Electrochemical Cells and Standard Electrode Potentials u2014 A-Level Chemistry | u7535u5316u5b66u7535u6c60u4e0eu6807u51c6u7535u6781u7535u52bf

Introduction to Electrochemical Cells

Electrochemical cells are devices that convert chemical energy into electrical energy (galvanic/voltaic cells) or use electrical energy to drive non-spontaneous chemical reactions (electrolytic cells). For A-Level Chemistry students, understanding how these cells work is fundamental to mastering the topic of electrochemistry.

电化学电池是将化学能转化为电能（原电池/伏打电池）或利用电能驱动非自发化学反应（电解池）的装置。对于A-Level化学学生来说，理解这些电池的工作原理是掌握电化学主题的基础。

Galvanic Cells: Producing Electricity from Chemical Reactions

A galvanic cell (also called a voltaic cell) consists of two half-cells connected by a salt bridge or porous barrier. Each half-cell contains an electrode immersed in an electrolyte solution. The spontaneous redox reaction drives electrons from the anode (oxidation site) through an external circuit to the cathode (reduction site), generating an electric current.

原电池（也称伏打电池）由两个通过盐桥或多孔隔膜连接的半电池组成。每个半电池包含一个浸在电解质溶液中的电极。自发的氧化还原反应驱动电子从阳极（氧化位点）通过外部电路流向阴极（还原位点），产生电流。

The key components of a galvanic cell include:

原电池的关键组成部分包括：

• Anode (负极/阳极): The electrode where oxidation occurs. Electrons are released here and flow out into the external circuit. 发生氧化的电极，电子在此释放并流入外部电路。
• Cathode (正极/阴极): The electrode where reduction occurs. Electrons from the external circuit are accepted here. 发生还原的电极，外部电路的电子在此被接收。
• Salt Bridge (盐桥): A U-shaped tube containing a concentrated electrolyte (usually KNO3 or KCl) that completes the circuit by allowing ion migration without mixing the two half-cell solutions. 含浓电解质（通常为KNO3或KCl）的U形管，通过允许离子迁移但不混合两个半电池溶液来完成电路。
• External Circuit (外部电路): A wire connecting the two electrodes through which electrons flow from anode to cathode. 连接两个电极的导线，电子通过它从阳极流向阴极。

Standard Electrode Potentials (E°)

The standard electrode potential (E°) of a half-cell measures the tendency of a species to gain electrons (be reduced) under standard conditions: 298 K (25°C), 1 mol dm⁻³ concentration for all solutions, and 100 kPa (1 atm) pressure for any gases involved. The half-cell is connected to a standard hydrogen electrode (SHE), which is assigned a potential of exactly 0.00 V.

标准电极电势（E°）衡量半电池中某物种在标准条件下获得电子（被还原）的趋势：298 K (25°C)，所有溶液浓度为1 mol dm⁻³，任何涉及的气体压力为100 kPa (1 atm)。半电池与标准氢电极（SHE）连接，SHE被赋予恰好0.00 V的电位。

The standard hydrogen electrode consists of a platinum electrode immersed in 1 mol dm⁻³ H+(aq) solution, with H2 gas bubbled through at 100 kPa pressure. Platinum is used because it is inert and provides a surface for the H+/H2 equilibrium:

标准氢电极由浸在1 mol dm⁻³ H+(aq)溶液中的铂电极组成，H2气体在100 kPa压力下通入。使用铂是因为它惰性且为H+/H2平衡提供表面：

2H+(aq) + 2e- ⇌ H2(g)    E° = 0.00 V

The Electrochemical Series

The electrochemical series arranges half-cells in order of their standard electrode potentials, from most negative (strongest reducing agents) to most positive (strongest oxidizing agents). This ordering allows chemists to predict the direction of redox reactions and calculate cell potentials.

电化学系列按标准电极电势从最负（最强还原剂）到最正（最强氧化剂）排列半电池。这种排序使化学家能够预测氧化还原反应的方向并计算电池电势。

Key half-cell equations every A-Level Chemistry student should memorise:

每位A-Level化学学生应记住的关键半电池方程式：

Half-equation | 半反应式	E° / V
K+ + e- ⇌ K	-2.93
Ca2+ + 2e- ⇌ Ca	-2.87
Mg2+ + 2e- ⇌ Mg	-2.37
Zn2+ + 2e- ⇌ Zn	-0.76
Fe2+ + 2e- ⇌ Fe	-0.44
2H+ + 2e- ⇌ H2	0.00
Cu2+ + 2e- ⇌ Cu	+0.34
I2 + 2e- ⇌ 2I-	+0.54
Fe3+ + e- ⇌ Fe2+	+0.77
Br2 + 2e- ⇌ 2Br-	+1.07
Cl2 + 2e- ⇌ 2Cl-	+1.36

Calculating Cell Potential (E°_cell)

The standard cell potential is calculated using the equation:

标准电池电势使用以下公式计算：

E°_cell = E°_reduction – E°_oxidation

A positive E°_cell indicates a spontaneous reaction (Delta G° is negative), meaning the cell can produce electricity. This is the fundamental principle behind all batteries. The more positive the E°_cell, the greater the driving force of the reaction.

正值的E°_cell表示自发反应（Delta G°为负），意味着该电池可以产生电能。这是所有电池背后的基本原理。E°_cell越正，反应的驱动力越大。

Worked Example: Zn/Cu Cell

Consider a galvanic cell with zinc and copper electrodes:

考虑一个带有锌和铜电极的原电池：

Step 1: Write the half-equations with their E° values.
Zn2+ + 2e- ⇌ Zn    E° = -0.76 V
Cu2+ + 2e- ⇌ Cu    E° = +0.34 V

Step 2: Identify which half-cell has the more positive E°. This will undergo reduction (cathode). The other will undergo oxidation (anode).
Cu2+/Cu has the more positive E° (+0.34 V), so Cu2+ is reduced at the cathode.

Step 3: Calculate E°_cell:
E°_cell = +0.34 – (-0.76) = +1.10 V

Step 4: Write the overall cell reaction. Reverse the oxidation half-equation and combine:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)    E°_cell = +1.10 V

This is the Daniell cell, one of the earliest practical batteries invented in 1836.

这就是丹尼尔电池，最早实用的电池之一，发明于1836年。

Factors Affecting Electrode Potentials: The Nernst Equation

When conditions deviate from standard, the Nernst equation allows calculation of the actual electrode potential:

当条件偏离标准时，能斯特方程允许计算实际电极电势：

E = E° – (RT/nF) ln Q

Where R = 8.314 J mol⁻¹ K⁻¹, T = temperature in Kelvin, n = number of electrons transferred, F = 96,500 C mol⁻¹, and Q = reaction quotient.

其中R = 8.314 J mol⁻¹ K⁻¹，T = 开尔文温度，n = 转移电子数，F = 96,500 C mol⁻¹，Q = 反应商。

At 298 K, this simplifies to the A-Level exam form:

在298 K时，简化为A-Level考试常用形式：

E = E° – (0.059/n) log10 Q

Key exam implications: increasing reactant concentration makes E more positive (stronger oxidizing agent); increasing product concentration makes E more negative (stronger reducing agent). Temperature and pressure changes also affect electrode potential.

考试关键含义：增加反应物浓度使E更正（更强氧化剂）；增加产物浓度使E更负（更强还原剂）。温度和压力变化也会影响电极电势。

Common A-Level Exam Questions and Tips

1. Predicting Reaction Feasibility

To determine if a redox reaction is feasible, calculate E°_cell. If positive, the reaction is thermodynamically feasible. However, a positive E° does not guarantee a measurable rate, as kinetic factors may prevent it.

要确定氧化还原反应是否可行，计算E°_cell。如果为正值，该反应热力学上可行。但正E°不保证可测量速率，动力学因素可能阻止反应发生。

2. Cell Diagrams (Cell Notation)

Cell diagrams use standard notation: oxidation half-cell on the left, reduction on the right, separated by || (salt bridge). Phase boundaries use |.

电池图使用标准符号：氧化半电池在左，还原半电池在右，用 || 分隔（盐桥），相界用 | 表示。

Example: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

3. Limitations of E° Values

Standard electrode potentials only apply under standard conditions. Real-world conditions rarely match, so actual potentials require the Nernst equation. E° values cannot predict reaction rates, only thermodynamic feasibility.

标准电极电势仅在标准条件下适用。现实条件很少匹配，实际电势需用能斯特方程计算。E°值无法预测反应速率，只能预测热力学可行性。

Types of Half-Cells and Electrode Systems | 半电池类型与电极系统

1. Metal/Metal Ion Electrodes

The simplest type of half-cell consists of a metal rod immersed in a solution of its own ions. Examples include Zn(s) | Zn2+(aq) and Cu(s) | Cu2+(aq). The metal acts as both the electrode surface and a participant in the redox equilibrium. These are the most straightforward to set up and analyse in the laboratory.

最简单的半电池类型是将金属棒浸入其自身离子的溶液中。例如 Zn(s) | Zn2+(aq) 和 Cu(s) | Cu2+(aq)。金属既充当电极表面又参与氧化还原平衡。这是实验室中最容易搭建和分析的类型。

2. Gas/Ion Electrodes

When one of the species in the half-equation is a gas, an inert electrode (usually platinum) is used to provide a surface for electron transfer. The standard hydrogen electrode (SHE) is the canonical example, but other gas electrodes include the chlorine electrode (Cl2 | Cl-) and oxygen electrode (O2 | H2O). The inert electrode must be coated with finely divided platinum (platinum black) to maximize surface area and catalyse the equilibrium.

当半反应式中某物种为气体时，使用惰性电极（通常为铂）提供电子转移的表面。标准氢电极（SHE）是典型例子，但其他气体电极包括氯电极（Cl2 | Cl-）和氧电极（O2 | H2O）。惰性电极必须涂有细分散的铂（铂黑）以最大化表面积并催化平衡。

3. Redox Electrodes (Ion/Ion Electrodes)

In redox electrodes, both the oxidized and reduced forms are in solution, and an inert electrode (platinum or graphite) simply transfers electrons. The Fe3+/Fe2+ system is a classic example: a platinum wire immersed in a solution containing both Fe3+ and Fe2+ ions. No metal is consumed or deposited – the electrode is purely an electron carrier.

在氧化还原电极中，氧化态和还原态都在溶液中，惰性电极（铂或石墨）仅传递电子。Fe3+/Fe2+系统是经典例子：将铂丝浸入含有Fe3+和Fe2+离子的溶液中。没有金属被消耗或沉积 – 电极纯粹是电子载体。

Electrolysis: Driving Non-Spontaneous Reactions | 电解：驱动非自发反应

While galvanic cells produce electricity from spontaneous redox reactions, electrolytic cells do the opposite – they use an external power supply to force non-spontaneous reactions to occur. This is the principle behind electroplating, aluminium extraction, and the chlor-alkali industry.

原电池通过自发氧化还原反应产生电能，而电解池则相反 – 它们使用外部电源迫使非自发反应发生。这是电镀、铝提取和氯碱工业背后的原理。

In an electrolytic cell, the anode is the positive electrode (connected to the positive terminal of the power supply) where oxidation occurs, and the cathode is the negative electrode where reduction occurs. This is the opposite assignment compared to a galvanic cell, which is a common source of confusion in exams.

在电解池中，阳极是正极（连接到电源的正极端子），发生氧化反应；阴极是负极，发生还原反应。这与原电池中的分配相反，是考试中常见的混淆点。

Key difference for exams: In a galvanic cell, the anode is negative and the cathode is positive. In an electrolytic cell, the anode is positive and the cathode is negative. Always identify the type of cell first before determining electrode polarity.

考试关键区别：在原电池中，阳极是负极，阴极是正极。在电解池中，阳极是正极，阴极是负极。在确定电极极性之前，始终先识别电池类型。

Comparing Half-Cell Potentials to Predict Reactions | 比较半电池电势预测反应

The electrochemical series is a powerful predictive tool. A species with a more positive E value will oxidize a species with a more negative E value. This rule allows us to predict:

电化学系列是一个强大的预测工具。具有更正E值的物种将氧化具有更负E值的物种。这条规则使我们能够预测：

• Displacement reactions: Will zinc displace copper from CuSO4 solution? Yes – Zn2+/Zn (E = -0.76 V) has a more negative potential than Cu2+/Cu (E = +0.34 V), so Zn will reduce Cu2+ ions.
  置换反应：锌能否从CuSO4溶液中置换铜？能 – Zn2+/Zn（E = -0.76 V）比Cu2+/Cu（E = +0.34 V）具有更负的电位，因此Zn会还原Cu2+离子。
• Reactivities of halogens: Will chlorine oxidize bromide ions? Yes – Cl2/Cl- (E = +1.36 V) is more positive than Br2/Br- (E = +1.07 V), so Cl2 can oxidize Br-. However, Br2 cannot oxidize Cl- because its E value is less positive.
  卤素反应活性：氯能否氧化溴离子？能 – Cl2/Cl-（E = +1.36 V）比Br2/Br-（E = +1.07 V）更正，因此Cl2可以氧化Br-。但Br2不能氧化Cl-，因为其E值较不正。
• Acid-metal reactions: Metals with negative E values react with acids (H+), while metals with positive E values (like copper and silver) do not dissolve in non-oxidizing acids.
  酸-金属反应：具有负E值的金属与酸（H+）反应，而具有正E值的金属（如铜和银）不会溶解在非氧化性酸中。

Quantitative Electrochemistry: Faraday’s Laws | 定量电化学：法拉第定律

Michael Faraday established the quantitative relationship between the amount of electricity passed through an electrolytic cell and the amount of substance produced or consumed at the electrodes.

迈克尔·法拉第建立了通过电解池的电量与电极上产生或消耗的物质数量之间的定量关系。

Faraday’s First Law: The mass of a substance produced at an electrode is directly proportional to the quantity of electricity (charge) passed through the cell.

法拉第第一定律：电极上产生的物质质量与通过电池的电量（电荷）成正比。

Faraday’s Second Law: When the same quantity of electricity is passed through different electrolytes, the masses of substances produced are proportional to their equivalent weights (molar mass divided by the number of electrons transferred).

法拉第第二定律：当相同的电量通过不同电解质时，产生的物质质量与其当量（摩尔质量除以转移的电子数）成正比。

The key equation linking charge, current, and time is:

连接电荷、电流和时间的关键方程式：

Q = I x t

Where Q = charge in coulombs (C), I = current in amperes (A), t = time in seconds (s).

其中Q = 电荷（库仑，C），I = 电流（安培，A），t = 时间（秒，s）。

The relationship between charge and amount of substance involves the Faraday constant:

电荷与物质量之间的关系涉及法拉第常数：

n(e-) = Q / F    where F = 96,500 C mol-1

Worked Example: A current of 2.00 A is passed through molten NaCl for 30 minutes. Calculate the mass of sodium produced.

例题：2.00 A的电流通过熔融NaCl 30分钟。计算生成的钠的质量。

Step 1: Q = 2.00 x (30 x 60) = 3,600 C

Step 2: n(e-) = 3,600 / 96,500 = 0.0373 mol

Step 3: Na+ + e- -> Na, so 1 mol e- produces 1 mol Na

Step 4: m(Na) = 0.0373 x 23.0 = 0.858 g

Commercial Applications of Electrochemistry | 电化学的商业应用

Lithium-Ion Batteries

The modern Li-ion battery operates through electrochemical intercalation – lithium ions move between a graphite anode and a lithium metal oxide cathode. The cell potential is typically 3.6-3.7 V, significantly higher than traditional aqueous cells, which is why they dominate portable electronics and electric vehicles.

现代锂离子电池通过电化学插层运作 – 锂离子在石墨阳极和锂金属氧化物阴极之间移动。电池电势通常为3.6-3.7 V，远高于传统水系电池，这就是它们主导便携式电子产品和电动汽车的原因。

Fuel Cells

Fuel cells convert chemical energy directly into electrical energy with high efficiency. Unlike batteries, they require a continuous supply of fuel and oxidant. The hydrogen-oxygen fuel cell is the most studied for A-Level:

燃料电池以高效率将化学能直接转化为电能。与电池不同，它们需要持续供应燃料和氧化剂。氢氧燃料电池是A-Level中研究最多的：

Alkaline Hydrogen Fuel Cell:

Anode: 2H2 + 4OH- -> 4H2O + 4e-

Cathode: O2 + 2H2O + 4e- -> 4OH-

Overall: 2H2 + O2 -> 2H2O    E(cell) = +1.23 V

Fuel cells produce only water as the waste product (in hydrogen-oxygen cells), making them environmentally attractive. However, challenges remain in hydrogen production, storage, and the cost of catalysts.

燃料电池（在氢氧电池中）仅产生水作为废物，使其在环境上具有吸引力。然而，氢气生产、储存和催化剂成本方面仍存在挑战。

Common Mistakes to Avoid | 常见错误提醒

1. Confusing E(cell) formula: The correct formula is E(cell) = E(reduction) – E(oxidation). Do NOT add the two potentials together – this is a common error.
   混淆电池电势公式：正确公式是E(cell) = E(reduction) – E(oxidation)。不要将两个电位相加 – 这是常见错误。
2. Forgetting sign conventions: When reversing a half-equation to represent oxidation, do NOT change the sign of E. The E value is always the reduction potential. The oxidation potential is simply the negative of the reduction potential, which is accounted for in the E(cell) subtraction formula.
   忘记符号惯例：反转半反应式以表示氧化时，不要改变E的符号。E值始终是还原电位。氧化电位仅仅是还原电位的负值，这在E(cell)减法公式中已被考虑。
3. Writing cell diagrams incorrectly: Always put the more negative half-cell on the left. The cell diagram for Zn/Cu should be Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s), not the reverse. The cell EMF should be positive when the cell is written this way.
   错误书写电池图：始终将较负的半电池放在左边。Zn/Cu的电池图应为Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)，而非相反。这样书写时电池电动势应为正值。
4. Ignoring non-standard conditions: Exam questions may specify non-standard concentrations or temperatures. If concentrations differ from 1 mol dm-3, the actual potential deviates from E, and the Nernst equation is required.
   忽略非标准条件：考试题目可能指定非标准浓度或温度。如果浓度不同于1 mol dm-3，实际电位偏离E，需要使用能斯特方程。
5. Misidentifying anode/cathode in electrolytic cells: Remember that in an electrolytic cell, oxidation still occurs at the anode and reduction at the cathode, but the polarity is reversed compared to a galvanic cell.
   错误识别电解池中的阳极/阴极：记住在电解池中，氧化仍发生在阳极，还原发生在阴极，但极性与原电池相反。

Key Equations Summary | 关键公式总结

Equation | 公式	Use | 用途
E(cell) = E(red) – E(ox)	Cell potential from half-cell potentials | 从半电池电势计算电池电势
E = E – (RT/nF) ln Q	Nernst equation (any temperature) | 能斯特方程（任意温度）
E = E – (0.059/n) log Q	Nernst equation at 298 K | 能斯特方程在298 K
Q = I x t	Charge from current and time | 从电流和时间计算电荷
n(e-) = Q / 96500	Moles of electrons from charge | 从电荷计算电子摩尔数
Delta G = -nFE(cell)	Gibbs free energy from cell potential | 从电池电势计算吉布斯自由能

Practice Questions for Exam Preparation | 考试准备练习题

Question 1: A galvanic cell is constructed using Mg | Mg2+(aq) and Ag | Ag+(aq) half-cells. Use the following data to calculate the standard cell EMF and write the overall cell reaction.

题目1：使用 Mg | Mg2+(aq) 和 Ag | Ag+(aq) 半电池构建原电池。使用以下数据计算标准电池电动势并写出总电池反应。

Mg2+ + 2e- = Mg    E = -2.37 V
Ag+ + e- = Ag    E = +0.80 V

Solution: Mg has the more negative E, so it undergoes oxidation (anode). Ag+ is reduced (cathode).

解答：Mg具有更负的E值，因此发生氧化（阳极）。Ag+被还原（阴极）。

Anode (oxidation): Mg -> Mg2+ + 2e-    E(ox) = +2.37 V

Cathode (reduction): Ag+ + e- -> Ag (x2)    E(red) = +0.80 V

E(cell) = +0.80 – (-2.37) = +3.17 V

Overall: Mg + 2Ag+ -> Mg2+ + 2Ag

Question 2: Explain why a magnesium electrode cannot be used to measure the standard electrode potential of the Mg2+/Mg half-cell using a simple voltmeter in aqueous solution, despite its very negative E value.

题目2：解释为什么尽管镁的E值非常负，但不能使用简单电压表在水溶液中测量Mg2+/Mg半电池的标准电极电势。

Answer: Magnesium reacts rapidly with water, producing hydrogen gas and magnesium hydroxide. This means the measured potential would not reflect the true Mg2+/Mg equilibrium – it would be a mixed potential involving both the Mg2+/Mg and H+/H2O couples. Even in apparently neutral water, Mg is thermodynamically unstable.

答案：镁与水迅速反应，产生氢气和氢氧化镁。这意味着测得的电位不会反映真正的Mg2+/Mg平衡 – 它将是一个涉及Mg2+/Mg和H+/H2O电对的混合电位。即使在看似中性的水中，镁在热力学上也是不稳定的。

Question 3: An electrochemical cell has a standard cell potential of +0.46 V. Write down the cell diagram using the correct notation, and identify the direction of electron flow in the external circuit. The half-cells are Cu2+/Cu and Fe3+/Fe2+.

题目3：某电化学电池的标准电池电势为+0.46 V。用正确符号写出电池图，并确定外部电路中电子流动的方向。半电池为Cu2+/Cu和Fe3+/Fe2+。

Data: Cu2+ + 2e- = Cu, E = +0.34 V; Fe3+ + e- = Fe2+, E = +0.77 V

Answer: Fe3+/Fe2+ has the more positive E (+0.77 V), so it is reduced at the cathode. Cu is oxidized at the anode. Electrons flow from Cu anode to Pt cathode through the external circuit.

答案：Fe3+/Fe2+具有更正值的E（+0.77 V），因此在阴极被还原。Cu在阳极被氧化。电子通过外部电路从Cu阳极流向Pt阴极。

Cell diagram: Cu(s) | Cu2+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s)

E(cell) = +0.77 – (+0.34) = +0.43 V (But the question states +0.46 V, which may indicate non-standard conditions are in play – a good discussion point!)

E(cell) = +0.77 – (+0.34) = +0.43 V（但题目说+0.46 V，这可能表明存在非标准条件 – 一个很好的讨论点！）

Connecting Electrochemistry to Other Topics | 电化学与其他主题的联系

Thermodynamics: Gibbs Free Energy

The relationship between cell potential and Gibbs free energy is given by Delta G = -nFE. A positive E(cell) corresponds to a negative Delta G, confirming that the reaction is spontaneous. This bridges electrochemistry with chemical thermodynamics – a favourite link in synoptic A-Level questions.

电池电势与吉布斯自由能之间的关系由Delta G = -nFE给出。正的E(cell)对应负的Delta G，确认反应是自发的。这将电化学与化学热力学联系起来 – 这是A-Level综合题中最受欢迎的连接点。

Equilibrium: The Nernst Equation at Equilibrium

When a galvanic cell reaches equilibrium (a “dead” battery), E(cell) = 0. At this point, the Nernst equation reduces to a relationship between E(cell) and the equilibrium constant K:

当原电池达到平衡（”耗尽”的电池），E(cell) = 0。此时，能斯特方程简化为E(cell)与平衡常数K之间的关系：

E(cell) = (RT/nF) ln K    or    E(cell) = (0.059/n) log K at 298 K

This equation allows calculation of equilibrium constants from electrochemical data – a powerful tool that links two major areas of physical chemistry.

该方程允许从电化学数据计算平衡常数 – 这是连接物理化学两个主要领域的强大工具。

Kinetics: Overpotential and Activation Energy

Even when a reaction is thermodynamically feasible (E(cell) > 0), it may occur very slowly due to high activation energy. The additional voltage required beyond the thermodynamic potential to drive a reaction at a practical rate is called the overpotential. This is why water electrolysis requires voltages above 1.23 V – the oxygen evolution reaction has a significant overpotential.

即使反应在热力学上可行（E(cell) > 0），由于高活化能，它可能进行得非常缓慢。超过热力学电位以实际速率驱动反应所需的额外电压称为过电位。这就是为什么水电解需要高于1.23 V的电压 – 氧析出反应具有显著的过电位。

Final Tips for A-Level Success | A-Level 成功终极提示

Electrochemistry questions consistently appear in A-Level Chemistry exams across all boards (CAIE, Edexcel, AQA, OCR). The most common question types combine E(cell) calculations with:
(1) writing overall redox equations,
(2) drawing and annotating cell diagrams,
(3) predicting reaction feasibility,
(4) applying the Nernst equation to non-standard conditions,
and (5) linking E(cell) to Delta G and equilibrium constants.

电化学题目始终出现在所有考试局（CAIE、Edexcel、AQA、OCR）的A-Level化学考试中。最常见的题型将E(cell)计算与以下内容结合：
（1）书写总氧化还原方程式，
（2）绘制并标注电池图，
（3）预测反应可行性，
（4）将能斯特方程应用于非标准条件，
（5）将E(cell)与Delta G和平衡常数联系起来。

Practice past paper questions extensively – the patterns are predictable once you understand the underlying principles. Always show your working for E(cell) calculations step by step, and double-check your sign conventions before writing your final answer.

大量练习历年真题 – 一旦理解基本原理，题型是可预测的。对于E(cell)计算，始终逐步展示你的解题过程，并在书写最终答案前仔细检查符号惯例。

Summary | 总结

Electrochemical cells and standard electrode potentials form the foundation of electrochemistry in A-Level Chemistry. Master these core concepts: half-cells, the electrochemical series, E°_cell calculations, and the Nernst equation, and you will be well-prepared for exam questions on this topic. Remember to practise drawing cell diagrams with correct notation and always check your sign conventions when calculating cell potentials.

电化学电池和标准电极电势构成了A-Level化学中电化学的基础。掌握这些核心概念：半电池、电化学系列、E°_cell计算和能斯特方程，你将能充分应对关于此主题的考试题目。记得练习用正确的符号画电池图，并在计算电池电势时始终检查符号惯例。

更多咨询请联系 16621398022（同微信）