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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Step 1: Convert time from hours to seconds. t = 2 hours × 3600 seconds1 hour = 7200 s Step 2: Calculate the total charge (Q) passed using the formula Q = I × t. Q = 3.0 A × 7200 s = 21600 C Step 3: Write the half-reaction for the deposition of aluminium. Aluminium ions are Al^3+. Al^3+ + 3e^- Al This reaction shows that 3 moles of electrons are required to deposit 1 mole of aluminium. Step 4: Calculate the moles of electrons passed. One Faraday (F) is the charge of one mole of electrons, given as 96500 C. Moles of e^- = (Q)/(F) = 21600 C96500 C/mol ≈ 0.22383 mol e^- Step 5: Calculate the moles of aluminium deposited using the stoichiometry from the half-reaction. Moles of Al = Moles of e^-3 = 0.22383 mol3 ≈ 0.07461 mol Al Step 6: Calculate the mass of aluminium deposited using its molar mass (Al = 27 g/mol). Mass of Al = Moles of Al × Molar mass of Al Mass of Al = 0.07461 mol × 27 g/mol ≈ 2.0145 g The calculated mass is approximately 2.01 g, which is closest to 2.02 g. The final answer is D 2.02 g. 6. In electrolytic purification processes, the impure metal to be purified is used as the anode. At the anode, the impure metal is oxidized to form metal ions, which then migrate to the cathode to be deposited as pure metal. The final answer is A anode. 7. One faraday is equal to the charge of one mole of electrons. Numerically, this value is approximately 96500 coulombs. The final answer is B 96 500 coulombs.