Appendix 2: Selected Standard Electrode Potentials | 附录二：精选标准电极电势

📚 Appendix 2: Selected Standard Electrode Potentials | 附录二：精选标准电极电势

The Cambridge International A-Level Chemistry syllabus provides an Appendix of selected standard electrode potentials (E⁰ values) that students are expected to use in electrochemical calculations. This appendix is not merely a data table—it is a powerful predictive tool for determining reaction feasibility, calculating cell potentials, and understanding redox chemistry across the entire course.

剑桥国际A-Level化学考纲提供了一个精选标准电极电势（E⁰值）附录，学生需要在电化学计算中使用这些数据。这个附录不仅仅是一个数据表——它是一个强大的预测工具，用于判断反应可行性、计算电池电势以及理解贯穿整个课程的氧化还原化学。

1. What Is a Standard Electrode Potential? | 什么是标准电极电势？

A standard electrode potential (E⁰) is the potential difference measured when a half-cell is connected to the standard hydrogen electrode (SHE) under standard conditions: 298 K, 1 atm pressure, and 1 mol dm⁻³ concentration for all aqueous species. It is measured in volts (V).

标准电极电势（E⁰）是在标准条件下，将半电池与标准氢电极（SHE）连接时所测得的电势差。标准条件为：298 K温度、1 atm压力、所有水溶液物种浓度为1 mol dm⁻³。其单位为伏特（V）。

By convention, all half-cell reactions are written as reduction reactions in the Appendix. For example:

按照惯例，附录中所有半电池反应均以还原反应的形式书写。例如：

Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E⁰ = −0.76 V

A more negative E⁰ value means the reduced form (Zn(s)) is a stronger reducing agent, while a more positive E⁰ value means the oxidised form (Zn²⁺) is a stronger oxidising agent.

越负的E⁰值意味着还原形态（Zn(s)）是更强的还原剂，而越正的E⁰值意味着氧化形态（Zn²⁺）是更强的氧化剂。

2. The Standard Hydrogen Electrode (SHE) | 标准氢电极（SHE）

The SHE is the reference electrode against which all other electrode potentials are measured. It consists of a platinum electrode in contact with H⁺(aq) at 1 mol dm⁻³ and H₂(g) at 1 atm pressure.

标准氢电极是测量所有其他电极电势的参考电极。它由铂电极组成，与浓度为1 mol dm⁻³的H⁺(aq)和1 atm压力的H₂(g)接触。

2H⁺(aq) + 2e⁻ ⇌ H₂(g) E⁰ = 0.00 V (by definition)

The platinum electrode is inert—it does not participate in the redox reaction but provides a surface for electron transfer. The SHE is assigned a potential of exactly 0.00 V by international convention.

铂电极是惰性的——它不参与氧化还原反应，但为电子转移提供表面。根据国际惯例，SHE被指定为恰好0.00 V的电势。

In practice, the SHE is difficult to set up in school laboratories; alternative reference electrodes such as silver/silver chloride or calomel electrodes are often used. However, exam questions typically assume the SHE is used directly.

在实践中，SHE在学校实验室中难以搭建；通常使用银/氯化银或甘汞电极等替代参考电极。然而，考试题目通常假设直接使用SHE。

3. Reading the Appendix: Key Conventions | 阅读附录：关键约定

The Appendix lists half-reactions in a specific order. Let us examine how to interpret the table correctly.

附录按特定顺序列出半反应。让我们研究如何正确解读该表。

Half-Reaction	E⁰ / V
F₂(g) + 2e⁻ ⇌ 2F⁻(aq)	+2.87
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l)	+1.52
Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq)	+0.77
2H⁺(aq) + 2e⁻ ⇌ H₂(g)	0.00
Fe²⁺(aq) + 2e⁻ ⇌ Fe(s)	−0.44
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)	−0.76

Reading from the bottom upward: the oxidised forms at the bottom are increasingly powerful reducing agents. Reading from the top downward: the oxidised forms at the top are increasingly powerful oxidising agents.

从下往上读：底部的氧化形态是越来越强的还原剂。从上往下读：顶部的氧化形态是越来越强的氧化剂。

The most positive E⁰ (F₂/E⁻ = +2.87 V) indicates that F₂ is the strongest oxidising agent in the table. Conversely, Li⁺/Li (E⁰ ≈ −3.04 V) would be at the very bottom, with Li(s) being an extremely powerful reducing agent.

最正的E⁰（F₂/F⁻ = +2.87 V）表明F₂是表中最强的氧化剂。相反，Li⁺/Li（E⁰ ≈ −3.04 V）将位于最底部，Li(s)是极强的还原剂。

4. Calculating Cell EMF | 计算电池电动势

The electromagnetic force (EMF) of an electrochemical cell is calculated from the two half-cell potentials using the following formula:

电化学电池的电动势（EMF）使用以下公式从两个半电池电势计算得出：

E⁰(cell) = E⁰(reduction at cathode) − E⁰(reduction at anode)

Alternatively, it can be expressed as:

或者可以表示为：

E⁰(cell) = E⁰(right-hand electrode) − E⁰(left-hand electrode)

The more positive half-cell undergoes reduction (cathode), while the more negative half-cell undergoes oxidation (anode). Electrons flow from the anode to the cathode through the external circuit.

更正的半电池发生还原反应（阴极），而更负的半电池发生氧化反应（阳极）。电子通过外部电路从阳极流向阴极。

Worked Example: Calculate the EMF of a cell made from Fe³⁺/Fe²⁺ (E⁰ = +0.77 V) and Zn²⁺/Zn (E⁰ = −0.76 V).

例题：计算由Fe³⁺/Fe²⁺（E⁰ = +0.77 V）和Zn²⁺/Zn（E⁰ = −0.76 V）组成的电池的电动势。

E⁰(cell) = +0.77 − (−0.76) = +1.53 V

Since E⁰(cell) is positive, the reaction is thermodynamically feasible under standard conditions.

由于E⁰(cell)为正，该反应在标准条件下是热力学可行的。

5. Predicting Reaction Feasibility | 预测反应可行性

For any redox reaction, we can predict spontaneity by comparing the E⁰ values of the two half-reactions. The oxidising agent from the half-cell with the higher (more positive) E⁰ will oxidise the reducing agent from the half-cell with the lower (more negative) E⁰.

对于任何氧化还原反应，我们可以通过比较两个半反应的E⁰值来预测自发性。来自较高（更正）E⁰半电池的氧化剂将氧化来自较低（更负）E⁰半电池的还原剂。

Rule of Thumb: A redox reaction is feasible if E⁰(cell) > 0.

经验法则：当E⁰(cell) > 0时，氧化还原反应是可行的。

Consider whether Cl₂(g) will oxidise Br⁻(aq) to Br₂(l). From the Appendix:

考虑Cl₂(g)是否会将Br⁻(aq)氧化为Br₂(l)。根据附录：

Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq) E⁰ = +1.36 V
Br₂(l) + 2e⁻ ⇌ 2Br⁻(aq) E⁰ = +1.07 V

Here, Cl₂ (E⁰ = +1.36 V) is a stronger oxidising agent than Br₂ (E⁰ = +1.07 V). Therefore:

在这里，Cl₂（E⁰ = +1.36 V）是比Br₂（E⁰ = +1.07 V）更强的氧化剂。因此：

E⁰(cell) = +1.36 − (+1.07) = +0.29 V > 0 → Feasible

Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(l)

This reaction is indeed observed experimentally—chlorine water turns bromide solutions orange-brown due to Br₂ formation.

该反应确实在实验中被观察到——氯水将溴化物溶液变为橙棕色，这是因为生成了Br₂。

6. The Electrochemical Series | 电化学系列

The standard electrode potentials arranged in descending order form the electrochemical series. This series allows chemists to compare the relative strengths of oxidising and reducing agents systematically.

按降序排列的标准电极电势构成了电化学系列。该系列允许化学家系统地比较氧化剂和还原剂的相对强度。

Key values students should memorise from the Cambridge Appendix include:

学生应记住的剑桥附录中的关键数值包括：

F₂/F⁻ : +2.87 V — strongest oxidising agent among common halogens
MnO₄⁻/Mn²⁺ in acid : +1.52 V — powerful oxidising agent used in titrations
Cr₂O₇²⁻/Cr³⁺ in acid : +1.33 V — used in redox titrations
I₂/I⁻ : +0.54 V — mild oxidising agent
Fe³⁺/Fe²⁺ : +0.77 V — important in transition metal chemistry
Zn²⁺/Zn : −0.76 V — common anode in batteries
Mg²⁺/Mg : −2.38 V — strong reducing agent

F₂/F⁻：+2.87 V — 常见卤素中最强的氧化剂
酸性条件下MnO₄⁻/Mn²⁺：+1.52 V — 用于滴定的强氧化剂
酸性条件下Cr₂O₇²⁻/Cr³⁺：+1.33 V — 用于氧化还原滴定
I₂/I⁻：+0.54 V — 温和氧化剂
Fe³⁺/Fe²⁺：+0.77 V — 过渡金属化学中的重要体系
Zn²⁺/Zn：−0.76 V — 电池中的常见阳极
Mg²⁺/Mg：−2.38 V — 强还原剂

Note that the electrochemical series is temperature-dependent; E⁰ values are quoted at 298 K. At different temperatures, the order may change.

注意电化学系列是温度依赖的；E⁰值在298 K下给出。在不同温度下，顺序可能发生变化。

7. Limitations of Electrode Potential Predictions | 电极电势预测的局限性

While E⁰ values are excellent thermodynamic predictors, they do not guarantee that a reaction will actually occur at a measurable rate. Several factors limit their predictive power:

虽然E⁰值是出色的热力学预测工具，但它们不能保证反应实际上以可测量的速率发生。有几个因素限制了其预测能力：

Kinetic limitations: A reaction may be thermodynamically feasible (E⁰(cell) > 0) but kinetically slow due to a high activation energy. For example, the reduction of MnO₄⁻ requires H⁺ ions and may be slow without acid.
Concentration effects: E⁰ values assume 1 mol dm⁻³. In real systems, concentrations differ. The Nernst equation describes how potential varies with concentration, but this is beyond A-Level scope—however, qualitative understanding is expected.
Formation of insoluble or gaseous products: If a product leaves the system (as a precipitate or gas), the reaction may proceed even when E⁰(cell) is slightly negative.
Overpotential: In electrolysis, extra voltage is needed to overcome kinetic barriers—this is why electrolysis of water requires more than the theoretical 1.23 V.

动力学限制：反应可能在热力学上可行（E⁰(cell) > 0），但由于活化能高而动力学缓慢。例如，MnO₄⁻的还原需要H⁺离子，在没有酸的情况下可能很慢。
浓度效应：E⁰值假设浓度为1 mol dm⁻³。在真实系统中，浓度不同。能斯特方程描述了电势如何随浓度变化，但这超出了A-Level范围——然而，定性理解是必需的。
不溶物或气体产物的形成：如果产物离开系统（作为沉淀或气体），即使E⁰(cell)略微为负，反应也可能进行。
过电势：在电解中，需要额外的电压来克服动力学障碍——这就是为什么水的电解需要超过理论值1.23 V的原因。

Exam questions frequently test this limitation—a reaction may have a positive E⁰(cell) but still not be observed because the reaction rate is negligible.

考试题目经常测试这一局限性——反应可能具有正的E⁰(cell)，但由于反应速率可忽略不计而仍然观察不到。

8. Applications: Batteries and Cells | 应用：电池和电化学电池

The Appendix values are used to predict the voltages of commercial batteries and to design new electrochemical cells.

附录值用于预测商业电池的电压以及设计新的电化学电池。

Example: The Zinc–Copper Cell

示例：锌铜电池

Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) E⁰ = +0.34 V
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) E⁰ = −0.76 V

E⁰(cell) = +0.34 − (−0.76) = +1.10 V

This is the classic Daniell cell. In the salt bridge, K⁺ ions migrate toward the copper half-cell and NO₃⁻ ions toward the zinc half-cell to maintain charge neutrality.

这是经典的丹尼尔电池。在盐桥中，K⁺离子向铜半电池迁移，NO₃⁻离子向锌半电池迁移，以维持电荷中性。

Example: The Hydrogen–Oxygen Fuel Cell

示例：氢氧燃料电池

O₂(g) + 4H⁺(aq) + 4e⁻ ⇌ 2H₂O(l) E⁰ = +1.23 V (acidic)
2H⁺(aq) + 2e⁻ ⇌ H₂(g) E⁰ = 0.00 V

E⁰(cell) = +1.23 − 0.00 = +1.23 V

Fuel cells are efficient because they convert chemical energy directly to electrical energy without combustion, and the only product is water—making them environmentally friendly.

燃料电池效率高，因为它们直接将化学能转化为电能而无需燃烧，唯一的产物是水——使其环保。

9. Electrolysis and the Appendix | 电解与附录

During electrolysis, the external voltage drives a non-spontaneous reaction. The Appendix helps predict which ions are discharged at each electrode.

在电解过程中，外部电压驱动非自发反应。附录有助于预测哪些离子在哪个电极被放电。

Consider the electrolysis of concentrated aqueous NaCl. The possible half-reactions at the cathode are:

考虑浓NaCl水溶液的电解。阴极可能的半反应为：

2H⁺(aq) + 2e⁻ ⇌ H₂(g) E⁰ = 0.00 V
Na⁺(aq) + e⁻ ⇌ Na(s) E⁰ = −2.71 V

Since H⁺ reduction has a much higher E⁰, hydrogen gas is preferentially evolved at the cathode—not sodium metal. This matches experimental observation.

由于H⁺还原具有高得多的E⁰，氢气在阴极优先析出——而不是钠金属。这符合实验观察结果。

However, concentration matters: in concentrated NaCl, the high Cl⁻ concentration makes chlorine gas the preferred product at the anode, even though the E⁰ for O₂ evolution (+1.23 V) is lower than that for Cl₂ evolution (+1.36 V). This is because the overpotential for O₂ evolution is very high.

然而，浓度很重要：在浓NaCl中，高Cl⁻浓度使得氯气成为阳极的优先产物，尽管O₂析出的E⁰（+1.23 V）低于Cl₂析出的E⁰（+1.36 V）。这是因为O₂析出的过电势非常高。

Students should note that in dilute NaCl, oxygen is produced at the anode because the low Cl⁻ concentration shifts the balance toward water oxidation.

学生应注意，在稀NaCl中，阳极产生氧气，因为低Cl⁻浓度将平衡转向水的氧化。

10. Worked Examples with the Appendix | 附录应用例题

Let us work through two typical exam-style questions that test the use of the Appendix.

让我们完成两道典型的考试风格题目，测试附录的使用。

Example 1: Use the Appendix to determine whether Fe³⁺(aq) can oxidise I⁻(aq) to I₂(aq).

例1：使用附录判断Fe³⁺(aq)能否将I⁻(aq)氧化为I₂(aq)。

Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) E⁰ = +0.77 V
I₂(aq) + 2e⁻ ⇌ 2I⁻(aq) E⁰ = +0.54 V

E⁰(cell) = +0.77 − (+0.54) = +0.23 V > 0

Since E⁰(cell) is positive, the reaction is feasible:

由于E⁰(cell)为正，该反应可行：

2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq)

This is why Fe³⁺ solutions turn iodine–starch paper blue-black.

这就是Fe³⁺溶液使碘-淀粉试纸变蓝黑色的原因。

Example 2: Two half-cells have E⁰ values of −0.44 V (Fe²⁺/Fe) and +1.52 V (MnO₄⁻/Mn²⁺ in acid). Write the overall cell reaction and calculate E⁰(cell).

例2：两个半电池的E⁰值分别为−0.44 V（Fe²⁺/Fe）和+1.52 V（酸性MnO₄⁻/Mn²⁺）。写出总电池反应并计算E⁰(cell)。

The MnO₄⁻ half-cell has the higher E⁰, so it is the cathode (reduction). Fe is oxidised:

MnO₄⁻半电池具有更高的E⁰，因此它是阴极（还原）。Fe被氧化：

Cathode: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l)
Anode: Fe(s) ⇌ Fe²⁺(aq) + 2e⁻

Balancing electrons (LCM = 10):

配平电子（最小公倍数 = 10）：

2MnO₄⁻(aq) + 16H⁺(aq) + 5Fe(s) → 2Mn²⁺(aq) + 8H₂O(l) + 5Fe²⁺(aq)

E⁰(cell) = +1.52 − (−0.44) = +1.96 V

11. Common Exam Pitfalls | 常见考试陷阱

The following mistakes are frequently observed in student responses to electrode potential questions:

以下是学生在电极电势题目中经常犯的错误：

Wrong sign convention: Always use the values as written in the Appendix (reduction potentials). Do not flip the sign “for oxidation”—the formula E⁰(cell) = E⁰(cathode) − E⁰(anode) handles this automatically. Flipping signs leads to double-counting errors.
Ignoring the anode/cathode distinction: The more positive E⁰ half-cell is always the cathode. Some students incorrectly assign the more negative value to the cathode.
Forgetting that E⁰ is intensive: Multiplying a half-reaction by a coefficient does not change its E⁰ value. The potential is an intensive property, like temperature.
Confusing feasibility with rate: A spontaneous reaction (E⁰(cell) > 0) may be extremely slow. Always mention kinetics when discussing whether a reaction “actually occurs”.
Incorrect salt bridge direction: In the salt bridge, cations flow toward the cathode (positive half-cell) and anions flow toward the anode (negative half-cell).

错误符号约定：始终使用附录中给出的值（还原电势）。不要为了氧化而翻转符号——公式E⁰(cell) = E⁰(阴极) − E⁰(阳极)会自动处理这一点。翻转符号会导致重复计算错误。
忽视阴/阳极区分：更正E⁰的半电池始终是阴极。一些学生错误地将更负的值分配给阴极。
忘记E⁰是强度性质：将半反应乘以系数不会改变其E⁰值。电势是强度性质，就像温度一样。
混淆可行性与速率：自发反应（E⁰(cell) > 0）可能极其缓慢。在讨论反应是否”实际发生”时，务必提及动力学因素。
盐桥方向错误：在盐桥中，阳离子向阴极（正半电池）流动，阴离子向阳极（负半电池）流动。

Understanding the difference between thermodynamic feasibility and kinetic reality is a hallmark of high-scoring A-Level answers.

理解热力学可行性与动力学现实之间的区别是高分段A-Level答案的标志。

12. Summary and Revision Strategy | 总结与复习策略

The Appendix of selected standard electrode potentials is one of the most versatile tools in your A-Level Chemistry arsenal. Mastery of this table enables you to:

精选标准电极电势附录是你A-Level化学工具箱中最通用的工具之一。掌握此表使你能：

Calculate cell potentials for any combination of half-cells
Predict whether a redox reaction is thermodynamically feasible
Compare the strengths of oxidising and reducing agents
Determine the products of electrolysis
Understand the principles behind commercial batteries and fuel cells

计算任意半电池组合的电池电势
预测氧化还原反应是否热力学可行
比较氧化剂和还原剂的强度
确定电解产物
理解商业电池和燃料电池背后的原理

For effective revision, create a condensed flashcard of the most frequently tested E⁰ values, practise writing half-reactions in both directions, and work through past-paper questions involving the Nernst-type calculations. Remember that the Appendix is provided in the exam—your task is not to memorise every value, but to know how to use them with precision and confidence.

为有效复习，创建最常考E⁰值的精简闪卡，练习双向书写半反应，并完成涉及计算类型的往年试题。记住附录在考试中会提供——你的任务不是记住每个值，而是知道如何精确且自信地使用它们。

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Appendix 2: Selected Standard Electrode Potentials | 附录二：精选标准电极电势