HOW TO SOLVE!!! HELP!!!
We electroplate a piece of metal submerged in a solution of copper (II) chloride with a current of 7.1 amperes and we accumulate a total of 22.9 grams of copper.Data: Cu 63.5 g/mol; Cl 35.5 g/mol ; 1 Faraday 96485 C/molHow many hours does it take for the copper to accumulate? (round to two decimals)
We electroplate a piece of metal submerged in a solution of copper (II) chloride with a current of ?
2020-04-12 3:18 pm
回答 (1)
2020-04-12 3:48 pm
Cu²⁺(aq) + 2e⁻ → Cu(s)
Moles of Cu accumulated = (22.9 g) / (63.5 g/mol) = 0.3606 mol
Moles of e⁻ passed = (0.3606 mol) × 2 = 0.7212 mol
Amount of electricity = (0.7212 mol) × (96485 C/mol) = 69580 C = 69580 A s
Time taken = [(69580 A s) / (7.1 A)] / (3600 s/h) = 2.72 h
====
OR:
(22.9 g Cu) × (1 mol Cu / 63.5 g Cu) × (2 mol e⁻ / 1 mol Cu) × (96485 A s / 1 mol e⁻) / (7.1 A) × (1 h / 3600 s)
= 2.72 h
Moles of Cu accumulated = (22.9 g) / (63.5 g/mol) = 0.3606 mol
Moles of e⁻ passed = (0.3606 mol) × 2 = 0.7212 mol
Amount of electricity = (0.7212 mol) × (96485 C/mol) = 69580 C = 69580 A s
Time taken = [(69580 A s) / (7.1 A)] / (3600 s/h) = 2.72 h
====
OR:
(22.9 g Cu) × (1 mol Cu / 63.5 g Cu) × (2 mol e⁻ / 1 mol Cu) × (96485 A s / 1 mol e⁻) / (7.1 A) × (1 h / 3600 s)
= 2.72 h
收錄日期: 2021-04-18 18:27:36
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