96.5 C of electricity is used to completely deposit silver metal (Ag) from an aqueous solution in which the silver ion is present as Ag⁺(aq) - VCE - SSCE Chemistry - Question 9 - 2004 - Paper 1

To determine the mass of silver and copper deposited through electrolysis, we can use Faraday's laws of electrolysis.

The mass of an element deposited is given by the formula: $m = \frac{Q}{F} \times \frac{M}{n}$ where:

• $m$ is the mass of the element deposited,
• $Q$ is the total charge (in coulombs),
• $F$ is Faraday's constant (approximately 96500 C/mol),
• $M$ is the molar mass of the element, and
• $n$ is the number of electrons transferred per atom of the element.

For silver (Ag), the molar mass is approximately 107.87 g/mol, and it requires 1 electron to deposit 1 mole of silver. Therefore, using 96.5 C of electricity: $m_{Ag} = \frac{96.5}{96500} \times \frac{107.87}{1}$ For copper (Cu), the molar mass is about 63.55 g/mol, and it requires 2 electrons to deposit 1 mole of copper. Thus, for copper with the same charge: $m_{Cu} = \frac{96.5}{96500} \times \frac{63.55}{2}$

Comparing both, we can show that the mass of silver deposited is half that of copper deposited.