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AQA A-Level Chemistry June 2018 Insert 1: Data Skills and Electrode Potentials | AQA A-level化学2018年6月卷1数据插页解析

2026-08-22 01:23 · 本文部分内容由AI辅助创作

📚 AQA A-Level Chemistry June 2018 Insert 1: Data Skills and Electrode Potentials | AQA A-level化学2018年6月卷1数据插页解析

In AQA A-level Chemistry Paper 1 (7405/1) for June 2018, students were given an insert labelled Insert 1. This insert supplied key chemical data, including selected standard electrode potentials and thermochemical values, so that questions could test data handling and application rather than simple recall. Many of the questions on redox equilibria, feasibility and energy changes depended directly on this data.

在AQA A-level化学2018年6月卷1（代码7405/1）中，考生拿到了一份标注为Insert 1的数据插页。该插页提供了关键的化学数据，包括选定的标准电极电势和热化学数值，因此考题能够考查数据处理与运用能力，而不是单纯的知识背诵。许多关于氧化还原平衡、反应可行性和能量变化的题目都直接依赖这些数据。

1. What Insert 1 Is and Why It Matters | 什么是数据插页以及为何重要

Insert 1 is a separate data sheet attached to the exam paper. It is not a question paper but a reference tool. In June 2018, it contained selected electrode potentials and enthalpy data. Students who knew how to read the table quickly could avoid losing marks on sign errors and incorrect half-equation selection.

Insert 1是试卷附带的一份独立数据表。它不是题目本身，而是一种参考工具。2018年6月卷中，它包含了选定的电极电势和焓变数据。能够快速读懂该表格的考生可以避免因符号错误和半反应选错而失分。

Because AQA expects students to use data as printed, you must not memorise the values. Instead, practise extracting the correct reduction half-equation and its E° value, then applying it to cell diagrams and feasibility calculations.

因为AQA希望考生直接使用插页给出的数据，所以不需要背诵这些数值。相反，应当练习从表中提取正确的还原半反应及其 E° 值，然后将它们用于电池图解和可行性计算中。

2. Reading the Standard Electrode Potential Table | 如何阅读标准电极电势表

The standard electrode potential table usually lists half-equations as reductions, with E° measured in volts under standard conditions: 298 K, 1.00 mol dm⁻³ ion concentrations and 100 kPa gas pressure. A more positive E° means the species on the left is more easily reduced, while a more negative E° means the species on the right is more easily oxidised.

标准电极电势表通常以还原半反应的形式列出，E° 的单位为伏特，标准条件为298 K、1.00 mol dm⁻³ 离子浓度和100 kPa 气体压力。E° 越正，表明左侧物种越容易被还原；E° 越负，表明右侧物种越容易被氧化。

Half-equation E° / V
Fe³⁺ + e⁻ → Fe²⁺ +0.77
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O +1.51
Cl₂ + 2e⁻ → 2Cl⁻ +1.36
Zn²⁺ + 2e⁻ → Zn −0.76
Cu²⁺ + 2e⁻ → Cu +0.34
Fe²⁺ + 2e⁻ → Fe −0.44

These representative values are the kind found in Insert 1. You must use them exactly as printed, including the sign and the number of electrons. Reversing a half-equation reverses the sign of E°, but the magnitude stays the same.

这些代表性数值与Insert 1中出现的类型一致。必须按插页原样使用这些数据，包括符号和电子数。将半反应反向书写时，E° 的符号反转，但数值大小保持不变。

3. Cell Diagrams and Conventional Representation | 电池图解与常规表示法

When an electrochemical cell is set up using two half-cells from Insert 1, the conventional cell diagram places the more positive E° half-cell on the right as the reduction site. The oxidation half-cell is placed on the left. Vertical lines show phase boundaries, and a double vertical line represents the salt bridge.

当使用Insert 1中的两个半电池组成一个电化学电池时，常规电池图解将 E° 更正的半电池放在右侧作为还原位点，氧化的半电池放在左侧。单竖线表示相界面，双竖线代表盐桥。

d MnO₄⁻/Mn²⁺ on the right written as MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The cell diagram is:

例如，由 Fe³⁺/Fe²⁺ 和 MnO₄⁻/Mn²⁺ 构成的电池，左侧为 Fe²⁺ → Fe³⁺ + e⁻，右侧为 MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O。其电池图解为：

Pt | Fe²⁺, Fe³⁺ || MnO₄⁻, Mn²⁺, H⁺ | Pt

A platinum electrode is used when the half-cell has no solid metal conductor, as both Fe³⁺/Fe²⁺ and MnO₄⁻/Mn²⁺ involve ions in solution. The salt bridge completes the circuit and allows ion movement without mixing the solutions.

当半电池中没有固态金属导体时，需要使用铂电极，因为 Fe³⁺/Fe²⁺ 和 MnO₄⁻/Mn²⁺ 都涉及溶液中的离子。盐桥构成完整回路，允许离子移动而不使两种溶液混合。

4. Calculating Standard Cell EMF | 计算标准电池电动势

The standard cell electromotive force, E°cell, is calculated by subtracting the oxidation half-cell potential from the reduction half-cell potential. This is often expressed as:

标准电池电动势 E°cell 的计算方法是用还原半电池电势减去氧化半电池电势。通常表示为：

E°cell = E°(cathode) − E°(anode)

Because the cathode is where reduction occurs and the anode is where oxidation occurs, you must identify which half-cell is reduced and which is oxidised. The half-cell with the more positive E° is reduced, so it becomes the cathode.

因为阴极发生还原反应，阳极发生氧化反应，所以必须先确定哪个半电池被还原、哪个被氧化。E° 更正的半电池被还原，因此成为阴极。

1 V and the Fe³⁺/Fe²⁺ half-cell has E° = +0.77 V. The MnO₄⁻/Mn²⁺ half-cell is reduced, so:

以Insert 1中的 MnO₄⁻/Mn²⁺ 与 Fe³⁺/Fe²⁺ 为例，MnO₄⁻/Mn²⁺ 半电池的 E° = +1.51 V，Fe³⁺/Fe²⁺ 半电池的 E° = +0.77 V。MnO₄⁻/Mn²⁺ 半电池被还原，因此：

E°cell = +1.51 V − +0.77 V = +0.74 V

A common mistake is to add the two values or to use the minus sign incorrectly after reversing the oxidation half-equation. Always subtract the anode potential, even if it is negative, and keep the signs exactly as given in the insert.

常见错误是将两个数值相加，或者在反向书写氧化半反应后错误使用负号。始终减去阳极电势，即使其为负值，也要完全按照插页中的符号处理。

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5. Predicting Feasibility from E°cell | 由电池电动势判断反应可行性

For a reaction to be feasible under standard conditions, the calculated E°cell must be positive. A positive E°cell corresponds to a negative free energy change, meaning the reaction is thermodynamically favoured. If E°cell is negative, the reaction is not feasible in the forward direction under standard conditions.

在标准条件下，一个反应要可行，计算得到的 E°cell 必须为正。正的 E°cell 对应负的自由能变化，说明反应在热力学上是有利的。如果 E°cell 为负，则在标准条件下正向反应不可行。

However, a positive E°cell does not guarantee a fast reaction. Activation energy barriers can make a thermodynamically feasible reaction kinetically slow, so E°cell predicts feasibility, not rate.

但是，正的 E°cell 并不保证反应速率快。活化能势垒可能使热力学上可行的反应在动力学上很慢，因此 E°cell 预测的是可行性，而不是速率。

on is feasible. This is the basis of using manganate(VII) to oxidise iron(II) in redox titrations, a key practical skill linked to this data.

对于 MnO₄⁻ 与 Fe²⁺ 的反应，E°cell = +0.74 V 为正值，因此反应可行。这正是高锰酸根氧化亚铁离子在氧化还原滴定中的应用基础，也是与此数据相关的关键实验技能。

6. Linking Electrode Potentials to ΔG and K | 电极电势与ΔG和K的联系

Insert 1 data can also be used to calculate the standard Gibbs free energy change using the relationship:

利用Insert 1的数据还可以通过以下关系式计算标准吉布斯自由能变化：

ΔG° = −nFE°cell

Here n is the number of moles of electrons transferred in the balanced overall equation, F is the Faraday constant (96 500 C mol⁻¹), and E°cell is the standard cell potential. A positive E°cell gives a negative ΔG°, confirming feasibility.

其中 n 是配平的总反应中转移电子的物质的量，F 是法拉第常数（96 500 C mol⁻¹），E°cell 是标准电池电动势。正的 E°cell 给出负的 ΔG°，从而证实反应可行。

A more positive E°cell corresponds to a larger K, meaning the equilibrium lies further to the right. This links electrode potentials to equilibrium position.

此外，ΔG° 与平衡常数 K 的关系为 ΔG° = −RT ln K。E°cell 越正，对应的 K 越大，说明平衡位置越偏向生成物。这就把电极电势与平衡位置联系了起来。

7. Redox Half-Equations and Balancing | 氧化还原半反应及配平

Insert 1 often includes half-equations that are already balanced as reductions. To construct the overall redox equation, you reverse the oxidation half-equation and balance the electrons transferred. The number of electrons lost must equal the number gained.

Insert 1通常已经以还原反应的形式给出了配平的半反应。要构建总氧化还原方程式，需要反向书写氧化半反应，并平衡转移的电子数。失去的电子数必须等于得到的电子数。

For the reaction between MnO₄⁻ and Fe²⁺, the MnO₄⁻ half-equation involves 5 electrons, while the Fe²⁺ oxidation involves 1 electron. Therefore the iron half-equation must be multiplied by 5:

对于 MnO₄⁻ 与 Fe²⁺ 的反应，MnO₄⁻ 半反应涉及5个电子，而 Fe²⁺ 的氧化涉及1个电子。因此铁的半反应必须乘以5：

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
5Fe²⁺ → 5Fe³⁺ + 5e⁻

Adding the two half-equations gives the overall equation:

将两个半反应相加，得到总方程式：

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

value used in ΔG° = −nFE°cell, so you must always balance the equation before selecting n.

在该配平方程中，n = 5 mol 电子。这就是在 ΔG° = −nFE°cell 中使用的数值，因此必须始终先配平方程式，再确定 n。

8. Common Data-Insert Pitfalls | 数据插页常见错误

Students often lose marks on data-insert questions because they read the wrong half-equation or treat all potentials as reductions when one must be reversed. The following list summarises the most frequent errors from June 2018-style questions.

考生在数据插页类题目中失分，通常是因为读错了半反应，或者把所有电势都当作还原电势处理，而其中一个需要反向。以下列表总结了2018年6月类型题中最常见的错误。