Nickel Electroplating: Current, Efficiency and Coating Quality
A small workshop electroplates steel spoons with nickel. The electrolyte is nickel(II) sulphate solution, the anodes are pure nickel, and each spoon is polished, degreased and weighed before plating. In a trial, one spoon is plated for 45.0 minutes at a constant current of 0.500 A, with the electrolyte at 50 °C. Its mass increases by 0.372 g. A black sludge forms at the anode after several days. The customer's specification requires a coating of at least 0.400 g per spoon. The workshop foreman suggests simply raising the current to 0.600 A for the same time; the rectifier can deliver up to 0.800 A. (1 F = 96 500 C mol⁻¹; Aᵣ Ni = 58.7.)
1. Calculate the theoretical mass of nickel that should have been deposited and the current efficiency. Show the electrode reaction.
Read the worked answer
Quantity of charge: Q = I × t = 0.500 A × (45.0 × 60 s) = 1350 C.
Moles of electrons: n(e⁻) = 1350 / 96 500 = 1.399 × 10⁻² mol.
The cathode reaction is Ni²⁺ + 2e⁻ → Ni, so 2 mol electrons deposit 1 mol Ni: n(Ni) = 1.399 × 10⁻² / 2 = 6.995 × 10⁻³ mol.
theoretical mass = 6.995 × 10⁻³ × 58.7 = 0.411 g (0.4106 g).
Current efficiency = (actual mass / theoretical mass) × 100 = (0.372 / 0.411) × 100 = 90.6%.
So about 9.4% of the charge did not produce deposited nickel. The specification needs 0.400 g, and the trial produced only 0.372 g, so the trial fails the specification even though the calculation shows the charge was theoretically sufficient.
2. Give two reasons why the actual mass is less than the theoretical mass, and explain one of them with an equation.
Read the worked answer
Hydrogen is evolved at the cathode. Water (or H⁺) is reduced in competition with Ni²⁺: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) (or 2H⁺ + 2e⁻ → H₂). This uses charge that the calculation assumed was all used for nickel.
Some current is lost through side reactions or electrical leakage - for example, reduction of dissolved oxygen, reduction of impurity metal ions in the bath, or current passing through the jig/contacts rather than the spoon.
Either reason lowers the mass deposited for a given charge without breaking Faraday's laws; the laws predict the maximum, not the actual, mass. The 90.6% efficiency is a measure of how much charge was wasted.
3. The foreman proposes raising the current to 0.600 A for the same 45.0 minutes. Evaluate this proposal and recommend a better change.
Read the worked answer
At 0.600 A for 45.0 min, Q = 0.600 × 2700 = 1620 C.
n(e⁻) = 1620 / 96 500 = 1.679 × 10⁻² mol; n(Ni) = 8.394 × 10⁻³ mol; theoretical mass = 0.493 g.
At the observed 90.6% efficiency, expected deposit = 0.493 × 0.906 = 0.447 g, which is above the 0.400 g specification. So the arithmetic says the proposal could work.
However, raising the current density can make the deposit rough, powdery or poorly adherent, and it usually increases hydrogen evolution, so the efficiency may fall. A rough coating may not meet the customer's quality requirement even if the mass is high. The safer recommendation is to extend the time at 0.500 A to about 48–49 minutes (needed theoretical mass = 0.400/0.906 = 0.442 g; time = 45 × 0.442/0.411 ≈ 48.4 min), or to improve the bath (concentration, temperature, agitation, anode bags) and then re-test the coating. Any change should be checked on a test coupon for thickness and adhesion, not just mass.