To determine the time required for a current of 0.3A to deposit 0.8g of copper, we need to calculate the amount of charge needed to deposit this amount of copper.
The charge (Q) can be calculated using the equation Q = I * t, where Q is the charge, I is the current, and t is the time in seconds.
1. Determine the amount of charge needed to deposit 0.8g of copper.
The molar mass of copper (Cu) is 63.55 g/mol.
0.8g of copper = 0.8g / (63.55 g/mol) ≈ 0.0126 mol of copper
To deposit 1 mole of copper, 2 moles of electrons are needed.
Therefore, to deposit 0.0126 mol of copper, (0.0126 mol) * (2 mol e⁻ / 1 mol of Cu) ≈ 0.0252 mol of electrons are needed.
2. Calculate the charge required to deposit the electrons.
The charge of one electron is approximately 1.6 * 10^-19 C (Coulombs).
The total charge required can be calculated as (0.0252 mol) * (6.023*10^23 electrons/mol) * (1.6 * 10^-19 C/electron) ≈ 2.42 C
3. Determine the time required.
Since Q = I * t, rearranging the equation gives t = Q / I
t = 2.42 C / 0.3A ≈ 8.07 seconds
Therefore, a current of 0.3A would need to pass through a solution of copper (II) sulfate for approximately 8.07 seconds to deposit 0.8g of copper.
How long does a current of 0•3A need to pass through a solution of copper (ii) sulphate to deposit 0•8g of copper
1 answer
1 answer