U9.10 · Electrolysis and Faraday's LawPractice · 电解与法拉第定律
Lessons covered:U9.10 Electrolysis and Faraday's Law
- 1Numeric
A spoon is silver-plated by making it the cathode in AgNO₃ solution and passing 0.500 A for 30.0 min. What is the greatest mass of silver that can deposit? M(Ag) = 107.87 g/mol, F = 96485 C/mol.
gExplanationt = 30.0 × 60 = 1800 s, q = 0.500 × 1800 = 900 C, n(e⁻) = 900 ÷ 96485 = 0.00933 mol. Ag⁺ + e⁻ → Ag, so n(Ag) = n(e⁻) and m = 0.00933 × 107.87 = 1.01 g.
- 2Numeric
Steel is zinc-plated by electrolyzing ZnSO₄ solution at 3.00 A. How many minutes are needed to deposit 5.00 g of zinc? M(Zn) = 65.38 g/mol, F = 96485 C/mol.
minExplanationn(Zn) = 5.00 ÷ 65.38 = 0.07648 mol; Zn²⁺ + 2e⁻ → Zn, so n(e⁻) = 0.15295 mol; q = 0.15295 × 96485 = 14758 C; t = q ÷ I = 4919 s = 82.0 min.
- 3Multiple choice
When molten MgCl₂ is electrolyzed, what forms at each electrode?
ExplanationMolten salt contains only Mg²⁺ and Cl⁻, with no water. Reduction at the cathode: Mg²⁺ + 2e⁻ → Mg; oxidation at the anode: 2Cl⁻ → Cl₂ + 2e⁻. This is how magnesium is made industrially; in aqueous solution, water would be reduced at the cathode instead, giving H₂.
- 4Multiple choice
1 M NaBr solution is electrolyzed with inert electrodes. What are the main products at the two electrodes? E°: Br₂(l) + 2e⁻ → 2Br⁻(aq), E° = +1.09 V; O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l), E° = +1.23 V; 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq), E° = −0.83 V; Na⁺(aq) + e⁻ → Na(s), E° = −2.71 V.
ExplanationCathode: comparing reduction of Na⁺ (−2.71 V) and H₂O (−0.83 V), water has the higher E° and is reduced more easily, giving H₂ and OH⁻. Anode: oxidizing Br⁻ needs only 1.09 V, easier than oxidizing water (1.23 V), so Br₂ forms.
- 5Numeric
1 M NiCl₂ solution is electrolyzed with inert electrodes, giving Ni at the cathode and Cl₂ at the anode. E°: Cl₂(g) + 2e⁻ → 2Cl⁻(aq), E° = +1.36 V; Ni²⁺(aq) + 2e⁻ → Ni(s), E° = −0.26 V. Under standard conditions, what is the minimum applied voltage?
VExplanationThe reaction to drive is Ni²⁺ + 2Cl⁻ → Ni + Cl₂, with E°cell = E°(cathode) − E°(anode) = −0.26 − 1.36 = −1.62 V < 0, so it isn’t spontaneous. The applied voltage must at least cancel it: 1.62 V (in practice somewhat more, to overcome resistance and extra resistance at the electrodes).
- 6Multiple choice
Solutions of and are each electrolyzed, and the mass of metal deposited at the cathode is plotted against the charge passed. Why is the silver line about 3.4 times as steep as the copper line?
ExplanationA charge q supplies n(e⁻) = q/F. The mass deposited is (q/F) ÷ (electrons per ion) × molar mass, so the slope is M/(nF): silver 107.87/(1 × 96,485) ≈ 1.118 mg/C and copper 63.55/(2 × 96,485) ≈ 0.329 mg/C, a ratio of (107.87 ÷ 63.55) × 2 ≈ 3.4. On the graph, at 1000 C there is about 1.12 g of silver and 0.33 g of copper. The slope depends only on electrons per ion and molar mass, not on reactivity or how fast ions move; copper’s E° is in fact lower than silver’s.
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