Copper plating a steel key: calculation and deposit quality
In a school electroplating project, a steel key of total surface area 18.0 cm2 is to be copper-plated from an acid copper sulfate bath (CuSO4 plus dilute H2SO4). The group uses a copper anode and a current of 0.500 A for 40.0 min. The cathode current efficiency is found to be 90%. Copper has M = 63.55 g/mol and density 8.96 g/cm3; the Faraday constant is 96 500 C/mol. The key is first degreased and pickled, but the group forgets to rinse one area, and the deposit there is patchy. After the run the deposit on the rest of the key is dark and powdery, especially near the edge closest to the anode. A more experienced student says the current was too high and that a graphite anode would have been cheaper. The teacher wants a quantitative check and a decision on the anode.
1. Calculate the mass of copper deposited and the average coating thickness. Is a 25 µm target met?
Read the worked answer
Q = It = 0.500 × 2400 = 1200 C. n(e-) = 1200/96 500 = 0.01244 mol; n(Cu) = 0.00622 mol. Mass at 100% = 0.00622 × 63.55 = 0.395 g. At 90% efficiency: 0.356 g. Volume = 0.356/8.96 = 0.0397 cm3. Thickness = 0.0397/18.0 = 0.00221 cm = 22.1 µm. The 25 µm target is not met; the coating is about 3 µm short, roughly 12% low. To reach 25 µm at the same efficiency would require about 45 min, or a higher current with better deposit control.
2. Diagnose the dark, powdery deposit and the patchy area. What changes would you recommend?
Read the worked answer
Average current density is 0.500/18 = 0.0278 A/cm2 (27.8 mA/cm2), not extreme for an acid sulfate bath, but the edge nearest the anode has a much higher local current density because of primary current distribution. That locally depletes Cu2+ and co-evolves H2, giving a dark, powdery, poorly adherent deposit.
The patchy area points to incomplete degreasing or rinsing, leaving an oily or passive spot. Remedies: improve cleaning and rinsing, shield or reposition the edges, lower the current or extend the time, stir or filter the bath, and control temperature and additives.
3. The student proposes replacing the copper anode with graphite to save money. Give a reasoned judgement.
Read the worked answer
Graphite would not dissolve; it would evolve O2 at the anode: 2H2O → O2 + 4H+ + 4e-. This acidifies the bath and depletes Cu2+ unless the bath is continually replenished, and the oxygen can oxidize bath additives.
A copper anode dissolves as Cu → Cu2+ + 2e-, maintaining the copper concentration and giving a stable deposit. The graphite 'saving' is false economy here; keep the copper anode and monitor Cu2+ and acid concentration.