WHAT QUANTITY OF SILVER IS DEPOSITED WHEN 96500C OF ELECTRICITY IS PASSED THROUGH A SOLUTION CONTAINING SILVER IONS

To determine the quantity of silver deposited when 96500C of electricity is passed through a solution containing silver ions, we need to consider Faraday’s law of electrolysis. According to this law, the amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the solution.

The formula for calculating the quantity of substance deposited during electrolysis is:

Quantity (in grams) = (Current (in amperes) × Time (in seconds) × Atomic mass) / Faraday constant

In this case, we are interested in the quantity of silver deposited. The atomic mass of silver (Ag) is 108 g/mol, and the Faraday constant (F) is 96500 C/mol. Therefore, we can calculate the quantity of silver deposited as follows:

Quantity of silver = (Current × Time × Atomic mass of silver) / Faraday constant

Quantity of silver = (96500C × Atomic mass of silver) / Faraday constant

Substituting the values, we get:

Quantity of silver = (96500C × 108g/mol) / 96500C/mol

Quantity of silver = 108g

Therefore, when 96500C of electricity is passed through a solution containing silver ions, a quantity of 108 grams of silver will be deposited.