Zinc Coating in a 45-Minute Slot
A small plating cell is used to zinc-coat a steel plate of exposed area 0.020 m2. The target coating thickness is 15 micrometres. The bath runs at 2.5 A and has a current efficiency of 85%. Zinc density is 7140 kg/m3, molar mass 65.4 g/mol, and the deposition reaction is Zn2+ + 2e- -> Zn. Faraday's constant is 96500 C/mol. The workshop has a 45-minute slot for plating, excluding 10 minutes for rinsing and setup. The bath supplier warns that the current density should not exceed 150 A/m2, otherwise the deposit becomes rough and poorly adherent. The trainee wants to know whether the existing current is enough, and if not, whether raising the current is a safe solution. The plate is flat and both sides are masked except one face of area 0.020 m2.
1. Calculate the mass of zinc required and the theoretical charge. Then find the actual plating time at 2.5 A and 85% efficiency. Does it fit the 45-minute slot?
Read the worked answer
Coating volume = area x thickness = 0.020 m2 x 15 x 10^-6 m = 3.0 x 10^-7 m3. Mass = density x volume = 7140 x 3.0 x 10^-7 = 2.142 x 10^-3 kg = 2.142 g. Moles of Zn = 2.142 / 65.4 = 0.03275 mol.
The reaction needs 2 mol electrons per mol Zn, so theoretical charge: Q = n F x moles = 2 x 96500 x 0.03275 = 6320 C.
At 85% efficiency, actual charge = 6320 / 0.85 = 7435 C. Time = Q / I = 7435 / 2.5 = 2974 s = 49.6 min.
This exceeds the 45-minute slot. At 2.5 A the existing current is not enough.
2. Can raising the current solve the problem within 45 minutes? Calculate the maximum allowed current from the current-density limit and the corresponding time.
Read the worked answer
The current-density limit is 150 A/m2 on the exposed area of 0.020 m2: maximum current = 150 x 0.020 = 3.0 A.
At 3.0 A, assuming the same 85% efficiency: time = 7435 / 3.0 = 2478 s = 41.3 min.
This is inside the 45-minute slot, with about 3.7 minutes of margin. So raising the current to 3.0 A is arithmetically sufficient and stays exactly at the supplier's current-density limit. It is acceptable only if the efficiency remains at 85%; if efficiency falls at the higher current, the time will be longer and the margin may disappear.
3. Explain why setting the bath to 3.5 A would be a poor judgement even though it would shorten the calculated time to about 35 minutes.
Read the worked answer
At 3.5 A the current density is 3.5 / 0.020 = 175 A/m2, which is above the supplier's 150 A/m2 limit. The deposit may become rough, porous, or poorly adherent. A rough deposit can also trap bath chemicals and fail a corrosion or adhesion check even if the average thickness is correct.
In addition, higher current can reduce current efficiency because more charge goes into side reactions such as hydrogen evolution. That would increase the required time and could cause hydrogen embrittlement or pitting. The correct judgement is to use 3.0 A or less, or to increase the exposed area or reduce the target thickness. The arithmetic alone is not sufficient; the current-density constraint must be respected.