CH12 Unit 8: Electrochemical & Electrolytic Cells — 原电池与电解池 完整解答¶
状态: ✅ 已完成
创建日期: 2026-02-18
最后更新: 2026-02-18
8.1 The Electrochemical Cell — 原电池¶
Q1¶
EN (Original): What does ULIRBLRWIO stand for?
中文翻译： ULIRBLRWIO 代表什么？
Answer 答案:
EN: This is a mnemonic for building an electrochemical cell:
- Universal — understand the setup
- Lowest — the lowest (most negative) E° is oxidized
- Is — is
- Reversed — reversed (written as oxidation)
- Biggest — the biggest (most positive) E° is reduced
- Left — left as a reduction
- Redox — combine redox half-reactions
- Write — write the net reaction
- Interpret — interpret the E° value
- Outcome — determine if spontaneous
In practice: "Understand: Lowest Is Reversed, Biggest Left as Reduction. Write, Interpret, Outcome." — i.e., identify the half-reaction with the lowest E° (reverse it for oxidation), keep the highest E° as reduction, combine, calculate net E°, and determine spontaneity.
中文： 这是构建原电池的助记词：找到最低（最负）的 E° 将其反转为氧化反应，保留最高（最正）的 E° 为还原反应，组合写出净反应，计算净 E°，判断自发性。
Q2¶
EN (Original): Will you ever double or triple an E° value? Explain.
中文翻译： 你会将 E° 值加倍或三倍吗？解释。
Answer 答案:
EN: No. Standard reduction potentials (E°) are intensive properties — they do not depend on the amount of substance. Even if you multiply a half-reaction by 2 or 3 to balance electrons, the E° value stays the same. This is because E° is measured per electron transferred, not per mole of reaction.
中文： 不会。 标准还原电位（E°）是强度性质——不取决于物质的量。即使将半反应乘以2或3来平衡电子，E° 值保持不变。这是因为 E° 是按每个电子转移测量的，而不是按摩尔反应。
Q3¶
EN (Original): What cell was used to measure the E° values of all other cells? What is the E° value for this cell?
中文翻译： 什么电池被用来测量所有其他电池的 E° 值？该电池的 E° 值是多少？
Answer 答案:
EN: The Standard Hydrogen Electrode (SHE), also called the Standard Hydrogen Half-Cell:
$$2\text{H}^+(aq,\;1\text{M}) + 2e^- \rightarrow \text{H}_2(g,\;1\text{atm})$$
Its E° value is defined as 0.00 V by convention. All other E° values are measured relative to this reference.
中文： 标准氢电极（SHE），也称标准氢半电池。其 E° 值被定义为 0.00 V，所有其他 E° 值都以此为参考。
✅ Answer Key 答案: E° = 0.00 V — 一致 ✓
Q4¶
EN (Original): If a reduction half reaction had E° = +4.52, what would be the E° value of the same reaction in reverse (oxidation)?
中文翻译： 如果还原半反应的 E° = +4.52，反向（氧化）反应的 E° 值是多少？
Answer 答案:
EN: When you reverse a half-reaction (from reduction to oxidation), you simply change the sign of E°.
$$E°{\text{oxidation}} = -E°$$}} = -(+4.52) = \boxed{-4.52\;\text{V}
中文： 将半反应从还原反转为氧化时，只需改变 E° 的符号。答案为 −4.52 V。
✅ Answer Key 答案: E° = −4.52 V — 一致 ✓
Q5¶
EN (Original): Determine if the reaction below is spontaneous by calculating the net E°.
3 Hg(l) + 2 NO₃⁻ + 8 H⁺ → 3 Hg²⁺ + 2 NO(g) + 4 H₂O
中文翻译： 通过计算净 E° 判断以下反应是否自发。
Answer 答案:
EN:
Step 1 — Identify half-reactions:
Reduction (NO₃⁻ → NO):
$$\text{NO}_3^-(aq) + 4\text{H}^+(aq) + 3e^- \rightarrow \text{NO}(g) + 2\text{H}_2\text{O}(l) \quad E° = +0.96\;\text{V}$$
Oxidation (Hg → Hg²⁺):
$$\text{Hg}(l) \rightarrow \frac{1}{2}\text{Hg}_2^{2+}?$$
Actually, looking at the product Hg²⁺, the relevant half-reaction from the data table:
$$\text{Hg}^{2+}(aq) + 2e^- \rightarrow \text{Hg}(l) \quad E°_{\text{red}} = +0.85\;\text{V}$$
Reversed for oxidation: E°_ox = −0.85 V
Step 2 — Calculate net E°:
$$E°{\text{net}} = E°$$}} + E°_{\text{ox}} = (+0.96) + (-0.85) = \boxed{+0.11\;\text{V}
Step 3 — Conclusion: Since E°_net > 0, the reaction is spontaneous (ΔG < 0).
中文：
还原半反应：NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O，E° = +0.96 V
氧化半反应：Hg → Hg²⁺ + 2e⁻，E°_ox = −0.85 V
净 E° = +0.96 + (−0.85) = +0.11 V > 0，反应自发。
✅ Answer Key 答案: Net E° = +0.11 V, spontaneous — 一致 ✓
Q6¶
EN (Original): What would the net E° be for the reverse reaction of the one shown in question 5. Would it be spontaneous?
中文翻译： 第5题反应的逆反应净 E° 是多少？是否自发？
Answer 答案:
EN: The reverse reaction simply reverses the sign of the net E°:
$$E°_{\text{reverse}} = -(+0.11) = \boxed{-0.11\;\text{V}}$$
Since E°_net < 0, the reverse reaction is not spontaneous.
中文： 逆反应的净 E° = −(+0.11) = −0.11 V < 0，反应不自发。
Q7¶
EN (Original): What process (oxidation or reduction) occurs at the anode in an electrochemical cell?
中文翻译： 在原电池的阳极发生什么过程（氧化还是还原）？
Answer 答案:
EN: Oxidation occurs at the anode. Remember: Anode = Oxidation ("An Ox").
中文： 阳极发生氧化反应。记忆法："An Ox"——阳极（Anode）= 氧化（Oxidation）。
Q8¶
EN (Original): What is the purpose of the salt bridge in an electrochemical cell?
中文翻译： 原电池中盐桥的作用是什么？
Answer 答案:
EN: The salt bridge serves two purposes:
- Maintains electrical neutrality — As electrons flow through the wire from anode to cathode, ions must flow between the half-cells to balance the charge. Cations from the salt bridge migrate toward the cathode half-cell (which is becoming negative), and anions migrate toward the anode half-cell (which is becoming positive).
- Completes the circuit — Without the salt bridge, charge would build up and the reaction would stop almost immediately.
中文： 盐桥的作用：
- 保持电中性 — 电子通过导线从阳极流向阴极时，离子必须在两个半电池间流动以平衡电荷。盐桥中的阳离子移向阴极（变负），阴离子移向阳极（变正）。
- 完成电路 — 没有盐桥，电荷积累会导致反应立即停止。
Q9¶
EN (Original): Will Hg²⁺ ions react with water to produce Hg metal? Explain.
中文翻译： Hg²⁺ 离子会与水反应生成 Hg 金属吗？解释。
Answer 答案:
EN: No. For Hg²⁺ to be reduced to Hg metal by water, water would need to act as a reducing agent (be oxidized):
- Reduction: Hg²⁺ + 2e⁻ → Hg(l), E° = +0.85 V
- Oxidation: H₂O → ½O₂ + 2H⁺ + 2e⁻, E°_ox = −1.23 V (standard conditions)
Net E° = +0.85 + (−1.23) = −0.38 V
Since E° is negative, this reaction is non-spontaneous under standard conditions. Water is too weak a reducing agent — Hg²⁺ cannot oxidize water to O₂, so no Hg metal will be produced.
On the Standard Reduction Table, Hg²⁺ sits below O₂/H₂O, so water (as a reducing agent on the right side) cannot reduce Hg²⁺ — the "backward Z" confirms no spontaneous reaction.
中文： 不会。 要使 Hg²⁺ 被水还原为 Hg 金属，水需要作为还原剂被氧化。使用标准还原电位：E°_net = +0.85 + (−1.23) = −0.38 V，为负值，反应不自发。水是非常弱的还原剂，Hg²⁺ 不能有效氧化水产生 O₂，因此不会生成 Hg 金属。
Q10¶
EN (Original): What are two common terms for the oxidation of a metal?
中文翻译： 金属氧化的两个常用术语是什么？
Answer 答案:
EN: Corrosion and tarnishing.
- Corrosion — typically refers to the gradual oxidative degradation of metals (especially iron → rust)
- Tarnishing — the oxidation that forms a thin discolored layer on the metal surface (especially silver, copper)
中文： 腐蚀（corrosion）和失去光泽/变色（tarnishing）。腐蚀通常指金属的氧化降解（如铁锈），变色指金属表面形成薄氧化层（如银、铜）。
Q11¶
EN (Original): What are the only two metals that can be used to cathodically protect Rubidium? Bonus: Why is this unlikely to work very well?
中文翻译： 哪两种金属可以用来阴极保护铷？额外题：为什么这不太可能有效？
Answer 答案:
EN: To cathodically protect rubidium, you need metals with a more negative E° than Rb.
Rb⁺ + e⁻ → Rb, E° = −2.98 V
Only metals with E° < −2.98 V qualify:
- Cs (Cs⁺ + e⁻ → Cs, E° = −3.03 V)
- Li (Li⁺ + e⁻ → Li, E° = −3.04 V)
Bonus: This is unlikely to work well because:
- Rubidium, cesium, and lithium are all extremely reactive alkali metals that react violently with water and air
- You cannot set up a conventional galvanic protection system with metals that spontaneously combust in the presence of moisture
- Rubidium itself would not be used as a structural metal, making cathodic protection pointless
中文： 阴极保护铷需要 E° 比 Rb（−2.98 V）更负的金属：
- 铯 Cs（E° = −3.03 V）
- 锂 Li（E° = −3.04 V）
额外题： 这不实际，因为铷、铯、锂都是极度活泼的碱金属，会与水和空气剧烈反应，无法建立常规的电化学保护系统。铷本身也不用作结构金属。
Q12¶
EN (Original): A very expensive electrochemical cell is set up with a silver electrode in a solution of silver nitrate in the first beaker and a gold electrode in a solution of gold nitrate in the second beaker. The electrodes are connected with a wire and a salt bridge is used to connect the two beakers.
a) Write the two half reactions with their correct E°. Indicate which is undergoing reduction and which is undergoing oxidation. Determine the net reaction and the net E°.
b) Identify the cathode and the anode in this cell.
c) Identify the species reduced and the species oxidized.
d) In this cell (as in all electrochemical cells) electrons move from the ___ to the ___
中文翻译： 用银电极浸在硝酸银溶液中，金电极浸在硝酸金溶液中，用导线和盐桥连接，建立一个昂贵的原电池。
Answer 答案:
(a) Half-reactions and net reaction:
From the Standard Reduction Potentials table:
- Au³⁺(aq) + 3e⁻ → Au(s), E° = +1.50 V
- Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V
Higher E° undergoes reduction, lower E° is reversed for oxidation:
Reduction: Au³⁺(aq) + 3e⁻ → Au(s), E° = +1.50 V
Oxidation: Ag(s) → Ag⁺(aq) + e⁻, E°_ox = −0.80 V (×3 to balance electrons, but E° unchanged)
Net reaction (multiply Ag by 3):
$$\boxed{\text{Au}^{3+}(aq) + 3\text{Ag}(s) \rightarrow \text{Au}(s) + 3\text{Ag}^+(aq)}$$
$$E°_{\text{net}} = (+1.50) + (-0.80) = \boxed{+0.70\;\text{V}}$$
(b) Cathode = Gold electrode (reduction occurs here). Anode = Silver electrode (oxidation occurs here).
(c) Species reduced: Au³⁺ (gains electrons → Au). Species oxidized: Ag (loses electrons → Ag⁺).
(d) Electrons move from the anode (silver electrode) to the cathode (gold electrode).
中文：
(a) 还原：Au³⁺ + 3e⁻ → Au，E° = +1.50 V；氧化：Ag → Ag⁺ + e⁻，E°_ox = −0.80 V
净反应：Au³⁺ + 3Ag → Au + 3Ag⁺，净 E° = +0.70 V
(b) 阴极 = 金电极，阳极 = 银电极
(c) 被还原：Au³⁺；被氧化：Ag
(d) 电子从阳极（银电极）流向阴极（金电极）
✅ Answer Key 答案: Au³⁺ + 3Ag → Au + 3Ag⁺, Net E° = +0.70 V — 一致 ✓
Q13¶
EN (Original): Consider the electrochemical cell shown below.
Two beakers connected by a salt bridge (1M NaNO₃) and a voltmeter (V). Left beaker: Pb electrode in 1M Pb(NO₃)₂ solution. Right beaker: Mn electrode in 1M Mn(NO₃)₂ solution.
a) Write the two half reactions with their correct E°. Indicate which is undergoing reduction and which is undergoing oxidation. Determine the net reaction and the net E°.
b) As the reaction proceeds which ion will decrease in concentration and which will increase?
c) Which electrode will increase in mass as the reaction proceeds, the cathode or the anode?
中文翻译： 如图所示的原电池：左烧杯有 Pb 电极在 Pb(NO₃)₂ 溶液中，右烧杯有 Mn 电极在 Mn(NO₃)₂ 溶液中，用盐桥连接。
Answer 答案:
(a) Half-reactions and net reaction:
From the Standard Reduction Potentials table:
- Pb²⁺(aq) + 2e⁻ → Pb(s), E° = −0.13 V
- Mn²⁺(aq) + 2e⁻ → Mn(s), E° = −1.19 V
Higher E° (less negative) undergoes reduction:
Reduction: Pb²⁺(aq) + 2e⁻ → Pb(s), E° = −0.13 V
Oxidation: Mn(s) → Mn²⁺(aq) + 2e⁻, E°_ox = +1.19 V
Net reaction:
$$\boxed{\text{Pb}^{2+}(aq) + \text{Mn}(s) \rightarrow \text{Pb}(s) + \text{Mn}^{2+}(aq)}$$
$$E°_{\text{net}} = (-0.13) + (+1.19) = \boxed{+1.06\;\text{V}}$$
(b) As the reaction proceeds:
- Pb²⁺ decreases in concentration (being reduced to Pb metal at the cathode)
- Mn²⁺ increases in concentration (Mn metal is oxidized to Mn²⁺ at the anode)
(c) The cathode (Pb electrode) increases in mass as Pb²⁺ ions are reduced and deposited as Pb metal onto it. The anode (Mn electrode) loses mass as Mn dissolves into solution.
中文：
(a) 还原：Pb²⁺ + 2e⁻ → Pb，E° = −0.13 V；氧化：Mn → Mn²⁺ + 2e⁻，E°_ox = +1.19 V
净反应：Pb²⁺ + Mn → Pb + Mn²⁺，净 E° = +1.06 V
(b) Pb²⁺ 浓度减少（被还原为金属），Mn²⁺ 浓度增加（金属被氧化）
(c) 阴极（Pb电极）质量增加——Pb²⁺ 被还原沉积为金属铅。
✅ Answer Key 答案: Pb²⁺ + Mn → Pb + Mn²⁺, Net E° = +1.06 V — 一致 ✓
8.2 Electrolytic Cells — 电解池¶
Q1¶
EN (Original): What are the two most common materials used to make inert electrodes?
中文翻译： 制造惰性电极最常用的两种材料是什么？
Answer 答案:
EN: Platinum (Pt) and graphite (carbon, C).
These materials are "inert" because they do not participate in the redox reactions — they merely conduct electrons.
中文： 铂（Pt）和石墨（碳，C）。这些材料是"惰性的"，不参与氧化还原反应，仅导电。
Q2¶
EN (Original): Platinum electrodes are used in a molten solution of AlCl₃ in an electrolytic cell. Write the two half reactions that occur with their correct E°. Indicate which is undergoing reduction and which is undergoing oxidation. Determine the net reaction and the net E°.
中文翻译： 在电解池中用铂电极电解熔融 AlCl₃。写出两个半反应及其 E°，指出哪个是还原、哪个是氧化，求净反应和净 E°。
Answer 答案:
EN:
This is a Type I electrolytic cell (molten salt, no water present).
In molten AlCl₃, the ions are Al³⁺ and Cl⁻.
Determine which is reduced and which is oxidized:
- The cation (Al³⁺) is reduced at the cathode
- The anion (Cl⁻) is oxidized at the anode
Reduction (cathode): Al³⁺(l) + 3e⁻ → Al(s), E° = −1.66 V
Oxidation (anode): 2Cl⁻(l) → Cl₂(g) + 2e⁻, E°_ox = −1.36 V
Balance electrons (×2 for Al, ×3 for Cl):
$$2\text{Al}^{3+}(l) + 6\text{Cl}^-(l) \rightarrow 2\text{Al}(s) + 3\text{Cl}_2(g)$$
$$E°_{\text{net}} = (-1.66) + (-1.36) = \boxed{-3.02\;\text{V}}$$
The negative net E° confirms this is non-spontaneous — requires external electrical energy (electrolytic cell).
中文：
这是I型电解池（熔融盐，无水）。
还原（阴极）：Al³⁺ + 3e⁻ → Al，E° = −1.66 V
氧化（阳极）：2Cl⁻ → Cl₂ + 2e⁻，E°_ox = −1.36 V
净反应：2Al³⁺ + 6Cl⁻ → 2Al + 3Cl₂
净 E° = −3.02 V（负值，非自发，需电能驱动）
✅ Answer Key 答案: Net E° = −3.02 V — 一致 ✓
Q3¶
EN (Original): What is the complication added when comparing a type II electrolytic cell to a type I electrolytic cell?
中文翻译： II型电解池与I型电解池相比增加了什么复杂性？
Answer 答案:
EN: In a Type II electrolytic cell, water is present as a solvent (aqueous solution instead of molten salt). This means that water itself can be oxidized or reduced, competing with the dissolved ions. You must determine which species is actually reduced at the cathode (the cation or water?) and which is actually oxidized at the anode (the anion or water?) by comparing their E° values.
中文： II型电解池中有水作为溶剂（水溶液而非熔融盐），水本身可以被氧化或还原，与溶解的离子竞争。需要通过比较 E° 值来确定阴极实际还原哪种物质（阳离子还是水），阳极实际氧化哪种物质（阴离子还是水）。
Q4¶
EN (Original): How do you determine which species will be reduced in an electrolytic cell?
中文翻译： 在电解池中如何确定哪个物质会被还原？
Answer 答案:
EN: The species with the higher (more positive / less negative) standard reduction potential (E°) will be preferentially reduced at the cathode. It is "easier" to reduce — requires less energy.
In a Type II cell, compare E° of the cation's reduction to E° of water's reduction:
- H₂O + e⁻ → ½H₂ + OH⁻, E° = −0.41 V (neutral water, 10⁻⁷ M)
Whichever has the higher E° is reduced.
中文： 具有较高（更正/更不负的）标准还原电位 E° 的物质将优先在阴极被还原。在II型电解池中，比较阳离子还原的 E° 与水还原的 E°（−0.41 V），E° 较高者被还原。
Q5¶
EN (Original): How do you determine which species will be oxidized in an electrolytic cell?
中文翻译： 在电解池中如何确定哪个物质会被氧化？
Answer 答案:
EN: The species with the lower (more negative / less positive) standard reduction potential (E°) will be preferentially oxidized at the anode. When reversed for oxidation, the lower E° gives a higher (more positive) E°_ox, meaning it is "easier" to oxidize.
In a Type II cell, compare the E° of the anion's reduction to E° of water's oxidation:
- H₂O → ½O₂ + 2H⁺ + 2e⁻, E°_ox = −0.82 V (from E°_red = +0.82 V for neutral water)
The species whose reduction E° is lower (more negative) is oxidized preferentially.
中文： 具有较低（更负/更不正的）标准还原电位 E° 的物质将优先在阳极被氧化。反转为氧化反应后，较低 E° 给出较高的 E°_ox。在II型电解池中，比较阴离子的 E° 与水氧化的 E°。
Q6¶
EN (Original): Direct current is passed through inert electrodes in a 1.00 M solution of CuF₂. Identify which species will be oxidized and reduced and then write the two half reactions and the net reaction for the events which occur. Also calculate the net E° value.
中文翻译： 用惰性电极在 1.00 M CuF₂ 溶液中通直流电。确定哪个物质被氧化和还原，写出两个半反应和净反应，并计算净 E°。
Answer 答案:
EN:
This is a Type II electrolytic cell (aqueous CuF₂). Species in solution: Cu²⁺, F⁻, and H₂O.
At the cathode (reduction) — compare Cu²⁺ vs H₂O:
- Cu²⁺ + 2e⁻ → Cu(s), E° = +0.34 V
- H₂O + e⁻ → ½H₂ + OH⁻, E° = −0.41 V
Cu²⁺ has the higher E° → Cu²⁺ is reduced
At the anode (oxidation) — compare F⁻ vs H₂O:
- F₂ + 2e⁻ → 2F⁻, E° = +2.87 V (very high — F⁻ is extremely hard to oxidize)
- ½O₂ + 2H⁺(10⁻⁷) + 2e⁻ → H₂O, E° = +0.82 V
F⁻ has a much higher reduction E° than water → water is easier to oxidize → H₂O is oxidized
Reduction (cathode): Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V
Oxidation (anode): H₂O(l) → ½O₂(g) + 2H⁺(aq) + 2e⁻, E°_ox = −0.82 V
Net reaction:
$$\text{Cu}^{2+}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{Cu}(s) + \frac{1}{2}\text{O}_2(g) + 2\text{H}^+(aq)$$
$$E°_{\text{net}} = (+0.34) + (-0.82) = \boxed{-0.48\;\text{V}}$$
Negative → non-spontaneous → requires electrical energy (as expected for electrolytic cell).
中文：
II型电解池（CuF₂ 水溶液）。溶液中有 Cu²⁺、F⁻ 和 H₂O。
阴极：Cu²⁺（E° = +0.34 V）vs H₂O（E° = −0.41 V），Cu²⁺ 的 E° 更高 → Cu²⁺ 被还原
阳极：F⁻（E°_red = +2.87 V）vs H₂O（E°_red = +0.82 V），F⁻ 极难氧化 → H₂O 被氧化
还原：Cu²⁺ + 2e⁻ → Cu，E° = +0.34 V
氧化：H₂O → ½O₂ + 2H⁺ + 2e⁻，E°_ox = −0.82 V
净反应：Cu²⁺ + H₂O → Cu + ½O₂ + 2H⁺
净 E° = −0.48 V（非自发，需外加电能）
✅ Answer Key 答案: Net E° = −0.48 V — 一致 ✓
Q7¶
EN (Original): Consider the following hypothetical cell:
Silver | Silver ion || Tin ion | Tin
Ag | Ag⁺ || Sn²⁺ | Sn
a) Write the two half reactions and the balanced REDOX reaction. (Water will not react in this cell). Also calculate the net E° value.
b) Identify the oxidizing agent.
c) Identify the reducing agent.
d) Identify this as an electrolytic or an electrochemical cell based on your Net E° calculation.
e) Would this be a spontaneous reaction?
中文翻译： 考虑以下电池：Ag | Ag⁺ || Sn²⁺ | Sn
Answer 答案:
EN:
In cell notation: Anode | Anode solution || Cathode solution | Cathode
So: Ag is the anode (oxidized), Sn is the cathode (Sn²⁺ is reduced).
(a) Half-reactions:
Oxidation (anode): 2Ag(s) → 2Ag⁺(aq) + 2e⁻, E°_ox = −0.80 V
Reduction (cathode): Sn²⁺(aq) + 2e⁻ → Sn(s), E° = −0.14 V
Net reaction:
$$\boxed{2\text{Ag}(s) + \text{Sn}^{2+}(aq) \rightarrow 2\text{Ag}^+(aq) + \text{Sn}(s)}$$
$$E°_{\text{net}} = (-0.14) + (-0.80) = \boxed{-0.94\;\text{V}}$$
(b) Oxidizing agent: Sn²⁺ — it gets reduced (gains electrons), causing Ag to be oxidized.
(c) Reducing agent: Ag — it gets oxidized (loses electrons), causing Sn²⁺ to be reduced.
(d) Since E°_net = −0.94 V < 0, this is an electrolytic cell (requires external energy).
(e) No, this is not spontaneous (negative E° means ΔG > 0).
中文：
(a) 氧化（阳极）：2Ag → 2Ag⁺ + 2e⁻，E°_ox = −0.80 V
还原（阴极）：Sn²⁺ + 2e⁻ → Sn，E° = −0.14 V
净反应：2Ag + Sn²⁺ → 2Ag⁺ + Sn，净 E° = −0.94 V
(b) 氧化剂：Sn²⁺（被还原）
(c) 还原剂：Ag（被氧化）
(d) 电解池（E° < 0，非自发）
(e) 不自发
✅ Answer Key 答案: Net E° = −0.94 V — 一致 ✓
Q8¶
EN (Original): What are the two common purposes type III electrolytic cells are used for?
中文翻译： III型电解池的两个常见用途是什么？
Answer 答案:
EN: The two common purposes are:
- Electroplating — coating an object with a thin layer of metal (e.g., chrome plating, gold plating) for appearance or corrosion protection
- Electrorefining (purification) — purifying impure metals by dissolving the impure metal at the anode and depositing pure metal at the cathode (e.g., copper refining)
中文：
- 电镀 — 在物体表面镀上一层薄金属（如镀铬、镀金），用于美观或防腐
- 电解精炼 — 将不纯金属在阳极溶解，在阴极沉积纯金属（如铜的精炼）
Q9¶
EN (Original): If you want to plate something with a metal, does it need to be the anode or the cathode? Explain.
中文翻译： 如果要在物体上电镀金属，该物体应作为阳极还是阴极？解释。
Answer 答案:
EN: The object to be plated must be the cathode. At the cathode, reduction occurs — metal ions in solution gain electrons and deposit as solid metal onto the cathode surface. The metal you want to plate with is used as the anode — it oxidizes and dissolves, replenishing the metal ions in solution.
中文： 要电镀的物体必须是阴极。在阴极发生还原反应——溶液中的金属离子获得电子，沉积为固体金属到阴极表面。要镀的金属用作阳极，氧化溶解以补充溶液中的金属离子。
术语表 | Terminology¶
| English 英文 | Chinese 中文 | Definition 定义 |
|---|---|---|
| Electrochemical Cell | 原电池（伏打电池） | Spontaneous redox reaction → electrical energy (E° > 0) |
| Electrolytic Cell | 电解池 | Electrical energy → non-spontaneous redox reaction (E° < 0) |
| Anode | 阳极 | Electrode where oxidation occurs |
| Cathode | 阴极 | Electrode where reduction occurs |
| Salt Bridge | 盐桥 | Connects half-cells; maintains electrical neutrality |
| Standard Reduction Potential (E°) | 标准还原电位 | Voltage measured vs SHE at standard conditions |
| Standard Hydrogen Electrode (SHE) | 标准氢电极 | Reference electrode; E° = 0.00 V by definition |
| Intensive Property | 强度性质 | Property independent of amount (E° is not multiplied) |
| Cathodic Protection | 阴极保护 | Using a more reactive metal to protect another from corrosion |
| Inert Electrode | 惰性电极 | Electrode that doesn't react (Pt, graphite) |
| Molten Salt (Type I) | 熔融盐（I型） | Electrolysis of pure molten ionic compound |
| Aqueous (Type II) | 水溶液（II型） | Electrolysis with water present — must consider water redox |
| Electroplating (Type III) | 电镀（III型） | Coating object with metal via electrolysis |
| Electrorefining | 电解精炼 | Purifying metals via electrolysis |
答案核对总表 | Answer Key Verification Summary¶
| Question | Our Answer | Answer Key | Status |
|---|---|---|---|
| Q3 | SHE, E° = 0.00 V | E° = 0.00 V | ✅ |
| Q4 | E° = −4.52 V | E° = −4.52 V | ✅ |
| Q5 | Net E° = +0.11 V, spontaneous | Net E° = +0.11 V, spontaneous | ✅ |
| Q12a | Au³⁺ + 3Ag → Au + 3Ag⁺, E° = +0.70 V | Same, E° = +0.70 | ✅ |
| Q13a | Pb²⁺ + Mn → Pb + Mn²⁺, E° = +1.06 V | Same, E° = +1.06 | ✅ |
| 8.2 Q2 | Net E° = −3.02 V | Net E° = −3.02 V | ✅ |
| 8.2 Q6 | Net E° = −0.48 V | Net E° = −0.48 V | ✅ |
| 8.2 Q7a | Net E° = −0.94 V | Net E° = −0.94 V | ✅ |
Result: 8/8 answers verified — 100% match ✅