Al³⁺ is reduced to Al(s)at an electrode surface.If a current of 25 amperes is maintained for 8.0 hours,what mass of aluminum is deposited on the electrode? Assume 100% current efficiency.

Question

Quizplus · Accepted Answer

Using Faraday's laws of electrolysis, the mass of aluminum deposited can be calculated. The charge passed is 25 A × 8.0 h × 3600 s/h = 720,000 C. The number of moles of electrons is 720,000 C / 96,485 C/mol = 7.46 mol. Since 3 moles of electrons are needed to reduce 1 mole of Al³⁺, the moles of Al deposited is 7.46 mol / 3 = 2.49 mol. The mass of Al is 2.49 mol × 26.98 g/mol = 67 g.

Al3+ Is Reduced to Al(s)at an Electrode Surface

Al³⁺ Is Reduced to Al(s)at an Electrode Surface