Nickel Plating That Fails Thickness by One Micrometre
A small workshop nickel-plates steel components to a specified minimum coating thickness of 25 µm. The bath is a Watts-type nickel bath with nickel anodes, and the rectifier is set at 0.85 A. Each component has 30 cm² of platable surface. A batch is run for 45 minutes, and the current efficiency is measured at 92 %; the remaining charge goes to hydrogen evolution at the cathode. Useful data: electrochemical equivalent of nickel = 1.095 g per ampere-hour; density of nickel = 8.9 g/cm³; atomic mass of Ni = 58.7; valency = 2; Faraday constant = 96 500 C/mol. The inspector rejects the batch for under-thickness. The operator proposes to 'just turn the current up to 1.2 A', since the rectifier can deliver it and the plating time would then stay at 45 minutes. The workshop owner wants the batch passed without buying new equipment.
1. Calculate the coating thickness actually deposited and decide whether the inspector's rejection is justified.
Read the worked answer
Route 1 - electrochemical equivalent. m = ECE × I × t × efficiency = 1.095 g/Ah × 0.85 A × 0.75 h × 0.92 = 1.095 × 0.6375 × 0.92 = 0.642 g
Route 2 - Faraday's laws (cross-check). Q = It = 0.85 × 2700 = 2295 C moles of electrons = 2295/96 500 = 0.02378 mol Ni²⁺ + 2e⁻ → Ni, so moles Ni = 0.01189 mol mass = 0.01189 × 58.7 = 0.698 g; × 0.92 = 0.642 g ✔
Thickness. Volume = 0.642/8.9 = 0.0721 cm³ Thickness = volume/area = 0.0721/30 = 2.40 × 10⁻³ cm = 24.0 µm
Judgement. The specification is 25 µm minimum; 24.0 µm is about 4 % short, so the rejection is technically correct, but the batch is marginal rather than grossly defective. The shortfall is fully explained by the arithmetic, so no fault (contact resistance, anode passivation) needs to be assumed.
2. At the existing 0.85 A, how long should the batch have run? Show the scaling argument rather than redoing the full Faraday calculation.
Read the worked answer
Thickness is directly proportional to time when current and efficiency are unchanged:
t₂ = t₁ × (target/actual) = 45 min × (25/24.0) = 46.9 min
So a run time of 47 minutes (say 48 minutes for a safety margin) would have met the specification at the existing current with no change to the bath or rectifier.
Sanity check. 47 min gives mass = 0.642 × 47/45 = 0.671 g → volume 0.0754 cm³ → thickness 25.1 µm ✔
This is the cheapest compliant fix and it does not change the current density, so deposit quality is unaffected.
3. Evaluate the operator's proposal to raise the current to 1.2 A for 45 minutes. Will it pass, and what plating-quality risks does it introduce?
Read the worked answer
Arithmetic first. Predicted thickness if efficiency stayed at 92 %: 24.0 × (1.2/0.85) = 33.9 µm - comfortably above 25 µm, so on paper it works.
Why the prediction is optimistic.
- Current density rises from 0.85/30 = 28 mA/cm² to 40 mA/cm². That is still inside the usual Watts-bath window but at its upper end, where the hydrogen-evolution side reaction takes a larger share of the charge; the true efficiency will fall below 92 %, so the gain is less than proportional.
- Higher current density promotes rough, dull, nodular or burnt deposits, internal stress and poor adhesion - a component can pass a thickness check and still fail a bend or salt-spray test.
- Thickness distribution worsens: edges and corners (current-density 'highs') build up while recessed areas stay thin, so a nominal 34 µm average may still hide thin patches below 25 µm.
Judgement. Raising the current is not wrong in principle, but it trades a small, certain time penalty for a real quality risk. The defensible choice is to extend the time to 47–48 minutes (or raise the current modestly with agitation and re-measure efficiency on a coupon) rather than to change one variable and assume the efficiency is unchanged.