Two cells are connected in series. One contain AlCl3, and the other contains AgNO3 as the electrolytes .what mass of Ag is deposited when 18g of Al is deposited at cathode?
96,485 coulombs will deposit 27/3 = 9 grams Al. You have 18 g; therefore,
96,485 coulombs x (18 g/9 g) = 96,485 x 2 = 192,970 coulombs.
96,485 coulombs will deposit 107.9 g Ag so how much Ag will be deosited with 192,970 coulombs? That's 107.9 g Ag x (192,970/96,495) = ?
To calculate the mass of Ag deposited when 18g of Al is deposited, we need to use the concept of Faraday's laws of electrolysis.
1. Write down the balanced chemical equation for the reaction at the cathode:
3Ag⁺ + 3e⁻ → 3Ag (reduction)
2. Determine the molar mass of Ag:
The molar mass of Ag is 107.87 g/mol.
3. Determine the charge (in coulombs) required to deposit 18g of Al:
The charge required can be determined using Faraday's laws:
Faraday's constant (F) = 96,485 C/mol (charge of 1 mole of electrons)
The molar mass of Al is 26.98 g/mol.
Therefore, the number of moles of Al is given by:
moles of Al = mass of Al / molar mass of Al
= 18g / 26.98 g/mol
= 0.668 mol
The charge required can be determined by multiplying the number of moles of Al by the charge of one mole of electrons:
charge = moles of Al × F
= 0.668 mol × 96,485 C/mol
= 64,535 C
4. Determine the number of moles of Ag deposited at the cathode:
For every mole of electrons transferred (charging of the cathode), three moles of Ag are deposited.
moles of Ag = (charge ÷ Faraday's constant) × (1 mole of Ag / 3 moles of electrons)
= (64,535 C ÷ 96,485 C/mol) × (1 mole of Ag / 3 moles of electrons)
= 0.669 mol
5. Calculate the mass of Ag deposited at the cathode:
mass of Ag = moles of Ag × molar mass of Ag
= 0.669 mol × 107.87 g/mol
= 72.14 g
Therefore, approximately 72.14 grams of Ag will be deposited at the cathode.
To find out the mass of Ag deposited when 18g of Al is deposited at the cathode, we need to consider the stoichiometry and the Faraday's law of electrolysis.
First, let's calculate the number of moles of Al deposited at the cathode. We know that the molar mass of aluminum (Al) is approximately 27 g/mol. Therefore, the number of moles of Al deposited can be calculated as follows:
Moles of Al = Mass of Al ÷ Molar mass of Al
Moles of Al = 18g ÷ 27 g/mol
Moles of Al = 0.67 mol
Now, we need to determine the stoichiometry of the electrolysis reaction. Looking at the balanced equation:
2 Al³⁺ + 6 e⁻ → 2 Al
3 Ag⁺ + 3 e⁻ → 3 Ag
From the stoichiometry, we can see that 6 moles of electrons are required to deposit 2 moles of Al and 3 moles of Ag.
Since 2 moles of Al requires 6 moles of electrons, we can say that 0.67 moles of Al will require:
Electrons = Moles of Al × (6 moles of electrons ÷ 2 moles of Al)
Electrons = 0.67 mol × (6 ÷ 2)
Electrons = 2.01 mol
Now, according to Faraday's law of electrolysis, we can calculate the charge (in Coulombs) involved:
Charge (in Coulombs) = Electrons × Faraday's constant
Charge (in Coulombs) = 2.01 mol × 96485 C/mol
Charge (in Coulombs) = 194,197 C
Since Ag⁺ is reduced at the cathode to form Ag, we can determine the number of moles of Ag deposited using the same stoichiometry as before:
Moles of Ag = Charge (in Coulombs) ÷ (Faraday's constant × Number of electrons involved in the reduction of Ag⁺)
Moles of Ag = 194,197 C ÷ (96485 C/mol × 3 moles of electrons/1 mole of Ag⁺)
Moles of Ag = 0.067 mol
Finally, we can calculate the mass of Ag deposited using the molar mass of silver (Ag), which is approximately 107.87 g/mol:
Mass of Ag = Moles of Ag × Molar mass of Ag
Mass of Ag = 0.067 mol × 107.87 g/mol
Mass of Ag = 7.22 g
Therefore, approximately 7.22 grams of Ag will be deposited when 18 grams of Al is deposited at the cathode.