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Galvanic Cells and Chemical Power Sources | 原电池与化学电源

2026-08-30 13:30 · 本文部分内容由AI辅助创作

📚 Galvanic Cells and Chemical Power Sources | 原电池与化学电源

In this revision guide, we will explore the principles of galvanic (voltaic) cells and modern chemical power sources. The CIE A-Level Chemistry syllabus requires a clear understanding of how redox reactions can be harnessed to produce electricity, how electrode potentials are measured and applied, and how this theory underpins primary cells, secondary cells and fuel cells.

本复习指南将带你系统梳理原电池(伏打电池)的基本原理以及现代化学电源。CIE A-Level 化学考纲要求同学们清晰理解:如何利用氧化还原反应产生电能、如何测定和应用电极电势,以及这些理论如何支撑一次电池、二次电池和燃料电池。


1. The Core Principles of Galvanic Cells | 原电池的核心原理

A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy through a spontaneous redox reaction. It is also called a voltaic cell. In the classic zinc-copper sulphate cell, zinc loses electrons and is oxidised, while copper(II) ions gain electrons and are reduced.

原电池是一种通过自发氧化还原反应将化学能转化为电能的电化学装置,又称伏打电池。在经典的锌–硫酸铜电池中,锌失去电子被氧化,而铜(II)离子获得电子被还原。

Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

Because the zinc half-cell has a stronger tendency to lose electrons than the copper half-cell, electrons are forced through the external wire from the zinc electrode to the copper electrode. This movement of electrons is the electric current while the chemical reaction proceeds.

由于锌半电池失去电子的倾向比铜半电池更强,电子被迫通过外电路从锌电极流向铜电极。电子的移动即形成电流,同时化学反应持续进行。


2. Structure of a Galvanic Cell | 原电池的结构

A galvanic cell is made up of two half-cells connected by an external circuit and a salt bridge. Each half-cell contains an electrode immersed in an electrolyte containing ions of the same metal.

原电池由两个半电池组成,通过外电路和盐桥相连。每个半电池都包含一个浸入含有同种金属离子电解质中的电极。

KNO₃ is chosen for the salt bridge because both K⁺ and NO₃⁻ are inert towards the electrode reactions. During cell operation, anions move towards the anode half-cell and cations move towards the cathode half-cell to balance the excess charges caused by oxidation and reduction.

盐桥选择 KNO₃,是因为 K⁺ 和 NO₃⁻ 对电极反应都是惰性的。电池工作时,阴离子向阳极半电池移动,阳离子向阴极半电池移动,以平衡氧化和还原引起的多余电荷。


3. The Standard Hydrogen Electrode | 标准氢电极

To compare the electrode potentials of different half-cells, we need one reference electrode. The standard hydrogen electrode (SHE) is chosen as the universal reference, and its potential is defined as zero at all temperatures.

为了比较不同半电池的电极电势,我们需要一个参考电极。标准氢电极(SHE)被选作通用参考,其电极电势在所有温度下都被定义为零。

2H⁺(aq) + 2e⁻ ⇌ H₂(g)   E⁰ = 0.00 V

In the SHE, hydrogen gas at 100 kPa (1 atm) is bubbled over a platinum electrode coated with finely divided platinum black. The platinum provides an inert surface for the half-reaction and does not take part chemically.

在标准氢电极中,压强为 100 kPa(1 atm)的氢气通过铂电极表面,铂电极上覆盖有粉末状的铂黑。铂提供惰性表面以进行半反应,本身不参与化学反应。

The conditions for a standard electrode potential E⁰ are: all aqueous solutions at 1 mol dm⁻³, all gases at 100 kPa, temperature usually 298 K, and all substances in their standard physical states.

标准电极电势 E⁰ 的条件是:所有水溶液浓度为 1 mol dm⁻³,所有气体压强为 100 kPa,温度通常为 298 K,所有物质均处于标准物理状态。


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4. Standard Electrode Potentials | 标准电极电势

The standard electrode potential E⁰ of a half-cell is measured by connecting it to the SHE and recording the voltage under standard conditions. The sign of E⁰ tells us whether the half-cell is more easily reduced or oxidised relative to the SHE.

半电池的标准电极电势 E⁰ 通过将其与标准氢电极连接,并在标准条件下记录电压来测定。E⁰ 的正负号告诉我们该半电池相对于标准氢电极是更容易被还原还是被氧化。

The following table lists some standard electrode potentials from the electrochemical series.

下表中列出了电化学系列中的一些标准电极电势。

Half-reaction E⁰ / V
Mg²⁺(aq) + 2e⁻ ⇌ Mg(s) -2.38
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) -0.76
2H⁺(aq) + 2e⁻ ⇌ H₂(g) 0.00
Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) +0.34
Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) +0.77
Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq) +1.36

5. Calculating Cell EMF and Cell Notation | 计算电池电动势与电池符号

The electromotive force (EMF) of a galvanic cell is the maximum voltage the cell can produce. Under standard conditions it is equal to the difference between the two electrode potentials.

电池电动势(EMF)是原电池能够产生的最大电压。在标准条件下,它等于两个电极电势之差。

E⁰ cell = E⁰ (cathode) − E⁰ (anode)

The half-cell with the more positive E⁰ is always reduced and acts as the cathode; the half-cell with the more negative E⁰ is oxidised and acts as the anode.

E⁰ 更正

Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com

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