📚 AQA A Level Chemistry June 2018 Mark Scheme 5: Thermodynamics, Redox and Transition Metals | AQA A Level化学 2018年6月评分方案5：热力学、氧化还原与过渡金属
The June 2018 AQA A Level Chemistry Unit 5 (CHEM5) mark scheme rewards precise definitions, correct sign conventions, logical Born-Haber cycles, and clear explanations of redox feasibility. This article breaks down the highest-yield areas from the mark scheme so you can turn mark-scheme language into full marks.
2018年6月AQA A Level化学Unit 5（CHEM5）评分方案重视精准定义、正确的符号规则、清晰的Born-Haber循环以及对氧化还原可行性的解释。本文拆解评分方案中最高频的考点，帮助你用评分标准语言拿到满分。
1. What the June 2018 Mark Scheme 5 Prioritises | 2018年6月评分方案5的考查重点
The June 2018 AQA A2 Chemistry Unit 5 mark scheme prioritises sign conventions, precise definitions, calculation layout, and evaluation of feasibility. Examiners do not simply look for the final answer; they award marks for Born-Haber cycle labels, correct arrows, and clear statements about entropy or electrode potential.
2018年6月AQA A2化学Unit 5评分方案重点考查符号规则、精确定义、计算步骤以及对可行性的评价。考官不仅看最终答案，还会根据Born-Haber循环标注、正确箭头以及关于熵或电极电位的清晰表述给分。
Many candidates lose marks because they use the wrong sign for lattice dissociation or forget that standard conditions are required for standard electrode potentials. This guide works through the main mark-scheme points from the June 2018 series. If you are on the 7405 specification, these same skills appear in Paper 1 and Paper 3 required practicals.
很多考生因混淆晶格解离的符号，或忘记标准电极电位必须写标准条件而失分。本指南将逐一讲解2018年6月系列中的主要评分点。如果你学习的是7405新大纲，同样的技能也会出现在Paper 1和Paper 3的必修实验考查中。
2. Born-Haber Cycles and Lattice Enthalpy | Born-Haber循环与晶格焓
Born-Haber cycles apply Hess's law to ionic compounds. Lattice formation enthalpy is the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions; it is always exothermic and written as a negative value.
Born-Haber循环将Hess定律应用于离子化合物。晶格生成焓是指由气态离子生成1摩尔固态离子化合物时的焓变，它始终放热，因此写作负值。
Lattice dissociation enthalpy is the reverse process: one mole of solid ionic compound is separated into gaseous ions. It has the same magnitude but the opposite sign, so it is endothermic and positive.
晶格解离焓是逆过程：1摩尔固态离子化合物解离为气态离子。其数值相同但符号相反，因此是吸热且为正值。
ΔHf = ΔHat(metal) + ΔHie(metal) + ½ΔHat(halogen₂) + ΔHea(halogen) + ΔHlattice
In the mark scheme, you must label each step clearly: atomisation, ionisation, electron affinity, and lattice enthalpy. Using a single arrow without a label may lose a labelling mark.
评分方案要求你清楚地标注每一步：原子化、电离、电子亲和和晶格焓。只用箭头而不标注可能丢失标注分。
The mark scheme often asks why experimental lattice enthalpy differs from the perfect ionic model. The perfect ionic model assumes ions are perfect spheres with only electrostatic attraction. When there is covalent character, the experimental lattice enthalpy is more exothermic than the theoretical value.
评分方案常问为什么实验晶格焓与理想离子模型不同。理想离子模型假设离子是完美球体，只存在静电引力。当存在共价性时，实验晶格焓比理论值更放热。
3. Enthalpy Changes in Solution | 溶解过程中的焓变
Enthalpy of hydration is the enthalpy change when one mole of gaseous ions is dissolved in water to form aqueous ions. It is always exothermic because ion-dipole attractions release energy.
水合焓是指1摩尔气态离子溶于水形成水合离子时的焓变。它始终放热，因为离子-偶极吸引力会释放能量。
Enthalpy of solution is the enthalpy change when one mole of an ionic solid dissolves in water to form aqueous ions. It can be endothermic or exothermic, depending on the balance between lattice dissociation and hydration.
溶解焓是指1摩尔离子固体溶于水形成水合离子时的焓变。它可以是吸热也可以是放热，取决于晶格解离和水合过程之间的平衡。
ΔHsolution = ΔHlattice dissociation + ΣΔHhydration
In the June 2018 style questions, you may be given a cycle or values and asked to calculate one missing enthalpy change. Always check the direction: lattice formation is negative, lattice dissociation is positive, and hydration enthalpies are negative.
在2018年6月风格的题目中，可能会给出循环图或数值，并要求计算一个缺失的焓变。务必检查方向：晶格生成是负值，晶格解离是正值，水合焓是负值。
4. Entropy and Total Entropy Changes | 熵与总熵变
Entropy is a measure of the disorder or dispersal of energy in a system. It has units J K⁻¹ mol⁻¹. Moving from solid to liquid to gas increases entropy, and producing more gas moles in a reaction also increases entropy.
熵是系统混乱度或能量分散程度的量度，单位是J K⁻¹ mol⁻¹。从固体到液体再到气体，熵增大；反应中生成更多气体摩尔也会增大熵。
The total entropy change determines whether a reaction is feasible. It is the sum of the entropy change of the system and the entropy change of the surroundings.
总熵变决定反应是否可行。它等于系统的熵变与环境的熵变之和。
ΔStotal = ΔSsystem + ΔSsurroundings
ΔSsurroundings = -ΔH / T
In calculations, temperature must be in kelvin. If ΔH is in kJ mol⁻¹ and entropy is in J K⁻¹ mol⁻¹, convert ΔH to J mol⁻¹ before dividing by T. Many students lose a unit mark here.
计算时温度必须使用开尔文。如果ΔH的单位是kJ mol⁻¹，熵的单位是J K⁻¹ mol⁻¹，则在除以T之前要把ΔH转换为J mol⁻¹。许多学生在这里丢失单位分。
5. Gibbs Free Energy and Feasibility | Gibbs自由能与可行性
The Gibbs free energy equation links enthalpy, entropy, and temperature to predict feasibility. A reaction is feasible when ΔG is negative.
Gibbs自由能公式将焓变、熵变和温度联系起来以预测可行性。当ΔG为负值时，反应可行。
ΔG = ΔH - TΔS
For the June 2018 mark scheme, you must convert ΔS into kJ K⁻¹ mol⁻¹ if ΔH is in kJ mol⁻¹, or convert ΔH into J mol⁻¹ if you keep ΔS in J K⁻¹ mol⁻¹. Same units throughout are essential.
在2018年6月的评分方案中，如果ΔH用kJ mol⁻¹，则必须把ΔS转换为kJ K⁻¹ mol⁻¹；或者若保持ΔS为J K⁻¹ mol⁻¹，则要把ΔH转换为J mol⁻¹。整个计算必须使用一致的单位。
/ 1000) = -100 + 59.6 = -40.4 kJ mol⁻¹. The negative value means the reaction is feasible.
例如：若ΔH = -100 kJ mol⁻¹，ΔS = -200 J K⁻¹ mol⁻¹，T = 298 K，则ΔG = -100 - 298 × (-200 / 1000) = -100 + 59.6 = -40.4 kJ mol⁻¹。负值表示该反应可行。
6. Standard Electrode Potentials and Cell EMF | 标准电极电位与电池电动势
Standard electrode potential is measured relative to the standard hydrogen electrode (SHE). Standard conditions are: 298 K, 100 kPa, and 1 mol dm⁻³ solutions. The SHE has an assigned potential of 0.00 V.
标准电极电位是相对于标准氢电极（SHE）测定的。标准条件为：298 K、100 kPa以及1 mol dm⁻³溶液。标准氢电极的指定电位为0.00 V。
A salt bridge connects the two half-cells and allows ions to move while preventing the solutions from mixing. A high-resistance voltmeter measures the cell EMF without drawing significant current.
盐桥连接两个半电池，允许离子迁移，同时防止溶液混合。高电阻伏特计测量电池电动势时不会产生明显电流。
Ecell = Eright - Eleft
In the mark scheme, the more positive electrode potential belongs to the stronger oxidising agent. The more negative electrode potential belongs to the stronger reducing agent. Always subtract the left-hand electrode from the right-hand electrode as written in the cell diagram.
在评分方案中，电极电位更正的一方是较强的氧化剂。电极电位更负的一方是较强的还原剂。务必按电池图式中所写的右半电池电位减去左半电池电位。
7. Predicting Redox Feasibility | 预测氧化还原可行性
A redox reaction is feasible in the forward direction if the calculated Ecell is positive. A negative Ecell means the forward reaction is not thermodynamically feasible.
如果计算得到的Ecell为正值，则正向氧化还原反应可行。Ecell为负值意味着正向反应在热力学上不可行。
However, the June 2018 mark scheme expects you to mention limitations. A feasible reaction may not occur if the
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