How many grams of silver are deposited at a platinum cathode in the electrolysis of AgNO₃ (aq) by 5.30 amps of electric current in 4.0 hours?

Question

Quizplus · Accepted Answer

The amount of silver deposited can be calculated using Faraday's laws of electrolysis. The molar mass of silver (Ag) is 107.8682 g/mol, and it has a valency of 1, meaning 1 mole of electrons is required to deposit 1 mole of silver. The charge (Q) passed through the electrolyte is given by Q = It, where I is the current in amperes and t is the time in seconds. Here, I = 5.30 A and t = 4.0 hours = 4 * 3600 s = 14400 s. So, Q = 5.30 * 14400 = 76320 C. The amount of silver deposited is given by the equation m = (Q * M) / (n * F), where m is the mass of the deposit, M is the molar mass of the depositing substance, n is the valency, and F is Faraday's constant (96485 C/mol). Substituting the values, we get m = (76320 * 107.8682) / (1 * 96485) ≈ 85.3 g.

How Many Grams of Silver Are Deposited at a Platinum