📚 Quantitative Calculations in Electrolysis | 电解的定量计算
Electrolysis is a cornerstone of A-Level electrochemistry. Quantitative calculations allow us to predict exactly how much product is formed from a given current and time, or how long a current must flow to deposit a desired mass. These calculations rest on Faraday’s laws and the simple relationship between charge, current and time.
电解是 A-Level 电化学的核心内容。定量计算帮助我们准确预测在给定电流与时间下能生成多少产物，或者为了获得指定质量的产物需要通多久的电。这类计算的基础是法拉第定律以及电荷、电流与时间之间的简单关系。
1. The Mole of Electrons | 电子的摩尔计量
The fundamental idea is that electrolysis is an electron-transfer process. An electric current is a flow of charge, and in an electrolytic cell this charge is carried by electrons in the external circuit and by ions moving through the electrolyte.
电解的本质是电子转移过程。电流是电荷的流动，在电解池中，电荷由外电路中的电子以及电解质中定向移动的离子共同传递。
Each mole of electrons carries a fixed amount of charge. This quantity is called the Faraday constant, F, and its value can be obtained from the Avogadro constant and the charge on a single electron.
每摩尔电子携带的电荷量是固定的，这个常数叫做法拉第常数 F。它的数值可以通过阿伏加德罗常数和单个电子所带电荷求得。
F = L × e ≈ (6.02 × 10²³) × (1.60 × 10⁻¹⁹) ≈ 96,500 C mol⁻¹
In A-Level calculations, F is usually taken as 96,500 C mol⁻¹. This means that when one mole of electrons is transferred in an electrolytic cell, 96,500 coulombs of charge have passed.
在 A-Level 计算中，通常取 F = 96,500 C mol⁻¹。这意味着当电解池中转移了 1 摩尔电子时，通过的电量就是 96,500 库仑。
Faraday’s first law states that the mass of a substance liberated at an electrode is directly proportional to the quantity of electricity passed. Faraday’s second law states that when the same quantity of electricity is passed through different electrolytes, the masses of substances liberated are proportional to their chemical equivalents, M/z.
法拉第第一定律指出，电极上析出物质的质量与通过的电量成正比。法拉第第二定律指出，当相同电量通过不同电解质时，析出物质的质量与其化学当量 M/z 成正比。
2. Key Equation Q = It | 核心公式 Q = It
The quantity of charge, Q, delivered by a steady current, I, in time, t, is given by:
恒定电流 I 在时间 t 内输送的电荷量 Q 可以表示为：
Q = I × t
Here Q is measured in coulombs (C), I in amperes (A), and t in seconds (s). If time is given in minutes, it must be converted to seconds by multiplying by 60.
其中 Q 的单位是库仑 (C)，I 的单位是安培 (A)，t 的单位是秒 (s)。如果时间以分钟给出，必须乘以 60 换算成秒。
For example, a current of 1.50 A flowing for 25.0 minutes delivers:
例如，1.50 A 的电流通过 25.0 分钟时，输送的电量为：
Q = 1.50 × 25.0 × 60 = 2,250 C
The equation can be rearranged to find current or time: I = Q/t and t = Q/I.
该公式可以变形为 I = Q/t 和 t = Q/I，用于求电流或时间。
3. Electrode Half-Reactions and Stoichiometry | 电极半反应与化学计量
Every electrolysis calculation must begin with the correct electrode half-equation. The half-equation tells us how many moles of electrons are needed to produce one mole of product.
任何电解计算都必须从正确的电极半反应开始。半反应告诉我们生成 1 摩尔产物需要多少摩尔电子。
| Half-reaction | 半反应 | Electrons per mole of product | 每摩尔产物所需电子数 |
| Ag⁺ + e⁻ → Ag | 1 |
| Cu²⁺ + 2e⁻ → Cu | 2 |
| Al³⁺ + 3e⁻ → Al | 3 |
| 2H⁺ + 2e⁻ → H₂ | 2 |
| 2Cl⁻ → Cl₂ + 2e⁻ | 2 |
| 4OH⁻ → O₂ + 2H₂O + 4e⁻ | 4 |
Notice that hydrogen, chlorine and oxygen are diatomic molecules. One mole of H₂, Cl₂ or O₂ requires two, two and four moles of electrons respectively, according to the half-equations above.
注意氢气、氯气和氧气都是双原子分子。根据上述半反应，生成 1 摩尔 H₂、Cl₂、O₂ 分别需要 2、2、4 摩尔电子。
4. Calculating Mass of Product | 产物质量的计算
To calculate the mass of a solid deposited at the cathode, use the following structured method:
要计算阴极上析出固体的质量，可以按以下步骤进行：
-
Write the balanced half-reaction for the product.
写出生成产物的配平半反应。
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Calculate Q from current and time.
由电流和时间计算 Q。
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Find moles of electrons: n(e⁻) = Q/F.
求电子的物质的量：n(e⁻) = Q/F。
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Use the half-reaction ratio to find moles of product.
利用半反应中的计量关系求产物的物质的量。
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Multiply by molar mass to find mass.
乘以摩尔质量得到质量。
Worked example | 例题： A current of 2.50 A is passed through a solution of copper(II) sulfate for 30.0 minutes. What mass of copper is deposited?
例题： 将 2.50 A 的电流通过硫酸铜溶液 30.0 分钟，求阴极上析出铜的质量。
Cu²⁺ + 2e⁻ → Cu
Q = 2.50 × 30.0 × 60 = 4,500 C
n(e⁻) = 4,500 / 96,500 = 0.0466 mol
n(Cu) = 0.0466 / 2 = 0.0233 mol
m(Cu) = 0.0233 × 63.5 = 1.48 g
The theoretical mass of copper deposited is 1.48 g.
理论上析出铜的质量为 1.48 g。
5. Calculating Gas Volumes at RTP | 室温常压下的气体体积计算
If the product is a gas, its volume can be calculated using the molar gas volume. At room temperature and pressure (r.t.p.), one mole of any gas occupies 24.0 dm³, which is 24,000 cm³.
如果生成物是气体，可以利用气体摩尔体积计算其体积。在室温常压 (r.t.p.) 下，1 摩尔任何气体体积为 24.0 dm³，即 24,000 cm³。
Volume at r.t.p. = moles of gas × 24.0 dm³ mol⁻¹
Worked example | 例题： A current of 0.50 A is passed through dilute sulfuric acid for 40.0 minutes. What volume of hydrogen gas is produced at the cathode?
例题： 将 0.50 A 的电流通过稀硫酸 40.0 分钟，在阴极上生成氢气的体积是多少？
2H⁺ + 2e⁻ → H₂
Q = 0.50 × 40.0 × 60 = 1,200 C
n(e⁻) = 1,200 / 96,500 = 0.0124 mol
n(H₂) = 0.0124 / 2 = 0.00622 mol
V(H₂) = 0.00622 × 24,000 = 149 cm³
Thus about 149 cm³ of hydrogen gas is produced.
因此大约生成 149 cm³ 氢气。
6. Calculating Time or Current Required | 所需时间或电流的计算
Sometimes the mass of product is fixed and we need to find the time or current. The same equations are used in reverse.
有时产物质量已经给定，需要求通电时间或电流。这时只需反向使用这些公式。
t = n(e⁻) × F / I
I = n(e⁻) × F / t
Worked example | 例题： How long must a current of 0.75 A flow to deposit 0.500 g of silver from a silver nitrate solution? M(Ag) = 108 g mol⁻¹.
例题： 在硝酸银溶液中，用 0.75 A 的电流沉积 0.500 g 银，需要多长时间？M(Ag) = 108 g mol⁻¹。
Ag⁺ + e⁻ → Ag
n(Ag) = 0.500 / 108 = 0.00463 mol
n(e⁻) = 0.00463 mol
Q = 0.00463 × 96,500 = 447 C
t = 447 / 0.75 = 596 s ≈ 9.9 min
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