This AP Chemistry lesson on electrochemistry notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on electrochemistry and check yourself with the quiz.
Quick answer: Electrochemistry links redox reactions to electrical energy. In a galvanic cell a spontaneous reaction pushes electrons through a wire, and the voltage is E°cell = E°cathode − E°anode. In an electrolytic cell a power supply forces a non-spontaneous reaction. ΔG° = −nFE° and Faraday’s law connect voltage, charge and the amount of product.
Level: AP Chemistry · Curriculum: AP Chem Unit 9
Electrochemistry notes: the key ideas
What is a galvanic cell and how does it work?
A galvanic cell (also called a voltaic cell) uses a spontaneous redox reaction to push electrons through an external circuit. Oxidation happens at the anode and reduction happens at the cathode, in every kind of cell.
Put a zinc strip in zinc sulfate solution and a copper strip in copper(II) sulfate solution. Zinc atoms give up electrons more readily than copper atoms, so the zinc half-cell is the anode: Zn → Zn²⁺ + 2e⁻. The electrons travel through the wire to the copper strip, where Cu²⁺ ions pick them up: Cu²⁺ + 2e⁻ → Cu. A salt bridge lets ions drift between the beakers so that neither side builds up charge. Anions move toward the anode, cations toward the cathode.
Remember “an ox and a red cat”: anode with oxidation, reduction at the cathode. Electrons always travel from anode to cathode through the wire.
How do you calculate a standard cell potential?
Each half-reaction has a standard reduction potential, E°, measured against the standard hydrogen electrode (1.0 mol/dm³ H⁺, 1 atm H₂, platinum, 25 °C), which is defined as 0.00 V. The half-cell with the higher E° is reduced, so it is the cathode.
E°cell = E°cathode − E°anode
E°cell is in volts (V). Both values are reduction potentials taken from the table. If E°cell is positive the reaction is spontaneous as written.
| Half-reaction (reduction) | E° (V) |
|---|---|
| Ag⁺ + e⁻ → Ag | +0.80 |
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| 2H⁺ + 2e⁻ → H₂ | 0.00 |
| Sn²⁺ + 2e⁻ → Sn | −0.14 |
| Zn²⁺ + 2e⁻ → Zn | −0.76 |
| Al³⁺ + 3e⁻ → Al | −1.66 |
| Mg²⁺ + 2e⁻ → Mg | −2.37 |
A cell joins Zn²⁺/Zn and Ag⁺/Ag half-cells. Find E°cell.
Silver has the higher E°, so silver is the cathode and zinc is the anode. E°cell = 0.80 − (−0.76) = +1.56 V. The overall equation is Zn + 2Ag⁺ → Zn²⁺ + 2Ag.
Never multiply E° by the stoichiometric coefficient. Potentials are intensive: doubling the silver half-reaction does not double its voltage. The subtraction already handles the anode sign, so do not change it again.
How are E°cell, ΔG° and K connected?
ΔG° = −nFE°cell, and ΔG° = −RT ln K, so log K = nE°cell / 0.0592 at 25 °C
n = moles of electrons transferred in the balanced equation; F = 96 485 C/mol; ΔG° in J/mol; R = 8.314 J/(mol·K).
Find ΔG° and K for the zinc-copper cell, E°cell = 0.34 − (−0.76) = 1.10 V, n = 2.
ΔG° = −2 × 96 485 × 1.10 = −2.12 × 10⁵ J/mol = −212 kJ/mol. log K = 2 × 1.10 / 0.0592 = 37.2, so K ≈ 10³⁷: the reaction goes essentially to completion.
Away from standard conditions the voltage changes with the reaction quotient Q. The Nernst equation at 25 °C is E = E° − (0.0592 / n) log Q. When the cell runs, Q grows until Q = K, at which point E = 0 and the battery is flat.
Zn/Cu cell with [Zn²⁺] = 1.0 mol/dm³ and [Cu²⁺] = 0.010 mol/dm³. Q = [Zn²⁺]/[Cu²⁺] = 100, log Q = 2.
E = 1.10 − (0.0592 / 2)(2) = 1.10 − 0.059 = 1.04 V. Diluting the reactant Cu²⁺ lowers the driving force, so the voltage drops.
What happens in an electrolytic cell, and what does Faraday’s law say?
An electrolytic cell uses an external power source to force a non-spontaneous reaction. The anode is still where oxidation happens, but it is now the positive electrode, and the cathode is negative.
In aqueous solutions water can also react, so the product at each electrode is whichever half-reaction needs the smaller applied voltage. That is why aqueous sodium chloride gives hydrogen at the cathode, not sodium.
Faraday’s law: the amount of substance changed at an electrode is proportional to the charge passed. Q = I t, moles of electrons = Q / F, and moles of product = moles of electrons / z, where z is the number of electrons per ion or atom.
A current of 2.50 A plates copper from Cu²⁺ for 40.0 min. Find the mass of copper.
t = 40.0 × 60 = 2400 s, so Q = 2.50 × 2400 = 6000 C. Moles of e⁻ = 6000 / 96 485 = 0.0622 mol. Cu²⁺ + 2e⁻ → Cu, so moles of Cu = 0.0311 mol. Mass = 0.0311 × 63.55 = 1.98 g.
- Oxidation is at the anode and reduction is at the cathode, in every cell.
- E°cell = E°cathode − E°anode, in volts; a positive value means spontaneous.
- ΔG° = −nFE°cell, with F = 96 485 C/mol; log K = nE°/0.0592 at 25 °C.
- Nernst: E = E° − (0.0592/n) log Q; E = 0 when Q = K.
- Faraday: Q = It, n(e⁻) = Q/F, then use the mole ratio to find the mass.
Cells, cell potential and electrolysis: frequently asked questions
What is the difference between a galvanic cell and an electrolytic cell?
A galvanic cell turns a spontaneous reaction into electrical energy, with a negative anode and a positive cathode. An electrolytic cell uses electrical energy to drive a non-spontaneous reaction, with a positive anode and a negative cathode. Oxidation is at the anode in both.
How do you decide which electrode is the cathode?
Compare the standard reduction potentials. The half-reaction with the higher (more positive) E° is reduced, so it is the cathode in a galvanic cell. The other half-cell is the anode and runs as an oxidation.
Why is E° not multiplied when a half-reaction is doubled?
Potential is energy per unit of charge, so it does not depend on how many moles react. Only n, the electrons in the balanced equation, changes. It matters in ΔG° = −nFE° but not in E°cell.
How is cell potential related to the equilibrium constant?
Combining ΔG° = −nFE° with ΔG° = −RT ln K gives log K = nE°/0.0592 at 25 °C. A large positive E° means a very large K, and E°cell = 0 means K = 1.
How do you use Faraday’s law to find a mass deposited?
Find the charge with Q = It, divide by 96 485 C/mol to get moles of electrons, divide by the electrons per ion to get moles of metal, then multiply by the molar mass.
