High School Chemistry

Redox and Electrochemistry氧化还原与电化学

Every battery you charge, every piece of aluminium foil you use, every rusted nail you see — all are driven by the gain and loss of electrons. Redox chemistry governs which metals dissolve in acid, which half-cells generate voltage, and how industrial electrolysis refines metals. This guide builds from the ground up: assigning oxidation numbers, recognising redox reactions by OIL RIG, balancing equations by the half-reaction method, ranking metals by the activity series, understanding galvanic (voltaic) cells and their components, calculating cell EMF from standard reduction potentials ($E^\circ_{cell}$), and finally understanding electrolysis and its applications. Worked examples and KaTeX formulas throughout.你充电的每一块电池、你使用的每一张铝箔、你见到的每一根锈钉——都由电子的得与失驱动。氧化还原（redox，氧化还原）化学决定哪些金属溶于酸、哪些半电池产生电压，以及工业电解如何精炼金属。本指南从基础开始：分配氧化数（oxidation number，氧化数）、用 OIL RIG 识别氧化还原反应、用半反应法（half-reaction，半反应）配平方程式、用活动性顺序（activity series，活动性顺序）对金属排序、理解原电池（galvanic cell，原电池）及其组件、从标准还原电势（standard reduction potential，标准还原电势）计算电池电动势（$E^\circ_{cell}$），最后理解电解（electrolysis，电解）及其应用。全程使用例题与公式。

7 sections7 节内容 ON SCH4U · BC Chem 12 · AB Chem 30 · US noteON SCH4U · BC Chem 12 · AB Chem 30 · 美国说明 Grade 12 honors depth (ON/BC/AB); NGSS divergence noted12 年级荣誉深度（ON/BC/AB）；注明 NGSS 差异
Read me first先读这里

How to use this guide如何使用本指南

Redox and electrochemistry is a Grade-12 topic in all three Canadian provincial curricula (Ontario SCH4U, BC Chemistry 12, Alberta Chemistry 30) and has no dedicated NGSS performance expectation for US students (see the syllabus-map note). The three provincial curricula agree on a core scope: oxidation numbers, identifying redox by electron transfer, balancing by half-reactions, the activity series, galvanic cells and their components, standard reduction potentials, and electrolysis. They differ on quantitative depth — Ontario and Alberta add Faraday's-law stoichiometry at Grade 12; BC Chemistry 12 adds quantitative redox titration. The table below tells you which sections are core for you.氧化还原与电化学是三个加拿大省级大纲（安大略 SCH4U、BC Chemistry 12、阿尔伯塔 Chemistry 30）中的 12 年级主题，对美国学生而言没有专门的 NGSS 表现期望（见大纲对照说明）。三个省级大纲在核心范围上一致：氧化数、通过电子转移识别氧化还原、半反应配平、活动性顺序、原电池及其组件、标准还原电势与电解。它们在定量深度上有所不同——安大略和阿尔伯塔在 12 年级加入法拉第定律化学计量；BC Chemistry 12 加入定量氧化还原滴定。下表告诉你哪些节是你的核心。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Chemistry美国 NGSS 高中化学 §1–§2 (oxidation numbers, OIL RIG identification) as enrichment — electron transfer language appears in HS-PS1-2 (reaction outcomes) and HS-PS3-5 (electric-field energy). Use this guide to prepare for AP Chemistry Unit 9.§1–§2（氧化数、OIL RIG 识别）作为拓展——电子转移语言出现在 HS-PS1-2（反应结果）和 HS-PS3-5（电场能）中。使用本指南为 AP Chemistry Unit 9 做准备。 §5–§7 (galvanic cells, cell EMF, electrolysis): above the NGSS assessed floor; treat as AP Chemistry preparation.§5–§7（原电池、电池电动势、电解）：超出 NGSS 评估范围；作为 AP Chemistry 准备。 NGSS HS-PS1 (Chemistry) — Unit 13 divergence note— 第 13 单元差异说明
🇨🇦 ON SCH4U (Grade 12)安大略 SCH4U（12 年级） §1–§7 in full. F2.1–F2.6 and F3.1–F3.4 are all assessed. Know oxidation numbers, half-reaction balancing, galvanic cells, standard reduction potentials, and electrolysis applications (F3.5, F3.6).§1–§7 完整学习。F2.1–F2.6 和 F3.1–F3.4 均被评估。掌握氧化数、半反应配平、原电池、标准还原电势和电解应用（F3.5、F3.6）。 Nothing — SCH4U Strand F covers the full scope.无 — SCH4U F 单元覆盖完整范围。 Ontario SCH3U/4U Chemistry — SCH4U Strand F F2.1–F2.6, F3.1–F3.6— SCH4U F 单元 F2.1–F2.6、F3.1–F3.6
🇨🇦 BC Chemistry 12BC Chemistry 12 §1–§7 in full. Big Idea: "Oxidation and reduction are complementary processes that involve the gain or loss of electrons." Content: oxidation-reduction process, electrochemical cells, electrolytic cells, quantitative relationships (including $E^\circ$, redox titration mole calculations).§1–§7 完整学习。大概念："氧化与还原是涉及电子得失的互补过程。"内容：氧化还原过程、电化学电池、电解槽、定量关系（包括 $E^\circ$、氧化还原滴定摩尔计算）。 Nothing — BC Chemistry 12 assesses the full scope including quantitative $E^\circ$ and redox titration.无 — BC Chemistry 12 评估完整范围，包括定量 $E^\circ$ 和氧化还原滴定。 BC Chemistry 11/12 — Chemistry 12 Big Idea + Content + Elaborations for redox/electrochemistry— Chemistry 12 大概念 + 内容 + 细化（氧化还原/电化学）
🇨🇦 AB Chemistry 30阿尔伯塔 Chemistry 30 §1–§7 in full. Unit B GO1 and GO2 require: oxidation/reduction definitions, oxidation numbers, half-reaction balancing, spontaneity predictions, standard cell potential calculations, and Faraday's-law stoichiometry (mass, amount, current, time).§1–§7 完整学习。B 单元 GO1 和 GO2 要求：氧化/还原定义、氧化数、半反应配平、自发性预测、标准电池电势计算和法拉第定律化学计量（质量、物质的量、电流、时间）。 Nothing — AB Chemistry 30 Diploma exam builds directly on all seven sections, including Faraday's law.无 — AB Chemistry 30 文凭考试直接建立在全部七节内容上，包括法拉第定律。 Alberta Chemistry 20/30 — Chemistry 30 Unit B GO1/GO2 knowledge outcomes verbatim— Chemistry 30 B 单元 GO1/GO2 知识结果原文
🇺🇸 IB / AP feeder trackIB / AP 衔接轨道 All seven sections plus every going-deeper box. IB Chemistry HL Reactivity 1 (electron transfer and energetics) and AP Chemistry Unit 9 both require fluent half-reaction writing and $E^\circ_{cell}$ calculation from day one.全部 7 节及每个"深入"框。IB Chemistry HL Reactivity 1（电子转移与能量）和 AP Chemistry Unit 9 第一天就要求熟练书写半反应和计算 $E^\circ_{cell}$。 Nothing — this unit is the foundation for all IB/AP electrochemistry, including Nernst-equation extensions and Faraday quantitative work.无 — 本单元是所有 IB/AP 电化学的基础，包括能斯特方程拓展和法拉第定量内容。 NGSS HS-PS1 (Chemistry) — see "What This Feeds Into" for IB Chemistry HL feeder links— 见"本单元的去向"中的 IB Chemistry HL 衔接链接

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise five things: OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons); oxidation number rules (fluorine is always −1, hydrogen is +1, oxygen is −2 except peroxides); $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$; electrons flow from anode (oxidation) to cathode (reduction) in a galvanic cell; and electrolysis reverses spontaneous redox by applying external voltage. Read every cram-cheat box. Skip the Faraday-law derivations if time is short.背熟五件事：OIL RIG（氧化是失去电子，还原是得到电子）；氧化数规则（氟始终为 −1，氢为 +1，氧为 −2，过氧化物除外）；$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$；在原电池中电子从阳极（氧化）流向阴极（还原）；电解通过施加外部电压来逆转自发的氧化还原反应。读每个速记框。若时间紧，可跳过法拉第定律推导。

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If you are going for the top mark如果你目标顶分

Be precise about electron flow, ion flow, and the role of each cell component; correctly identify anode/cathode in both galvanic and electrolytic cells (they switch polarity); balance redox equations in acidic solution using the full half-reaction method (add H$_2$O for O, add H$^+$ for H, add e$^-$ to balance charge); calculate $E^\circ_{cell}$ and interpret its sign for spontaneity; and apply Faraday's law ($Q = nF$, $m = \frac{MIt}{nF}$) for electrolysis calculations. ON SCH4U F2.6 and AB Chemistry 30 GO2 both require spontaneity prediction from $E^\circ_{cell}$.精确理解电子流动、离子流动和每个电池组件的作用；在原电池和电解槽中正确识别阳极/阴极（极性互换）；用完整的半反应法在酸性溶液中配平氧化还原方程式（加 H$_2$O 平衡 O，加 H$^+$ 平衡 H，加 e$^-$ 平衡电荷）；计算 $E^\circ_{cell}$ 并解读其符号的自发性含义；将法拉第定律（$Q = nF$，$m = \frac{MIt}{nF}$）应用于电解计算。ON SCH4U F2.6 和 AB Chemistry 30 GO2 都要求从 $E^\circ_{cell}$ 预测自发性。

Honors flag.荣誉级标记。 This entire unit carries the Honors chip for US NGSS students, where there is no assessed redox or electrochemistry PE at the high-school level. For ON SCH3U students (Grade 11), redox and electrochemistry enter only at SCH4U (Grade 12) — treat all seven sections as SCH4U honors content. BC Chemistry 12 and AB Chemistry 30 students: all seven sections are core assessed content, not honors.对于美国 NGSS 学生，整个单元标 Honors，因为高中阶段没有被评估的氧化还原或电化学表现期望。对于 ON SCH3U 学生（11 年级），氧化还原与电化学仅在 SCH4U（12 年级）引入——将全部 7 节视为 SCH4U 荣誉内容。BC Chemistry 12 和 AB Chemistry 30 学生：全部 7 节均为核心被评估内容，而非荣誉。
Section 1 · Oxidation numbers第 1 节 · 氧化数

Oxidation Numbers氧化数

Six rules that assign electron bookkeeping to every atom in a formula.六条规则——为化学式中每个原子分配电子记账。
  • Rule 1规则 1 — pure element: oxidation number = 0. Examples: Na(s), O$_2$(g), Fe(s) all have oxidation number 0.— 纯元素：氧化数 = 0。例：Na(s)、O$_2$(g)、Fe(s) 的氧化数均为 0。
  • Rule 2规则 2 — monatomic ion: oxidation number = charge. Example: Na$^+$ is $+1$; Cl$^-$ is $-1$; Fe$^{3+}$ is $+3$.— 单原子离子：氧化数 = 电荷。例：Na$^+$ 为 $+1$；Cl$^-$ 为 $-1$；Fe$^{3+}$ 为 $+3$。
  • Rule 3规则 3 — fluorine: always $-1$ in compounds (most electronegative element).— 氟：在化合物中始终为 $-1$（电负性最强的元素）。
  • Rule 4规则 4 — hydrogen: $+1$ in most compounds; $-1$ in metal hydrides (NaH, CaH$_2$).— 氢：在大多数化合物中为 $+1$；在金属氢化物（NaH、CaH$_2$）中为 $-1$。
  • Rule 5规则 5 — oxygen: $-2$ in most compounds; $-1$ in peroxides (H$_2$O$_2$, Na$_2$O$_2$); $+2$ in OF$_2$ (fluorine overrides).— 氧：在大多数化合物中为 $-2$；在过氧化物（H$_2$O$_2$、Na$_2$O$_2$）中为 $-1$；在 OF$_2$ 中为 $+2$（氟优先级更高）。
  • Rule 6规则 6 — the sum of all oxidation numbers in a neutral molecule = 0; in a polyatomic ion = the ion's charge.— 中性分子中所有氧化数之和 = 0；多原子离子中 = 该离子的电荷。
ON SCH4U F2.1 lists "oxidation number" as assessed terminology. AB Chemistry 30 GO1 (verbatim): "define oxidizing agent, reducing agent, oxidation number, half-reaction." BC Chemistry 12 Content: "the oxidation-reduction process: — oxidation number — balancing redox reactions."ON SCH4U F2.1 将"氧化数"列为被评估术语。AB Chemistry 30 GO1（原文）："定义氧化剂、还原剂、氧化数、半反应。"BC Chemistry 12 内容："氧化还原过程：——氧化数——配平氧化还原反应。"
Worked Example 1 · Assigning oxidation numbers in Cr$_2$O$_7^{2-}$例题 1 · 在 Cr$_2$O$_7^{2-}$ 中分配氧化数

Find the oxidation number of chromium in the dichromate ion, Cr$_2$O$_7^{2-}$.求重铬酸根离子 Cr$_2$O$_7^{2-}$ 中铬的氧化数。

Set up the equation.建立方程。 Let the oxidation number of Cr be $x$. There are 2 Cr atoms and 7 O atoms. Oxygen is $-2$ (Rule 5; no peroxide here). Ion charge is $2-$:设 Cr 的氧化数为 $x$。有 2 个 Cr 原子和 7 个 O 原子。氧为 $-2$（规则 5；无过氧化物）。离子电荷为 $2-$：

$$ 2x + 7(-2) = -2 $$ $$ 2x - 14 = -2 \implies 2x = 12 \implies x = +6 $$

Chromium is in the $+6$ oxidation state in Cr$_2$O$_7^{2-}$.Cr$_2$O$_7^{2-}$ 中铬的氧化态为 $+6$。 This is one of the most powerful oxidising agents in common use (e.g. acidified dichromate in alcohol breathalyser reactions).这是常见的最强氧化剂之一（如酸化重铬酸钾用于酒精呼气检测反应）。

What is the oxidation number of sulfur in H$_2$SO$_4$?H$_2$SO$_4$ 中硫的氧化数是多少？
§1 · Q1
$+2$$+2$
$+4$$+4$
$+6$$+6$
$-2$$-2$
H is $+1$ (×2 = $+2$); O is $-2$ (×4 = $-8$). Sum must be 0: $2 + x + (-8) = 0 \Rightarrow x = +6$. Sulfur is $+6$ in sulfuric acid — its highest oxidation state.H 为 $+1$（×2 = $+2$）；O 为 $-2$（×4 = $-8$）。总和必须为 0：$2 + x + (-8) = 0 \Rightarrow x = +6$。硫在硫酸中为 $+6$——其最高氧化态。
Use Rule 6: $2(+1) + x + 4(-2) = 0$, so $2 + x - 8 = 0$, giving $x = +6$. Sulfur ranges from $-2$ (H$_2$S) to $+6$ (H$_2$SO$_4$).使用规则 6：$2(+1) + x + 4(-2) = 0$，即 $2 + x - 8 = 0$，得 $x = +6$。硫的氧化数范围从 $-2$（H$_2$S）到 $+6$（H$_2$SO$_4$）。
What is the oxidation number of nitrogen in NH$_3$?NH$_3$ 中氮的氧化数是多少？
§1 · Q2
$-3$$-3$
$0$$0$
$+3$$+3$
$+5$$+5$
H is $+1$ (×3 = $+3$). Sum = 0: $x + 3 = 0 \Rightarrow x = -3$. Nitrogen is at its most reduced state ($-3$) in ammonia. It ranges from $-3$ (NH$_3$) to $+5$ (HNO$_3$).H 为 $+1$（×3 = $+3$）。总和 = 0：$x + 3 = 0 \Rightarrow x = -3$。氮在氨中处于最低氧化态（$-3$）。其范围从 $-3$（NH$_3$）到 $+5$（HNO$_3$）。
Each H contributes $+1$; three H atoms give $+3$. For the molecule sum to be 0, N must be $-3$. This is nitrogen's lowest (most reduced) common oxidation state.每个 H 贡献 $+1$；三个 H 原子共 $+3$。为使分子总和为 0，N 必须为 $-3$。这是氮最低（最还原）的常见氧化态。
Going deeper — why oxidation numbers are bookkeeping, not reality深入 — 为何氧化数是记账而非现实

An oxidation number is a formal charge assignment based on the assumption that all bonds are ionic (electrons assigned entirely to the more electronegative atom). In reality, covalent bonds share electrons. For example, in CH$_4$, carbon is assigned $-4$ (all four C–H electrons counted as carbon's since C is slightly more electronegative than H). This does not mean carbon literally carries four extra electrons; it is a bookkeeping convention that lets us track electron flow in redox reactions consistently. The convention works because oxidation-number changes in a balanced redox equation always sum to zero overall — a direct consequence of electron conservation. SCH4U F2.3 and AB Chemistry 30 GO1 both assess the ability to write half-reactions and balance equations using this formalism.氧化数是基于所有化学键均为离子键的假设（电子完全分配给电负性更强的原子）而做出的形式上的电荷分配。实际上，共价键是共享电子的。例如，在 CH$_4$ 中，碳被分配 $-4$（因为 C 比 H 电负性略强，所有四个 C-H 电子都算作碳的）。这并不意味着碳真的携带四个额外电子；这是一种记账惯例，让我们能够一致地追踪氧化还原反应中的电子流动。这个惯例之所以有效，是因为配平的氧化还原方程式中氧化数变化的总和始终为零——这是电子守恒的直接结果。SCH4U F2.3 和 AB Chemistry 30 GO1 都评估用这种形式主义书写半反应和配平方程式的能力。

Section 2 · Identifying redox reactions (OIL RIG)第 2 节 · 识别氧化还原反应（OIL RIG）

Identifying Redox Reactions: OIL RIG识别氧化还原反应：OIL RIG

OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).OIL RIG：氧化是失去电子，还原是得到电子。
  • Oxidation氧化 — loss of electrons; oxidation number increases (becomes more positive). The species losing electrons is the reducing agent.— 失去电子；氧化数升高（变得更正）。失去电子的物质是还原剂（自身被氧化，同时使其他物质被还原）。
  • Reduction还原 — gain of electrons; oxidation number decreases (becomes more negative). The species gaining electrons is the oxidising agent.— 得到电子；氧化数降低（变得更负）。得到电子的物质是氧化剂（自身被还原，同时使其他物质被氧化）。
  • Identifying redox识别氧化还原 : assign oxidation numbers to all atoms (§1), then compare reactant to product. If any atom's oxidation number changes, the reaction is a redox reaction. If no oxidation numbers change (e.g. acid–base neutralisation, precipitation), it is not redox.：为所有原子分配氧化数（§1），然后对比反应物与生成物。如果任何原子的氧化数发生变化，则该反应是氧化还原反应。如果氧化数不变（如酸碱中和、沉淀），则不是氧化还原反应。

Key relationship:关键关系：

$$ \text{electrons lost by reducing agent} = \text{electrons gained by oxidising agent} $$ ON SCH4U F3.1 (verbatim): "explain redox reactions in terms of the loss and gain of electrons and the associated change in oxidation number." AB Chemistry 30 GO1: "differentiate between redox reactions and other reactions, using half-reactions and/or oxidation numbers."ON SCH4U F3.1（原文）："从电子得失及氧化数变化的角度解释氧化还原反应。"AB Chemistry 30 GO1："用半反应和/或氧化数区分氧化还原反应与其他反应。"
Worked Example 2 · Identifying oxidised/reduced species in a reaction例题 2 · 识别反应中被氧化/被还原的物质

In the reaction $\text{Fe} + \text{CuSO}_4 \to \text{FeSO}_4 + \text{Cu}$, identify (a) what is oxidised, (b) what is reduced, (c) the oxidising agent, and (d) the reducing agent.在反应 $\text{Fe} + \text{CuSO}_4 \to \text{FeSO}_4 + \text{Cu}$ 中，识别 (a) 被氧化的物质，(b) 被还原的物质，(c) 氧化剂，(d) 还原剂。

Assign oxidation numbers.分配氧化数。

Fe: $0$ (pure element) $\to$ Fe in FeSO$_4$: $+2$ (increases). Cu in CuSO$_4$: $+2$ $\to$ Cu: $0$ (decreases). SO$_4^{2-}$ does not change ($+6$ throughout — spectator).Fe：$0$（纯元素）→ FeSO$_4$ 中 Fe：$+2$（升高）。CuSO$_4$ 中 Cu：$+2$ → Cu：$0$（降低）。SO$_4^{2-}$ 不变（全程 $+6$——旁观者）。

(a) Iron (Fe) is oxidised(a) 铁（Fe）被氧化 — its oxidation number goes from $0$ to $+2$ (loses 2 electrons).——氧化数从 $0$ 升至 $+2$（失去 2 个电子）。

(b) Copper (Cu) is reduced(b) 铜（Cu）被还原 — its oxidation number goes from $+2$ to $0$ (gains 2 electrons).——氧化数从 $+2$ 降至 $0$（得到 2 个电子）。

(c) Oxidising agent: CuSO$_4$(c) 氧化剂：CuSO$_4$ (the species that is itself reduced — gains electrons).（自身被还原——得到电子的物质）。

(d) Reducing agent: Fe(d) 还原剂：Fe (the species that is itself oxidised — loses electrons). ✓（自身被氧化——失去电子的物质）。✓

In the reaction $2\text{Mg} + \text{O}_2 \to 2\text{MgO}$, which species is the reducing agent?在反应 $2\text{Mg} + \text{O}_2 \to 2\text{MgO}$ 中，哪种物质是还原剂？
§2 · Q1
O$_2$, because it gains electronsO$_2$，因为它得到电子
Mg, because it loses electrons (is oxidised)Mg，因为它失去电子（被氧化）
MgO, because it is formedMgO，因为它是生成物
Mg, because it gains electronsMg，因为它得到电子
Mg goes from $0$ to $+2$ (oxidised — loses electrons). The reducing agent is the species that is itself oxidised. O$_2$ goes from $0$ to $-2$ (reduced — gains electrons), so O$_2$ is the oxidising agent.Mg 从 $0$ 升至 $+2$（被氧化——失去电子）。还原剂是自身被氧化的物质。O$_2$ 从 $0$ 降至 $-2$（被还原——得到电子），故 O$_2$ 是氧化剂。
OIL RIG: the reducing agent is oxidised (loses electrons). Mg loses electrons ($0 \to +2$) — it IS the reducing agent. O$_2$ gains electrons ($0 \to -2$) — it is the oxidising agent.OIL RIG：还原剂被氧化（失去电子）。Mg 失去电子（$0 \to +2$）——它就是还原剂。O$_2$ 得到电子（$0 \to -2$）——它是氧化剂。
Which of the following is NOT a redox reaction?以下哪项不是氧化还原反应？
§2 · Q2
$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$
$\text{Zn} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2$$\text{Zn} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2$
$2\text{Fe} + 3\text{Cl}_2 \to 2\text{FeCl}_3$$2\text{Fe} + 3\text{Cl}_2 \to 2\text{FeCl}_3$
$\text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O}$$\text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_2\text{O}$
Acid–base neutralisation (NaOH + HCl): Na is $+1$ throughout, O is $-2$ throughout, H is $+1$ throughout, Cl is $-1$ throughout — no oxidation-number change. It is a double-displacement / ion-exchange reaction, not redox. All other options involve oxidation-number changes.酸碱中和（NaOH + HCl）：Na 全程 $+1$，O 全程 $-2$，H 全程 $+1$，Cl 全程 $-1$——无氧化数变化。这是复分解/离子交换反应，不是氧化还原。其他选项均涉及氧化数变化。
Check oxidation numbers: if any atom's oxidation number changes, it's redox. NaOH + HCl has no oxidation-number changes at all — it's an acid-base neutralisation (ion exchange).检查氧化数：如果任何原子的氧化数发生变化，则是氧化还原。NaOH + HCl 完全没有氧化数变化——这是酸碱中和（离子交换）。
Section 3 · Balancing redox by half-reactions第 3 节 · 半反应法配平氧化还原方程式

Balancing Redox by the Half-Reaction Method半反应法配平氧化还原方程式

Split the reaction into two half-reactions, balance each, then recombine.将反应分为两个半反应，分别配平，再合并。

Steps for acidic solution:酸性溶液中的步骤：

  1. Write the two unbalanced half-reactions (one oxidation, one reduction).写出两个未配平的半反应（一个氧化，一个还原）。
  2. Balance all atoms other than H and O.配平除 H 和 O 以外的所有原子。
  3. Balance O by adding H$_2$O to the deficient side.通过向缺少 O 的一侧加 H$_2$O 来配平 O。
  4. Balance H by adding H$^+$ to the deficient side.通过向缺少 H 的一侧加 H$^+$ 来配平 H。
  5. Balance charge by adding electrons (e$^-$) to the more-positive side.通过向电荷更正的一侧加电子（e$^-$）来配平电荷。
  6. Multiply the half-reactions so the electrons cancel, then add them.将两个半反应乘以适当系数使电子消去，然后相加。
  7. Simplify (cancel common species). Check: atoms and charge balanced.化简（消去相同物种）。检查：原子和电荷均已配平。
ON SCH4U F2.3 (verbatim): "write balanced chemical equations for oxidation-reduction reactions, using various methods including oxidation numbers of atoms and the half-reaction method of balancing." AB Chemistry 30 GO1: "write and balance equations for redox reactions in acidic and neutral solutions by … developing simple half-reaction equations … and assigning oxidation numbers."ON SCH4U F2.3（原文）："用氧化数法和半反应配平法等方法书写氧化还原反应的配平化学方程式。"AB Chemistry 30 GO1："通过建立简单的半反应方程式……和分配氧化数，书写和配平酸性和中性溶液中氧化还原反应的方程式。"
Worked Example 3 · Balancing MnO$_4^-$ + Fe$^{2+}$ in acidic solution例题 3 · 在酸性溶液中配平 MnO$_4^-$ + Fe$^{2+}$

Balance the redox reaction: $\text{MnO}_4^- + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+}$ in acidic solution.在酸性溶液中配平氧化还原反应：$\text{MnO}_4^- + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+}$。

Reduction half-reaction (Mn):还原半反应（Mn）：

$$ \text{MnO}_4^- \to \text{Mn}^{2+} $$

Balance O: add 4H$_2$O right. Balance H: add 8H$^+$ left. Balance charge: left = $-1+8=+7$; right = $+2$. Add 5e$^-$ left:配平 O：右侧加 4H$_2$O。配平 H：左侧加 8H$^+$。配平电荷：左侧 = $-1+8=+7$；右侧 = $+2$。左侧加 5e$^-$：

$$ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \to \text{Mn}^{2+} + 4\text{H}_2\text{O} $$

Oxidation half-reaction (Fe):氧化半反应（Fe）：

$$ \text{Fe}^{2+} \to \text{Fe}^{3+} + e^- $$

Equalise electrons: multiply Fe half-reaction ×5:均衡电子：Fe 半反应 ×5：

$$ 5\text{Fe}^{2+} \to 5\text{Fe}^{3+} + 5e^- $$

Add and cancel 5e$^-$:相加并消去 5e$^-$：

$$ \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+} $$

Check atoms and charge: Mn ✓, O ✓, H ✓, Fe ✓. Left charge: $-1+8+10 = +17$. Right charge: $+2 + 0 + 15 = +17$. ✓检查原子和电荷：Mn ✓，O ✓，H ✓，Fe ✓。左侧电荷：$-1+8+10 = +17$。右侧电荷：$+2 + 0 + 15 = +17$。✓

When balancing a redox half-reaction in acidic solution, what is added to balance oxygen atoms?在酸性溶液中配平氧化还原半反应时，加入什么来配平氧原子？
§3 · Q1
H$_2$O moleculesH$_2$O 分子
O$_2$ gasO$_2$ 气体
OH$^-$ ionsOH$^-$ 离子
Electrons电子
In acidic solution: H$_2$O balances O; H$^+$ balances H; e$^-$ balances charge. (In basic solution: H$_2$O and OH$^-$ are used instead, but that is beyond this guide's scope.)在酸性溶液中：H$_2$O 配平 O；H$^+$ 配平 H；e$^-$ 配平电荷。（在碱性溶液中改用 H$_2$O 和 OH$^-$，但超出本指南范围。）
The four species added in acidic solution balancing are: H$_2$O (for O), H$^+$ (for H), and e$^-$ (for charge). O$_2$ and OH$^-$ are not added in acidic-solution balancing.酸性溶液配平中加入的四种物质是：H$_2$O（配平 O）、H$^+$（配平 H）和 e$^-$（配平电荷）。O$_2$ 和 OH$^-$ 不在酸性溶液配平中加入。
In the balanced half-reaction for the reduction of Cr$_2$O$_7^{2-}$ to Cr$^{3+}$ in acid, how many electrons are transferred per Cr$_2$O$_7^{2-}$?在酸性溶液中 Cr$_2$O$_7^{2-}$ 还原为 Cr$^{3+}$ 的配平半反应中，每个 Cr$_2$O$_7^{2-}$ 转移多少个电子？
§3 · Q2
33
44
66
1212
Each Cr goes from $+6$ to $+3$: a change of $-3$ per Cr atom. There are 2 Cr atoms in Cr$_2$O$_7^{2-}$, so $2 \times 3 = 6$ electrons total. Full half-reaction: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$.每个 Cr 从 $+6$ 降至 $+3$：每个 Cr 原子变化 $-3$。Cr$_2$O$_7^{2-}$ 中有 2 个 Cr 原子，故共 $2 \times 3 = 6$ 个电子。完整半反应：$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$。
Electrons per ion = (change in oxidation number per atom) × (number of atoms). Cr: $+6 \to +3$ is $\Delta = 3$ per Cr; two Cr atoms give $2 \times 3 = 6$ electrons.每个离子的电子数 = （每个原子的氧化数变化）×（原子数）。Cr：$+6 \to +3$ 即 $\Delta = 3$/个 Cr；两个 Cr 原子得 $2 \times 3 = 6$ 个电子。
Section 4 · The activity series第 4 节 · 活动性顺序

The Activity Series活动性顺序

A ranked list of metals (and H$_2$) by their tendency to be oxidised — to lose electrons.金属（及 H$_2$）按失去电子倾向（被氧化）从强到弱的排列。
  • More active metal活性更强的金属 — higher in the series (e.g. Li, K, Ca, Na, Mg, Al, Zn, Fe, Ni, Sn, Pb, H$_2$, Cu, Ag, Au). Metals above H$_2$ dissolve in acid and displace H$_2$ gas.— 在顺序中排名较高（如 Li、K、Ca、Na、Mg、Al、Zn、Fe、Ni、Sn、Pb、H$_2$、Cu、Ag、Au）。H$_2$ 以上的金属溶于酸并置换出 H$_2$ 气体。
  • Displacement reaction置换反应 — a more active metal (higher in the series) displaces a less active metal ion from solution: $\text{A} + \text{B}^{n+} \to \text{A}^{n+} + \text{B}$, where A is more active than B.— 活性更强的金属（在顺序中更高）将活性较弱的金属离子从溶液中置换出来：$\text{A} + \text{B}^{n+} \to \text{A}^{n+} + \text{B}$，其中 A 比 B 活性更强。
  • Prediction rule预测规则 — a reaction is spontaneous (goes as written) only if the metal being oxidised is higher in the series than the ion being reduced. If the metal is lower, no reaction occurs.— 只有当被氧化的金属在顺序中高于被还原的离子时，反应才是自发的（按书写方向进行）。如果金属排名较低，则不发生反应。
ON SCH4U F: the metal activity series is referenced in the SCH3U C2.5 context ("predict the products of single displacement reactions, using the metal activity series") and the full electrochemical application is in SCH4U F. AB Chemistry 30 GO1: "compare the relative strengths of oxidizing and reducing agents, using empirical data" — the activity series is this empirical data.ON SCH4U F：金属活动性顺序在 SCH3U C2.5 背景下被引用（"用金属活动性顺序预测单置换反应的产物"），完整的电化学应用在 SCH4U F 中。AB Chemistry 30 GO1："用实验数据比较氧化剂和还原剂的相对强度"——活动性顺序就是这些实验数据。
Worked Example 4 · Predicting displacement reactions例题 4 · 预测置换反应

Predict whether the following reactions occur. If yes, write the balanced equation. Activity series order (partial, most to least active): Mg, Al, Zn, Fe, Pb, H$_2$, Cu, Ag. (a) Fe(s) + CuSO$_4$(aq) → ? (b) Cu(s) + ZnSO$_4$(aq) → ?预测以下反应是否发生。若发生，写出配平方程式。活动性顺序（部分，从强到弱）：Mg、Al、Zn、Fe、Pb、H$_2$、Cu、Ag。(a) Fe(s) + CuSO$_4$(aq) → ？(b) Cu(s) + ZnSO$_4$(aq) → ？

(a) Fe is more active than Cu (Fe is above Cu in the series).(a) Fe 比 Cu 活性更强（Fe 在顺序中排在 Cu 上方）。 Reaction occurs: Fe displaces Cu$^{2+}$ from solution.反应发生：Fe 将 Cu$^{2+}$ 从溶液中置换出来。

$$ \text{Fe(s)} + \text{CuSO}_4\text{(aq)} \to \text{FeSO}_4\text{(aq)} + \text{Cu(s)} $$

(b) Cu is less active than Zn (Cu is below Zn in the series).(b) Cu 比 Zn 活性更弱（Cu 在顺序中排在 Zn 下方）。 Cu cannot displace Zn$^{2+}$ from solution. No reaction occurs.Cu 无法将 Zn$^{2+}$ 从溶液中置换出来。不发生反应。

$$ \text{Cu(s)} + \text{ZnSO}_4\text{(aq)} \to \text{no reaction} $$
Which metal will NOT dissolve in dilute hydrochloric acid (HCl)?哪种金属不会溶于稀盐酸（HCl）？
§4 · Q1
Zinc (Zn)锌（Zn）
Iron (Fe)铁（Fe）
Magnesium (Mg)镁（Mg）
Copper (Cu)铜（Cu）
Metals above H$_2$ in the activity series dissolve in acid (they displace H$_2$ gas). Cu is below H$_2$ — it cannot reduce H$^+$ to H$_2$, so it does not dissolve in dilute HCl. Zn, Fe, and Mg are all above H$_2$ and do dissolve.活动性顺序中 H$_2$ 以上的金属溶于酸（它们置换出 H$_2$ 气体）。Cu 在 H$_2$ 以下——它无法将 H$^+$ 还原为 H$_2$，故不溶于稀盐酸。Zn、Fe 和 Mg 均在 H$_2$ 以上，均能溶解。
The rule: metals above H$_2$ in the activity series react with acid to produce H$_2$ gas. Cu is below H$_2$ (less active than hydrogen) and therefore does NOT dissolve in dilute HCl or H$_2$SO$_4$.规则：活动性顺序中 H$_2$ 以上的金属与酸反应产生 H$_2$ 气体。Cu 在 H$_2$ 以下（活性弱于氢），因此不溶于稀 HCl 或 H$_2$SO$_4$。
Silver (Ag) is placed in a solution of copper(II) sulfate (CuSO$_4$). What happens?将银（Ag）放入硫酸铜（CuSO$_4$）溶液中，会发生什么？
§4 · Q2
Ag dissolves and copper metal depositsAg 溶解，铜金属沉积
No reaction — Cu is more active than Ag, so Cu$^{2+}$ is not displaced不发生反应——Cu 比 Ag 活性更强，Cu$^{2+}$ 不会被置换
Silver dissolves and silver sulfate formsAg 溶解，硫酸银生成
Both metals dissolve两种金属均溶解
Activity series: Cu is above Ag (Cu is more active). For displacement to occur, the metal placed in solution must be MORE active than the metal ion in solution. Ag is LESS active than Cu — so Ag cannot displace Cu$^{2+}$. No reaction occurs.活动性顺序：Cu 在 Ag 上方（Cu 更活泼）。要发生置换，放入溶液的金属必须比溶液中的金属离子活性更强。Ag 比 Cu 活性更弱——故 Ag 无法置换 Cu$^{2+}$。不发生反应。
Displacement requires: metal in (solid) > metal ion (in solution) in activity. Cu > Ag in activity, so Ag cannot displace Cu$^{2+}$. The reaction would go in the other direction: Cu(s) + 2Ag$^+$(aq) → Cu$^{2+}$(aq) + 2Ag(s).置换需要：固体金属的活性 > 溶液中金属离子的活性。Cu 的活性 > Ag，故 Ag 无法置换 Cu$^{2+}$。反应向另一个方向进行：Cu(s) + 2Ag$^+$(aq) → Cu$^{2+}$(aq) + 2Ag(s)。
Section 5 · Galvanic (voltaic) cells第 5 节 · 原电池（伏打电池）

Galvanic (Voltaic) Cells原电池（伏打电池）

A galvanic cell converts spontaneous redox energy into electrical work.原电池将自发的氧化还原能量转化为电功。
  • Anode阳极 — where oxidation occurs. The anode is the negative electrode in a galvanic cell (it loses mass as metal dissolves). Electrons flow out of the anode through the external circuit.— 氧化发生的地方。阳极是原电池中的负极（金属溶解时质量减少）。电子通过外部电路从阳极流出。
  • Cathode阴极 — where reduction occurs. The cathode is the positive electrode in a galvanic cell (metal deposits, gains mass). Electrons flow into the cathode from the external circuit.— 还原发生的地方。阴极是原电池中的正极（金属沉积，质量增加）。电子通过外部电路流入阴极。
  • Salt bridge盐桥 — allows ions to migrate between the two half-cells to maintain electrical neutrality. Without a salt bridge, charge would build up and the cell would stop. Anions migrate to the anode half-cell; cations migrate to the cathode half-cell.— 允许离子在两个半电池之间迁移以维持电中性。没有盐桥，电荷会积聚，电池会停止工作。阴离子迁移到阳极半电池；阳离子迁移到阴极半电池。
  • Memory aid: An Ox, Red Cat记忆口诀：阳氧阴还 (Anode = Oxidation; Cathode = Reduction).（阳极 = 氧化；阴极 = 还原）。
ON SCH4U F3.2 (verbatim): "identify the components of a galvanic cell, and explain how each component functions in a redox reaction." F2.5: "analyse the processes in galvanic cells … (electron flow, electrode polarity, cell potential, ion movement)." AB Chemistry 30 GO2: "define anode, cathode, anion, cation, salt bridge/porous cup, electrolyte, external circuit, power supply, voltaic cell and electrolytic cell."ON SCH4U F3.2（原文）："识别原电池的组件，并解释每个组件在氧化还原反应中的功能。"F2.5："分析原电池中的过程（电子流动、电极极性、电池电势、离子运动）。"AB Chemistry 30 GO2："定义阳极、阴极、阴离子、阳离子、盐桥/多孔杯、电解质、外部电路、电源、伏打电池和电解槽。"
Worked Example 5 · Describing the Zn–Cu galvanic cell例题 5 · 描述 Zn-Cu 原电池

A galvanic cell is built with Zn(s)/Zn$^{2+}$(aq) in one half-cell and Cu(s)/Cu$^{2+}$(aq) in the other, connected by a salt bridge. (a) Identify the anode and cathode. (b) Write the half-reactions. (c) Describe the electron and ion flow.用 Zn(s)/Zn$^{2+}$(aq) 和 Cu(s)/Cu$^{2+}$(aq) 各一个半电池构建原电池，用盐桥连接。(a) 确定阳极和阴极。(b) 写出半反应。(c) 描述电子和离子的流动。

(a) Activity series: Zn is more active than Cu (Zn is above Cu).(a) 活动性顺序：Zn 比 Cu 更活泼（Zn 在 Cu 上方）。 Zn is more easily oxidised — it is the anode (negative electrode). Cu$^{2+}$ is more easily reduced — Cu is the cathode (positive electrode).Zn 更容易被氧化——它是阳极（负极）。Cu$^{2+}$ 更容易被还原——Cu 是阴极（正极）。

(b) Half-reactions:(b) 半反应：

$$ \text{Anode (oxidation): } \text{Zn(s)} \to \text{Zn}^{2+}\text{(aq)} + 2e^- $$ $$ \text{Cathode (reduction): } \text{Cu}^{2+}\text{(aq)} + 2e^- \to \text{Cu(s)} $$ $$ \text{Overall: } \text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \to \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)} $$

(c) Electron and ion flow.(c) 电子和离子流动。 Electrons flow from Zn anode (negative) through the external wire to Cu cathode (positive). In the salt bridge, anions (e.g. SO$_4^{2-}$ or Cl$^-$) migrate toward the anode half-cell; cations migrate toward the cathode half-cell — maintaining electrical neutrality.电子从 Zn 阳极（负极）通过外部导线流向 Cu 阴极（正极）。在盐桥中，阴离子（如 SO$_4^{2-}$ 或 Cl$^-$）向阳极半电池迁移；阳离子向阴极半电池迁移——维持电中性。

In a galvanic cell, electrons flow through the external circuit from:在原电池中，电子通过外部电路从以下方向流动：
§5 · Q1
Cathode to anode阴极流向阳极
Positive terminal to negative terminal正极流向负极
Anode (negative) to cathode (positive)阳极（负极）流向阴极（正极）
Through the salt bridge only仅通过盐桥
Electrons are produced at the anode (where oxidation occurs, the negative electrode) and consumed at the cathode (where reduction occurs, the positive electrode). Electron flow: anode → external wire → cathode. Ions flow through the salt bridge (not electrons).电子在阳极（氧化发生的地方，负极）产生，在阴极（还原发生的地方，正极）消耗。电子流动：阳极 → 外部导线 → 阴极。离子通过盐桥流动（不是电子）。
Electrons flow from where they are produced (anode, oxidation, negative) to where they are consumed (cathode, reduction, positive). The salt bridge carries ions, not electrons — it completes the circuit internally.电子从产生处（阳极，氧化，负极）流向消耗处（阴极，还原，正极）。盐桥携带的是离子，不是电子——它在内部完成电路。
What is the purpose of the salt bridge in a galvanic cell?原电池中盐桥的作用是什么？
§5 · Q2
Allow ions to flow between half-cells to maintain electrical neutrality允许离子在半电池间流动以维持电中性
Provide a path for electrons between the two electrodes为两电极间的电子提供通路
Increase the voltage of the cell提高电池的电压
Prevent any chemical reaction from occurring阻止任何化学反应发生
As oxidation proceeds at the anode, the anode solution builds up positive charge (Zn$^{2+}$ accumulates). As reduction proceeds at the cathode, the cathode solution loses positive charge (Cu$^{2+}$ is consumed). The salt bridge allows anions to flow to the anode and cations to flow to the cathode, restoring electrical neutrality and allowing the cell to continue operating.随着阳极处氧化的进行，阳极溶液积累正电荷（Zn$^{2+}$ 积累）。随着阴极处还原的进行，阴极溶液失去正电荷（Cu$^{2+}$ 被消耗）。盐桥允许阴离子流向阳极、阳离子流向阴极，恢复电中性，使电池继续工作。
Electrons do NOT flow through the salt bridge — they flow through the external wire. The salt bridge allows ion migration to maintain charge balance. Without it, the anode solution becomes too positive and the cathode solution too negative — the cell stops.电子不通过盐桥——它们通过外部导线流动。盐桥允许离子迁移以维持电荷平衡。没有盐桥，阳极溶液变得过正，阴极溶液过负——电池停止工作。
Section 6 · Standard reduction potentials and cell EMF第 6 节 · 标准还原电势与电池电动势

Standard Reduction Potentials and Cell EMF标准还原电势与电池电动势

Standard reduction potentials rank half-reactions by their tendency to be reduced.标准还原电势按被还原的倾向对半反应进行排名。
  • Standard reduction potential $E^\circ$标准还原电势 $E^\circ$ — measured at 25 °C, 1 atm, 1 mol/L concentrations, relative to the standard hydrogen electrode (SHE, $E^\circ = 0.00\ \mathrm{V}$). A more positive $E^\circ$ means a greater tendency to be reduced (stronger oxidising agent). A more negative $E^\circ$ means a greater tendency to be oxidised (stronger reducing agent).— 在 25 °C、1 atm、1 mol/L 浓度下，相对于标准氢电极（SHE，$E^\circ = 0.00\ \mathrm{V}$）测量。$E^\circ$ 越正，被还原的倾向越大（更强的氧化剂）。$E^\circ$ 越负，被氧化的倾向越大（更强的还原剂）。
  • Cell EMF formula:电池电动势公式：
$$ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} $$
  • Positive $E^\circ_\text{cell}$正的 $E^\circ_\text{cell}$ $\Rightarrow$ spontaneous reaction (galvanic cell). Negative $E^\circ_\text{cell}$ $\Rightarrow$ non-spontaneous (requires external voltage — electrolysis).$\Rightarrow$ 自发反应（原电池）。负的 $E^\circ_\text{cell}$ $\Rightarrow$ 非自发（需要外部电压——电解）。
  • Reading the table读表规则 : the half-reaction with the more positive $E^\circ$ is the cathode (reduced). The half-reaction with the more negative $E^\circ$ is the anode (reversed to write oxidation). Do NOT multiply $E^\circ$ by stoichiometric coefficients — it is an intensive property.：$E^\circ$ 较正的半反应是阴极（被还原）。$E^\circ$ 较负的半反应是阳极（反向写为氧化）。不要将 $E^\circ$ 乘以化学计量系数——它是强度性质。
ON SCH4U F3.3 (verbatim): "describe galvanic cells in terms of oxidation and reduction half-cells whose voltages can be used to determine overall cell potential." F2.6: "predict the spontaneity of redox reactions, based on overall cell potential as determined using a table of standard reduction potentials for redox half-reactions." F3.4: "explain how the hydrogen half-cell is used as a standard reference." AB Chemistry 30 GO2: "explain that the values of standard reduction potential are all relative to volts, as set for the hydrogen electrode at standard conditions … calculate the standard cell potential for electrochemical cells." BC Chemistry 12 Elaboration: "electrochemical cells: half-reactions, cell voltage (E0)."ON SCH4U F3.3（原文）："从氧化和还原半电池的角度描述原电池，其电压可用于确定总电池电势。"F2.6："根据标准还原电势表确定的总电池电势，预测氧化还原反应的自发性。"F3.4："解释氢半电池如何用作标准参考。"AB Chemistry 30 GO2："解释标准还原电势的值均相对于氢电极在标准条件下设定的伏特……计算电化学电池的标准电池电势。"BC Chemistry 12 细化："电化学电池：半反应，电池电压（E0）。"
Worked Example 6 · Calculating $E^\circ_\text{cell}$ for the Zn–Cu cell例题 6 · 计算 Zn-Cu 电池的 $E^\circ_\text{cell}$

Given: $\text{Cu}^{2+} + 2e^- \to \text{Cu}$, $E^\circ = +0.34\ \mathrm{V}$; $\text{Zn}^{2+} + 2e^- \to \text{Zn}$, $E^\circ = -0.76\ \mathrm{V}$. Calculate $E^\circ_\text{cell}$ and state whether the reaction is spontaneous as written: Zn(s) + Cu$^{2+}$(aq) $\to$ Zn$^{2+}$(aq) + Cu(s).已知：$\text{Cu}^{2+} + 2e^- \to \text{Cu}$，$E^\circ = +0.34\ \mathrm{V}$；$\text{Zn}^{2+} + 2e^- \to \text{Zn}$，$E^\circ = -0.76\ \mathrm{V}$。计算 $E^\circ_\text{cell}$ 并说明按书写方向反应是否自发：Zn(s) + Cu$^{2+}$(aq) → Zn$^{2+}$(aq) + Cu(s)。

Identify cathode and anode.确定阴极和阳极。 Cu$^{2+}$ has the higher (more positive) $E^\circ = +0.34\ \mathrm{V}$, so Cu$^{2+}$ is reduced at the cathode. Zn has the lower $E^\circ = -0.76\ \mathrm{V}$, so Zn is oxidised at the anode.Cu$^{2+}$ 具有更高（更正）的 $E^\circ = +0.34\ \mathrm{V}$，故 Cu$^{2+}$ 在阴极被还原。Zn 的 $E^\circ = -0.76\ \mathrm{V}$ 较低，故 Zn 在阳极被氧化。

Apply the formula.应用公式。

$$ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} = (+0.34) - (-0.76) = +1.10\ \mathrm{V} $$

Interpret.解读。 $E^\circ_\text{cell} = +1.10\ \mathrm{V} > 0$, so the reaction is spontaneous as written (consistent with Zn being more active than Cu). The Daniell cell — one of the first practical batteries — is this exact cell. ✓$E^\circ_\text{cell} = +1.10\ \mathrm{V} > 0$，故该反应按书写方向是自发的（与 Zn 比 Cu 活性更强一致）。丹尼尔电池——最早的实用电池之一——正是这个电池。✓

Given $E^\circ(\text{Ag}^+ / \text{Ag}) = +0.80\ \mathrm{V}$ and $E^\circ(\text{Fe}^{2+} / \text{Fe}) = -0.44\ \mathrm{V}$, what is $E^\circ_\text{cell}$ for the spontaneous reaction?已知 $E^\circ(\text{Ag}^+ / \text{Ag}) = +0.80\ \mathrm{V}$，$E^\circ(\text{Fe}^{2+} / \text{Fe}) = -0.44\ \mathrm{V}$，自发反应的 $E^\circ_\text{cell}$ 是多少？
§6 · Q1
$+0.36\ \mathrm{V}$
$+1.24\ \mathrm{V}$
$-1.24\ \mathrm{V}$
$-0.36\ \mathrm{V}$
Ag$^+$ has the higher $E^\circ$ (+0.80 V) — cathode. Fe has the lower $E^\circ$ (−0.44 V) — anode. $E^\circ_\text{cell} = (+0.80) - (-0.44) = +1.24\ \mathrm{V}$. Positive, confirming spontaneity. Overall: Fe + 2Ag$^+$ → Fe$^{2+}$ + 2Ag.Ag$^+$ 的 $E^\circ$ 较高（+0.80 V）——阴极。Fe 的 $E^\circ$ 较低（−0.44 V）——阳极。$E^\circ_\text{cell} = (+0.80) - (-0.44) = +1.24\ \mathrm{V}$。为正，确认自发性。总反应：Fe + 2Ag$^+$ → Fe$^{2+}$ + 2Ag。
$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$. The cathode is the more positive half-reaction (Ag$^+$/Ag, +0.80 V); the anode is the more negative (Fe$^{2+}$/Fe, −0.44 V). Do NOT multiply the potentials by stoichiometric coefficients.$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$。阴极是较正的半反应（Ag$^+$/Ag，+0.80 V）；阳极是较负的（Fe$^{2+}$/Fe，−0.44 V）。不要将电势乘以化学计量系数。
A cell has $E^\circ_\text{cell} = -0.45\ \mathrm{V}$. What does this tell you?某电池的 $E^\circ_\text{cell} = -0.45\ \mathrm{V}$。这说明什么？
§6 · Q2
The reaction is spontaneous and will run as a galvanic cell该反应是自发的，可作为原电池运行
The cell voltage is below 1 V but still spontaneous电池电压低于 1 V 但仍然自发
The reaction as written is non-spontaneous; it requires external voltage (electrolysis)按书写方向反应是非自发的；需要外部电压（电解）
The anode and cathode have been assigned incorrectly阳极和阴极分配错误
$E^\circ_\text{cell} < 0$ means the redox reaction as written is non-spontaneous ($\Delta G > 0$). The reverse reaction would be spontaneous. To drive a non-spontaneous reaction forward, external electrical energy must be supplied — this is the principle of electrolysis (§7).$E^\circ_\text{cell} < 0$ 意味着按书写方向的氧化还原反应是非自发的（$\Delta G > 0$）。逆反应将是自发的。要驱动非自发反应向前进行，必须提供外部电能——这是电解的原理（§7）。
The sign of $E^\circ_\text{cell}$ tells you about spontaneity: positive = spontaneous (galvanic); negative = non-spontaneous (requires electrolysis). It has nothing to do with anode/cathode assignment errors — the formula $E^\circ_\text{cathode} - E^\circ_\text{anode}$ is defined consistently.$E^\circ_\text{cell}$ 的符号告诉你自发性：正 = 自发（原电池）；负 = 非自发（需要电解）。这与阳极/阴极分配错误无关——公式 $E^\circ_\text{cathode} - E^\circ_\text{anode}$ 定义一致。
Section 7 · Electrolysis and applications第 7 节 · 电解及应用

Electrolysis and Applications电解及应用

Electrolysis uses external electrical energy to force non-spontaneous redox reactions.电解利用外部电能驱动非自发的氧化还原反应。
  • Electrolytic cell vs galvanic cell电解槽与原电池的对比 — in a galvanic cell, spontaneous redox produces electricity. In an electrolytic cell, electricity forces non-spontaneous redox. Both have anode (oxidation) and cathode (reduction), but polarity reverses: the anode of an electrolytic cell is connected to the positive terminal of the power supply; the cathode to the negative terminal.— 在原电池中，自发氧化还原产生电能。在电解槽中，电能驱动非自发氧化还原。两者都有阳极（氧化）和阴极（还原），但极性反转：电解槽的阳极连接到电源的正极；阴极连接到负极。
  • Faraday's law法拉第定律 — the mass of substance deposited or dissolved at an electrode is proportional to the quantity of charge passed:— 在电极处沉积或溶解的物质质量与通过的电荷量成正比：
$$ Q = It \qquad m = \frac{M \cdot I \cdot t}{n \cdot F} $$
  • $Q$ = charge (C), $I$ = current (A), $t$ = time (s), $m$ = mass (g), $M$ = molar mass (g/mol), $n$ = moles of electrons per mole of substance, $F$ = Faraday constant $= 96\,485\ \mathrm{C/mol}$.$Q$ = 电荷量（C），$I$ = 电流（A），$t$ = 时间（s），$m$ = 质量（g），$M$ = 摩尔质量（g/mol），$n$ = 每摩尔物质的摩尔电子数，$F$ = 法拉第常数 $= 96\,485\ \mathrm{C/mol}$。
  • Applications应用 : electroplating (coating metal surfaces), metal refining (pure copper, zinc, aluminium), chlor-alkali process (Cl$_2$ and NaOH from brine), and cathodic protection (preventing corrosion by making the protected metal the cathode).：电镀（金属表面镀层）、金属精炼（纯铜、锌、铝）、氯碱工业（从盐水制备 Cl$_2$ 和 NaOH）和阴极保护（通过使受保护金属成为阴极来防止腐蚀）。
ON SCH4U F3.5 (verbatim): "explain some applications of electrochemistry in common industrial processes." F3.6: "explain the corrosion of metals in terms of an electrochemical process." AB Chemistry 30 GO2: "calculate mass, amounts, current and time in single voltaic and electrolytic cells by applying Faraday's law and stoichiometry." BC Chemistry 12 Elaboration: "electrolytic cells: half-reactions, minimum voltage to operate, applications including metal refining (e.g. zinc, aluminum), preventing metal corrosion (cathodic protection)."ON SCH4U F3.5（原文）："解释电化学在常见工业过程中的一些应用。"F3.6："从电化学过程的角度解释金属腐蚀。"AB Chemistry 30 GO2："用法拉第定律和化学计量学计算单个伏打电池和电解槽中的质量、物质的量、电流和时间。"BC Chemistry 12 细化："电解槽：半反应，运行所需最低电压，应用包括金属精炼（如锌、铝）、防止金属腐蚀（阴极保护）。"
Worked Example 7 · Faraday's law — mass of copper deposited例题 7 · 法拉第定律——沉积铜的质量

A current of $2.00\ \mathrm{A}$ is passed through a CuSO$_4$ electrolytic cell for $30.0\ \mathrm{min}$. Copper deposits at the cathode: Cu$^{2+}$ + 2e$^-$ $\to$ Cu. Calculate the mass of copper deposited. ($M_\text{Cu} = 63.55\ \mathrm{g/mol}$, $F = 96\,485\ \mathrm{C/mol}$)将 $2.00\ \mathrm{A}$ 的电流通过 CuSO$_4$ 电解槽 $30.0\ \mathrm{min}$。铜在阴极沉积：Cu$^{2+}$ + 2e$^-$ → Cu。计算沉积铜的质量。（$M_\text{Cu} = 63.55\ \mathrm{g/mol}$，$F = 96\,485\ \mathrm{C/mol}$）

Calculate charge $Q$.计算电荷量 $Q$。

$$ Q = I \times t = 2.00\ \mathrm{A} \times (30.0 \times 60)\ \mathrm{s} = 2.00 \times 1800 = 3600\ \mathrm{C} $$

Apply Faraday's law.应用法拉第定律。 Cu$^{2+}$ requires $n = 2$ electrons per Cu atom.Cu$^{2+}$ 每个 Cu 原子需要 $n = 2$ 个电子。

$$ m = \frac{M \cdot Q}{n \cdot F} = \frac{63.55 \times 3600}{2 \times 96\,485} = \frac{228\,780}{192\,970} = 1.186\ \mathrm{g} $$

Result: $1.19\ \mathrm{g}$ of copper deposits at the cathode. ✓结果：$1.19\ \mathrm{g}$ 铜在阴极沉积。✓

In an electrolytic cell, which electrode is connected to the positive terminal of the power supply?在电解槽中，哪个电极与电源的正极相连？
§7 · Q1
The cathode, where reduction occurs阴极，还原发生的地方
Neither electrode — polarity does not apply to electrolytic cells两个电极均不——极性不适用于电解槽
Both electrodes simultaneously两个电极同时
The anode, where oxidation occurs阳极，氧化发生的地方
In an electrolytic cell, the power supply forces current in reverse. The anode is connected to the positive (+) terminal of the power supply — it forces electrons out of the electrode, driving oxidation. The cathode is connected to the negative (−) terminal — it supplies electrons, driving reduction. This is the opposite polarity to a galvanic cell.在电解槽中，电源强制电流反向流动。阳极连接到电源的正（+）极——它强制电子离开电极，驱动氧化。阴极连接到电源的负（−）极——它提供电子，驱动还原。这与原电池的极性相反。
Anode = oxidation in BOTH galvanic and electrolytic cells. But the polarity of the anode differs: in a galvanic cell, the anode is negative (it produces electrons); in an electrolytic cell, the anode is connected to the positive terminal (the power supply forces electrons out).阳极 = 氧化，在原电池和电解槽中均如此。但阳极的极性不同：在原电池中，阳极是负极（产生电子）；在电解槽中，阳极连接到正极（电源强制电子离开）。
A $3.00\ \mathrm{A}$ current deposits silver from AgNO$_3$ solution (Ag$^+$ + e$^-$ → Ag, $M_\text{Ag} = 107.87\ \mathrm{g/mol}$) for $20.0\ \mathrm{min}$. What mass of silver deposits?$3.00\ \mathrm{A}$ 电流从 AgNO$_3$ 溶液中沉积银（Ag$^+$ + e$^-$ → Ag，$M_\text{Ag} = 107.87\ \mathrm{g/mol}$）$20.0\ \mathrm{min}$。沉积的银的质量是多少？
§7 · Q2
$4.02\ \mathrm{g}$
$2.01\ \mathrm{g}$
$8.04\ \mathrm{g}$
$1.34\ \mathrm{g}$
$Q = 3.00 \times (20.0 \times 60) = 3600\ \mathrm{C}$. $n = 1$ (Ag$^+$ + 1e$^-$ → Ag). $m = \frac{107.87 \times 3600}{1 \times 96\,485} = \frac{388\,332}{96\,485} = 4.02\ \mathrm{g}$.$Q = 3.00 \times (20.0 \times 60) = 3600\ \mathrm{C}$。$n = 1$（Ag$^+$ + 1e$^-$ → Ag）。$m = \frac{107.87 \times 3600}{1 \times 96\,485} = \frac{388\,332}{96\,485} = 4.02\ \mathrm{g}$。
Steps: $Q = It$; moles of e$^-$ = $Q/F$; moles of Ag = moles of e$^-$ (since $n=1$); mass = moles × $M$. Or use the formula directly: $m = MIt/(nF)$. Here $n=1$ for Ag.步骤：$Q = It$；e$^-$ 摩尔数 = $Q/F$；Ag 摩尔数 = e$^-$ 摩尔数（因为 $n=1$）；质量 = 摩尔数 × $M$。或直接用公式：$m = MIt/(nF)$。这里 Ag 的 $n=1$。
Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline for every redox question每道氧化还原题的解题纪律
  • Assign oxidation numbers first.先分配氧化数。 Write each atom's oxidation number above the formula before deciding what changed. Mark which atoms increase (oxidation) and which decrease (reduction). This prevents the most common error: misidentifying the oxidising and reducing agents.在判断变化之前，先在化学式上方写出每个原子的氧化数。标记哪些原子升高（氧化）哪些降低（还原）。这可以防止最常见的错误：错误识别氧化剂和还原剂。
  • Don't mix up OIL RIG and the agent labels.不要混淆 OIL RIG 和试剂标签。 The species that is oxidised is the reducing agent (it causes reduction in something else). The species that is reduced is the oxidising agent. This reversal trips many students.被氧化的物质是还原剂（它使其他物质被还原）。被还原的物质是氧化剂。这种反转让很多学生出错。
  • Check atom and charge balance after every half-reaction.每个半反应后检查原子和电荷守恒。 Write the total charge on each side. The electrons you add must exactly balance the charge difference. In the final equation, verify that electrons cancel completely.在每侧写出总电荷。加入的电子必须精确平衡电荷差。在最终方程中，验证电子完全消去。
Galvanic cells and cell EMF (§5, §6)原电池与电池电动势（§5、§6）
  • Cathode is always higher $E^\circ$.阴极始终具有更高的 $E^\circ$。 Look up the two half-reactions. The one with the more positive (or less negative) $E^\circ$ is the cathode (is reduced). Reverse the other for the anode (oxidation). Apply $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$. A positive result confirms spontaneity.查找两个半反应。$E^\circ$ 较正（或较不负）的是阴极（被还原）。将另一个反向作为阳极（氧化）。应用 $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$。正结果确认自发性。
  • Never multiply $E^\circ$ by coefficients.永远不要将 $E^\circ$ 乘以系数。 Even if you multiply a half-reaction by 2 (to equalise electrons), the $E^\circ$ value stays the same — it is an intensive property (like temperature or concentration), not extensive.即使将半反应乘以 2（以均衡电子），$E^\circ$ 值也保持不变——它是强度性质（如温度或浓度），而非广度性质。
Electrolysis and Faraday's law (§7)电解与法拉第定律（§7）
  • Unit check before computing.计算前检查单位。 Time must be in seconds for $Q = It$. Convert minutes to seconds (multiply by 60). Keep at least 3 significant figures throughout.$Q = It$ 中时间必须以秒为单位。将分钟转换为秒（乘以 60）。全程保留至少 3 位有效数字。
  • Identify $n$ from the half-reaction.从半反应中确定 $n$。 $n$ is the number of electrons in the balanced half-reaction for one mole of substance. Cu$^{2+}$ + 2e$^-$ $\to$ Cu gives $n = 2$. Ag$^+$ + e$^-$ $\to$ Ag gives $n = 1$. Al$^{3+}$ + 3e$^-$ $\to$ Al gives $n = 3$. This is the most common source of error in Faraday calculations.$n$ 是一摩尔物质的配平半反应中的电子数。Cu$^{2+}$ + 2e$^-$ → Cu 给出 $n = 2$。Ag$^+$ + e$^-$ → Ag 给出 $n = 1$。Al$^{3+}$ + 3e$^-$ → Al 给出 $n = 3$。这是法拉第计算中最常见的错误来源。
  • Polarity reversal in electrolytic vs galvanic cells.电解槽与原电池中的极性反转。 In both: anode = oxidation, cathode = reduction. But in a galvanic cell the anode is negative; in an electrolytic cell the anode is connected to the positive terminal. Exam questions test this distinction directly.在两者中：阳极 = 氧化，阴极 = 还原。但在原电池中阳极是负极；在电解槽中阳极连接到正极。考题直接测试这一区别。
Oxidation numbers and identification (§1, §2)氧化数与识别（§1、§2）
  • Know the priority order for the six rules.了解六条规则的优先顺序。 Fluorine overrides oxygen (OF$_2$: O is $+2$). Oxygen overrides hydrogen in most compounds (H$_2$O: H is $+1$). For metal hydrides, hydrogen is $-1$. Always apply the rules in order: pure element → monatomic ion → F → H → O → sum = charge.氟优先于氧（OF$_2$：O 为 $+2$）。在大多数化合物中氧优先于氢（H$_2$O：H 为 $+1$）。金属氢化物中氢为 $-1$。始终按顺序应用规则：纯元素 → 单原子离子 → F → H → O → 总和 = 电荷。
  • Confirm OIL RIG with numbers, not guessing.用数字而非猜测确认 OIL RIG。 Always calculate the oxidation-number change numerically. Write "Fe: $0 \to +2$, loss of 2 e$^-$, oxidised" rather than relying on memory about which metal reacts with which ion.始终用数字计算氧化数变化。写出"Fe：$0 \to +2$，失去 2 个 e$^-$，被氧化"，而不是靠记忆哪种金属与哪种离子反应。
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OIL RIG?OIL RIG？
Oxidation Is Loss (of electrons); Reduction Is Gain (of electrons). The reducing agent is oxidised; the oxidising agent is reduced.氧化是失去电子；还原是得到电子。还原剂被氧化；氧化剂被还原。
Oxidation number of O (general)?O 的氧化数（一般情况）？
$-2$ in most compounds; $-1$ in peroxides (H$_2$O$_2$, Na$_2$O$_2$); $+2$ in OF$_2$ (F has higher priority).大多数化合物中为 $-2$；过氧化物（H$_2$O$_2$、Na$_2$O$_2$）中为 $-1$；OF$_2$ 中为 $+2$（F 优先级更高）。
Half-reaction method: steps in acid?半反应法：酸性中的步骤？
1. Split into two half-reactions. 2. Balance atoms other than O and H. 3. Add H$_2$O to balance O. 4. Add H$^+$ to balance H. 5. Add e$^-$ to balance charge. 6. Multiply to cancel e$^-$, then add.1. 分为两个半反应。2. 配平 O 和 H 以外的原子。3. 加 H$_2$O 配平 O。4. 加 H$^+$ 配平 H。5. 加 e$^-$ 配平电荷。6. 乘以系数消去 e$^-$，然后相加。
Activity series: what does it predict?活动性顺序：它预测什么？
A more active metal (higher in the series) spontaneously displaces a less active metal ion from solution. Metals above H$_2$ dissolve in acid.活性更强的金属（顺序中较高）自发地从溶液中置换活性较弱的金属离子。H$_2$ 以上的金属溶于酸。
Galvanic cell: anode and cathode?原电池：阳极和阴极？
Anode (negative): oxidation occurs, metal dissolves, electrons flow out. Cathode (positive): reduction occurs, metal deposits, electrons flow in. Salt bridge maintains electrical neutrality.阳极（负极）：氧化发生，金属溶解，电子流出。阴极（正极）：还原发生，金属沉积，电子流入。盐桥维持电中性。
Cell EMF formula?电池电动势公式？
$$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$$ Positive: spontaneous. Negative: non-spontaneous (requires electrolysis).正值：自发。负值：非自发（需要电解）。
Standard hydrogen electrode (SHE)?标准氢电极（SHE）？
Reference half-cell: $2\text{H}^+ + 2e^- \to \text{H}_2$, $E^\circ = 0.00\ \mathrm{V}$ by definition. Measured at 25 °C, 1 atm H$_2$, [H$^+$] = 1 mol/L. All other $E^\circ$ values are measured relative to this.参考半电池：$2\text{H}^+ + 2e^- \to \text{H}_2$，$E^\circ = 0.00\ \mathrm{V}$（定义）。在 25 °C、1 atm H$_2$、[H$^+$] = 1 mol/L 下测量。所有其他 $E^\circ$ 值均相对于此测量。
Electrolysis: galvanic vs electrolytic?电解：原电池与电解槽的区别？
Galvanic: spontaneous redox → electricity. Electrolytic: electricity → non-spontaneous redox. Both: anode = oxidation, cathode = reduction. Polarity reversal: electrolytic anode is connected to + terminal.原电池：自发氧化还原 → 电能。电解槽：电能 → 非自发氧化还原。两者：阳极 = 氧化，阴极 = 还原。极性反转：电解槽阳极连接 + 极。
Faraday's law formula?法拉第定律公式？
$$m = \frac{M \cdot I \cdot t}{n \cdot F}$$ $M$ = molar mass, $I$ = current (A), $t$ = time (s), $n$ = e$^-$ per formula unit, $F = 96\,485\ \mathrm{C/mol}$.$M$ = 摩尔质量，$I$ = 电流（A），$t$ = 时间（s），$n$ = 每式量的 e$^-$ 数，$F = 96\,485\ \mathrm{C/mol}$。
Oxidising agent vs reducing agent?氧化剂与还原剂？
Oxidising agent: gains electrons (is itself reduced). Reducing agent: loses electrons (is itself oxidised). Higher (more positive) $E^\circ$ = stronger oxidising agent.氧化剂：得到电子（自身被还原）。还原剂：失去电子（自身被氧化）。$E^\circ$ 越高（越正）= 氧化剂越强。
Corrosion as electrochemistry?腐蚀的电化学本质？
Iron rusting is a galvanic process: Fe is the anode (oxidised: Fe $\to$ Fe$^{2+}$ + 2e$^-$); O$_2$ and H$_2$O are reduced at the cathode. Cathodic protection makes the metal the cathode by connecting it to a more active metal (sacrificial anode).铁锈蚀是原电池过程：Fe 是阳极（被氧化：Fe → Fe$^{2+}$ + 2e$^-$）；O$_2$ 和 H$_2$O 在阴极被还原。阴极保护通过将金属连接到活性更强的金属（牺牲阳极）使其成为阴极。
Spontaneity from $E^\circ_\text{cell}$?从 $E^\circ_\text{cell}$ 判断自发性？
$E^\circ_\text{cell} > 0$: spontaneous, $\Delta G < 0$, works as galvanic cell.
$E^\circ_\text{cell} < 0$: non-spontaneous, $\Delta G > 0$, requires electrolysis.
$E^\circ_\text{cell} = 0$: equilibrium.$E^\circ_\text{cell} > 0$：自发，$\Delta G < 0$，作为原电池工作。
$E^\circ_\text{cell} < 0$：非自发，$\Delta G > 0$，需要电解。
$E^\circ_\text{cell} = 0$：平衡。
Mixed Practice综合练习

Practice Quiz综合测验

What is the oxidation number of Mn in KMnO$_4$? 🇨🇦 SCH4U F2.1KMnO$_4$ 中 Mn 的氧化数是多少？🇨🇦 SCH4U F2.1
Q1
$+2$$+2$
$+4$$+4$
$+3$$+3$
$+7$$+7$
K = $+1$; O = $-2$ (×4 = $-8$). Sum = 0: $1 + x + (-8) = 0 \Rightarrow x = +7$. KMnO$_4$ is potassium permanganate, one of the strongest common oxidising agents. Mn ranges from $-1$ to $+7$.K = $+1$；O = $-2$（×4 = $-8$）。总和 = 0：$1 + x + (-8) = 0 \Rightarrow x = +7$。KMnO$_4$ 是高锰酸钾，最常见的最强氧化剂之一。Mn 的氧化数范围为 $-1$ 到 $+7$。
Apply Rule 6: $1 + x + 4(-2) = 0$. Solve: $x = +7$. Remember K contributes $+1$ (it's a Group 1 metal in a compound) and each O contributes $-2$.应用规则 6：$1 + x + 4(-2) = 0$。求解：$x = +7$。记住 K 贡献 $+1$（化合物中的第 1 族金属），每个 O 贡献 $-2$。
In the reaction $\text{Zn} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2$, which species is the oxidising agent? 🇨🇦 SCH4U F3.1 / AB Chem 30 GO1在反应 $\text{Zn} + 2\text{HCl} \to \text{ZnCl}_2 + \text{H}_2$ 中，哪种物质是氧化剂？🇨🇦 SCH4U F3.1 / AB Chem 30 GO1
Q2
H$^+$ (from HCl), because it is reduced ($+1 \to 0$)H$^+$（来自 HCl），因为它被还原（$+1 \to 0$）
Zn, because it is oxidised ($0 \to +2$)Zn，因为它被氧化（$0 \to +2$）
Cl$^-$, because it is a spectator ionCl$^-$，因为它是旁观者离子
ZnCl$_2$, because it is formedZnCl$_2$，因为它是生成物
H in HCl has oxidation number $+1$; H in H$_2$ has oxidation number $0$. H gains electrons (oxidation number decreases): H$^+$ is reduced. The species that is reduced is the oxidising agent. Zn goes from $0 \to +2$ (loses electrons = oxidised), so Zn is the reducing agent.HCl 中 H 的氧化数为 $+1$；H$_2$ 中 H 的氧化数为 $0$。H 得到电子（氧化数降低）：H$^+$ 被还原。被还原的物质是氧化剂。Zn 从 $0 \to +2$（失去电子 = 被氧化），故 Zn 是还原剂。
The oxidising agent is the species that gets reduced (gains electrons). H$^+$ goes from $+1$ to $0$ (gains one electron per H atom) — it is reduced, making it the oxidising agent. Zn is oxidised ($0 \to +2$) — it is the reducing agent. Cl$^-$ does not change oxidation number ($-1$ throughout).氧化剂是被还原（得到电子）的物质。H$^+$ 从 $+1$ 降至 $0$（每个 H 原子得到一个电子）——它被还原，使其成为氧化剂。Zn 被氧化（$0 \to +2$）——它是还原剂。Cl$^-$ 的氧化数不变（全程 $-1$）。
When balancing the half-reaction $\text{NO}_3^- \to \text{NO}$ in acidic solution, what must be added to balance oxygen atoms? 🇨🇦 SCH4U F2.3在酸性溶液中配平半反应 $\text{NO}_3^- \to \text{NO}$ 时，需要加什么来配平氧原子？🇨🇦 SCH4U F2.3
Q3
OH$^-$ ions to the left左侧加 OH$^-$ 离子
2H$_2$O to the right (NO$_3^-$ has 3 O, NO has 1 O; excess 2 O appear as H$_2$O)右侧加 2H$_2$O（NO$_3^-$ 有 3 个 O，NO 有 1 个 O；多余 2 个 O 以 H$_2$O 形式出现）
O$_2$ gas to the left左侧加 O$_2$ 气体
Nothing — oxygen is already balanced无需加任何东西——氧已经平衡
NO$_3^-$ has 3 O; NO has 1 O. Excess = 2 O on the left. Balance by adding 2H$_2$O to the right: $\text{NO}_3^- \to \text{NO} + 2\text{H}_2\text{O}$. Then balance H with 4H$^+$ on the left. In acidic solution: always use H$_2$O and H$^+$ (not OH$^-$).NO$_3^-$ 有 3 个 O；NO 有 1 个 O。左侧多出 2 个 O。通过在右侧加 2H$_2$O 配平：$\text{NO}_3^- \to \text{NO} + 2\text{H}_2\text{O}$。然后在左侧加 4H$^+$ 配平 H。在酸性溶液中：始终使用 H$_2$O 和 H$^+$（不用 OH$^-$）。
In acidic solution, the O-balancing agent is H$_2$O (added to the side that is short of O). The side short of O here is the right side: NO has 1 O but NO$_3^-$ has 3, so add 2H$_2$O to the right. Then add H$^+$ to the left to balance the 4 H atoms introduced.在酸性溶液中，配平 O 的试剂是 H$_2$O（加到 O 不足的一侧）。此处 O 不足的是右侧：NO 有 1 个 O，但 NO$_3^-$ 有 3 个，故在右侧加 2H$_2$O。然后在左侧加 H$^+$ 以配平引入的 4 个 H 原子。
Copper metal is placed in a silver nitrate solution. Which observation is expected? 🇨🇦 SCH3U C2.5 / AB Chem 30 GO1将铜金属放入硝酸银溶液中。预期会观察到什么？🇨🇦 SCH3U C2.5 / AB Chem 30 GO1
Q4
No reaction — Cu is above Ag in the activity series不发生反应——Cu 在活动性顺序中排在 Ag 上方
No reaction — Cu is below Ag in the activity series不发生反应——Cu 在活动性顺序中排在 Ag 下方
Cu dissolves, silver metal deposits on the copper (Cu displaces Ag$^+$)Cu 溶解，银金属在铜上沉积（Cu 置换 Ag$^+$）
Silver dissolves and copper deposits银溶解，铜沉积
Activity series: Cu is above Ag (Cu is more active). Cu can displace Ag$^+$ from solution. Reaction: Cu + 2Ag$^+$ → Cu$^{2+}$ + 2Ag. Observation: Cu metal dissolves (solution turns blue from Cu$^{2+}$); silver crystals deposit on the copper surface.活动性顺序：Cu 在 Ag 上方（Cu 更活泼）。Cu 能从溶液中置换 Ag$^+$。反应：Cu + 2Ag$^+$ → Cu$^{2+}$ + 2Ag。观察：Cu 金属溶解（溶液因 Cu$^{2+}$ 变蓝）；银晶体在铜表面沉积。
Cu is ABOVE Ag in the activity series (more active), so Cu spontaneously reduces Ag$^+$ to Ag metal while being oxidised to Cu$^{2+}$. This is a single-displacement redox reaction. The solution turns blue (Cu$^{2+}$) and silver dendrites form.Cu 在活动性顺序中排在 Ag 上方（更活泼），故 Cu 自发地将 Ag$^+$ 还原为 Ag 金属，同时被氧化为 Cu$^{2+}$。这是单置换氧化还原反应。溶液变蓝（Cu$^{2+}$），银树枝晶生成。
In a Zn–Cu galvanic cell, which electrode loses mass during operation? 🇨🇦 SCH4U F2.4 / F2.5在 Zn-Cu 原电池运行过程中，哪个电极质量减少？🇨🇦 SCH4U F2.4 / F2.5
Q5
The zinc anode, because Zn is oxidised and dissolves into solution锌阳极，因为 Zn 被氧化后溶解进入溶液
The copper cathode, because Cu$^{2+}$ enters the solution铜阴极，因为 Cu$^{2+}$ 进入溶液
Both electrodes lose mass equally两个电极质量均等减少
Neither electrode — mass is conserved in the cell overall两个电极均不减少——电池整体质量守恒
At the Zn anode: Zn(s) → Zn$^{2+}$(aq) + 2e$^-$. Solid Zn dissolves into the solution as Zn$^{2+}$ ions — the anode electrode loses mass. At the Cu cathode: Cu$^{2+}$(aq) + 2e$^-$ → Cu(s). Cu$^{2+}$ from solution deposits as solid Cu — the cathode electrode gains mass.在 Zn 阳极：Zn(s) → Zn$^{2+}$(aq) + 2e$^-$。固体 Zn 以 Zn$^{2+}$ 离子形式溶解进入溶液——阳极质量减少。在 Cu 阴极：Cu$^{2+}$(aq) + 2e$^-$ → Cu(s)。溶液中的 Cu$^{2+}$ 以固体 Cu 形式沉积——阴极质量增加。
At the anode (oxidation), solid metal dissolves — mass decreases. At the cathode (reduction), metal ions deposit — mass increases. The Zn anode loses mass; the Cu cathode gains mass.在阳极（氧化），固体金属溶解——质量减少。在阴极（还原），金属离子沉积——质量增加。Zn 阳极质量减少；Cu 阴极质量增加。
Given $E^\circ(\text{Cl}_2/\text{Cl}^-) = +1.36\ \mathrm{V}$ and $E^\circ(\text{Br}_2/\text{Br}^-) = +1.07\ \mathrm{V}$, which is the stronger oxidising agent? 🇨🇦 SCH4U F2.6 / AB Chem 30 GO1已知 $E^\circ(\text{Cl}_2/\text{Cl}^-) = +1.36\ \mathrm{V}$，$E^\circ(\text{Br}_2/\text{Br}^-) = +1.07\ \mathrm{V}$，哪种是更强的氧化剂？🇨🇦 SCH4U F2.6 / AB Chem 30 GO1
Q6
Br$^-$, because it has a lower $E^\circ$Br$^-$，因为它的 $E^\circ$ 较低
Br$_2$, because it has a lower $E^\circ$Br$_2$，因为它的 $E^\circ$ 较低
Cl$^-$, because it has a higher $E^\circ$Cl$^-$，因为它的 $E^\circ$ 较高
Cl$_2$, because it has a higher $E^\circ$ (stronger tendency to be reduced)Cl$_2$，因为它的 $E^\circ$ 较高（被还原的倾向更强）
A more positive $E^\circ$ for the reduction half-reaction means a greater tendency to be reduced — that is, a stronger oxidising agent. Cl$_2$ has $E^\circ = +1.36\ \mathrm{V}$ (more positive than Br$_2$'s $+1.07\ \mathrm{V}$), so Cl$_2$ is the stronger oxidising agent. In fact, Cl$_2$ can displace Br$^-$ from solution (Cl$_2$ + 2Br$^-$ → 2Cl$^-$ + Br$_2$).还原半反应的 $E^\circ$ 越正，被还原的倾向越大——即氧化剂越强。Cl$_2$ 的 $E^\circ = +1.36\ \mathrm{V}$（比 Br$_2$ 的 $+1.07\ \mathrm{V}$ 更正），故 Cl$_2$ 是更强的氧化剂。事实上，Cl$_2$ 能从溶液中置换 Br$^-$（Cl$_2$ + 2Br$^-$ → 2Cl$^-$ + Br$_2$）。
Higher $E^\circ$ = stronger oxidising agent (more easily reduced). Cl$_2$ (+1.36 V) has a higher $E^\circ$ than Br$_2$ (+1.07 V), so Cl$_2$ is the stronger oxidising agent. The corresponding reducing agents (Cl$^-$ and Br$^-$) are NOT oxidising agents — they are the reduced forms.$E^\circ$ 越高 = 氧化剂越强（越容易被还原）。Cl$_2$（+1.36 V）的 $E^\circ$ 高于 Br$_2$（+1.07 V），故 Cl$_2$ 是更强的氧化剂。相应的还原剂（Cl$^-$ 和 Br$^-$）不是氧化剂——它们是被还原的形式。
In an electrolytic cell, which electrode is the anode? 🇨🇦 SCH4U F / BC Chemistry 12在电解槽中，哪个电极是阳极？🇨🇦 SCH4U F / BC Chemistry 12
Q7
The negative electrode, where reduction occurs负极，还原发生的地方
The positive electrode, where oxidation occurs正极，氧化发生的地方
The negative electrode, where oxidation occurs负极，氧化发生的地方
Either electrode — it depends on the reaction任一电极——取决于反应
In an electrolytic cell, the anode is connected to the positive (+) terminal of the power supply. The power supply drives electrons out of the anode, forcing oxidation. The cathode (negative terminal) accepts electrons, driving reduction. Mnemonic: An Ox (Anode = Oxidation), Red Cat (Cathode = Reduction) — true for BOTH cell types.在电解槽中，阳极连接到电源的正（+）极。电源将电子从阳极驱出，强制氧化。阴极（负极）接受电子，驱动还原。助记：阳极=氧化，阴极=还原——对两种电池类型均成立。
An Ox, Red Cat: Anode = Oxidation, Cathode = Reduction — always. In an electrolytic cell, the anode is the positive electrode (connected to +). Note the contrast with a galvanic cell where the anode is the negative electrode — the process (oxidation) is the same, but the polarity is externally imposed.阳极=氧化，阴极=还原——始终如此。在电解槽中，阳极是正极（连接+）。注意与原电池的对比——原电池中阳极是负极——过程（氧化）相同，但极性是外部施加的。
A $5.00\ \mathrm{A}$ current electrolyses molten Al$_2$O$_3$ (Al$^{3+}$ + 3e$^-$ → Al, $M_\text{Al} = 26.98\ \mathrm{g/mol}$) for $1.00\ \mathrm{h}$. What mass of aluminium is produced? 🇨🇦 AB Chem 30 GO2$5.00\ \mathrm{A}$ 电流电解熔融 Al$_2$O$_3$（Al$^{3+}$ + 3e$^-$ → Al，$M_\text{Al} = 26.98\ \mathrm{g/mol}$）$1.00\ \mathrm{h}$。产生铝的质量是多少？🇨🇦 AB Chem 30 GO2
Q8
$3.36\ \mathrm{g}$
$10.1\ \mathrm{g}$
$1.68\ \mathrm{g}$
$5.03\ \mathrm{g}$
$Q = 5.00 \times 3600 = 18\,000\ \mathrm{C}$. $n = 3$ (Al$^{3+}$ + 3e$^-$). $m = \frac{26.98 \times 18\,000}{3 \times 96\,485} = \frac{485\,640}{289\,455} = 1.68\ \mathrm{g}$.$Q = 5.00 \times 3600 = 18\,000\ \mathrm{C}$。$n = 3$（Al$^{3+}$ + 3e$^-$）。$m = \frac{26.98 \times 18\,000}{3 \times 96\,485} = \frac{485\,640}{289\,455} = 1.68\ \mathrm{g}$。
Formula: $m = MIt/(nF)$. Here $I = 5.00\ \mathrm{A}$, $t = 3600\ \mathrm{s}$ (1 h), $M = 26.98$, $n = 3$, $F = 96\,485$. The key is $n = 3$ for Al (not 2). Using $n=2$ would give $2.52\ \mathrm{g}$, not $1.68\ \mathrm{g}$.公式：$m = MIt/(nF)$。这里 $I = 5.00\ \mathrm{A}$，$t = 3600\ \mathrm{s}$（1 h），$M = 26.98$，$n = 3$，$F = 96\,485$。关键是 Al 的 $n = 3$（不是 2）。用 $n=2$ 会得到 $2.52\ \mathrm{g}$，而非 $1.68\ \mathrm{g}$。
A cell is constructed with the half-reactions: Cu$^{2+}$ + 2e$^-$ → Cu, $E^\circ = +0.34\ \mathrm{V}$; and Ni$^{2+}$ + 2e$^-$ → Ni, $E^\circ = -0.25\ \mathrm{V}$. What is $E^\circ_\text{cell}$ for the spontaneous reaction? 🇨🇦 SCH4U F2.6 / BC Chemistry 12用以下半反应构建电池：Cu$^{2+}$ + 2e$^-$ → Cu，$E^\circ = +0.34\ \mathrm{V}$；Ni$^{2+}$ + 2e$^-$ → Ni，$E^\circ = -0.25\ \mathrm{V}$。自发反应的 $E^\circ_\text{cell}$ 是多少？🇨🇦 SCH4U F2.6 / BC Chemistry 12
Q9
$+0.59\ \mathrm{V}$
$+0.09\ \mathrm{V}$
$-0.59\ \mathrm{V}$
$-0.09\ \mathrm{V}$
Cu$^{2+}$ has the higher $E^\circ$ (+0.34 V) → cathode. Ni has the lower $E^\circ$ (−0.25 V) → anode (Ni is oxidised). $E^\circ_\text{cell} = (+0.34) - (-0.25) = +0.59\ \mathrm{V}$. Positive: spontaneous. Overall: Ni(s) + Cu$^{2+}$(aq) → Ni$^{2+}$(aq) + Cu(s).Cu$^{2+}$ 的 $E^\circ$ 较高（+0.34 V）→ 阴极。Ni 的 $E^\circ$ 较低（−0.25 V）→ 阳极（Ni 被氧化）。$E^\circ_\text{cell} = (+0.34) - (-0.25) = +0.59\ \mathrm{V}$。正值：自发。总反应：Ni(s) + Cu$^{2+}$(aq) → Ni$^{2+}$(aq) + Cu(s)。
$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$. Cathode = more positive half-reaction (Cu$^{2+}$/Cu, +0.34 V). Anode = less positive (Ni$^{2+}$/Ni, −0.25 V). Subtraction: $0.34 - (-0.25) = +0.59\ \mathrm{V}$. Never add the two potentials.$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}$。阴极 = 较正的半反应（Cu$^{2+}$/Cu，+0.34 V）。阳极 = 较不正的（Ni$^{2+}$/Ni，−0.25 V）。相减：$0.34 - (-0.25) = +0.59\ \mathrm{V}$。不要将两个电势相加。
Which industrial process uses electrolysis to produce both Cl$_2$ gas and NaOH from brine? 🇨🇦 SCH4U F3.5 / BC Chemistry 12哪种工业过程利用电解从盐水生产 Cl$_2$ 气体和 NaOH？🇨🇦 SCH4U F3.5 / BC Chemistry 12
Q10
The Haber process哈伯法
The chlor-alkali process氯碱工业
The Hall-Heroult process霍尔-埃鲁法
The contact process接触法
The chlor-alkali process electrolyses concentrated NaCl(aq) (brine): at the anode, Cl$^-$ is oxidised to Cl$_2$ gas; at the cathode, H$_2$O is reduced to H$_2$ and OH$^-$ (NaOH). This is a major industrial process producing plastics raw materials (Cl$_2$ for PVC) and caustic soda (NaOH). The Hall-Heroult process is for aluminium refining; Haber makes NH$_3$; contact makes H$_2$SO$_4$.氯碱工业电解浓缩 NaCl(aq)（盐水）：阳极处 Cl$^-$ 被氧化为 Cl$_2$ 气体；阴极处 H$_2$O 被还原为 H$_2$ 和 OH$^-$（NaOH）。这是生产塑料原料（PVC 用 Cl$_2$）和烧碱（NaOH）的重要工业过程。霍尔-埃鲁法用于铝精炼；哈伯法生产 NH$_3$；接触法生产 H$_2$SO$_4$。
Chlor-alkali = brine electrolysis = Cl$_2$ + NaOH + H$_2$. Hall-Heroult = aluminium from molten Al$_2$O$_3$. Haber = N$_2$ + H$_2$ → NH$_3$ (not electrolysis). Contact = SO$_2$ → SO$_3$ → H$_2$SO$_4$ (not electrolysis).氯碱工业 = 盐水电解 = Cl$_2$ + NaOH + H$_2$。霍尔-埃鲁法 = 从熔融 Al$_2$O$_3$ 制铝。哈伯法 = N$_2$ + H$_2$ → NH$_3$（非电解）。接触法 = SO$_2$ → SO$_3$ → H$_2$SO$_4$（非电解）。
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Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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Next Steps后续单元

What This Feeds Into本单元的去向

Redox chemistry is one of the two pillars of Grade-12 chemistry (the other is equilibrium). Every electrochemical device — from the lithium-ion battery in your phone to the fuel cell in a hydrogen car — is grounded in the half-reaction and cell-potential framework you built here. The cross-references below point at the IB college-credit feeder and the broader HS Chemistry context.氧化还原化学是 12 年级化学的两大支柱之一（另一个是平衡）。每一个电化学装置——从手机中的锂离子电池到氢能汽车中的燃料电池——都建立在你在这里构建的半反应和电池电势框架上。以下链接指向 IB 大学学分衔接课程和更广泛的高中化学背景。

Within High School Chemistry.在 HS Chemistry 内部。

Redox builds on Chemical Reactions (Unit 6 — reaction types and balancing), the Activity Series from single-displacement reactions (Unit 6), and Solution Stoichiometry (Unit 5 — mole calculations) which underpin the Faraday's-law mass calculations in §7. Chemical Equilibrium (Unit 12) connects directly to redox: the Nernst equation (a Grade-12 AP/IB extension) relates cell potential to concentration using $E = E^\circ - (RT/nF)\ln Q$ — the foundation for advanced electrochemistry.氧化还原建立在化学反应（第 6 单元——反应类型与配平）、单置换反应中的活动性顺序（第 6 单元）以及溶液化学计量（第 5 单元——摩尔计算）之上，这些支撑了 §7 中的法拉第定律质量计算。化学平衡（第 12 单元）与氧化还原直接相关：能斯特方程（12 年级 AP/IB 拓展）用 $E = E^\circ - (RT/nF)\ln Q$ 将电池电势与浓度联系起来——高级电化学的基础。

College-credit feeder and cross-subject links.大学学分衔接与跨学科链接。

IB Chemistry HL · Reactivity 1: What Drives Chemical Reactions? (electron transfer, oxidation state, and the energetics/spontaneity of redox — the primary IB college-credit feeder for this unit)IB Chemistry HL · Reactivity 1：是什么驱动化学反应？（电子转移、氧化态以及氧化还原的能量/自发性——本单元的主要 IB 大学学分衔接） IB Chemistry HL · Reactivity 2: How Much, How Fast, How Far? (electrochemical cells, cell potential, Faraday's law at IB depth)IB Chemistry HL · Reactivity 2：多少、多快、多远？（IB 深度的电化学电池、电池电势、法拉第定律）

If you are aiming for IB Chemistry HL or AP Chemistry (Unit 9 — Electrochemistry), the half-reaction method, standard reduction potentials, and galvanic/electrolytic cell diagrams here are assumed from day one of the college-credit course. IB Chemistry HL Reactivity 1 extends this with the relationship between $\Delta G^\circ$, $E^\circ_\text{cell}$, and $K$ (the thermodynamic triad). IB Reactivity 2 adds the Nernst equation and quantitative Faraday calculations. AP Chemistry Unit 9 adds concentration cells and electrolysis of solutions vs melts. Corrosion chemistry (SCH4U F3.6) links directly to galvanic-cell theory — rusting is a spontaneous galvanic process.如果你目标是 IB Chemistry HL 或 AP Chemistry（Unit 9——电化学），这里的半反应法、标准还原电势以及原电池/电解槽图从大学学分课程的第一天就被默认掌握。IB Chemistry HL Reactivity 1 通过 $\Delta G^\circ$、$E^\circ_\text{cell}$ 和 $K$ 的关系（热力学三合一）来延伸这部分内容。IB Reactivity 2 加入能斯特方程和定量法拉第计算。AP Chemistry Unit 9 加入浓差电池和溶液与熔融态的电解比较。腐蚀化学（SCH4U F3.6）直接链接到原电池理论——锈蚀是一个自发的原电池过程。

DINGRUI SCHOLARS

High School Chemistry · Redox and Electrochemistry · v1高中化学 · 氧化还原与电化学 · v1