9.11 Electrolysis

Electroplating

Run an electroplating cell and watch the anode dissolve as glowing ions drift across and coat the cathode. Progress bar tracks plating completion.

Cathode (−)Anode (+)Ion Migration4 Metals
Topic 9.11

Electrolysis and Faraday’s Law

Calculate the amount of charge flow based on changes in the amounts of reactants and products in an electrochemical cell.

Faraday's laws relate the stoichiometry of the redox reaction in an electrochemical cell to the number of electrons transferred, the mass of material deposited or removed from an electrode (as in electroplating), the current, the time elapsed, and the charge of the ionic species.

The equation on the sheet is deceptively simple:

I = q/t, so q = I × t

The full conversion chain is just dimensional analysis:

I (amperes) × t (seconds) → q (coulombs) → ÷ F → mol e⁻ → half-reaction ratio → mol substance → × M → mass

with F = 96 485 C per mole of electrons (Faraday's constant, given on the sheet).

The step everyone skips is the half-reaction ratio. You must know how many electrons each ion requires:

  • Ag⁺ + e⁻ → Ag: 1 mol e⁻ per mol Ag
  • Cu²⁺ + 2 e⁻ → Cu: 2 mol e⁻ per mol Cu
  • Al³⁺ + 3 e⁻ → Al: 3 mol e⁻ per mol Al

Ignoring this is the single largest source of error in electrolysis problems — and a favorite exam distractor, since two cells wired in series pass identical charge but deposit very different numbers of moles.

Units discipline. Time must be in seconds. One ampere is one coulomb per second, so 1 A × 1 s = 1 C. Convert minutes and hours before you begin.

Applications: electroplating, refining copper to high purity, producing aluminum from bauxite (the Hall–Héroult process), and generating chlorine and sodium hydroxide from brine.

Key points

  • q = It, with time in seconds.
  • Divide charge by 96 485 C/mol to get moles of electrons.
  • The half-reaction gives moles of electrons per mole of substance — never assume 1:1.
  • Cells in series pass the same charge but deposit different moles of different metals.

Equations

  • on the exam sheet
    • current (amperes)
    • charge (coulombs)
    • time (seconds)
  • not on the sheetF = 96 485 C mol⁻¹ is given on the sheet.

Common mistakes

  • Time in minutes. Convert to seconds first.
  • Assuming 1 mol e⁻ per mol of metal. Read the half-reaction.
  • Using the wrong Faraday value. 96 485 C/mol, printed on the sheet.
  • Forgetting to convert moles to grams when the question asks for mass.

Worked example

A current of 2.50 A is passed through a solution of CuSO₄ for 45.0 minutes. (a) Calculate the mass of copper deposited. (b) If the same current were passed for the same time through AgNO₃, what mass of silver would be deposited? (c) Explain why the two masses differ.

Charge passed (common to both):
t = 45.0 min × 60 s/min = 2700 s
q = It = (2.50 A)(2700 s) = 6750 C

Moles of electrons:
n(e⁻) = 6750 C ÷ 96 485 C/mol = 0.06996 mol e⁻

(a) Copper. Cu²⁺ + 2 e⁻ → Cu, so 2 mol e⁻ per mol Cu:
n(Cu) = 0.06996 ÷ 2 = 0.03498 mol
m(Cu) = 0.03498 × 63.55 g/mol = 2.22 g

(b) Silver. Ag⁺ + e⁻ → Ag, so 1 mol e⁻ per mol Ag:
n(Ag) = 0.06996 mol
m(Ag) = 0.06996 × 107.87 g/mol = 7.55 g

(c) Why they differ. Two independent reasons compound. First, each Cu²⁺ ion requires two electrons to be reduced while each Ag⁺ requires only one, so the same charge deposits half as many moles of copper as of silver. Second, silver's molar mass (107.87) is substantially larger than copper's (63.55). Together these give more than three times the mass of silver.

This is precisely why the half-reaction step cannot be skipped.

Full notes for topic 9.11 →