Electrolysis in IGCSE Edexcel Chemistry | IGCSE Edexcel 化学：电解考点精讲

📚 Electrolysis in IGCSE Edexcel Chemistry | IGCSE Edexcel 化学：电解考点精讲

Electrolysis is a process that uses a direct electric current to drive a non-spontaneous chemical reaction. In IGCSE Chemistry, it is a core topic that connects the abstract concepts of ions, electrons and oxidation-reduction to real-world applications like aluminium extraction and electroplating. Understanding electrolysis means mastering the layout of an electrolytic cell, predicting products at inert and active electrodes, and writing balanced half-equations for both molten and aqueous electrolytes.

电解是利用直流电驱动非自发化学反应的过程。在IGCSE化学中，它是连接离子、电子和氧化还原等抽象概念与实际应用（如铝的提取和电镀）的核心主题。理解电解意味着掌握电解池的构造，预测惰性电极和活性电极上的产物，并为熔融态和水溶液电解质书写配平的半反应方程式。

1. Defining Electrolysis and Key Components | 电解的定义与关键组成

Electrolysis is the decomposition of an ionic compound, either molten or in aqueous solution, by the passage of an electric current. The substance that is broken down is called the electrolyte. The two rods or plates that carry the current into the electrolyte are called electrodes. The positive electrode is the anode and the negative electrode is the cathode.

电解是指通过电流使熔融或水溶液中的离子化合物分解的过程。被分解的物质称为电解质。将电流导入电解质的两根棒或板称为电极。正极是阳极，负极是阴极。

In an electrolytic cell, the power source (battery or DC supply) pushes electrons onto the cathode, giving it a negative charge. At the same time, electrons are drawn away from the anode, making it positive. This creates an electric field that forces ions in the electrolyte to move towards the oppositely charged electrode. Cations (positive ions) migrate to the cathode, and anions (negative ions) migrate to the anode.

在电解池中，电源（电池或直流电源）将电子推向阴极，使其带负电。同时，电子从阳极被抽走，使其带正电。这样就产生了一个电场，迫使电解质中的离子向带相反电荷的电极移动。阳离子（正离子）向阴极迁移，阴离子（负离子）向阳极迁移。

The key difference between a chemical cell and an electrolytic cell is that a chemical cell uses spontaneous redox reactions to generate electricity, while an electrolytic cell uses electricity to force a non-spontaneous redox reaction to occur. In IGCSE, the focus is firmly on the electrolytic cell.

化学电池与电解池的主要区别在于：化学电池利用自发的氧化还原反应产生电能，而电解池则利用电能迫使非自发的氧化还原反应发生。在IGCSE中，重点明确放在电解池上。

2. Movement of Ions and Electron Flow | 离子移动与电子流动

During electrolysis, current is carried through the wires and electrodes by the flow of electrons, but through the electrolyte itself the current is carried by the movement of mobile ions. In the external circuit, electrons flow from the negative terminal of the power source to the cathode, and then from the anode back to the positive terminal. This direction of electron flow is from negative to positive.

在电解过程中，电流在导线和电极中通过电子的流动来传导，而在电解质中则通过可移动离子的运动来传导。在外部电路中，电子从电源的负极流向阴极，然后从阳极流回正极。电子流动的方向是从负极到正极。

Inside the electrolyte, cations are attracted to the cathode and gain electrons there, a process called reduction. Anions are attracted to the anode and lose electrons there, a process called oxidation. You can remember this with the mnemonic “RED CAT, AN OX”: Reduction occurs at the Cathode, and Oxidation occurs at the Anode.

在电解质内部，阳离子被吸引到阴极并在那里获得电子，这称为还原。阴离子被吸引到阳极并在那里失去电子，这称为氧化。你可以用 “RED CAT, AN OX” 这个记忆口诀：还原发生在阴极 (Reduction at Cathode)，氧化发生在阳极 (Oxidation at Anode)。

For any electrolysis experiment, you must be able to sketch a simple cell, label the electrodes (anode as +, cathode as -), show the electrolyte, the external d.c. supply, and arrows indicating the direction of electron flow in the wires and the movement of ions in the electrolyte.

对于任何电解实验，你必须能够画出简单的电解池示意图，标出电极（阳极为+，阴极为-）、电解质、外部直流电源，并用箭头标出导线中的电子流动方向和电解质中的离子移动方向。

3. Electrolysis of Molten Ionic Compounds | 熔融离子化合物的电解

When a solid ionic compound is melted, the ionic lattice breaks down and the ions become free to move. This is a necessary condition for electrolysis, because solid ionic compounds do not conduct electricity. Molten lead(II) bromide (PbBr₂) is a classic example used in school laboratories to demonstrate electrolysis.

当固体离子化合物熔化时，离子晶格被打破，离子可以自由移动。这是电解的必要条件，因为固体离子化合物不导电。熔融溴化铅 (PbBr₂) 是学校实验室常用来演示电解的经典例子。

In molten lead(II) bromide, only two types of ion are present: Pb²⁺ and Br⁻. The Pb²⁺ ions move to the cathode, gain electrons and are reduced to form a molten lead layer at the bottom of the crucible. This can be observed as a grey, shiny metal. The Br⁻ ions move to the anode, lose electrons and are oxidised to form bromine molecules, which appear as a reddish-brown gas.

在熔融溴化铅中，只存在两种离子：Pb²⁺ 和 Br⁻。Pb²⁺ 离子移向阴极，获得电子并被还原，在坩埚底部形成一层熔融的铅。可以观察到这是一种灰色、闪亮的金属。Br⁻ 离子移向阳极，失去电子并被氧化，生成溴分子，表现为红棕色的气体。

The half-equations for this process are: at the cathode, reduction:
Pb²⁺ + 2e⁻ → Pb;
at the anode, oxidation:
2Br⁻ → Br₂ + 2e⁻.
The products of electrolysing a molten binary compound are always the element itself at the cathode and the non-metal element at the anode.

该过程的半反应方程式为：在阴极，还原反应：Pb²⁺ + 2e⁻ → Pb；在阳极，氧化反应：2Br⁻ → Br₂ + 2e⁻。电解熔融二元化合物的产物总是：阴极得到金属单质，阳极得到非金属单质。

4. Electrolysis of Aqueous Solutions – The Role of Water | 水溶液电解 – 水的角色

When an ionic compound is dissolved in water, the solution contains not only the ions from the solute but also H⁺ and OH⁻ ions from the partial ionisation of water:
H₂O ⇌ H⁺ + OH⁻.
This makes predicting products more complex, because more than one type of cation and anion is present.

当离子化合物溶于水时，溶液中不仅含有溶质解离出的离子，还含有水部分电离产生的 H⁺ 和 OH⁻ 离子：H₂O ⇌ H⁺ + OH⁻。这使得产物预测更加复杂，因为存在不止一种阳离子和阴离子。

At the cathode, two reduction reactions are possible: the cation from the ionic compound can gain electrons, or the H⁺ from water can gain electrons to form hydrogen gas. At the anode, two oxidation reactions are possible: the anion from the ionic compound can lose electrons, or the OH⁻ from water can lose electrons to form oxygen gas and water.

在阴极，可能发生两种还原反应：来自离子化合物的阳离子获得电子，或者来自水的 H⁺ 获得电子生成氢气。在阳极，可能发生两种氧化反应：来自离子化合物的阴离子失去电子，或者来自水的 OH⁻ 失去电子生成氧气和水。

The actual products depend on the selective discharge rules, which are based on the reactivity series for cations and a set of anion preferences for anions. Understanding these rules is a core IGCSE exam requirement.

实际产物取决于选择性放电规则：对阳离子而言，取决于金属活动性顺序；对阴离子而言，则有一套优先顺序。掌握这些规则是IGCSE考试的核心要求。

5. Selective Discharge at the Cathode | 阴极的选择性放电

At the cathode, the cation that is preferentially discharged is the one that is least reactive in the reactivity series. This is because less reactive metals have a greater tendency to gain electrons and form neutral atoms. Hydrogen (H⁺) is treated as if it were a metal in this context and is considered more reactive than copper (Cu), silver (Ag) and gold (Au), but less reactive than metals above lead (Pb) in the reactivity series.

在阴极，优先放电的阳离子是金属活动性顺序中最不活泼的离子。这是因为较不活泼的金属更容易获得电子形成中性原子。在这个语境中，氢离子 H⁺ 被视为一种“金属”，其活泼性在铜 (Cu)、银 (Ag) 和金 (Au) 之上，但在铅 (Pb) 以上的金属之下。

The rule for cathode discharge in aqueous solution is:

If the metal cation is from potassium (K⁺) to aluminium (Al³⁺) inclusive, or is a Group 1 or Group 2 metal, then hydrogen gas (H₂) is produced from H⁺ ions, because these metals are more reactive than hydrogen.
If the metal cation is from zinc (Zn²⁺) to lead (Pb²⁺) inclusive, both the metal and hydrogen can be produced, but in IGCSE Edexcel you are usually told to expect hydrogen gas unless the solution is very concentrated – but check the specific syllabus past papers.
If the metal cation is copper (Cu²⁺), silver (Ag⁺) or gold (Au³⁺), the pure metal itself is deposited at the cathode because these metals are less reactive than hydrogen.

对水溶液中阴极放电的规则是：

如果金属阳离子是钾 (K⁺) 到铝 (Al³⁺)（包含），或者属于第1族或第2族金属，则产物为氢气 (H₂)，由 H⁺ 离子还原得到，因为这些金属比氢更活泼。
如果金属阳离子是锌 (Zn²⁺) 到铅 (Pb²⁺)（包含），则金属和氢气都可能生成，但在IGCSE Edexcel中通常要求预期产物为氢气，除非溶液非常浓——但应查阅具体大纲的历年真题确认。
如果金属阳离子是铜 (Cu²⁺)、银 (Ag⁺) 或金 (Au³⁺)，则纯金属本身会沉积在阴极上，因为这些金属的活泼性比氢弱。

This can be summarised in the half-equations: for reactive metal solutions, cathode reaction: 2H⁺ + 2e⁻ → H₂. For less reactive metals, cathode reaction: e.g. Cu²⁺ + 2e⁻ → Cu.

这可以用半反应方程式来总结：对于活泼金属的溶液，阴极反应为：2H⁺ + 2e⁻ → H₂。对于较不活泼的金属，阴极反应为：例如 Cu²⁺ + 2e⁻ → Cu。

6. Selective Discharge at the Anode | 阳极的选择性放电

At the anode, the anion that is discharged is the one that is most easily oxidised. The order of ease of discharge is:

Halide ions (Cl⁻, Br⁻, I⁻) are discharged in preference to OH⁻ when present in concentrated solution. In dilute solution, OH⁻ may be discharged if the halide concentration is very low, but for IGCSE Edexcel, if a halide is present, the halogen is usually formed at the anode.
Hydroxide ions (OH⁻) from water are discharged to give oxygen gas and water: 4OH⁻ → O₂ + 2H₂O + 4e⁻.
Sulfate ions (SO₄²⁻) and nitrate ions (NO₃⁻) are never discharged in aqueous solution because OH⁻ is oxidised more easily. If the anion is sulfate or nitrate, oxygen gas is produced at the anode.

在阳极，放电的阴离子是最容易被氧化的那种。放电容易程度的顺序为：

卤化物离子 (Cl⁻, Br⁻, I⁻) 在浓溶液中优先于 OH⁻ 放电。在稀溶液中，如果卤化物浓度非常低，可能放电 OH⁻，但对于IGCSE Edexcel而言，如果存在卤化物，通常会在阳极生成卤素单质。
来自水的氢氧根离子 (OH⁻) 放电生成氧气和水：4OH⁻ → O₂ + 2H₂O + 4e⁻。
硫酸根离子 (SO₄²⁻) 和硝酸根离子 (NO₃⁻) 在水溶液中永远不会放电，因为 OH⁻ 更容易被氧化。如果阴离子是硫酸根或硝酸根，阳极产物为氧气。

To predict the anode product: if the electrolyte contains a halide ion (chloride, bromide, iodide), you will see the halogen (chlorine, bromine, iodine) produced. If the anion is sulfate, nitrate, or something else that does not discharge easily, oxygen is produced. The half-equation for chloride ions: 2Cl⁻ → Cl₂ + 2e⁻; for bromide: 2Br⁻ → Br₂ + 2e⁻; for hydroxide: 4OH⁻ → O₂ + 2H₂O + 4e⁻.

预测阳极产物：如果电解质中含有卤离子（氯离子、溴离子、碘离子），就会产生卤素（氯气、溴蒸气、碘）。如果阴离子是硫酸根、硝酸根或其他不易放电的离子，则产生氧气。氯离子的半反应方程式为：2Cl⁻ → Cl₂ + 2e⁻；溴离子：2Br⁻ → Br₂ + 2e⁻；氢氧根：4OH⁻ → O₂ + 2H₂O + 4e⁻。

7. Worked Example: Electrolysis of Aqueous Sodium Chloride | 例题：氯化钠水溶液的电解

Consider the electrolysis of concentrated aqueous sodium chloride (brine) using inert carbon electrodes. The ions present are: Na⁺, Cl⁻, H⁺ and OH⁻. At the cathode, we compare Na⁺ and H⁺. Sodium is more reactive than hydrogen (Na is high in the reactivity series), so H⁺ is discharged. Bubbles of colourless hydrogen gas are seen. Cathode half-equation: 2H⁺ + 2e⁻ → H₂.

以使用惰性碳电极电解浓氯化钠水溶液（盐水）为例。溶液中存在 Na⁺、Cl⁻、H⁺ 和 OH⁻ 离子。在阴极，我们比较 Na⁺ 和 H⁺。钠比氢更活泼（钠在活动性顺序中位置很高），所以 H⁺ 放电，可看到无色氢气气泡产生。阴极半反应方程式：2H⁺ + 2e⁻ → H₂。

At the anode, we compare Cl⁻ and OH⁻. In a concentrated solution, Cl⁻ ions are discharged in preference to OH⁻. Pale green chlorine gas is seen bubbling off. Anode half-equation: 2Cl⁻ → Cl₂ + 2e⁻. The solution remaining around the cathode gradually becomes sodium hydroxide (NaOH) because Na⁺ and OH⁻ remain. This is an important industrial process for producing chlorine, hydrogen and sodium hydroxide, known as the chlor-alkali industry.

在阳极，我们比较 Cl⁻ 和 OH⁻。在浓溶液中，Cl⁻ 优先于 OH⁻ 放电，可以看到黄绿色氯气气泡冒出。阳极半反应方程式：2Cl⁻ → Cl₂ + 2e⁻。阴极周围剩余的溶液逐渐变为氢氧化钠 (NaOH)，因为 Na⁺ 和 OH⁻ 留在溶液中。这是一个生产氯气、氢气和氢氧化钠的重要工业过程，称为氯碱工业。

8. Electrolysis Using Active Electrodes – Copper Purification | 活性电极的电解 – 铜的精炼

When the electrodes themselves take part in the reaction, they are called active electrodes. A crucial example is the purification of blister copper using an electrolytic cell. The impure copper is made the anode, and a thin sheet of pure copper acts as the cathode. The electrolyte is copper(II) sulfate solution acidified with a little sulfuric acid.

当电极本身参与反应时，它们被称为活性电极。一个重要的例子是利用电解池精炼粗铜。将不纯的铜用作阳极，一片薄的高纯度铜板用作阴极。电解质是加了少量硫酸酸化的硫酸铜溶液。

During electrolysis, at the anode, copper atoms lose electrons and enter the solution as Cu²⁺ ions: Cu(s) → Cu²⁺(aq) + 2e⁻. Impurities such as gold, silver and platinum do not oxidise and fall to the bottom as “anode sludge”. At the cathode, Cu²⁺ ions from the solution are reduced and deposited as pure copper on the cathode: Cu²⁺(aq) + 2e⁻ → Cu(s). The pure copper gradually gets thicker, and the concentration of Cu²⁺ stays approximately constant. This process yields copper of 99.99% purity.

在电解过程中，阳极的铜原子失去电子，以 Cu²⁺ 离子的形式进入溶液：Cu(s) → Cu²⁺(aq) + 2e⁻。金、银、铂等杂质不发生氧化，沉到底部成为“阳极泥”。在阴极，溶液中的 Cu²⁺ 离子获得电子，以纯铜的形式沉积在阴极上：Cu²⁺(aq) + 2e⁻ → Cu(s)。纯铜板逐渐增厚，而 Cu²⁺ 的浓度大致保持不变。此过程可得到纯度高达 99.99% 的铜。

9. Electroplating | 电镀

Electroplating is the process of coating a metal object with a thin layer of another metal using electrolysis. This is usually done to improve corrosion resistance (e.g., chromium plating on steel) or for decorative purposes (e.g., silver plating on cutlery). The object to be plated is made the cathode, the plating metal is made the anode, and the electrolyte contains ions of the plating metal.

电镀是利用电解在金属物体表面镀上一层薄薄的另一种金属的过程。这样做通常是为了提高耐腐蚀性（例如钢材上镀铬）或为了装饰目的（例如餐具上镀银）。待镀物件作为阴极，镀层金属作为阳极，电解质含有镀层金属的离子。

For silver-plating a steel spoon, the spoon is the cathode, a bar of pure silver is the anode, and the electrolyte is a solution of silver nitrate (or another soluble silver salt). At the anode: Ag(s) → Ag⁺(aq) + e⁻, silver dissolves. At the cathode: Ag⁺(aq) + e⁻ → Ag(s), silver forms a shiny layer on the spoon. The net effect is the transfer of silver from the anode to the cathode without changing the concentration of the electrolyte.

以给钢勺镀银为例，勺子作为阴极，纯银棒作为阳极，电解质为硝酸银（或其他可溶性银盐）溶液。阳极反应：Ag(s) → Ag⁺(aq) + e⁻，银溶解；阴极反应：Ag⁺(aq) + e⁻ → Ag(s)，银在勺子表面形成闪亮的镀层。净效果是银从阳极转移到了阴极，电解液的浓度保持不变。

10. Electrolysis of Aluminium Oxide – The Hall-Héroult Process | 氧化铝的电解 – 霍尔-埃鲁法

Aluminium is extracted from its ore bauxite (Al₂O₃) by electrolysis, because it is too reactive to be obtained by reduction with carbon. Aluminium oxide has a very high melting point (over 2000 °C), so it is dissolved in molten cryolite (Na₃AlF₆) to lower the melting point to around 950 °C and reduce energy costs. The electrolytic cell has a steel tank lined with carbon, which acts as the cathode. Large carbon blocks serve as the anodes.

铝是通过电解从铝土矿 (Al₂O₃) 中提取的，因为铝过于活泼，无法用碳还原获得。氧化铝的熔点极高（超过 2000 °C），因此将其溶解在熔融的冰晶石 (Na₃AlF₆) 中，使熔点降至约 950 °C，从而降低能耗。电解池使用碳衬里钢槽作为阴极，大块碳块作为阳极。

In the molten mixture, Al³⁺ and O²⁻ ions are present. At the cathode, Al³⁺ ions are reduced to molten aluminium metal, which sinks to the bottom of the cell and is tapped off: Al³⁺ + 3e⁻ → Al(l). At the anode, O²⁻ ions are oxidised to oxygen gas: 2O²⁻ → O₂ + 4e⁻. The oxygen reacts with the hot carbon anodes to produce carbon dioxide, so the anodes are gradually consumed and need periodic replacement: C(s) + O₂(g) → CO₂(g).

在熔融混合物中，含有 Al³⁺ 和 O²⁻ 离子。在阴极，Al³⁺ 离子被还原为熔融的液态铝，沉到电解池底部，定期排出：Al³⁺ + 3e⁻ → Al(l)。在阳极，O²⁻ 离子被氧化为氧气：2O²⁻ → O₂ + 4e⁻。氧气与炽热的碳阳极反应生成二氧化碳，因此阳极会逐渐消耗，需要定期更换：C(s) + O₂(g) → CO₂(g)。

The overall reaction is often summarised as: 2Al₂O₃(l) → 4Al(l) + 3O₂(g). This process is hugely important but also very energy-intensive.

总反应通常概括为：2Al₂O₃(l) → 4Al(l) + 3O₂(g)。这一过程非常重要，但也是高耗能的。

11. Quantitative Electrolysis and Faraday’s Laws | 定量的电解与法拉第定律

IGCSE Edexcel may require simple calculations linking current, time and the mass of metal deposited. The key concept is that the amount of substance produced at an electrode is directly proportional to the quantity of electric charge passed. The charge (Q) is measured in coulombs (C) and is calculated by: Q = I × t, where I is the current in amperes and t is the time in seconds.

IGCSE Edexcel 可能要求进行简单的计算，将电流、时间与沉积的金属质量联系起来。核心概念是：电极上产生的物质量与通过的电量成正比。电量 (Q) 以库仑 (C) 为单位，计算公式为：Q = I × t，其中 I 为电流（安培），t 为时间（秒）。

One faraday (F) is the charge on one mole of electrons, approximately 96 500 C mol⁻¹. The mass of a substance discharged is determined by the number of moles of electrons transferred in the half-equation. For example, to deposit 1 mole of copper (Cu²⁺ + 2e⁻ → Cu) requires 2 moles of electrons, i.e., 2F of charge. To calculate mass: work out moles of electrons (Q ÷ F), then use the stoichiometric ratio from the half-equation to find moles of the metal, and finally multiply by the relative atomic mass (Aᵣ).

1 法拉第 (F) 是 1 摩尔电子所带的电量，约等于 96 500 C mol⁻¹。物质放电的质量由半反应方程式中转移的电子摩尔数决定。例如，沉积 1 摩尔铜 (Cu²⁺ + 2e⁻ → Cu) 需要 2 摩尔电子，即 2F 的电量。计算质量时：先求出电子的摩尔数（Q ÷ F），然后利用半反应方程式中的化学计量比求出金属的摩尔数，最后乘以相对原子质量 (Aᵣ)。

Example: calculate the mass of silver deposited when a current of 0.5 A is passed through silver nitrate solution for 30 minutes.
Step 1: t = 30 × 60 = 1800 s.
Step 2: Q = 0.5 A × 1800 s = 900 C.
Step 3: moles of electrons = 900 C ÷ 96 500 C mol⁻¹ ≈ 0.00933 mol.
Step 4: Ag⁺ + e⁻ → Ag, so 1 mole of electrons gives 1 mole of silver. Moles of Ag = 0.00933 mol.
Step 5: mass of Ag = 0.00933 mol × 108 g mol⁻¹ ≈ 1.01 g.

例题：计算当 0.5 A 的电流通过硝酸银溶液 30 分钟时所沉积的银的质量。
步骤 1：t = 30 × 60 = 1800 秒。
步骤 2：Q = 0.5 A × 1800 s = 900 C。
步骤 3：电子的摩尔数 = 900 C ÷ 96 500 C mol⁻¹ ≈ 0.00933 mol。
步骤 4：Ag⁺ + e⁻ → Ag，所以 1 摩尔电子沉积 1 摩尔银，银的摩尔数 = 0.00933 mol。
步骤 5：银的质量 = 0.00933 mol × 108 g mol⁻¹ ≈ 1.01 g。

12. Common Exam Pitfalls and Summary | 常见考试失分点与总结

Many IGCSE candidates lose marks by confusing the signs of the electrodes: in electrolysis, cathode is always negative and anode is always positive, regardless of cell type. Others forget to state that ions must be free to move, which only happens in the molten or aqueous state, not in solid ionic compounds. Also, always write full half-equations with correct charges and balancing; missing state symbols may be penalised depending on the mark scheme.

许多IGCSE考生因混淆电极的符号而失分：在电解中，阴极总是负极，阳极总是正极，这与电池类型无关。还有人忘记说明离子必须能自由移动，这仅在熔融或水溶液状态下发生，固体离子化合物不导电。此外，一定要书写正确的、配平的半反应方程式，并标清电荷；漏写状态符号可能因评分方案而被扣分。

In aqueous electrolysis, candidates often forget to consider the water-derived ions and therefore predict the wrong product. Always list all four ions (or more) present, apply the selective discharge rules, and then write both half-equations. For reactive electrodes, remember that the anode itself can oxidise, so the product at the anode is different from what you would see with inert graphite electrodes.

在水溶液电解中，考生经常忘记考虑水产生的离子，从而预测出错误的产物。务必先列出所有存在的离子（四个或更多），应用选择性放电规则，然后写出两个半反应方程式。对于活性电极，要记住阳极本身可能会被氧化，因此阳极产物与使用惰性石墨电极时观察到的不同。

Mastery of electrolysis provides a strong foundation for understanding redox chemistry and many industrial processes. By systematically identifying the electrolyte’s state, the ions present and the electrode materials, you can confidently predict products and construct accurate half-equations.

掌握电解知识为了解氧化还原化学和许多工业过程奠定了坚实的基础。通过系统地识别电解质的状态、存在的离子和电极材料，你就能自信地预测产物并构建准确的半反应方程式。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Electrolysis in IGCSE Edexcel Chemistry | IGCSE Edexcel 化学：电解 考点精讲