Please help!:( What mass of copper would be produced by the reduction of copper(II) ions during pas- sage of 1 ampere of current through a solution of copper(II) sulfate for 89 minutes?

coulombs = A X S

convert 89 mins to seconds

____ c (1 c/1 mol e-) = ___ mol e-
____ mol e- (1 mol/2 e- cu 2+) = mol cu
mol cu x MW cu = g cu

I got 5340. Is that right?

To determine the mass of copper produced during the reduction of copper(II) ions, we need to use the principles of Faraday's laws of electrolysis.

1. Calculate the total charge passed through the solution:
Faraday's law states that the mass of a substance produced during electrolysis is directly proportional to the amount of charge passed through the solution. The formula is:

Q = I * t

Where Q is the total charge in Coulombs (C), I is the current in Amperes (A), and t is the time in seconds (s). To convert minutes to seconds, multiply the given time by 60:

t = 89 minutes * 60 seconds/minute = 5340 seconds

Substituting the given values:

Q = 1 A * 5340 s = 5340 C

2. Calculate the number of moles of electrons passed through the solution:
Since copper(II) ions (Cu^2+) are being reduced, the reaction involves the gain of two electrons (2e^-) per copper(II) ion reduced.

1 mole of electrons = Avogadro's number (6.02 * 10^23) electrons

So, the number of moles of electrons can be calculated as:

Moles of electrons = Q / (2 * Faraday's constant)

where Faraday's constant (F) = 96,485 C/mol e^-

Substituting the values:

Moles of electrons = 5340 C / (2 * 96485 C/mol e^-) ≈ 0.0276 mol

3. Calculate the mass of copper produced:
The balanced equation for the reduction of copper(II) ions is:

Cu^2+ (aq) + 2e^- → Cu(s)

The stoichiometry of the reaction shows that 1 mole of copper(II) ions (Cu^2+) produces 1 mole of copper (Cu).

Therefore, the mass of copper produced can be calculated using the molar mass of copper:

Molar mass of copper = 63.55 g/mol

Mass of copper = Moles of electrons * Molar mass of copper

Mass of copper = 0.0276 mol * 63.55 g/mol ≈ 1.756 g

So, approximately 1.756 grams of copper would be produced by the reduction of copper(II) ions in 89 minutes with a current of 1 ampere.