Using Faraday's law, how many grams of copper deposit when 96485 C of charge pass through?

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Prepare for the NANTeL Chemistry Certification and Engineering Fundamentals Test with multiple choice questions, detailed explanations, and key insights to boost your understanding and confidence. Ace your exam!

Multiple Choice

Using Faraday's law, how many grams of copper deposit when 96485 C of charge pass through?

Explanation:
Faraday's law links the charge that passes through the system to the mass of metal deposited, using the relation m = Q M / (n F), where M is the metal's molar mass, n is the number of electrons needed per atom, and F is the Faraday constant. For copper deposition, Cu2+ gains 2 electrons to become Cu, so n = 2. One Faraday (96485 C) corresponds to 1 mole of electrons, so the moles of copper deposited are (1 mole of electrons) / 2 = 0.5 mole. Using copper's molar mass of about 63.55 g/mol, the mass deposited is 0.5 × 63.55 ≈ 31.8 g.

Faraday's law links the charge that passes through the system to the mass of metal deposited, using the relation m = Q M / (n F), where M is the metal's molar mass, n is the number of electrons needed per atom, and F is the Faraday constant. For copper deposition, Cu2+ gains 2 electrons to become Cu, so n = 2. One Faraday (96485 C) corresponds to 1 mole of electrons, so the moles of copper deposited are (1 mole of electrons) / 2 = 0.5 mole. Using copper's molar mass of about 63.55 g/mol, the mass deposited is 0.5 × 63.55 ≈ 31.8 g.

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