 2H⁺ + 2e⁻ → H₂ at the cathode, 2H₂O → O₂ + 4H⁺ + 4e⁻ at the anode.
- inert electrodes: Cu²⁺ + 2e⁻ → Cu at the cathode, O₂ evolved at the anode and H⁺ produced.
- 浓 NaCl(aq)（卤水）：阴极 2H⁺ + 2e⁻ → H₂，阳极 2Cl⁻ → Cl₂ + 2e⁻；溶液中剩余 NaOH。
- 稀 H₂SO₄(aq)：阴极 2H⁺ + 2e⁻ → H₂，阳极 2H₂O → O₂ + 4H⁺ + 4e⁻。
- CuSO₄(aq) 配惰性电极：阴极 Cu²⁺ + 2e⁻ → Cu，阳极放出 O₂ 并生成 H⁺。
en is evolved. This is the basis of copper refining and electroplating, where the concentration of the electrolyte stays almost constant.
若在硫酸铜溶液中使用铜阳极，阳极自身溶解（Cu → Cu²⁺ + 2e⁻），不放出氧气。这就是铜的精炼和电镀的基础，此时电解液浓度几乎保持不变。
7. Worked Example 1: Mass of Copper Deposited | 例题一：沉积铜的质量
mass of copper deposited at the cathode. Take F = 96500 C mol⁻¹ and M(Cu) = 63.5 g mol⁻¹.
用惰性电极，将 2.00 A 的电流通过硫酸铜(II)溶液 30.0 分钟。计算阴极沉积铜的质量。取 F = 96500 C mol⁻¹，M(Cu) = 63.5 g mol⁻¹。
utes = 30.0 × 60 = 1800 s. Step 2: calculate the charge, Q = I × t = 2.00 × 1800 = 3600 C.
第一步：把时间换算成秒。30.0 分钟 = 30.0 × 60 = 1800 s。第二步：计算电荷，Q = I × t = 2.00 × 1800 = 3600 C。
l. Step 4: use the half-equation Cu²⁺ + 2e⁻ → Cu, so n(Cu) = n(e⁻) ÷ 2 = 1.865 × 10⁻² mol.
第三步：计算电子的物质的量，n(e⁻) = Q ÷ F = 3600 ÷ 96500 = 3.73 × 10⁻² mol。第四步：根据半反应 Cu²⁺ + 2e⁻ → Cu，得 n(Cu) = n(e⁻) ÷ 2 = 1.865 × 10⁻² mol。
The anode reaction is 2H₂O → O₂ + 4H⁺ + 4e⁻, so n(O₂) = 3.73 × 10⁻² ÷ 4 = 9.33 × 10⁻³ mol.
第五步：计算质量，m = n × M = 1.865 × 10⁻² × 63.5 = 1.18 g。阳极反应为 2H₂O → O₂ + 4H⁺ + 4e⁻，故 n(O₂) = 3.73 × 10⁻² ÷ 4 = 9.33 × 10⁻³ mol。
n(e⁻) = (I × t) ÷ F → n(substance) = n(e⁻) ÷ z → m = n × M
Rearranging the same relationships lets you answer reverse questions, such as finding the current needed to deposit 5.00 g of copper in 1.00 hour, or the time needed for a given mass.
把同样的关系式变形，就能回答逆向问题，例如求 1.00 小时内沉积 5.00 g 铜所需的电流，或沉积一定质量所需的时间。
8. Worked Example 2: Gas Volumes and the 2:1 Ratio | 例题二：气体体积与 2:1 之比
Suppose the same charge of 3600 C is passed through dilute sulfuric acid instead. Hydrogen is formed at the cathode and oxygen at the anode. Find the volume of each gas at room temperature and pressure, where the molar gas volume is 24 dm³ mol⁻¹.
假设同样的 3600 C 电荷改为通过稀硫酸。阴极生成氢气，阳极生成氧气。求室温常压下每种气体的体积，此时摩尔气体体积为 24 dm³ mol⁻¹。
3 × 10⁻² mol. For hydrogen, 2H⁺ + 2e⁻ → H₂, so n(H₂) = 3.73 × 10⁻² ÷ 2 = 1.865 × 10⁻² mol.
电子的物质的量不变：3.73 × 10⁻² mol。对于氢气，2H⁺ + 2e⁻ → H₂，故 n(H₂) = 3.73 × 10⁻² ÷ 2 = 1.865 × 10⁻² mol。
For oxygen, 2H₂O → O₂ + 4H
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