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Multiple Choice
Cu2+ is reduced to Cu(s) at an electrode. If a current of 1.25 A is passed for 72 hours, what mass of copper is deposited at the electrode? (MW of Cu: 63.55 g/mol)
A
91.5 g
B
55.8 g
C
83.1 g
D
110 g
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Verified step by step guidance
1
Identify the relevant electrochemical reaction: Cu²⁺ + 2e⁻ → Cu(s). This indicates that 2 moles of electrons are required to reduce 1 mole of Cu²⁺ to Cu(s).
Calculate the total charge (Q) passed through the electrode using the formula Q = I × t, where I is the current (1.25 A) and t is the time in seconds (72 hours converted to seconds).
Convert the total charge (Q) from coulombs to moles of electrons using Faraday's constant (approximately 96485 C/mol). This will give you the moles of electrons involved in the reaction.
Determine the moles of copper deposited using the stoichiometry of the reaction. Since 2 moles of electrons deposit 1 mole of copper, divide the moles of electrons by 2 to find the moles of copper.
Calculate the mass of copper deposited by multiplying the moles of copper by the molar mass of copper (63.55 g/mol).