Two cells are connected in series. One contains AlCl3 and the other contains AgNo3 as the electrolytes. What mass of silver is deposited when 18g of Al is deposited at cathode?
To determine the mass of silver deposited when 18g of Al is deposited at the cathode, we need to use Faraday's law of electrolysis.
1. First, let's calculate the moles of Al that is deposited at the cathode. The molar mass of Al is 27 g/mol, and we have 18g of Al, so:
moles of Al = 18g / 27 g/mol = 0.67 mol of Al
2. Since the cells are connected in series, we know that the same number of moles of Al will be deposited at the cathode as the number of moles of Ag deposited at the anode.
3. The balanced chemical equation for the reaction at the cathode is:
Al^3+ + 3e^- -> Al
4. From the balanced equation, we can see that 3 moles of electrons are required to deposit 1 mole of Al. Therefore, the number of electrons involved in the process is 0.67 mol x 3 mol e^-/1 mol Al = 2 mol of electrons.
5. Each mole of electron corresponds to a charge of 1 Faraday (F). Therefore, the total charge passed through the circuit can be calculated by multiplying the number of moles of electrons by the Faraday constant (F = 96,485 C/mol):
Total charge = 2 mol e^- x 96,485 C/mol = 192,970 C
6. According to Faraday's law, the mass of the substance deposited at the electrode can be calculated using the equation:
mass = (Total charge x molar mass of the substance) / (Faraday constant)
7. The molar mass of silver (Ag) is 107.87 g/mol. Plugging the values into the equation, we get:
mass of Ag = (192,970 C x 107.87 g/mol) / 96,485 C/mol = 214.85 g of Ag
Therefore, the mass of silver deposited when 18g of Al is deposited at the cathode is approximately 214.85 grams.