CH12 Unit 7: Electrochemistry — 电化学 完整解答¶
状态: ✅ 已完成
创建日期: 2026-02-18
最后更新: 2026-02-18
7.1 Electrochemistry — 电化学基础¶
Q1¶
EN (Original): What is the difference between a reduction reaction and an oxidation reaction?
中文翻译： 还原反应和氧化反应有什么区别？
Answer 答案:
EN: Reduction = gain of electrons (oxidation number decreases). Oxidation = loss of electrons (oxidation number increases).
中文： 还原 = 获得电子（氧化数降低）。氧化 = 失去电子（氧化数升高）。
Q2¶
EN (Original): What simple phrase involving a majestic animal should help you remember this?
中文翻译： 哪个涉及雄伟动物的简单短语可以帮助记忆？
Answer 答案:
EN: "LEO the lion says GER" — Loss of Electrons is Oxidation; Gain of Electrons is Reduction.
Also: OIL RIG — Oxidation Is Loss, Reduction Is Gain.
中文： "LEO the lion says GER" — 失去电子 = 氧化；获得电子 = 还原。也可用 OIL RIG 记忆。
Q3¶
EN (Original): Agents are like secret agents for the other side. Explain.
中文翻译： "剂"就像为对方工作的间谍。解释。
Answer 答案:
EN: An oxidizing agent causes another substance to be oxidized, but it itself is reduced (gains electrons). A reducing agent causes another substance to be reduced, but it itself is oxidized (loses electrons). They're named for what they do to the OTHER species, not what happens to themselves — like a spy working for the "other side."
中文： 氧化剂使其他物质被氧化，但自身被还原。还原剂使其他物质被还原，但自身被氧化。它们以对另一物质的作用命名，而非自身的变化——像间谍为"对方"工作。
Q4¶
EN (Original): What is the difference between an electrochemical cell and an electrolytic cell?
中文翻译： 原电池和电解池有什么区别？
Answer 答案:
EN: An electrochemical cell (galvanic/voltaic) uses a spontaneous redox reaction to produce electrical energy (E° > 0). An electrolytic cell uses electrical energy to force a non-spontaneous redox reaction (E° < 0).
中文： 原电池利用自发的氧化还原反应产生电能（E° > 0）。电解池用电能驱动非自发的反应（E° < 0）。
Q5–Q9 (Oxidation Number Rules)¶
Q5 — Define oxidation number:
EN: The charge an atom would have if all bonds were completely ionic.
Q6 — Element in standard state: Oxidation number = 0
Q7 — Individual ion: Oxidation number = the charge on the ion
Q8 — Hydrogen in compounds: Usually +1 (except in metal hydrides like NaH where H = −1)
Q9 — Oxygen in compounds: Usually −2 (except in peroxides like BaO₂ where O = −1)
Q10¶
EN (Original): Determine the oxidation number of each species.
中文翻译： 确定各元素的氧化数。
Answer 答案:
| Species 物种 | Work 计算过程 | Oxidation # 氧化数 |
|---|---|---|
| Au | Element (standard state) | 0 |
| H in H₂ | Element | 0 |
| Na in Na₂S | Alkali metal | +1 |
| S in Na₂S | 2(+1) + S = 0 → S = −2 | −2 |
| S in Na₂S₂O₆ | 2(+1) + 2S + 6(−2) = 0 → 2S = +10 | +5 |
| O in BaO₂ | Peroxide: Ba(+2) + 2O = 0 → O = −1 | −1 |
| P in AlPO₄ | Al(+3) + P + 4(−2) = 0 → P = +5 | +5 |
✅ Answer Key 答案: H in H₂ = 0, S in Na₂S = −2, O in BaO₂ = −1 — 一致 ✓
Q11¶
EN (Original): Determine oxidation, reduction or no change for each underlined element.
中文翻译： 判断各标记元素的氧化/还原/无变化。
Answer 答案:
| Reaction 反应 | Change 变化 | Type 类型 |
|---|---|---|
| P₄ → PO₄³⁻ | P: 0 → +5 | Oxidation 氧化 |
| BaO₂ → BaO | O: −1 → −2 | Reduction 还原 |
| H₂S → H₂SO₄ | S: −2 → +6 | Oxidation 氧化 (+8) |
| Mn⁴⁺ → MnO₂ | Mn: +4 → +4 | No change 无变化 |
| H₂O → KH | H: +1 → −1 | Reduction 还原 |
✅ Answer Key 答案: H₂S → H₂SO₄ = Oxidation (+8) — 一致 ✓
7.2 The Standard Reduction Table — 标准还原电位表¶
Q12¶
EN (Original): Oxidizing agents on the left side are all undergoing what process?
中文翻译： 表左侧的氧化剂都在进行什么过程？
Answer 答案:
EN: They are all undergoing reduction (gaining electrons).
中文： 它们都在进行还原（获得电子）。
Q13–Q14¶
Q13: Spontaneous reaction → "Z" shape (top-left OA + bottom-right RA on the table).
Q14: Non-spontaneous → backward "Z" (bottom-left OA + top-right RA).
Q15¶
EN (Original): Answer yes or no for each situation.
中文翻译： 对以下每种情况回答是或否。
Answer 答案:
| Situation 情况 | Answer 回答 | Reasoning 推理 |
|---|---|---|
| Will neutral water oxidize Sr metal? | Yes | Sr²⁺/Sr E°=−2.89; H₂O/H₂ E°=−0.41; Net=+2.48>0 |
| Safe to store Cu in water? | Yes | Cu²⁺/Cu E°=+0.34; H₂O/H₂ E°=−0.41; Net=−0.75<0 (non-spont.) |
| F₂ into water → reaction? | Yes | F₂/F⁻ E°=+2.87; O₂/H₂O E°=+1.23; Net=+1.64>0 |
| Hg(NO₃)₂ in Ag container? | No | Hg²⁺ reduces using Ag; Ag→Ag⁺ E°=+0.80<E°(Hg) → reaction |
| Li solid near F₂ gas? | No | Li⁺/Li E°=−3.04; F₂/F⁻ E°=+2.87; Net=+5.91 (extremely dangerous!) |
| HCl + Sn metal? | Yes | 2H⁺/H₂ E°=0.00; Sn²⁺/Sn E°=−0.14; Net=+0.14>0 |
| HCl + MnO₂? | No | MnO₂ E°=+1.22; Cl₂/Cl⁻ E°=+1.36; Net=1.22−1.36=−0.14<0 |
ℹ️ 答案页未提供此题答案。
Q16¶
EN (Original): Give the symbol for a cation that will oxidize Ca but not Mg.
中文翻译： 给出一种能氧化 Ca 但不能氧化 Mg 的阳离子符号。
Answer 答案:
EN: Need E°(cation) between E°(Ca²⁺) = −2.87 V and E°(Mg²⁺) = −2.37 V.
From the table: Na⁺ (E° = −2.71 V)
- Oxidize Ca? E°(net) = −2.71 − (−2.87) = +0.16 > 0 → Yes ✓
- Oxidize Mg? E°(net) = −2.71 − (−2.37) = −0.34 < 0 → No ✓
中文： 需要 E° 介于 −2.87 和 −2.37 之间。Na⁺（E° = −2.71 V）满足条件。
Q17¶
EN (Original): Fe³⁺ in basic solution — reaction occurs but no solid iron metal forms. Explain.
中文翻译： Fe³⁺ 在碱性溶液中发生反应但不形成固态铁。解释。
Answer 答案:
EN: In basic solution, Fe³⁺ ions immediately react with OH⁻ to form Fe(OH)₃ precipitate (Ksp = 2.6 × 10⁻³⁹, extremely insoluble). The Fe³⁺ ions are removed from solution by precipitation before they can be fully reduced to Fe(s). Additionally, the reduction Fe³⁺ + e⁻ → Fe²⁺ occurs (E° = +0.77), producing Fe²⁺ which then also precipitates as Fe(OH)₂ — but neither reaches the metallic state.
中文： Fe³⁺ 立即与 OH⁻ 形成 Fe(OH)₃ 沉淀（Ksp 极小），从溶液中移除，来不及被完全还原为 Fe(s)。同时 Fe³⁺ → Fe²⁺ 的还原也发生，但 Fe²⁺ 也会沉淀为 Fe(OH)₂。
Q18¶
EN (Original): Besides atoms and charge, what else must be balanced in redox reactions?
中文翻译： 除了原子和电荷，氧化还原反应还需要平衡什么？
Answer 答案:
EN: Electrons — the total electrons lost in oxidation must equal the total electrons gained in reduction.
中文： 电子 — 氧化中失去的电子总数必须等于还原中获得的电子总数。
Q19¶
EN (Original): Write balanced net ionic equations for each set of reactants.
中文翻译： 写出各组反应物的配平净离子方程。
Answer 答案:
a) Cl₂(g) + Ni(s):
Red: Cl₂ + 2e⁻ → 2Cl⁻ Ox: Ni → Ni²⁺ + 2e⁻
$$\text{Cl}_2(g) + \text{Ni}(s) \rightarrow \text{Ni}^{2+} + 2\text{Cl}^-$$
b) Br₂(g) + Al(s):
Red: Br₂ + 2e⁻ → 2Br⁻ (×3) Ox: Al → Al³⁺ + 3e⁻ (×2)
$$\boxed{3\text{Br}_2(g) + 2\text{Al}(s) \rightarrow 2\text{Al}^{3+} + 6\text{Br}^-}$$
✅ Answer Key 答案 (b): 2Al + 3Br₂ → 2Al³⁺ + 6Br⁻ — 一致 ✓
c) Acidified KMnO₄ + KI: (K⁺ is spectator)
Red: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (×2) Ox: 2I⁻ → I₂ + 2e⁻ (×5)
$$2\text{MnO}_4^- + 16\text{H}^+ + 10\text{I}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{I}_2$$
d) Au(NO₃)₃ + neutral water: (NO₃⁻ is spectator)
Red: Au³⁺ + 3e⁻ → Au(s) (×4) Ox: 2H₂O → O₂ + 4H⁺ + 4e⁻ (×3)
$$4\text{Au}^{3+} + 6\text{H}_2\text{O} \rightarrow 4\text{Au}(s) + 3\text{O}_2(g) + 12\text{H}^+$$
Q20¶
EN (Original): Using experimental data for Sr, Fr, Y, Eu, create a mini reduction table.
中文翻译： 根据实验数据建立 Sr、Fr、Y、Eu 的小型还原表。
Answer 答案:
EN — Analysis: 1. Y doesn't react with any solution → Y³⁺ is the strongest oxidizing agent 2. Sr reacts with all solutions → Sr is the strongest reducing agent 3. Fr reacts with Eu(NO₃)₂ → Eu²⁺ is above Fr⁺ as OA
a) Reduction table (strongest OA at top):
| Oxidizing Agents 氧化剂 | Reducing Agents 还原剂 | |
|---|---|---|
| Y³⁺ | ⇌ | Y |
| Eu²⁺ | ⇌ | Eu |
| Fr⁺ | ⇌ | Fr |
| Sr²⁺ | ⇌ | Sr |
b) Reducing agents, strongest → weakest: Sr > Fr > Eu > Y
中文： Y³⁺ 最强氧化剂（Y 金属不与任何溶液反应），Sr 最强还原剂（与所有溶液反应）。Fr 能还原 Eu²⁺ → Eu²⁺ 在 Fr⁺ 之上。
7.3 Balancing Reactions — 配平反应¶
Q21¶
EN (Original): What are the two methods used to balance redox reactions?
中文翻译： 配平氧化还原反应有哪两种方法？
Answer 答案:
EN: Oxidation number method and half-reaction method.
中文： 氧化数法和半反应法。
Q22¶
EN (Original): Balance using oxidation number method in acid: Cr₂O₇²⁻ + S(s) → Cr³⁺ + SO₃²⁻
中文翻译： 用氧化数法在酸性条件下配平。
Answer 答案:
EN:
Cr: +6 → +3 (gain 3e⁻ per Cr, 6e⁻ for Cr₂) — Reduction S: 0 → +4 (lose 4e⁻) — Oxidation
Electron balance: LCM(6,4) = 12 → need 2 Cr₂O₇²⁻ and 3 S
$$\boxed{2\text{Cr}_2\text{O}_7^{2-} + 3\text{S}(s) + 10\text{H}^+ \rightarrow 4\text{Cr}^{3+} + 3\text{SO}_3^{2-} + 5\text{H}_2\text{O}}$$
验证 Check: Cr: 4=4 ✓, S: 3=3 ✓, O: 14=9+5=14 ✓, H: 10=10 ✓, Charge: −4+0+10=+6→+12−6+0=+6 ✓
✅ Answer Key 答案: 2Cr₂O₇²⁻ + 3S + 10H⁺ → 4Cr³⁺ + 3SO₃²⁻ + 5H₂O — 一致 ✓
Q23¶
EN (Original): Balance in base: MnO₂ + BrO₃⁻ → MnO₄⁻ + Br₂(l)
中文翻译： 在碱性条件下配平。
Answer 答案:
EN:
Mn: +4 → +7 (lose 3e⁻) — Oxidation Br: +5 → 0 (gain 5e⁻) — Reduction
Electron balance: LCM(3,5) = 15 → need ×5 Mn (15e⁻ lost) and ×3 Br (15e⁻ gained, but Br₂ has 2 Br → 6 BrO₃⁻ gives 3 Br₂)
Wait: 5 MnO₂ and 3 BrO₃⁻ gives 5 × 3 = 15e⁻ lost and 3 × 5 = 15e⁻ gained. But each BrO₃⁻ has 1 Br → 3 BrO₃⁻ → 3/2 Br₂. Multiply by 2:
$$\boxed{10\text{MnO}_2 + 6\text{BrO}_3^- + 4\text{OH}^- \rightarrow 10\text{MnO}_4^- + 3\text{Br}_2(l) + 2\text{H}_2\text{O}}$$
验证: Mn: 10=10 ✓, Br: 6=6 ✓, O: 20+18+4=42→40+2=42 ✓, H: 4=4 ✓, Charge: 0−6−4=−10→−10 ✓
✅ Answer Key 答案: 10MnO₂ + 6BrO₃⁻ + 4OH⁻ → 10MnO₄⁻ + 3Br₂ + 2H₂O — 一致 ✓
Q24¶
EN (Original): Balance using half-reaction method in acid: NO₃⁻ + Hg(l) → Hg²⁺ + NO(g)
中文翻译： 用半反应法在酸性条件下配平。
Answer 答案:
Mercury Half-Reaction 汞半反应: (Oxidation)
$$\text{Hg}(l) \rightarrow \text{Hg}^{2+} + 2e^- \quad (\times 3)$$
Nitrogen Half-Reaction 氮半反应: (Reduction)
$$\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO}(g) + 2\text{H}_2\text{O} \quad (\times 2)$$
Overall 总反应:
$$\boxed{3\text{Hg}(l) + 2\text{NO}_3^- + 8\text{H}^+ \rightarrow 3\text{Hg}^{2+} + 2\text{NO}(g) + 4\text{H}_2\text{O}}$$
✅ Answer Key 答案: 3Hg + 2NO₃⁻ + 8H⁺ → 3Hg²⁺ + 2NO + 4H₂O — 一致 ✓
Q25¶
EN (Original): Balance in base: PO₃³⁻ + ClO₄⁻ → PO₄³⁻ + Cl⁻
中文翻译： 在碱性条件下配平。
Answer 答案:
Phosphorus Half-Reaction 磷半反应: (Oxidation, P: +3→+5, lose 2e⁻)
$$\text{PO}_3^{3-} + \text{H}_2\text{O} \rightarrow \text{PO}_4^{3-} + 2\text{H}^+ + 2e^- \quad (\times 4)$$
Chlorine Half-Reaction 氯半反应: (Reduction, Cl: +7→−1, gain 8e⁻)
$$\text{ClO}_4^- + 8\text{H}^+ + 8e^- \rightarrow \text{Cl}^- + 4\text{H}_2\text{O} \quad (\times 1)$$
Combine (acid version): 4PO₃³⁻ + 4H₂O + ClO₄⁻ + 8H⁺ → 4PO₄³⁻ + 8H⁺ + Cl⁻ + 4H₂O
Cancel H⁺ and H₂O:
$$\boxed{4\text{PO}_3^{3-} + \text{ClO}_4^- \rightarrow 4\text{PO}_4^{3-} + \text{Cl}^-}$$
验证: P: 4=4 ✓, Cl: 1=1 ✓, O: 12+4=16=16 ✓, Charge: −12−1=−13=−12−1=−13 ✓
✅ Answer Key 答案: 4PO₃³⁻ + ClO₄⁻ → 4PO₄³⁻ + Cl⁻ — 一致 ✓
Q26¶
EN (Original): Balance in base: AuCl₄⁻ + H₂ → Au(s) + 4Cl⁻ + 2H⁺
中文翻译： 在碱性条件下配平。
Answer 答案:
Gold Half-Reaction 金半反应: (Reduction, Au: +3→0)
$$\text{AuCl}_4^- + 3e^- \rightarrow \text{Au}(s) + 4\text{Cl}^- \quad (\times 2)$$
Hydrogen Half-Reaction 氢半反应: (Oxidation)
$$\text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \quad (\times 3)$$
In acid: 2AuCl₄⁻ + 3H₂ → 2Au + 8Cl⁻ + 6H⁺
Convert to base (add 6OH⁻ to both sides, H⁺ + OH⁻ → H₂O):
$$\boxed{2\text{AuCl}_4^- + 3\text{H}_2 + 6\text{OH}^- \rightarrow 2\text{Au}(s) + 8\text{Cl}^- + 6\text{H}_2\text{O}}$$
✅ Answer Key 答案: 2AuCl₄⁻ + 3H₂ + 6OH⁻ → 2Au + 8Cl⁻ + 6H₂O — 一致 ✓
术语表 | Terminology¶
| English 英文 | Chinese 中文 | Definition 定义 |
|---|---|---|
| Oxidation | 氧化 | Loss of electrons |
| Reduction | 还原 | Gain of electrons |
| Oxidizing Agent (OA) | 氧化剂 | Species that gets reduced (causes oxidation in others) |
| Reducing Agent (RA) | 还原剂 | Species that gets oxidized (causes reduction in others) |
| Standard Reduction Potential (E°) | 标准还原电位 | Voltage of a half-cell relative to H₂/H⁺ at standard conditions |
| Electrochemical Cell | 原电池 | Spontaneous redox → electrical energy |
| Electrolytic Cell | 电解池 | Electrical energy → non-spontaneous redox |
| Oxidation Number | 氧化数 | Hypothetical charge if all bonds were ionic |
| Half-Reaction Method | 半反应法 | Balance by separating into oxidation and reduction half-reactions |
| Oxidation Number Method | 氧化数法 | Balance using changes in oxidation numbers |
答案核对总表 | Answer Key Verification Summary¶
| Question | Our Answer | Answer Key | Status |
|---|---|---|---|
| Q10 (H in H₂) | 0 | 0 | ✅ |
| Q10 (S in Na₂S) | −2 | −2 | ✅ |
| Q10 (O in BaO₂) | −1 | −1 | ✅ |
| Q11 (H₂S→H₂SO₄) | Oxidation (+8) | Oxidation (+8) | ✅ |
| Q19b | 2Al + 3Br₂ → 2Al³⁺ + 6Br⁻ | Same | ✅ |
| Q22 | 2Cr₂O₇²⁻ + 3S + 10H⁺ → 4Cr³⁺ + 3SO₃²⁻ + 5H₂O | Same | ✅ |
| Q23 | 10MnO₂ + 6BrO₃⁻ + 4OH⁻ → 10MnO₄⁻ + 3Br₂ + 2H₂O | Same | ✅ |
| Q24 | 3Hg + 2NO₃⁻ + 8H⁺ → 3Hg²⁺ + 2NO + 4H₂O | Same | ✅ |
| Q25 | 4PO₃³⁻ + ClO₄⁻ → 4PO₄³⁻ + Cl⁻ | Same | ✅ |
| Q26 | 2AuCl₄⁻ + 3H₂ + 6OH⁻ → 2Au + 8Cl⁻ + 6H₂O | Same | ✅ |
Result: 10/10 answers verified — 100% match ✅