AP Chemistry · Lesson 3

Electrolysis and Faraday’s law

Know Know it

An electrolytic cell uses an outside power source to drive a redox reaction that is not favorable on its own (ΔG∘>0\Delta G^\circ > 0). Oxidation still happens at the anode and reduction at the cathode. The power source pulls electrons out of the anode and pushes them into the cathode.

The amount of reaction is set by the charge that passes. Charge in coulombs is current in amperes times time in seconds: q=Itq = It. Moles of electrons are q/Fq/F, with F = 96,485 C/mol. The half-reaction tells you how many moles of electrons each mole of product needs.

In electroplating, the object to coat is the cathode in a solution of the metal’s ions. Silver needs 1 mol of electrons per mole of silver (Ag++e−→Ag\text{Ag}^+ + e^- \rightarrow \text{Ag}), copper needs 2, and aluminum needs 3.

Comparing the two kinds of cell
Galvanic cellElectrolytic cell
Energyreleases energy as electricityuses energy from a power source
Sign of Ecell∘E^\circ_{\text{cell}} for the reactionpositivenegative
Sign of ΔG∘\Delta G^\circnegativepositive
Oxidation happens atanodeanode
Reduction happens atcathodecathode

Worked example Easy

A current of 1.50 A flows through an electrolytic cell for 45.0 minutes. How many moles of electrons pass through the cell? (F = 96,485 C/mol)

  1. 0.0420 molAnswer
  2. 0.0210 mol
  3. 7.00×10−47.00 \times 10^{-4} mol
  4. 4050 mol

How to solve it

  1. Time: 45.0 min × 60 = 2700 s.
  2. Charge: q=It=(1.50)(2700)=4050q = It = (1.50)(2700) = 4050 C.
  3. Moles of electrons: 4050/96,485=0.04204050/96{,}485 = 0.0420 mol.

Why each choice is right or wrong

  • A. Correct. The charge is (1.50)(2700 s)=4050(1.50)(2700 \text{ s}) = 4050 C, and 4050/96,485=0.04204050/96{,}485 = 0.0420 mol of electrons.
  • B. Incorrect. 0.0210 mol divides by 2 as if for a two-electron half-reaction; the question asks for moles of electrons, not of metal.
  • C. Incorrect. 7.00×10−47.00 \times 10^{-4} mol uses 45.0 as the time in seconds; 45.0 minutes is 2700 s.
  • D. Incorrect. 4050 is the charge in coulombs, not the moles of electrons; divide by F to convert.

Trap Seconds, and electrons per atom

Two slips cause most wrong answers here: leaving the time in minutes or hours (current is coulombs per second), and forgetting the electrons per atom. Copper(II) needs 2 mol of electrons per mole of copper, so the same charge plates half as many moles of copper as of silver.

Explain Explain it

Task verbs: calculate, explain

Work every electrolysis problem along one chain: current and time give charge (q=Itq = It); charge gives moles of electrons (q/Fq/F); the half-reaction gives moles of product; the molar mass gives grams. You can run the chain backward to find a time or a current.

Cells connected in series carry the same current for the same time, so they pass the same moles of electrons. The moles of metal plated in each then depend only on the charge of its ion.

Worked example Medium

How many minutes does it take to plate 0.500 g of silver onto a spoon from a solution of Ag+\text{Ag}^{+} using a current of 0.250 A? (Ag: 107.87 g/mol; F = 96,485 C/mol)

  1. 59.6 min
  2. 1790 min
  3. 29.8 minAnswer
  4. 1.86 min

How to solve it

  1. Moles of silver: 0.500/107.87=4.64×10−30.500/107.87 = 4.64 \times 10^{-3} mol, which needs the same moles of electrons.
  2. Charge: (4.64×10−3)(96,485)=447(4.64 \times 10^{-3})(96{,}485) = 447 C.
  3. Time: 447/0.250=1790447/0.250 = 1790 s, and 1790/60=29.81790/60 = 29.8 min.

Why each choice is right or wrong

  • A. Incorrect. 59.6 min uses 2 mol of electrons per mole of silver, but Ag+\text{Ag}^+ needs only one electron.
  • B. Incorrect. 1790 is the time in seconds; the question asks for minutes, so divide by 60.
  • C. Correct. Moles of Ag = 0.500/107.87 = 4.64×10−34.64 \times 10^{-3} mol, which needs 447 C; 447/0.250 = 1790 s, or 29.8 min.
  • D. Incorrect. 1.86 min multiplies the charge by the current instead of dividing by it.

Calculate Your turn

A student passes a steady current of 2.00 A through a solution of copper(II) sulfate for 30.0 minutes, plating copper metal onto the cathode. Calculate the mass of copper deposited. (Cu: 63.55 g/mol; F = 96,485 C/mol)

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Compare with a model answer

A full answer does this

  • Finds the moles of electrons from the charge, about 0.0373 mol
  • Uses 2 mol of electrons per mole of copper and finds the mass, about 1.19 g

Model answer

Charge: q=(2.00)(1800 s)=3600q = (2.00)(1800 \text{ s}) = 3600 C. Moles of electrons: 3600/96,485=0.03733600/96{,}485 = 0.0373 mol. Since Cu2++2e−→Cu\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}, moles of Cu = 0.0373/2 = 0.0187 mol. Mass: (0.0187)(63.55)=1.19(0.0187)(63.55) = 1.19 g of copper.

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