What mass of silver can be deposited by the passage of a constant current of 5.00 A through a AgNO₃(aq)solution for 2.00 hours?

Question

Quizplus · Accepted Answer

The mass of silver deposited can be calculated using Faraday's laws of electrolysis. The amount of substance deposited (in moles) is directly proportional to the quantity of electricity (charge) that flows through the cell. The charge (Q) can be calculated as current (I) multiplied by time (t), where I = 5.00 A and t = 2.00 hours (7200 seconds, since 1 hour = 3600 seconds). So, Q = 5.00 A * 7200 s = 36000 C. The molar mass of silver (Ag) is 107.8682 g/mol, and its valency is 1, meaning 1 mole of electrons (or 96500 C) will deposit 1 mole of Ag. Therefore, the mass (m) of Ag deposited can be calculated as: m = (Q / 96500) * Molar mass of Ag = (36000 / 96500) * 107.8682 g/mol ≈ 40.3 g.

What Mass of Silver Can Be Deposited by the Passage