Mistake Master
Home› Unit 9 · Applications of Thermodynamics› 9.1·9.2·9.3·9.4·9.5·9.6·9.7·9.8·9.9·9.10·9.11› Lesson
Skill Check 0 / 10 complete

Electrolysis and Faraday’s law

Electrolysis is chemistry paid for by the coulomb. Faraday's law is the exchange rate — a ladder from current and time down to grams of metal, one careful unit conversion at a time.

§1

The Faraday ladder.

▸

In electrolysis, the amount of substance produced is set by the charge passed. Faraday's law is a ladder of conversions from electrical measurements to grams.

The steps: charge (C) = current (A) × time (s); then moles of electrons = charge ÷ F (Faraday's constant, 96,485 C/mol e⁻); then moles of metal = moles of electrons ÷ (electrons per ion); and finally grams = moles × molar mass.

Each rung has a unit trap: time must be in seconds, and you must divide by the charge per ion (e.g. 2 for Cu²⁺, 3 for Al³⁺) before converting to mass.

UNIT 9 TOPIC 9.11 • ELECTROLYSIS AND FARADAY'S LAW FARADAY'S LAW Electrolytic cell power source (battery) + − e⁻ e⁻ electron flow from power source Ag⁺ NO₃⁻ AgNO₃ (aq) anode (+) oxidation cathode (−) reduction Cathode: Ag⁺ + e⁻ → Ag(s) Faraday's law: charge → moles → mass q = I · t charge = current × time mol e⁻ = q / F charge ÷ Faraday constant F = 96,485 C·mol⁻¹ (charge per mole e⁻) Worked example — silver plating Given: I = 2.00 A, t = 965 s q = I·t = (2.00)(965) = 1930 C mol e⁻ = 1930 / 96,485 = 0.0200 mol e⁻ Ag⁺ + e⁻ → Ag (1 mol e⁻ per mol Ag) mass = 0.0200 mol × 107.87 g·mol⁻¹ mass of Ag deposited = 2.16 g More charge (larger I or longer t) → more Ag deposited. CED ANCHOR Electrolysis stoichiometry links charge → moles of e⁻ → mass: q = I·t, mol e⁻ = q / F, m = mol × M. Anode (+) oxidation · cathode (−) reduction · more charge → more product · F = 96,485 C per mole e⁻. AP Chemistry · Unit 9 · Applications of Thermodynamics
Fig. 9.11.1 Faraday's law climbs from charge to mass. Charge = current × time (seconds); moles of electrons = charge ÷ F; moles of metal = moles of electrons ÷ (electrons per ion); grams = moles × molar mass.
§2

Climbing the ladder.

▸

One conversion per rung, watching units.

  1. Charge. Multiply current (amps) by time in seconds: q = I × t.
  2. Moles of electrons. Divide the charge by F = 96,485 C/mol.
  3. Moles of metal. Divide moles of electrons by the ion's charge (electrons per ion).
  4. Mass. Multiply moles of metal by its molar mass for grams.
§3

The pieces you'll meet.

▸

The rungs of Faraday's law.

q
Charge
= current × time (in seconds).
F
Faraday's constant
96,485 C per mole of electrons.
n(e⁻)
Moles of electrons
= charge ÷ F.
per ion
Electrons per ion
Divide by the ion charge (2 for Cu²⁺).
mass
Grams
= moles of metal × molar mass.
seconds
Seconds
Convert minutes/hours to seconds first.
§4

Worked example: plating copper.

▸

Given. 2.0 A for 30 minutes deposits Cu from Cu²⁺ (molar mass 63.5 g/mol).

Charge. t = 30 min = 1800 s; q = 2.0 × 1800 = 3600 C.

Electrons. mol e⁻ = 3600 ÷ 96,485 ≈ 0.0373 mol.

Metal, then mass. Cu²⁺ needs 2 e⁻, so mol Cu = 0.0373 ÷ 2 ≈ 0.0187 mol; mass = 0.0187 × 63.5 ≈ 1.19 g.

§5

Mistakes that cost real points.

▸
Pitfall · 01

"Moles of electrons equals moles of metal deposited."

You must divide moles of electrons by the ion's charge: Cu²⁺ needs 2 electrons per atom, Al³⁺ needs 3. Treating moles of electrons as moles of metal overcounts the metal by that factor.

Fix. Divide moles of electrons by the electrons-per-ion (the ion charge) to get moles of metal.

Pitfall · 02

"Charge in coulombs is the same as moles of electrons."

Charge must be divided by Faraday's constant (96,485 C/mol) to get moles of electrons. Using coulombs directly as moles skips a step and is off by a huge factor.

Fix. Convert charge to moles of electrons by dividing by F = 96,485 C/mol.

Pitfall · 03

"Plug the time in minutes straight into q = I × t."

Current is in amperes (coulombs per second), so time must be in seconds. Using minutes (or hours) without converting makes the charge — and everything downstream — wrong by 60 (or 3600).

Fix. Convert time to seconds before computing charge; and report grams if grams are asked, using molar mass.

§6

Skill Check.

▸

Ten scenarios. Pick the chips that match your answer, then check. A scenario marks complete the first time every part is right. Progress saves on this device.

0 of 10 scenarios complete