Calculate the mass of copper deposited by the electrolysis of copper (II) sulphate solution placed in series with a silver nitrate solution, when 0.108 g of silver is being deposited
The first step is to write the balanced equation for the electrolysis of copper (II) sulphate:
CuSO4(aq) + 2H2O(l) → Cu(s) + O2(g) + 4H+(aq) + SO42-(aq)
This equation tells us that for every mole of electrons transferred, one mole of copper is deposited at the cathode.
The second step is to use Faraday's law of electrolysis to relate the amount of silver deposited to the amount of electrons transferred:
Q = It
Where Q is the charge (in coulombs), I is the current (in amperes), and t is the time (in seconds).
Since one mole of electrons equals one Faraday of charge (96485 C), we can rearrange the equation to solve for the number of moles of electrons transferred:
n = Q / F = It / F
Where n is the number of moles of electrons transferred.
The third step is to use the number of moles of electrons transferred to calculate the mass of copper deposited. Since the molar mass of copper is 63.55 g/mol, we can use the following equation:
m = n x M
Where m is the mass of copper deposited (in grams), n is the number of moles of electrons transferred, and M is the molar mass of copper.
Putting it all together, we have:
n = (0.108 g / 107.87 g/mol) / 1 = 0.001001 mol e-
m = 0.001001 mol x 63.55 g/mol = 0.0638 g
Therefore, the mass of copper deposited is 0.0638 g.