# Sample problems electrolytic cells

1. Question: How many minutes will it take to deposit 2.15 grams of silver (Ag) from a solution of AgNO₃ using a current of 2.00 A? (Ag⁺ + e⁻ → Ag) Solution: First find the number of moles of Ag: 2.15 g / (107.87 g/mol) = 0.02 mol. This requires 0.02 mol of electrons, or 0.02 x 96485 C = 1929.7 C. Time required = charge / current = 1929.7 C / 2 A = 964.85 s ≈ 16.08 minutes.

2. Question: How many grams of aluminum (Al) can be deposited from a solution of AlCl₃ using a current of 3.00 A for 2.00 hours? (Al³⁺ + 3e⁻ → Al) Solution: The total charge is 3 A x 2 h x 3600 s/h = 21600 C. This is equal to 21600 C / 96485 C/mol = 0.224 mol of electrons. For every 3 mol of electrons, 1 mol of Al is deposited, so 0.224/3 = 0.075 mol of Al is deposited. This is 0.075 mol x 26.98 g/mol = 2.02 g Al.
3. Question: How much time is required to electroplate a silver spoon with 0.5 g of Ag using a current of 1 A? (Ag⁺ + e⁻ → Ag) Solution: The number of moles of Ag is 0.5 g / (107.87 g/mol) = 0.0046 mol. This requires 0.0046 mol of electrons, or 0.0046 x 96485 C = 445 C. Time required = charge / current = 445 C / 1 A = 445 s.
4. Question: What is the current required to electrolyze a solution of NaCl, producing 0.5 moles of Cl₂ in 30 minutes? (2Cl⁻ → Cl₂ + 2e⁻) Solution: The number of moles of electrons needed is 0.5 moles Cl₂ x 2 = 1 mole of electrons. The charge needed is 1 mol x 96485 C/mol = 96485 C. Current = charge / time = 96485 C / (30 x 60 s) = 53.6 A.
5. Question: How much time would be needed to deposit 5 g of Cu onto an electrode from a Cu²⁺ solution with a current of 2 A? (Cu²⁺ + 2e⁻ → Cu) Solution: The number of moles of Cu is 5 g / (63.55 g/mol) = 0.079 mol. This requires 0.079 x 2 = 0.158 mol of electrons, or 0.158 x 96485 C = 15245 C. Time required = charge / current = 15245 C / 2 A = 7623 s.
6. Question: How many grams of Na will be deposited if 2 Faradays of electricity is passed through a solution of Na⁺ ions? (Na⁺ + e⁻ → Na) Solution: One Faraday deposits 1 mole of a monovalent metal ion. Thus 2 Faradays will deposit 2 moles of Na. This is 2 mol x 22.99 g/mol = 45.98 g Na.
7. Question: How much charge is required to deposit 1.5 g of Mg from a Mg²⁺ solution? (Mg²⁺ + 2e⁻ → Mg) Solution: The number of moles of Mg is 1.5 g / (24.31 g/mol) = 0.0617 mol. This requires 0.0617 x 2 = 0.1234 mol of electrons, or 0.1234 x 96485 C = 11907 C.
8. Question: What current is required to deposit 3 g of Al from a Al³⁺ solution in 2 hours? (Al³⁺ + 3e⁻ → Al) Solution: The number of moles of Al is 3 g / (26.98 g/mol) = 0.111 mol. This requires 0.111 x 3 = 0.333 mol of electrons, or 0.333 x 96485 C = 32131 C. Current = charge / time = 32131 C / (2 x 3600 s) = 4.46 A.
9. Question: How much time is required to produce 22.4 L of Cl₂ gas at STP from a Cl⁻ solution with a current of 5 A? (2Cl⁻ → Cl₂ + 2e⁻) Solution: 22.4 L of gas at STP is 1 mol, and for every 1 mol of Cl₂, 2 mol of electrons is required. Thus, we need 2 x 96485 C = 192970 C. Time required = charge / current = 192970 C / 5 A = 38594 s.
10. Question: What current is needed to reduce 1 mol of H⁺ to H₂ in 30 minutes? (2H⁺ + 2e⁻ → H₂) Solution: The charge needed is 2 x 96485 C = 192970 C. Current = charge / time = 192970 C / (30 x 60 s) = 107.2 A.