Final answer:
The number of copper atoms produced by the electrolysis of copper nitrate solution with a 0.6 amp current for 20 minutes is approximately 2.24 × 10²¹ atoms.
Step-by-step explanation:
The question concerns the electrolysis of copper nitrate solution, specifically the amount of copper (Cu) produced after an electric current of 0.6 amps flows for 20 minutes. By applying Faraday's laws of electrolysis, we calculate the number of moles of electrons transferred to reduce the Cu2+ ions into copper atoms. The charge (Q) circulated during electrolysis is calculated as follows:
- Q = Current (I) × Time (t)
- Q = 0.60 A × 20 min × 60 sec/min = 720 C
Using the Faraday constant (F), which is approximately 96500 C/mol, to find the moles of electrons transferred:
- Moles of electrons (ne) = Q / F
- ne = 720 C / 96500 C/mol ≈ 7.46 × 10-3 mol
Since two electrons reduce one Cu2+ ion to one Cu atom, moles of Cu produced are:
- Moles of Cu = ne / 2 ≈ 3.73 × 10-3 mol
Last, the number of copper atoms is calculated using Avogadro's number (NA = 6.02 × 1023 atoms/mol):
- Number of Cu atoms = Moles of Cu × NA ≈ 3.73 × 10-3 mol × 6.02 × 1023 atoms/mol ≈ 2.24 × 1021 atoms