When during electrolysis of a solution of AgNO3 9650 coulombs of charge pa s through the electroplating bath, the mass of silver deposited on the cathode will be

To determine the mass of silver deposited on the cathode during the electrolysis of a silver nitrate (AgNO3) solution when 9650 coulombs of charge passes through, we can use Faraday's laws of electrolysis. Let's break this down step by step.

Understanding the Electrolysis Process

During electrolysis, ions in the solution are reduced or oxidized at the electrodes. In the case of silver nitrate, silver ions (Ag⁺) are reduced at the cathode to form solid silver (Ag). The relevant half-reaction can be represented as:

Ag⁺ + e^- → Ag

Calculating the Amount of Silver Deposited

To find out how much silver is deposited, we need to know the relationship between the charge passed, the number of moles of electrons involved, and the molar mass of silver. Here’s how we can do that:

• Faraday's Constant: The charge required to deposit one mole of a monovalent ion (like Ag⁺) is approximately 96500 coulombs. This is known as Faraday's constant.
• Charge Passed: In this case, we have 9650 coulombs of charge.
• Moles of Electrons: The number of moles of electrons can be calculated using the formula:

Number of moles of electrons = Charge (C) / Faraday's constant (C/mol)

Substituting the values:

Number of moles of electrons = 9650 C / 96500 C/mol = 0.1 mol

Relating Moles of Electrons to Moles of Silver

Since the reduction of one mole of silver ions requires one mole of electrons, the moles of silver deposited will also be 0.1 mol.

Calculating the Mass of Silver

Next, we need to convert the moles of silver to mass. The molar mass of silver (Ag) is approximately 107.87 g/mol. We can use the following formula:

Mass (g) = Moles × Molar Mass

Substituting the values:

Mass of silver = 0.1 mol × 107.87 g/mol = 10.787 g

Final Result

Thus, when 9650 coulombs of charge passes through the electroplating bath, approximately 10.79 grams of silver will be deposited on the cathode. This calculation illustrates the direct relationship between charge, moles of electrons, and the mass of metal deposited during electrolysis.