This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Alright ngwaemeldaayeng — let's do this. Here's the solution to your problem: Given: Current (I) = 2 A Time (t) = 10 minutes Atomic mass of Copper (Cu) = 64 g/mol Faraday's constant (F) = 96,500 C/mol e^- Part a) Calculate the quantity of electricity used: Step 1: Convert time from minutes to seconds. t = 10 min × 60 s/min t = 600 s Step 2: Calculate the quantity of electricity in Coulombs. The formula for quantity of electricity is Q = I × t. Q = 2 A × 600 s Q = 1200 C Step 3: Calculate the quantity of electricity in Faradays. We know that 1 Faraday (F) = 96,500 C. Quantity in Faradays = Quantity in Coulombs96,500 C/F Quantity in Faradays = 1200 C96,500 C/F Quantity in Faradays ≈ 0.0124 F Part b) Calculate the mass of copper deposited at the cathode: Step 4: Write the half-reaction for copper deposition and determine the moles of electrons required. The copper ions in CuSO_4 are Cu^2+. Cu^2+(aq) + 2e^- Cu(s) From the reaction, 2 moles of electrons are required to deposit 1 mole of copper. Step 5: Use Faraday's laws to calculate the mass of copper deposited. 2 moles of electrons correspond to 2 × 96,500 C or 2 Faradays. 1 mole of copper has a mass of 64 g. So, 2 Faradays deposit 64 g of copper. Using the quantity of electricity in Faradays from Step 3: Mass of Cu = ( Quantity in Faradays ) × Molar mass of CuMoles of electrons Mass of Cu = 0.0124 F × 64 g/mol2 F/mol Mass of Cu = 0.0124 × 32 g Mass of Cu ≈ 0.3968 g Send me the next one 📸