Nickel plating batch fails thickness specification
A small workshop in Mandalay nickel-plates steel rivets in batches. Each batch has a total exposed surface area of 0.045 m², and the customer requires an average coating thickness of 25 µm. The tank uses a nickel anode and a NiSO₄/NiCl₂ electrolyte. For one batch the operator sets the current at 1.80 A and plates for 40.0 min. The current efficiency under these conditions is 88%. Data: density of nickel = 8.90 g cm⁻³; molar mass Ni = 58.7 g mol⁻¹; Faraday constant = 96 500 C mol⁻¹; Ni²⁺ + 2e⁻ → Ni. After plating, the rivets look bright, but the workshop has no thickness gauge. The customer will reject the batch if the average coating is below 25 µm. The operator suggests simply raising the current to 15.0 A for the same 40.0 min to make the coating thicker faster.
1. Calculate the mass of nickel needed for the specified coating and the mass actually deposited in 40.0 min. Hence calculate the average thickness obtained and decide whether the batch meets the 25 µm specification.
Read the worked answer
Required volume: 0.045 m² = 450 cm²; 25 µm = 2.5 × 10⁻³ cm.
V = 450 × 2.5 × 10⁻³ = 1.125 cm³.
m = ρV = 8.90 × 1.125 = 10.0 g Ni required.
Actual charge: Q = It = 1.80 × (40.0 × 60) = 4320 C.
mol e⁻ = 4320 / 96 500 = 0.04477 mol.
mol Ni = 0.04477 / 2 = 0.02238 mol.
Theoretical Ni mass = 0.02238 × 58.7 = 1.314 g.
At 88% efficiency: actual mass = 1.314 × 0.88 = 1.156 g.
Average thickness = m / (ρA) = 1.156 / (8.90 × 450) = 2.89 × 10⁻⁴ cm = 2.89 µm.
This is far below 25 µm, so the batch fails the specification.
2. The operator proposes raising the current to 15.0 A for 40.0 min. Evaluate whether this will meet the specification, and explain two reasons why simply increasing current may worsen coating quality.
Read the worked answer
At 15.0 A for 2400 s: Q = 15.0 × 2400 = 36 000 C.
mol e⁻ = 36 000 / 96 500 = 0.3731 mol; mol Ni = 0.1865 mol.
Theoretical mass = 0.1865 × 58.7 = 10.95 g.
At 88% efficiency: actual mass = 9.64 g.
Thickness = 9.64 / (8.90 × 450) = 2.41 × 10⁻³ cm = 24.1 µm.
This is still slightly below 25 µm, and the efficiency is likely to fall at the higher current density.
Simply raising current is risky because: (1) high current density can cause rapid hydrogen evolution, making the deposit rough, powdery or poorly adherent; (2) the local pH may rise and precipitate nickel hydroxide, and hydrogen can embrittle the steel. Better control of agitation, temperature and current density, or a longer plating time, is needed.
3. Calculate how long the existing 1.80 A process would need to run to reach 25 µm at 88% efficiency, and comment on the practical implication.
Read the worked answer
Required Ni mass = 10.0 g.
mol Ni = 10.0 / 58.7 = 0.1704 mol.
mol e⁻ = 2 × 0.1704 = 0.3408 mol.
Q = 0.3408 × 96 500 = 32 890 C.
With 88% efficiency: t = Q / (I × efficiency) = 32 890 / (1.80 × 0.88) = 20 760 s ≈ 5.77 h.
So the existing 40 min cycle is only about one-eighth of the required time. Practically, the workshop must either accept a much longer cycle, use several plating tanks, or increase current while improving agitation and electrolyte control to keep the deposit sound.