Unit 9 · Topic 9.11 Beta

Electrolysis and Faraday's Law

In electrolysis a current drives an unfavored redox reaction.

Practice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

Every aluminum can starts as aluminum oxide, and the only practical way to get the metal out is to push enormous electric currents through the molten ore. Smelters run at hundreds of thousands of amperes, and their output in kilograms per hour is set by a single, simple chain of conversions: Faraday’s law.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. In an electrolytic cell, what does the power supply do?

  1. Drives a redox reaction that is not favored
  2. Lowers the activation energy
  3. Makes E°cell positive
  4. Keeps the solutions from mixing
Show the answer

An electrolytic cell uses outside energy to drive an unfavored reaction.

  • Correct: Drives a redox reaction that is not favored:
  • Lowers the activation energy:
  • Makes E°cell positive:
  • Keeps the solutions from mixing:

2. What is the Faraday constant?

  1. 96,485 C per mole of electrons
  2. 6.022 × 10²³ electrons
  3. 8.314 J/(mol·K)
  4. 1.602 × 10⁻¹⁹ C per mole
Show the answer

F is the charge on one mole of electrons.

  • Correct: 96,485 C per mole of electrons:
  • 6.022 × 10²³ electrons:
  • 8.314 J/(mol·K):
  • 1.602 × 10⁻¹⁹ C per mole:

3. How many moles of Cu form from 0.10 mol of electrons in Cu²⁺ + 2e⁻ → Cu?

  1. 0.050 mol
  2. 0.10 mol
  3. 0.20 mol
  4. 0.025 mol
Show the answer

Two electrons per Cu: 0.10 ÷ 2 = 0.050 mol.

  • Correct: 0.050 mol:
  • 0.10 mol:
  • 0.20 mol:
  • 0.025 mol:

Part 4 · See it

See it first

A beaker of copper(II) sulfate with a copper anode and a key as the cathode, joined to a power supply that moves electrons from the anode to the key. Copper dissolves at the anode and plates onto the key. Below, a chain: current times time gives charge; divide by F for moles of electrons; divide by 2 for moles of copper; multiply by 63.55 for grams of copper.
Copper plating: Cu²⁺ is reduced onto the key (cathode) and Cu dissolves from the anode. Below, the Faraday chain from current and time to grams. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A current I flows for a time ta charge q = It, in coulombs, passes through the cell
  2. Each mole of electrons carries 96,485 Cmoles of electrons = q ÷ F
  3. The half-reaction fixes how many electrons each ion needsmoles of product = mol e⁻ ÷ electrons per ion
  4. Each mole of product has a known massmass = moles × molar mass

Part 6 · Key ideas

Key ideas

  • Electrolysis uses an electric current to drive an unfavored redox reaction, such as plating a metal or splitting a molten salt.
  • Charge: q = It (coulombs = amperes × seconds). Convert minutes and hours to seconds.
  • Moles of electrons: q ÷ F, with F = 96,485 C/mol e⁻.
  • Use the half-reaction: Ag⁺ needs 1 e⁻, Cu²⁺ needs 2, Al³⁺ needs 3. Then moles × molar mass.
  • The same charge through cells in series gives the same mol e⁻ but different masses of metal.

Part 7 · Misconception

A common mistake

The wrong idea: One mole of electrons deposits one mole of any metal.

What actually happens: The half-reaction decides: one mole of electrons deposits 1 mol Ag (Ag⁺ + e⁻), ½ mol Cu (Cu²⁺ + 2e⁻) or ⅓ mol Al (Al³⁺ + 3e⁻). Always divide moles of electrons by the electrons per ion.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Experimental setup

Copper plating a key

A student plates copper onto a brass key. The key and a copper strip dip into 1.0 M CuSO₄(aq) and are connected to a power supply. A steady current of 2.50 A runs for 30.0 minutes. Cu²⁺(aq) + 2e⁻ → Cu(s). Molar mass of Cu = 63.55 g/mol; F = 96,485 C/mol e⁻.

1. How much charge, in coulombs, passes through the circuit?

Type a number and its unit.

Show the answer

q = It = (2.50 A)(30.0 min × 60 s/min) = (2.50 C/s)(1800 s) = 4.50 × 10³ C.

  • Answer: 4.50e+3 C

2. Calculate the mass of copper plated onto the key.

Type a number and its unit.

Show the answer

4.50 × 10³ C × (1 mol e⁻ / 96,485 C) = 0.04664 mol e⁻; × (1 mol Cu / 2 mol e⁻) = 0.02332 mol Cu; × 63.55 g/mol = 1.48 g Cu.

  • Answer: 1.48 g

3. Which electrode is the key, and what happens there?

  1. The anode, where Cu²⁺ ions gain electrons and copper plates on
  2. The cathode, where Cu²⁺ gains electrons and copper plates on
  3. The cathode, where copper atoms lose electrons and dissolve
  4. The anode, where copper atoms lose electrons and dissolve
Show the answer

The key gains copper, so Cu²⁺ is reduced on it: the key is the cathode.

  • The anode, where Cu²⁺ ions gain electrons and copper plates on: Gaining electrons is reduction, which happens at the cathode, not the anode.
  • Correct: The cathode, where Cu²⁺ gains electrons and copper plates on: Right: plating is reduction, which happens at the cathode.
  • The cathode, where copper atoms lose electrons and dissolve: Losing electrons is oxidation, which happens at the anode.
  • The anode, where copper atoms lose electrons and dissolve: That is what happens at the copper strip, not at the key.

4. How many minutes would it take to plate 2.00 g of copper at the same current?

Type a number and its unit.

Show the answer

2.00 g ÷ 63.55 g/mol = 0.03147 mol Cu; × 2 = 0.06294 mol e⁻; × 96,485 C/mol = 6073 C; t = q/I = 6073 C ÷ 2.50 A = 2429 s = 40.5 min.

  • Answer: 40.5 min

Data table

Three cells in series

Three electrolytic cells are connected one after another (in series), so the same charge, 1.93 × 10⁴ C, passes through each. F = 96,485 C/mol e⁻.

Metal plated in each cell
CellSolutionCathode half-reactionMolar mass of metal (g/mol)
1AgNO₃Ag⁺ + e⁻ → Ag107.87
2Cu(NO₃)₂Cu²⁺ + 2e⁻ → Cu63.55
3Al₂O₃ dissolved in molten cryoliteAl³⁺ + 3e⁻ → Al26.98

5. What mass of aluminum is produced in Cell 3?

Type a number and its unit.

Show the answer

0.2000 mol e⁻ × (1 mol Al / 3 mol e⁻) × 26.98 g/mol = 1.80 g Al.

  • Answer: 1.80 g

6. Which cell produces the greatest mass of metal, and why?

  1. Cell 3: aluminum ions carry the largest charge, so they attract the most electrons.
  2. The three masses are equal, because the same charge passes through each cell.
  3. Cell 2: copper is between silver and aluminum, so it balances charge and mass best.
  4. Cell 1: Ag⁺ needs one electron, giving the most moles, and Ag is heaviest.
Show the answer

Same mol e⁻ in each cell; mol metal = mol e⁻ ÷ (electrons per ion), then × molar mass. Silver wins on both counts.

  • Cell 3: aluminum ions carry the largest charge, so they attract the most electrons.: A larger charge means more electrons per atom, so fewer atoms form from the same charge.
  • The three masses are equal, because the same charge passes through each cell.: Equal charge gives equal moles of electrons, but different moles and masses of metal.
  • Cell 2: copper is between silver and aluminum, so it balances charge and mass best.: There is no balancing; work out each cell: Ag is largest in both moles and mass.
  • Correct: Cell 1: Ag⁺ needs one electron, giving the most moles, and Ag is heaviest.: Right: 0.200 mol Ag (21.6 g) versus 0.100 mol Cu (6.36 g) and 0.0667 mol Al (1.80 g).

7. A student plates copper and finds the key gained less mass than calculated from the current and time. Which is a plausible cause?

  1. Some current went into another reduction at the cathode, such as making H₂.
  2. The student used a larger key, which needs more copper.
  3. The copper anode lost mass during the run.
  4. The CuSO₄ solution was more concentrated than 1.0 M.
Show the answer

Faraday’s law assumes every electron reduces Cu²⁺. Side reactions use some electrons, so less copper forms.

  • Correct: Some current went into another reduction at the cathode, such as making H₂.: Right: if not every electron reduces Cu²⁺, less copper plates than Faraday’s law predicts.
  • The student used a larger key, which needs more copper.: The mass plated depends on the charge, not on the size of the key.
  • The copper anode lost mass during the run.: The anode is expected to lose mass as Cu dissolves; it does not reduce the mass plated.
  • The CuSO₄ solution was more concentrated than 1.0 M.: A more concentrated solution would not reduce the copper plated by a given charge.

Part 9 · Summary

Summary

In electrolysis a current drives an unfavored redox reaction. The amount of product follows a chain: charge q = It; moles of electrons = q/F; moles of product from the half-reaction’s electrons per ion; mass from the molar mass. The same chain run backward gives the current or time needed.

Part 10 · Up next

What comes next

This is the last topic in the course.

Part 11 · Connections

Connections