Copper Plating Efficiency and an Inhibitor Below Specification
A workshop student copper-plates a steel coupon in an acidified CuSO4 bath. The coupon is weighed before and after plating: mass gain 0.250 g. The plating runs at 0.500 A for 30.0 min. The student calculates the expected copper mass from Faraday’s law and finds a discrepancy, suspecting a competing cathode reaction. In a separate corrosion test, the student immerses 2.0 cm2 iron coupons in 1.0 M HCl at 25 °C with a commercial inhibitor at 0, 5, 10, and 20 ppm. The volumes of H2 collected in the first 5.0 min are 25.0, 18.0, 10.0, and 7.5 mL, respectively. The workshop specification says the inhibitor must give at least 80% inhibition at the recommended dose of 10 ppm. The student must decide whether the plating data are acceptable and whether the inhibitor meets the specification.
1. Calculate the current efficiency for copper deposition and identify the most likely competing cathode reaction. Is the observed mass consistent with pure copper?
Read the worked answer
Charge passed: Q = I × t = 0.500 A × (30.0 × 60 s) = 900 C.
Moles of electrons: n(e−) = Q/F = 900 C / 96485 C mol−1 = 9.33 × 10−3 mol.
Cathode reaction: Cu2+ + 2e− → Cu, so expected n(Cu) = 9.33 × 10−3 / 2 = 4.66 × 10−3 mol.
Expected mass of Cu = 4.66 × 10−3 mol × 63.55 g mol−1 = 0.296 g.
Observed mass gain = 0.250 g, so current efficiency = (0.250 / 0.296) × 100 = 84.4%.
About 15.6% of the charge did not deposit copper. In an acidified CuSO4 bath the likely competing cathode reaction is hydrogen evolution: 2H+ + 2e− → H2. Reduction of dissolved oxygen is also possible but is usually smaller in a deaerated plating bath. The mass gain is consistent with the deposit being copper; the discrepancy is in the charge efficiency, not in the identity of the deposit, assuming the coupon was dried and weighed without oxide or drag-out contamination.
2. Using the H2 volumes, calculate percentage inhibition at each inhibitor concentration and decide whether 10 ppm meets the 80% specification. State one experimental reason the 20 ppm result might underestimate inhibition.
Read the worked answer
For a fixed time and coupon area, the corrosion rate is proportional to the H2 volume. Let V0 = 25.0 mL at 0 ppm.
Percentage inhibition = [(V0 − V) / V0] × 100.
At 5 ppm: [(25.0 − 18.0) / 25.0] × 100 = 28%. At 10 ppm: [(25.0 − 10.0) / 25.0] × 100 = 60%. At 20 ppm: [(25.0 − 7.5) / 25.0] × 100 = 70%.
At the recommended 10 ppm, inhibition is only 60%, well below the 80% specification. Even 20 ppm gives 70%, so simply doubling the dose does not reach the target in the tested range. The inhibitor as used does not meet the specification; the workshop should test higher concentrations, a different inhibitor, or a formulation/pretreatment change.
One reason the 20 ppm result might underestimate inhibition: the 5 min measurement may include an initial burst of corrosion before the inhibitor has adsorbed and reached steady state. A pre-adsorption period or longer exposure would give a fairer steady-state inhibition value. Other reasons include temperature drift, coupon surface preparation, or loss of H2 by dissolution/leakage in the collection setup.
3. If the target coating is 20.0 μm on a 10.0 cm2 coupon (copper density 8.96 g cm−3), calculate the plating time at the same efficiency and current. Does the observed 0.250 g deposit correspond to the target?
Read the worked answer
Target coating volume = thickness × area = (20.0 × 10−4 cm) × 10.0 cm2 = 0.0200 cm3.
Target copper mass = density × volume = 8.96 g cm−3 × 0.0200 cm3 = 0.179 g.
At the observed rate, 0.250 g is deposited in 30.0 min. Time for 0.179 g = (0.179 g / 0.250 g) × 30.0 min = 21.5 min.
So the target 20.0 μm coating would require about 21.5 min at the same current and efficiency.
The observed 0.250 g in 30.0 min corresponds to a thicker coating: thickness = mass / (density × area) = 0.250 g / (8.96 g cm−3 × 10.0 cm2) = 2.79 × 10−3 cm = 27.9 μm. The run overplated relative to the 20.0 μm target. The plating data are therefore not automatically acceptable for the coating specification; the time should be reduced to about 21–22 min, or the current adjusted, and the coating thickness verified.