📚 Common Pitfalls in IGCSE CCEA Chemistry: Detailed Solutions | IGCSE CCEA 化学：易错题精讲
In IGCSE CCEA Chemistry, many students lose marks not because they lack knowledge, but because they fall into the same predictable traps. This article collects the most common mistakes made in exams – from mole calculations and electrolysis to organic naming and energy changes – and explains exactly how to avoid them. Each section presents a typical error, deconstructs the misconception behind it, and provides a step‑by‑step correct solution. Use this as a revision tool to sharpen your accuracy and boost your confidence before the final paper.
在 IGCSE CCEA 化学考试中，很多学生丢分不是因为知识欠缺，而是掉进了相同的、可预测的陷阱中。本文收集了考试中最常见的错误——从摩尔计算、电解到有机命名和能量变化——并详细解释了如何避免这些错误。每个小节都先展示典型错例，剖析背后的错误观念，再给出逐步正确的解法。请将此文作为复习工具，在最后冲刺阶段提高答题的准确性并增强自信。
1. Moles and Molar Calculations | 摩尔与摩尔计算
One of the most frequent errors occurs when students confuse the mass of a substance with the number of moles. A typical question asks: “Calculate the number of moles in 4.4 g of carbon dioxide (CO₂).” The common mistake is to divide the mass by something other than the molar mass, or to use incorrect units. Some students write: number of moles = 4.4 ÷ 44 = 0.1 mol – which is correct numerically – but they often forget to include the unit ‘mol’ or misread the relative formula mass of CO₂ as 28 instead of 44. Others mistakenly apply the formula for concentration instead of the simple mass‑mole relationship.
最常见的错误之一是将物质的质量与物质的量混淆。一道典型题目是：“计算4.4 g二氧化碳(CO₂)的物质的量。”常见错误是用错误的分母去除质量，或者单位使用不当。一些学生写：物质的量 = 4.4 ÷ 44 = 0.1 摩尔，数值正确，但经常忘记写上单位“mol”，或者把CO₂的相对分子质量读成28而不是44。另一些学生会误用与浓度有关的公式，而不是简单的质量‑物质的量关系。
The correct approach: First, determine the molar mass of CO₂: C (12) + O₂ (2 × 16) = 44 g mol⁻¹. Then apply the formula: amount (mol) = mass (g) ÷ molar mass (g mol⁻¹). So 4.4 g ÷ 44 g mol⁻¹ = 0.10 mol. Always write the unit. A further subtlety: in problems where the mass is given in kilograms, it must first be converted to grams (1 kg = 1000 g). Many candidates lose a mark by using 0.0044 kg directly in the formula, which gives a value 1000 times too small.
正确的做法：首先计算出CO₂的摩尔质量：C (12) + O₂ (2 × 16) = 44 g mol⁻¹。然后应用公式：物质的量(mol) = 质量(g) ÷ 摩尔质量(g mol⁻¹)。因此4.4 g ÷ 44 g mol⁻¹ = 0.10 mol。一定要写上单位。另一个容易忽略的细节：如果题目给出的质量单位是千克，必须先换算成克(1 kg = 1000 g)。很多考生直接用0.0044 kg代入公式，得到的结果小了1000倍，从而丢分。
2. Balancing Equations and State Symbols | 方程式配平与状态符号
Even when students correctly balance a chemical equation, they often lose marks for omitting state symbols. CCEA mark schemes consistently award one mark for correct state symbols in equations such as the thermal decomposition of calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g). A common mistake is to use (aq) for calcium oxide, or to leave state symbols out entirely. Another pitfall is forgetting that elements like hydrogen, oxygen and nitrogen must be written as diatomic molecules (H₂, O₂, N₂) in equations; writing O instead of O₂ unbalances the equation and misrepresents the reactant.
即使学生正确地配平了化学方程式，他们常常会因为遗漏状态符号而丢分。CCEA的评分方案一贯规定，像碳酸钙热分解这样的方程式：CaCO₃(s) → CaO(s) + CO₂(g)，状态符号占有1分。常见错误是把氧化钙的状态写成 (aq)，或者干脆不写状态符号。另一个陷阱是忘记氢气、氧气、氮气等元素在方程式中必须以双原子分子形式存在(H₂, O₂, N₂)；错写成 O 而不是 O₂ 不仅让方程式无法配平，还错误地表示了反应物。
How to get it right: First, learn the standard diatomic elements: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂. When writing an equation, always consider the physical states under the given conditions. Use (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous (dissolved in water). Ionic compounds that are not dissolved are usually (s). Acids and alkalis in solution are (aq). After balancing the numbers of atoms, check that the state symbol for each species matches the description in the question. For example, a reaction that occurs in solution demands (aq) for soluble salts and (l) for water.
如何做到正确：首先，记住标准双原子分子：H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂。书写方程式时，始终要根据给定条件考虑物理状态。(s) 表示固体，(l) 表示液体，(g) 表示气体，(aq) 表示水溶液（溶于水）。未溶解的离子化合物通常是 (s)。溶液中的酸和碱为 (aq)。配平原子数目之后，还要检查每种物质的状态符号是否与题目描述一致。例如，在溶液中发生的反应，可溶盐要求写 (aq)，水要求写 (l)。
3. Electrolysis of Aqueous Solutions | 水溶液的电解
A classic mistake arises when predicting the products of electrolysis for aqueous solutions. Students often blindly apply the reactivity series and assume that the metal ion is always discharged at the cathode. For a solution like aqueous copper(II) sulfate with inert electrodes, Cu²⁺ is indeed discharged at the cathode to give copper metal. However, for aqueous sodium chloride, the cation Na⁺ is less reactive than water, so hydrogen gas (from water) is produced at the cathode instead of sodium. At the anode, the halide ion (Cl⁻) is oxidised to chlorine gas because its concentration outweighs the tendency to discharge oxygen from water. The common error is to predict oxygen at the anode and sodium at the cathode.
在预测水溶液电解产物时，常会出现一个经典误解。学生往往生搬硬套金属活动性顺序，认为阴极总是析出金属离子。对于像硫酸铜水溶液（惰性电极）这样的例子，Cu²⁺ 确实在阴极放电生成铜。然而，对于氯化钠水溶液，阳离子 Na⁺ 的放电能力弱于水，所以阴极析出的是氢气（来自水）而非金属钠。在阳极，卤素离子(Cl⁻)被氧化成氯气，因为其浓度优势超过了水放电析出氧的趋势。常见的错误答案是：阳极生成氧气，阴极生成钠。
To avoid confusion, memorise the priority rules for discharge. At the cathode: cations with reduction potentials less than that of water (e.g., Na⁺, K⁺, Ca²⁺, Mg²⁺, Al³⁺) are not discharged; instead, water is reduced: 2H₂O + 2e⁻ → H₂ + 2OH⁻. For less reactive metals (Cu²⁺, Ag⁺), the metal ions are reduced. At the anode: if the solution contains a high concentration of halide ions (Cl⁻, Br⁻, I⁻), they are discharged in preference to OH⁻ from water. In dilute solutions, or with sulfates/nitrates, oxygen is produced from OH⁻: 4OH⁻ → O₂ + 2H₂O + 4e⁻. Always note electrode material: copper anode can dissolve (Cu → Cu²⁺ + 2e⁻), overriding normal halide discharge.
要避免混淆，必须记住放电的优先顺序。阴极：还原电势比水弱的阳离子（如 Na⁺, K⁺, Ca²⁺, Mg²⁺, Al³⁺）不会被放电；此时水被还原：2H₂O + 2e⁻ → H₂ + 2OH⁻。较不活泼的金属离子（Cu²⁺, Ag⁺）则优先还原。阳极：如果溶液中含有高浓度卤离子（Cl⁻, Br⁻, I⁻），它们会优先于水中的 OH⁻ 放电。在稀溶液中或存在硫酸根/硝酸根时，OH⁻ 被氧化生成氧气：4OH⁻ → O₂ + 2H₂O + 4e⁻。还要注意电极材料：铜阳极可能会溶解（Cu → Cu²⁺ + 2e⁻），这会改变通常的卤素放电顺序。
4. Rates of Reaction and Collision Theory | 反应速率与碰撞理论
When explaining why increasing the concentration or pressure increases the rate of reaction, students frequently give vague answers such as “particles move faster”, which is more relevant to temperature. The correct explanation must refer to the number of particles per unit volume and the resulting frequency of collisions. Another error involves catalysts: saying “a catalyst increases the rate of reaction by increasing the energy of the particles” is incorrect. A catalyst provides an alternative reaction pathway with a lower activation energy; it does not alter the energy of the reacting particles themselves.
在解释为什么增大浓度或压强会提高反应速率时，学生常常给出模糊的回答，如“粒子运动更快”，这其实更适合用于温度的影响。正确的解释必须提到单位体积内的粒子数增多了，从而碰撞频率增大。关于催化剂的另一个错误是：称“催化剂通过增大粒子能量来加快反应速率”，这是不正确的。催化剂提供了一条具有较低活化能的替代反应路径，它并不改变反应粒子本身的能量。
A precise answer for concentration: “Increasing the concentration means there are more reactant particles per unit volume, so the frequency of successful collisions increases, leading to a higher rate of reaction.” For pressure (gases): “Higher pressure compresses the gas, bringing particles closer together; more particles in a given volume leads to more frequent collisions.” Remember that a catalyst lowers the activation energy. The Maxwell‑Boltzmann distribution can be used to illustrate that, with a lower activation energy, a greater proportion of particles have energy equal to or exceeding the new activation energy, so a greater proportion of collisions are effective. Never state that a catalyst directly gives particles more energy.
浓度的精确答案：“增大浓度意味着单位体积内反应物的粒子数增多，因此有效碰撞的频率增加，导致反应速率提高。”对于压强（气体）：“增大压强压缩了气体，使粒子靠得更近；给定体积内的粒子数增多，碰撞更加频繁。”务必记住催化剂降低活化能。可用麦克斯韦‑玻尔兹曼分布来说明：由于活化能降低，更多比例的粒子具有等于或超过新活化能的能量，因此有效碰撞的比例增大。绝对不能说催化剂直接给予粒子更多能量。
5. Dynamic Equilibrium and Le Chatelier’s Principle | 动态平衡与勒夏特列原理
Many students misinterpret the effect of a catalyst on equilibrium position. A catalyst speeds up both the forward and reverse reactions equally, so it does not change the position of equilibrium; it only allows the system to reach equilibrium more quickly. Another common error is applying Le Chatelier’s principle to changes in concentration of solids or pure liquids – these are essentially constant and do not shift the equilibrium. Furthermore, when describing the effect of increasing temperature on an exothermic reaction (ΔH negative), students often say “equilibrium shifts to the right because the reaction is exothermic” instead of the proper reasoning: the system opposes the increase in temperature by favouring the endothermic direction (left), so the equilibrium shifts to the left.
许多学生对催化剂对平衡位置的影响存在误解。催化剂同等程度地加快正反应和逆反应的速率，因此它不会改变平衡位置，只是让体系更快地达到平衡。另一个常见错误是对固体或纯液体的浓度变化应用勒夏特列原理——这些物质的浓度基本不变，不会导致平衡移动。此外，当描述高温对放热反应（ΔH为负）的影响时，学生常说“平衡向右移动，因为反应放热”，而不是正确的推理：体系通过向吸热方向（左）移动来削弱温度的升高，因此平衡向左移动。
Le Chatelier’s principle states: if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the position of equilibrium shifts to oppose that change. For temperature: if the forward reaction is exothermic (ΔH = – x kJ mol⁻¹), increasing the temperature will shift equilibrium to the left (endothermic direction) to absorb the added heat. If the forward reaction is endothermic, the opposite occurs. For pressure: increasing pressure favours the side with fewer moles of gas. Do not use the catalyst argument for equilibrium yield. When exam questions ask “Explain why a higher temperature is not always used in industry even though it increases rate,” the answer must discuss the trade‑off between rate and equilibrium yield and the optimum conditions.
勒夏特列原理指出：如果一个处于平衡的体系受到浓度、压强或温度的改变，平衡位置将朝削弱这种改变的方向移动。对于温度：若正反应放热（ΔH = – x kJ mol⁻¹），升高温度将使平衡向左（吸热方向）移动以吸收额外的热量。若正反应吸热，则相反。对于压强：增大压强有利于气体分子总数较少的一侧。不要用催化剂解释平衡产率。当考题问及“为什么工业上不总是用高温，虽然高温能提高速率”，答案必须讨论速率与平衡产率的权衡以及最优条件。
6. Acid–Base Titration and Indicators | 酸碱滴定与指示剂
A recurring mistake involves the choice of indicator for a titration. Phenolphthalein is suitable for strong acid – strong base and strong acid – weak base titrations, but not for weak acid – strong base titrations? Actually, phenolphthalein changes colour in the pH range 8.3–10.0, so it is ideal for strong base versus any acid (strong or weak) because the equivalence point lies in the alkaline region for weak acid‑strong base. Methyl orange (pH 3.1–4.4) is used for strong acid versus weak base. Students frequently confuse these. Another error is in the calculation: forgetting to convert cm³ to dm³ when applying M₁V₁ = M₂V₂. If volumes are in cm³, the ratio can be used directly if units are consistent, but using a volume in dm³ in the formula with concentrations in mol dm⁻³ requires all volumes in dm³.
一个反复出现的错误是指示剂的选择。酚酞适用于强酸–强碱和强酸–弱碱滴定，实际上酚酞的变色范围是pH 8.3–10.0，因此它对于强碱与任何酸（强或弱）的滴定都非常理想，因为弱酸‑强碱的等当点位于碱性区域。甲基橙（pH 3.1–4.4）用于强酸与弱碱的滴定。学生经常混淆这点。另一类错误在于计算：应用 M₁V₁ = M₂V₂ 时忘记将 cm³ 换算成 dm³。如果体积单位都用 cm³，只要两者单位一致，比值可以直接使用；但如果公式中的浓度单位是 mol dm⁻³，则所有体积必须以 dm³ 为单位。
Correct approach: For a strong acid‑strong base titration, either indicator can be used because the vertical portion of the pH curve spans pH 3–10. For strong acid‑weak base, the equivalence point is below pH 7, so methyl orange is suitable. For weak acid‑strong base, the equivalence point is above pH 7, so phenolphthalein is suitable. Titration calculations: always check the equation stoichiometry first. For NaOH + HCl → NaCl + H₂O, the mole ratio is 1:1, so M₁V₁ = M₂V₂ holds. But for H₂SO₄ + 2NaOH, it is M₁V₁ (acid) × 2 = M₂V₂ (base) or M₁V₁ = M₂V₂ / 2. Common error: forgetting the factor of 2. Convert volumes: 25.0 cm³ = 0.0250 dm³. Use the relationship: moles = concentration × volume (in dm³).
正确的做法：强酸‑强碱滴定既可用酚酞也可用甲基橙，因为pH突跃范围涵盖pH 3–10。强酸‑弱碱滴定等当点pH低于7，适合甲基橙。弱酸‑强碱滴定等当点pH高于7，适合酚酞。滴定计算：始终先检查化学计量比。对于 NaOH + HCl → NaCl + H₂O，摩尔比为1:1，因此 M₁V₁ = M₂V₂ 成立。但对于 H₂SO₄ + 2NaOH，则为 M₁V₁（酸）× 2 = M₂V₂（碱），或 M₁V₁ = M₂V₂ / 2。常见错误：漏掉系数2。进行体积换算：25.0 cm³ = 0.0250 dm³。使用关系：摩尔数 = 浓度 × 体积（以 dm³ 计）。
7. Organic Chemistry: Naming and Functional Groups | 有机化学：命名与官能团
Naming organic compounds correctly is a minefield for many candidates. The most frequent mistakes include: numbering the carbon chain from the wrong end, miscounting the longest continuous chain, and misidentifying the functional group. For example, butan‑2‑ol is often named as butan‑3‑ol because students start numbering from the end closest to the –OH group incorrectly, or they fail to recognise that the alcohol functional group takes priority in numbering. Another error is confusing the suffixes: –ane (alkane), –ene (alkene), –anol (alcohol), –anoic acid (carboxylic acid), –yl –anoate (ester). Drawing structural isomers is also problematic: many draw the same structure twice or produce impossible bonding (e.g., pentavalent carbon).
对许多考生来说，正确命名有机化合物是一个雷区。最常见的错误包括：从错误的一端开始给碳链编号，数错最长的连续碳链，以及误认官能团。例如，butan‑2‑ol 常被命名为 butan‑3‑ol，因为学生没有从离 –OH 基团最近的一端开始编号，或者他们没有意识到醇的官能团应给予最小编号优先。另一个错误是混淆后缀：–ane（烷烃）、–ene（烯烃）、–anol（醇）、–anoic acid（羧酸）、–yl –anoate（酯）。绘制结构异构体也经常出错：很多人重复画出相同的结构，或画出不可能的键（如五价碳）。
To name a compound: (1) identify the functional group and its suffix. (2) Find the longest continuous carbon chain containing that group. (3) Number the chain so that the functional group gets the lowest possible number; if it is an alkene, the double bond must have the lowest number. (4) Name any alkyl side chains as prefixes (methyl, ethyl) with their position numbers. (5) Put everything together: numbers separated by commas, with hyphens between numbers and words. Example: CH₃CH₂CH(CH₃)CH₂OH is 2‑methylbutan‑1‑ol. Common wrong name: 3‑methylbutan‑4‑ol (wrong numbering direction). For esters, the alcohol part comes first (alkyl), then the carboxylic acid part (alkanoate): e.g., methyl ethanoate, not ethyl methanoate. Remember that isomers must have the same molecular formula but different structural arrangements; count atoms carefully.
命名步骤：(1) 识别官能团及其后缀。(2) 找出含该官能团的最长连续碳链。(3) 给碳链编号，使官能团获得最小的位次号；如果是烯烃，双键也必须获得最小的位次号。(4) 把烷基侧链作为前缀（甲基、乙基），并标明其位次。(5) 组合在一起：数字间用逗号，数字与名称间用连字符。示例：CH₃CH₂CH(CH₃)CH₂OH 应为 2‑methylbutan‑1‑ol。常见错误名：3‑methylbutan‑4‑ol（编号方向错误）。对于酯，醇部分在前（烷基），然后是酸部分（烷酸酯）：例如 methyl ethanoate，不是 ethyl methanoate。注意异构体必须具有相同的分子式但不同的结构排列，仔细数原子。
8. Energetics: Exothermic and Endothermic Reactions | 能量学：放热与吸热反应
A subtle error appears in energy profile diagrams and bond‑energy calculations. Students often label the enthalpy change (ΔH) as the difference between reactants and the activation energy, rather than the difference between products and reactants. They also misinterpret breaking bonds as exothermic and making bonds as endothermic. In reality, breaking bonds absorbs energy (endothermic) and making bonds releases energy (exothermic). This confusion leads to an inverted sign for ΔH when using bond energies. For example, for H₂ + Cl₂ → 2HCl, many calculate ΔH = bonds broken – bonds formed correctly, but then give the wrong sign (+ or –), thinking energy released is positive ΔH.
在能量分布图和键能计算中，一个隐蔽的错误经常出现。学生经常把焓变(ΔH)标为反应物与活化能之差，而非产物与反应物之差。他们也误解了键的断裂与形成：认为断键是放热，成键是吸热。实际上，断键吸收能量（吸热），成键释放能量（放热）。这种混淆导致用键能计算 ΔH 时符号错乱。例如，对于反应 H₂ + Cl₂ → 2HCl，许多人会正确地计算 ΔH = 断键吸收能量 – 成键释放能量，但结果却漏掉或写错符号（+ 或 –），以为释放能量对应正的 ΔH。
The correct method: ΔH = sum of bond energies of bonds broken (reactants) – sum of bond energies of bonds formed (products). In H₂ + Cl₂, bonds broken: one H–H (436 kJ mol⁻¹) and one Cl–Cl (243 kJ mol⁻¹), total = 679 kJ. Bonds formed: two H–Cl bonds (2 × 431 = 862 kJ). ΔH = 679 – 862 = –183 kJ mol⁻¹, so the reaction is exothermic. Students who reverse the subtraction get +183 kJ mol⁻¹, which incorrectly suggests endothermic. Also, when drawing energy profiles, ensure the curve for exothermic reactions shows products at a lower energy than reactants, with ΔH indicated as a downward arrow (negative). For endothermic, products are higher. Activation energy is always the energy from reactants to the peak of the curve; label it clearly. Don’t confuse it with ΔH.
正确的做法：ΔH = 反应物断裂的所有键的键能之和 – 产物形成所有键的键能之和。在 H₂ + Cl₂ 中，断裂的键：一个 H–H (436 kJ mol⁻¹) 和一个 Cl–Cl (243 kJ mol⁻¹)，总计 679 kJ。形成的键：两个 H–Cl 键 (2 × 431 = 862 kJ)。ΔH = 679 – 862 = –183 kJ mol⁻¹，因此反应放热。做相反减法的学生得到 +183 kJ mol⁻¹，错误地表明为吸热。此外，绘制能量分布图时，确保放热反应的曲线显示产物的能量比反应物低，ΔH 以向下箭头表示（负值）。吸热反应则产物能量更高。活化能总是从反应物到曲线峰顶的能量差值，应清晰标出，切勿与 ΔH 混淆。
9. Ionic and Covalent Bonding | 离子键与共价键
Students very frequently lose marks when drawing dot‑and‑cross diagrams, especially for ionic compounds. One common mistake is failing to use different symbols (dots and crosses) for electrons from different atoms, or not putting brackets and charges around the ions. For example, the drawing for magnesium oxide (MgO) should show Mg with no outer electrons (having lost its two outer electrons) and the oxide ion with a full octet, surrounded by brackets with a 2– charge, while the Mg²⁺ ion is shown without brackets but with the 2+ charge. Many candidates draw the transferred electrons still around the magnesium, or they omit the charges entirely. Another error is drawing covalent bonds as the transfer of electrons, rather than sharing.
学生在画电子点叉图时，尤其是离子化合物，经常丢分。一个常见错误是没有用不同的符号（点和叉）来表示来自不同原子的电子，或没有在离子周围加上方括号和电荷。例如，氧化镁 (MgO) 的图应显示 Mg 没有外层电子（失去了它的两个外层电子），氧离子具有完整的八电子结构，外加方括号和 2– 电荷；而 Mg²⁺ 离子则不加括号但标注 2+ 电荷。许多考生的图仍把转移出去的电子画在镁周围，或完全漏掉电荷。另一个错误是将共价键画成电子的转移，而不是共用。
To draw an ionic diagram correctly: (a) Represent the metal atom with its outer electrons (e.g., using dots). (b) Represent the non‑metal atom with its outer electrons (using crosses). (c) Show the transfer of electron(s) from metal to non‑metal by moving the dot(s) to the non‑metal. (d) Draw the resulting ions: the non‑metal more often needs brackets, with its full octet, and the negative charge written as superscript outside the bracket; the metal ion is drawn without outer electrons, with a positive charge. The ions should be drawn side by side with a clear ionic formula. For covalent molecules (like H₂O), show shared pairs between O and each H, with O’s original electrons as dots and H’s as crosses, to demonstrate the shared origin. Always fulfil the octet rule for Period 2 elements (except for H, which needs 2 electrons).
正确绘制离子图的步骤：(a) 用外层电子（如点）表示金属原子。(b) 用外层电子（如叉）表示非金属原子。(c) 通过将点（金属电子）移到非金属一侧，展示电子转移。(d) 画出生成的离子：非金属通常需要方括号，内部为完整的八电子结构，负电荷作为上标写在括号外；金属离子则不画外层电子，标注正电荷。离子应并排绘制，并清晰写出离子式。对于共价分子（如 H₂O），在 O 和各 H 之间画出共用电子对，O 原有的电子用点，H 的用叉，以体现共用来源。始终满足第二周期元素的八隅体规则（H 只需 2 个电子）。
10. Redox Reactions and Oxidation States | 氧化还原反应与氧化态
Many IGCSE students struggle to identify the oxidising and reducing agents in a redox equation, often confusing the concepts. A very common misconception is: “The species that gets oxidised is the oxidising agent.” That is wrong. The oxidising agent is the species that causes oxidation by accepting electrons, and therefore itself gets reduced. Similarly, the reducing agent is oxidised. For example, in the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂, iron oxide is reduced to iron, so it is the oxidising agent. Carbon monoxide is oxidised to carbon dioxide, so it is the reducing agent. Students who swap the agents will lose easy marks. Another pitfall: assigning oxidation numbers without following the rules, especially to oxygen in peroxides (–1 rather than –2) and hydrogen in metal hydrides (–1).
许多IGCSE学生在氧化还原方程中识别氧化剂和还原剂时感到困难，经常混淆概念。一个非常普遍的误解是：“被氧化的物质就是氧化剂。”这是错误的。氧化剂是通过接受电子而造成氧化的物质，因此它自身被还原。同理，还原剂则自身被氧化。例如，在反应 Fe₂O₃ + 3CO → 2Fe + 3CO₂ 中，氧化铁被还原成铁，因此它是氧化剂；一氧化碳被氧化成二氧化碳，因此它是还原剂。把二者颠倒的学生会丢掉容易拿到的分。另一个陷阱：不遵循规则指定氧化数，尤其是在过氧化物中氧为 –1 而非 –2，以及在金属氢化物中氢为 –1。
Mnemonic to remember: OIL RIG – Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). The oxidising agent gains electrons (is reduced), the reducing agent loses electrons (is oxidised). To work out oxidation states: (1) free elements = 0; (2) simple ions = charge on ion; (3) oxygen usually –2 (except in peroxides –1, in OF₂ +2); (4) hydrogen usually +1 (except in metal hydrides –1); (5) sum of oxidation states in a neutral compound = 0, in an ion = charge on ion. Once oxidation states are assigned, identify which atoms’ oxidation states increase (oxidation) and decrease (reduction). Then state the agent accordingly. Practice with a range of equations, including disproportionation where the same element is both oxidised and reduced (e.g., Cl₂ + 2NaOH → NaCl + NaClO + H₂O).
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