Nickel Plating versus Galvanising for Steel Brackets in a Marine Yard
Steel brackets for a seaside installation must be protected against corrosion. The workshop can either nickel-plate the brackets or hot-dip galvanise them. Standard reduction potentials: E°(Ni²⁺/Ni) = −0.25 V, E°(Fe²⁺/Fe) = −0.44 V, E°(Zn²⁺/Zn) = −0.76 V.
A trial nickel plating is run on a bracket of total exposed area 0.0200 m² in 250.0 mL of 0.500 mol L⁻¹ NiSO4(aq), using a pure nickel anode and a constant current of 0.500 A for 60.0 min (F = 96 485 C mol⁻¹, M(Ni) = 58.69 g mol⁻¹). The bracket gains 0.482 g. The plater proposes to double the layer thickness by raising the current to 1.00 A for the same time, arguing that 'more current means more metal, so the job is twice as fast'.
The client also asks what will happen when a coated bracket is deliberately scratched through to the steel and left in salt spray. In trials, the scratch on a nickel-plated bracket rusts heavily within days, while the steel next to a scratch on a galvanised bracket stays bright for months.
1. Calculate the theoretical mass of nickel that 0.500 A for 60.0 min should deposit, determine the current efficiency, identify the most likely side reaction, and estimate how much the nickel-ion concentration of the bath fell.
Read the worked answer
Charge passed: Q = I·t = 0.500 A × 3600 s = 1800 C.
Moles of electrons: 1800 / 96 485 = 1.866×10⁻² mol e⁻.
Cathode reaction: Ni²⁺ + 2e⁻ → Ni, so n(Ni) = 1.866×10⁻² / 2 = 9.33×10⁻³ mol.
Theoretical mass: 9.33×10⁻³ × 58.69 = 0.547 g.
Actual: 0.482 g → n = 0.482 / 58.69 = 8.21×10⁻³ mol.
Current efficiency: 0.482 / 0.547 = 88.1% (or 8.21/9.33 = 88.0%).
The missing 1.12×10⁻³ mol of electrons went into a competing cathode process. In an aqueous sulfate bath the realistic competitor is hydrogen evolution: 2H2O + 2e⁻ → H2 + 2OH⁻ (equivalently 2H⁺ + 2e⁻ → H2), which produces 1.12×10⁻³ mol of H2 and slowly raises the local pH. This is why plating baths are buffered and why very high current densities give porous, hydrogen-contaminated deposits.
Bath depletion: the bath started with 0.250 L × 0.500 mol L⁻¹ = 0.125 mol Ni²⁺. After plating, 0.125 − 0.0082 = 0.1168 mol remains → 0.467 mol L⁻¹, a fall of about 6.6%. Small per run, but over many brackets the bath must be replenished and the anode dissolves to replace it (Ni → Ni²⁺ + 2e⁻ at the anode).
2. Using the standard potentials, explain why a scratch in the nickel coating makes the steel corrode faster, whereas a scratch in the zinc coating protects the steel. Support your answer with half-equations and E°cell values, and name the protection mechanism.
Read the worked answer
Nickel on steel - the coating is cathodic to iron. E°(Ni²⁺/Ni) = −0.25 V is higher than E°(Fe²⁺/Fe) = −0.44 V, so iron is the more easily oxidised metal. Once the coating is breached, salt water connects the two metals and a galvanic cell forms:
anode (Fe): Fe → Fe²⁺ + 2e⁻ E° = −0.44 V cathode (Ni): Ni²⁺ + 2e⁻ → Ni E° = −0.25 V
E°cell = (−0.25) − (−0.44) = +0.19 V, spontaneous. The small scratched area of iron becomes the anode and is consumed faster than it would be unprotected, because the large surrounding nickel area supplies a big cathodic current. This is the classic failure of a 'noble' coating: it protects only while it is continuous.
Zinc on steel - the coating is anodic to iron. E°(Zn²⁺/Zn) = −0.76 V is lower than that of iron, so zinc is oxidised preferentially:
anode (Zn): Zn → Zn²⁺ + 2e⁻ E° = −0.76 V cathode (Fe): 2H2O + O2 + 4e⁻ → 4OH⁻ (or H2 evolution)
E°cell = (−0.44) − (−0.76) = +0.32 V, and the zinc, not the steel, is consumed. The steel is held at a potential where it cannot oxidise; the mechanism is cathodic protection (sacrificial anode). Galvanising therefore protects even at a scratch, and the protection extends a short distance across bare steel.
Judgement: for a marine site with inevitable mechanical damage, galvanising is the correct choice; nickel plating is decorative and, once damaged, actively harmful.
3. Evaluate the plater's proposal to double the thickness by doubling the current to 1.00 A for the same 60.0 min. Is the arithmetic right, and what practical chemistry argues against it?
Read the worked answer
The arithmetic is right in principle. Faraday's law gives m = (M·I·t)/(n·F), so mass deposited is proportional to charge. Doubling I at constant t doubles Q, so if the current efficiency stayed at 88% the deposit would double to about 0.96 g. On paper the proposal works.
The chemistry argues against it:
- Current efficiency falls with current density. Doubling the current over the same 0.0200 m² doubles the cathode current density. The nickel deposition and hydrogen evolution reactions are in competition, and the H2 branch is far more sensitive to potential. Expect the efficiency to drop below 88%, so the layer will be less than double and hydrogen production will rise.
- Deposit quality. Metal deposited faster than the ions can be supplied and arranged tends to be powdery, dendritic, stressed or 'burnt' at the edges - poor adhesion and worse corrosion resistance, exactly the opposite of what a marine bracket needs.
- Mass-transport limits. The bath is only 0.500 mol L⁻¹ Ni²⁺ and is unstirred; the cathode diffusion layer can be depleted locally, which again favours H2 evolution and hydroxide precipitation.
- Anode and bath constraints. The nickel anode must dissolve fast enough to keep the ion concentration up; otherwise repeated fast runs progressively deplete the bath.
Better judgement: keep the current at 0.500 A (a proven 88% efficiency and sound deposit) and double the time to 120 min, or increase the cathode area, rather than doubling the current density. Verify with a thickness measurement on a coupon instead of trusting the charge calculation.