AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/9/9-11)
Unit 9 · Topic 9.11
9.11 Electrolysis and Faraday's Law
Current is charge per unit time (I = q/t), and one mole of electrons carries 96,485 coulombs (the Faraday constant). From the current and the time, you can find the moles of electrons that flowed and then use the half-reaction to work out how much metal plates out, how much gas forms, or the charge on an ion.
Key terms
- electrolysis
- current (I = q/t)
- coulomb
- Faraday constant
- electroplating
Charge, current and time
Electric charge, q, is measured in coulombs (C). Current, I, is the rate of charge flow, in amperes (A), where 1 A = 1 C/s. So q = I × t, with t in seconds.
The Faraday constant, F = 96,485 C/mol e⁻, is the charge on one mole of electrons. Dividing a charge by F gives moles of electrons: mol e⁻ = q / F.
Using the half-reaction
The half-reaction tells you how many electrons each atom or ion needs. Ag⁺ + e⁻ → Ag needs 1 mol of electrons per mole of silver. Cu²⁺ + 2e⁻ → Cu needs 2. Al³⁺ + 3e⁻ → Al needs 3. At the anode, 2Cl⁻ → Cl₂ + 2e⁻ releases 2 mol of electrons per mole of Cl₂ gas.
The standard path
- Convert time to seconds.
- Find the charge: q = I × t.
- Find moles of electrons: q ÷ 96,485 C/mol e⁻.
- Use the half-reaction's mole ratio to find moles of the substance.
- Convert to grams (× molar mass) or, for a gas, to volume if needed.
- To go backward (find time or current from a mass), run the same steps in reverse.
What these calculations can tell you
Faraday's laws connect five things: electrons transferred, mass deposited or removed at an electrode, current, time, and the charge of the ions involved. If you know all but one, you can find the last.
This is how electroplating is controlled: a jeweler sets the current and the time to put a precise mass of silver or gold on an object. It also explains why producing aluminum takes so much electricity: each Al³⁺ needs three electrons.
Gases at the electrodes
Gases are handled the same way. In the electrolysis of water, the cathode reaction 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) needs 2 mol of electrons per mole of H₂. The anode reaction 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ releases 4 mol of electrons per mole of O₂. The same electrons pass through both electrodes, so twice as many moles of H₂ form as O₂, and at the same temperature and pressure the H₂ takes up twice the volume.
After finding moles of gas, use PV = nRT (topic 3.4) if you need a volume.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Mass of silver plated
A current of 2.00 A passes through a solution of AgNO₃ for 30.0 minutes. What mass of silver (molar mass 107.87 g/mol) plates out at the cathode?
Show the solutionHide the solution
- Step 1: Time: 30.0 min × 60 s/min = 1800 s.
- Step 2: Charge: q = It = (2.00 A)(1800 s) = 3600 C.
- Step 3: Moles of electrons: 3600 C ÷ 96,485 C/mol = 0.03731 mol e⁻.
- Step 4: Ag⁺ + e⁻ → Ag, so moles of Ag = 0.03731 mol.
- Step 5: Mass = 0.03731 mol × 107.87 g/mol = 4.02 g.
Answer: 4.02 g of Ag
- Example 2Calculator allowed
Trap: two electrons per copper ion
How long, in minutes, must a current of 0.500 A flow to plate 1.00 g of copper (molar mass 63.55 g/mol) from a Cu²⁺ solution?
Show the solutionHide the solution
- Step 1: Moles of Cu: 1.00 g ÷ 63.55 g/mol = 0.01574 mol.
- Step 2: Cu²⁺ + 2e⁻ → Cu, so moles of electrons = 2 × 0.01574 = 0.03147 mol. Forgetting this factor of 2 would give half the correct time.
- Step 3: Charge: 0.03147 mol × 96,485 C/mol = 3037 C.
- Step 4: Time: t = q / I = 3037 C ÷ 0.500 A = 6070 s.
- Step 5: In minutes: 6070 s ÷ 60 = 101 min.
Answer: About 101 minutes (6.07 × 10³ s)
- Example 3Calculator allowed
Finding an ion's charge
A current of 1.00 A flows for 965 s through a solution of a zinc salt, and 0.327 g of zinc (molar mass 65.38 g/mol) plates out. What is the charge on the zinc ion?
Show the solutionHide the solution
- Step 1: Charge: q = (1.00 A)(965 s) = 965 C.
- Step 2: Moles of electrons: 965 ÷ 96,485 = 0.01000 mol e⁻.
- Step 3: Moles of Zn deposited: 0.327 ÷ 65.38 = 0.00500 mol.
- Step 4: Electrons per zinc atom: 0.01000 ÷ 0.00500 = 2.00.
- Step 5: Each zinc ion gained 2 electrons to become a neutral atom, so the ion is Zn²⁺.
Answer: The ion is Zn²⁺ (charge 2+).
Common mistakes
- Leaving time in minutes or hours. Amperes are coulombs per second.
- Forgetting the electron ratio from the half-reaction (2 for Cu²⁺, 3 for Al³⁺).
- Multiplying by F when you should divide (or the reverse). Check units: C ÷ (C/mol) = mol.
- Putting the metal deposit at the anode. Metal ions are reduced and plate out at the cathode.
On the exam
- Electrolysis calculations appear in both multiple-choice and free-response questions. Lay out the unit chain (A → C → mol e⁻ → mol substance → g) so each step can earn credit.
- You may be asked to work backward to find time, current, or the charge of an ion from experimental data.
Connected topics
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Check yourself
4 questions on 9.11 Electrolysis and Faraday's Law. Pick an answer to see if you got it, and why.
A current of 2.00 A passes through a solution of CuSO₄ for 30.0 minutes, plating copper onto the cathode: Cu²⁺(aq) + 2e⁻ → Cu(s). What mass of copper (molar mass 63.55 g/mol) is deposited? (F = 96,485 C/mol e⁻)
How long must a current of 0.500 A flow through a solution of AgNO₃ to deposit 1.08 g of silver (molar mass 107.87 g/mol)? (Ag⁺(aq) + e⁻ → Ag(s); F = 96,485 C/mol e⁻)
A current of 10.0 A passes through Al₂O₃ dissolved in molten cryolite for 1.00 hour, producing aluminum at the cathode: Al³⁺ + 3e⁻ → Al. What mass of aluminum (molar mass 26.98 g/mol) forms? (F = 96,485 C/mol e⁻)
In the electrolysis of water, hydrogen forms at the cathode: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq). What current must flow for 965 s to produce 0.0250 mol of H₂? (F = 96,485 C/mol e⁻)
0 of 4 answered