Plating a Steel Fitting from a Mixed Copper–Nickel Bath
A workshop plates a steel (iron) fitting in a bath that is 0.80 M in NiSO₄ and 0.020 M in CuSO₄, held at pH 3.0 and 25 °C. The bath volume is 250 mL and is used once. Standard reduction potentials: E°(Cu²⁺/Cu) = +0.34 V, E°(Ni²⁺/Ni) = −0.25 V, E°(Fe²⁺/Fe) = −0.44 V. A current of 2.50 A is passed for 45.0 min. The cathode current efficiency for metal deposition is measured as 96 %; the remaining charge goes into hydrogen evolution. The foreman wants to know what will actually deposit, whether the copper in the bath will run out, and whether the coating will protect the steel where it is scratched. F = 96 485 C mol⁻¹, Ar(Cu) = 63.55.
1. Decide which metal deposits first at the cathode and show, by calculation, the cathode potential at which nickel codeposition would begin.
Read the worked answer
Compute the actual (Nernst) potentials under the bath conditions:
E(Cu) = 0.34 + (0.0592/2)log(0.020) = 0.34 − 0.050 = +0.290 V E(Ni) = −0.25 + (0.0592/2)log(0.80) = −0.25 − 0.003 = −0.253 V
Copper is far more easily reduced, so copper plates first and the cathode sits near +0.29 V. Nickel can only begin when the cathode potential falls to −0.253 V, i.e. when E(Cu) has dropped to that value: 0.34 + 0.0296 log[Cu²⁺] = −0.253 → log[Cu²⁺] = −0.593/0.0296 = −20.0 → [Cu²⁺] ≈ 1 × 10⁻²⁰ M.
That is effectively zero: copper plates out quantitatively before nickel starts, so the deposit is layered - copper first, then nickel. Note also that at pH 3 the H⁺/H₂ couple is at −0.0592 × 3 = −0.178 V, which is above E(Ni); hydrogen evolution is thermodynamically favoured over nickel and only the high overpotential of H₂ on nickel allows nickel to deposit at all. That is exactly the 4 % loss recorded as current efficiency.
2. Calculate how much copper the 250 mL bath can supply and whether the full charge passed is available for copper deposition.
Read the worked answer
Total charge: Q = I t = 2.50 A × (45.0 × 60 s) = 6750 C. Moles of electrons = 6750/96 485 = 0.06996 mol.
But copper is a limiting reagent: n(Cu²⁺) = 0.250 L × 0.020 M = 5.00 × 10⁻³ mol, which requires 2 × 5.00 × 10⁻³ = 0.0100 mol of electrons = 965 C. That is only about 14 % of the charge passed, so copper runs out after roughly 965/2.50 ≈ 386 s (about 6.4 min).
Mass of copper available = 5.00 × 10⁻³ × 63.55 = 0.318 g; with 96 % efficiency, 0.305 g is deposited and about 0.013 g is lost to hydrogen evolution during the copper phase.
The remaining 5785 C is spent first on hydrogen and then on nickel plating (with a mixed efficiency). Recovering the nickel from the spent bath is worth considering: 0.80 × 0.250 = 0.200 mol Ni²⁺ is far in excess of what 45 minutes at 2.5 A could deposit.
3. Explain what happens at the cathode after the copper is exhausted, and advise the foreman on whether a nickel coating protects scratched steel. Compare with a zinc coating.
Read the worked answer
After copper is exhausted, the cathode potential falls: no Cu²⁺ remains to hold it at +0.29 V. Hydrogen evolution (E ≈ −0.18 V at pH 3) takes an increasing share of the current, and nickel begins to deposit once the potential reaches about −0.25 V. The practical consequences are (i) bubbling and pitting of the plate, (ii) a drop in current efficiency, and (iii) hydrogen embrittlement of the steel, which is a reason to bake the part afterwards.
Protection. Nickel (E° = −0.25 V) is more noble than iron (E° = −0.44 V). Where the coating is intact it is just a barrier, but a scratch makes the exposed iron the anode of a galvanic cell and it corrodes faster than it would unprotected; the nickel, as cathode, is protected. A zinc coating (E° = −0.76 V) is the opposite case: zinc is the anode, so it dissolves preferentially and protects the steel even at a scratch - sacrificial protection. If the part is to be used outdoors and damaged in service, zinc (or a zinc-rich primer) is the better engineering choice; nickel is chosen for appearance and wear resistance, not for sacrificial protection.