Final answer:
To find the grams of silver deposited in the electrolysis, the number of moles of electrons corresponding to the total charge passed through the solution is calculated. This, in turn, gives the moles of silver, and multiplied by the molar mass of silver, provides the mass deposited. The answer, 85.33 grams, indicates the correct response is 'None of the above.'
Step-by-step explanation:
To calculate 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, we must first use Faraday's laws of electrolysis. The amount of substance (silver in this case) that is deposited or dissolved at an electrode is directly proportional to the amount of electricity that passes through the electrolyte. Here's how the calculation works:
- Determine the total charge (Q) passed through the electrolyte, using Q = I x t, where I is current in amperes and t is time in seconds. So, Q = 5.30 A x (4.0 h x 3600 s/h) = 76,320 C
- Next, calculate the number of moles of electrons using the charge and Faraday's constant (96,485 C/mol). Moles of electrons (n) = Q / (96,485 C/mol) = 76,320 C / 96,485 C/mol = 0.791 mol
- Since the cathode reaction for the electrolysis of AgNO₃ is Ag+ (aq) + e¯ → Ag(s), 1 mole of electrons will deposit 1 mole of silver. Therefore, 0.791 mol of electrons will deposit 0.791 mol of silver.
- Finally, calculate the mass of silver deposited using the molar mass of silver (107.87 g/mol). Mass = moles x molar mass = 0.791 mol x 107.87 g/mol = 85.33 g
Therefore, 85.33 grams of silver are deposited at the cathode, which means that the correct answer is None of the above.