Question: How many moles of electrons are needed to reduce 1 mole of Ag⁺ to Ag? (Ag⁺ + e⁻ → Ag) Solution: From the reaction, it’s clear that 1 mole of Ag⁺ ions requires 1 mole of electrons for reduction.
- Question: What is the charge in coulombs required to reduce 3 moles of Cu²⁺ to Cu? (Cu²⁺ + 2e⁻ → Cu) Solution: From the reaction, 2 moles of electrons are needed for 1 mole of Cu²⁺ ions. So, for 3 moles of Cu²⁺, 3 x 2 = 6 moles of electrons are needed. Using Faraday’s constant (1 mole of electrons = 96485 C), 6 moles of electrons = 6 x 96485 = 578910 C.
- Question: How many moles of Al³⁺ can be reduced to Al by a current of 2 A flowing for 3 hours? (Al³⁺ + 3e⁻ → Al) Solution: Total charge = current x time = 2 A x 3 x 3600 s = 21600 C. Using Faraday’s constant, this corresponds to 21600/96485 ≈ 0.224 moles of electrons. As 3 moles of electrons are needed to reduce 1 mole of Al³⁺, we can reduce 0.224/3 = 0.075 moles of Al³⁺.
- Question: How long in seconds would it take to deposit 0.5 moles of Au from Au³⁺ by a current of 1 A? (Au³⁺ + 3e⁻ → Au) Solution: 1.5 moles of electrons are needed to deposit 0.5 moles of Au. The charge associated with 1.5 moles of electrons is 1.5 x 96485 = 144728 C. So, time = total charge / current = 144728 / 1 = 144728 s.
- Question: What current in amperes is needed to deposit 2 moles of Zn from Zn²⁺ in 1 hour? (Zn²⁺ + 2e⁻ → Zn) Solution: 4 moles of electrons are needed to deposit 2 moles of Zn. The charge associated with 4 moles of electrons is 4 x 96485 = 385940 C. So, current = total charge / time = 385940 / 3600 = 107.2 A.
- Question: How many moles of Pb²⁺ can be reduced to Pb by a charge of 50000 C? (Pb²⁺ + 2e⁻ → Pb) Solution: The number of moles of electrons corresponding to 50000 C is 50000/96485 ≈ 0.518 moles. As 2 moles of electrons reduce 1 mole of Pb²⁺, we can reduce 0.518/2 = 0.259 moles of Pb²⁺.
- Question: How much time in hours would it take to deposit 1 mole of Ag from Ag⁺ by a current of 2 A? (Ag⁺ + e⁻ → Ag) Solution: The charge associated with 1 mole of electrons is 1 x 96485 = 96485 C. So, time = total charge / current = 96485 / 2 = 48242.5 s = 13.4 hours.
- Question: What current in amperes is needed to deposit 0.5 moles of Cu from Cu²⁺ in 2 hours? (Cu²⁺ + 2e⁻ → Cu) Solution: 1 mole of electrons are needed to deposit 0.5 moles of Cu. The charge associated with 1 mole of electrons is 1 x 96485 = 96485 C. So, current = total charge / time = 96485 / (2 x 3600) = 13.4 A.
- Question: How many moles of Al³⁺ can be reduced to Al by a current of 3 A flowing for 4 hours? (Al³⁺ + 3e⁻ → Al) Solution: Total charge = current x time = 3 A x 4 x 3600 s = 43200 C. Using Faraday’s constant, this corresponds to 43200/96485 ≈ 0.448 moles of electrons. As 3 moles of electrons are needed to reduce 1 mole of Al³⁺, we can reduce 0.448/3 = 0.149 moles of Al³⁺.
Question: How long in seconds would it take to deposit 1 mole of Au from Au³⁺ by a current of 2 A? (Au³⁺ + 3e⁻ → Au) Solution: 3 moles of electrons are needed to deposit 1 mole of Au. The charge associated with 3 moles of electrons is 3 x 96485 = 289455 C. So, time = total charge / current = 289455 / 2 = 144728 s.
