How long does a current of 0•3A need to pass through a solution of copper (ii) sulphate to deposit 0•8g of copper
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.
To determine the time required for a current of 0.3A to deposit 0.8g of copper from a solution of copper (II) sulfate, we can use Faraday's law of electrolysis.
1. Find the molar mass of copper (Cu):
The molar mass of copper is approximately 63.55 g/mol.
2. Calculate the number of moles of copper deposited using the molar mass:
Moles of copper = Mass of copper / Molar mass
Moles of copper = 0.8g / 63.55 g/mol ≈ 0.0126 mol
3. Determine the number of electrons required to deposit one mole of copper:
According to the balanced equation for the copper deposition reaction, 2 moles of electrons are required to produce 1 mole of copper.
Therefore, 1 mole of copper requires 2 moles of electrons.
4. Calculate the total number of electrons required to deposit 0.0126 mol of copper:
Number of electrons = Moles of copper × 2
Number of electrons = 0.0126 mol × 2 ≈ 0.0252 mol e^-
5. Use Faraday's constant to convert the number of electrons to coulombs:
Faraday's constant (F) is approximately 96485 C/mol e^-.
Charge (Coulombs) = Number of electrons × Faraday's constant
Charge (Coulombs) = 0.0252 mol e^- × 96485 C/mol e^- ≈ 2.43 C
6. Apply Ohm's law to find the time (t) required for the specified current (I):
Ohm's Law: I = Q / t, where I is the current, Q is the charge, and t is the time.
Rearranging the equation, t = Q / I
t = 2.43 C / 0.3 A ≈ 8.1 seconds
Therefore, a current of 0.3A would need approximately 8.1 seconds to pass through the solution of copper (II) sulfate to deposit 0.8g of copper.