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6.2.15Additional lessons

Electroplating — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 146 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Part 1 of 5How Electroplating Works

6.2.15 Electroplating

Electroplating coats an object with a thin metal layer by electrolysis.

Cathode
  • The object to be plated is made the negative electrode so that metal ions are reduced onto its surface.
Anode
  • The plating metal is the positive electrode and dissolves to replace the ions removed from solution.
Electrolyte
  • A solution containing ions of the plating metal, for example silver nitrate for plating with silver.
Silver example
  • Cathode
    • Ag⁺(aq) + e⁻ → Ag(s);
      • anode: Ag(s) → Ag⁺(aq) + e⁻, so the concentration stays constant.

Part 2 of 5Why Plate Objects

1
Plating can protect iron or steel from corrosion by covering it with a more resistant metal such as chromium or zinc.
2
Plating can improve appearance, giving jewellery and cutlery a bright finish of silver or gold.
3
A thin layer of a metal such as tin is cheaper than making the whole object from that metal.
4
The quality of the coating depends on a steady, low current and a clean surface on the object.
5
Fast deposition at high current produces a rough or powdery layer that does not stick well.

Part 3 of 5Quantitative Electrolysis

Quantitative Electrolysis
Charge
  • Q = I × t, where Q is in coulombs, I is in amperes and t is in seconds.
Moles of electrons
  • n(e⁻) = Q ÷ 96 500, using the charge on one mole of electrons.
Moles of metal
  • Divide the moles of electrons by the number of electrons needed to deposit one atom, which is the ion charge.
Mass
  • Multiply moles of metal by its molar mass to find the mass deposited.
IonElectrons per atomMass deposited per mole of electrons
Ag⁺1107.87 g
Cu²⁺231.8 g
Al³⁺39.0 g

Part 4 of 5Worked Example

Question

  • A current of 2.00 A passes for 10.0 minutes through silver nitrate solution.
  • Calculate the mass of silver deposited on the cathode.
  • (Ar of Ag = 107.87;
    • F = 96 500 C mol⁻¹)

  1. Time = 10.0 × 60 = 600 s, so Q = I × t = 2.00 × 600 = 1200 C.
  2. n(e⁻) = 1200 ÷ 96 500 = 0.01244 mol, and because Ag⁺ + e⁻ → Ag, n(Ag) = 0.01244 mol.
  3. Mass = 0.01244 mol × 107.87 g mol⁻¹ = 1.34 g.

Answer 1.34 g of silver

Part 5 of 5Exam Checklist & Tips

6.2.15 – Key Points for the Exam

  • Electroplating Coating an object with a thin layer of metal using electrolysis.
  • Faraday constant The charge on one mole of electrons, 96 500 C mol⁻¹.
  • Coulomb The unit of electric charge, equal to one ampere flowing for one second.
  • Plating metal The metal that is deposited and acts as the anode.

Common Mistakes

  • Using minutes instead of seconds when calculating charge.
  • Forgetting to divide by the ion charge when converting moles of electrons to moles of metal.
  • Making the object the anode, which would dissolve it instead of plating it.
  • Claiming the electrolyte concentration falls in plating with a matching metal anode.

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