Determine the amount of silver liberated at the cathode if a current of 0.500 A flows through an AgNO ₃ electrolyte for 1 hour. The atomic weight of silver is 107.9 g mol ⁻¹ .

Step-by-step solution

Using Faraday's first law of electrolysis, the mass of the substance deposited at the electrode is given by: m = ZIt where Z is the electrochemical equivalent, I is the current, and t is the time. Z = M nF For the deposition of silver from AgNO ₃ , the reaction is Ag ^+ + e^- Ag . Thus, n = 1 . Given: M = 107.9 g mol ⁻¹ I = 0.500 A t = 1 hour = 3600 s F = 96500 C mol ⁻¹ Substituting the values: m = 107.9 0.500 3600 1 96500 m = 107.9 1800 96500 m = 1942.2 96500 2.01 g