BSEB Intermediate Chemistry study materials
BSEB Intermediate Chemistry study resources

BSEB Intermediate Chemistry: Electrochemistry

BSEB Intermediate Chemistry · Work through your exam, one topic at a time.

All guide sections

Electrochemistry

Electrochemistry in this paper means three linked skills: conductance and Kohlrausch's law, the Nernst equation applied to a cell, and Faraday's laws with commercial cells and corrosion.

Conductance

  • Conductivity κ (S cm⁻¹) falls on dilution because the number of ions per unit volume drops; molar conductivity Λm = 1000κ/c rises and levels off at infinite dilution.
  • Kohlrausch's law: Λ°m = ν₊λ°₊ + ν₋λ°₋. It gives Λ° for weak electrolytes and, by difference, for sparingly soluble salts such as AgCl.
  • Degree of dissociation α = Λm/Λ°m, and Ka = cα²/(1 − α).

Cells and the Nernst equation

  • E°cell = E°cathode − E°anode; ΔG° = −nFE°; at equilibrium E°cell = (0.059/n) log K.
  • At 298 K: E = E° − (0.059/n) log Q, with Q written as products over reactants, pure solids and liquids omitted.
  • Notation: anode (oxidation) on the left, salt bridge as ‖, cathode on the right.

Worked example

Zn | Zn²⁺(0.1 M) ‖ Cu²⁺(0.01 M) | Cu, with E° = 1.10 V. Reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, so n = 2 and Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10. E = 1.10 − (0.059/2) log 10 = 1.10 − 0.0295 = 1.07 V. The voltage drops below E° because the reactant ion is depleted and the product ion accumulates relative to standard conditions.

Traps that cost marks

  • Inverting Q so the sign of the correction flips.
  • Forgetting that 1 S cm² mol⁻¹ = 10⁻⁴ S m² mol⁻¹.
  • Faraday calculation slips: mass = (I·t·M)/(n·F) with F = 96500 C.
  • Naming the electrode with the higher reduction potential as the anode; it is the cathode.
  • Assuming a catalyst alters E° - it does not.