An electrochemical cell is a device with two electrodes (anode and cathode), half-cell solutions, and a salt bridge that interconverts chemical and electrical energy through redox reactions; galvanic cells run thermodynamically favored reactions (E° > 0), while electrolytic cells force unfavored ones with an external power source.
An electrochemical cell physically separates a redox reaction into two half-reactions. Oxidation happens at the anode, reduction happens at the cathode, and the electrons travel through an external wire instead of jumping directly between particles. That electron flow is the whole point. It either generates a voltage you can use (galvanic/voltaic cell) or consumes an applied voltage to force a reaction that wouldn't happen on its own (electrolytic cell).
The CED (9.8.A.1) expects you to know the job of every component. The electrodes are where the half-reactions occur, the half-cell solutions supply the active ions, the salt bridge lets ions migrate to keep each half-cell electrically neutral, and a voltmeter or ammeter measures the cell's output. You also need to describe what's happening at both the macroscopic level (the anode loses mass, gas bubbles form at an electrode) and the particulate level (electrons leave Zn atoms, Cu²⁺ ions plate onto the cathode). One memory hook works in every cell type. Oxidation always happens at the anode and reduction always happens at the cathode, whether the cell is galvanic or electrolytic.
The electrochemical cell is the central object of the back half of Unit 9 (Thermodynamics and Electrochemistry). Three learning objectives are built around it. AP Chem 9.8.A asks you to connect the cell's physical parts to how it operates. AP Chem 9.9.A asks you to judge whether a cell is thermodynamically favored using E°cell and standard reduction potentials. AP Chem 9.10.A asks you to predict how the cell potential shifts under nonstandard concentrations. This is also where Unit 9's two big ideas fuse together. The equation ΔG° = -nFE° turns a measurable voltage into a thermodynamic verdict, so a cell diagram is really a free-energy problem in disguise.
Keep studying AP Chemistry Unit 9
Galvanic Cell (Unit 9)
A galvanic (voltaic) cell is the specific type of electrochemical cell that runs a thermodynamically favored reaction and produces a positive voltage. It's a battery in its simplest form. Per 9.8.A.2, 'galvanic' and 'voltaic' mean the same thing on the exam.
Cell Potential and Free Energy (Unit 9)
E°cell and ΔG° are two readings of the same dial. A positive E°cell means a negative ΔG°, so the cell reaction is thermodynamically favored. This link (Topic 9.9) is the single most tested relationship involving electrochemical cells.
Nonstandard Conditions and Q (Unit 9)
Real cells rarely sit at standard conditions where Q = 1. Topic 9.10 treats cell potential as a driving force toward equilibrium, so the farther Q is from K, the bigger the voltage. As the cell discharges and approaches equilibrium, the voltage shrinks toward zero. And be careful, Le Châtelier arguments don't apply because the cell isn't at equilibrium.
Faraday's Law and Electrolysis (Unit 9)
Electrolytic cells are electrochemical cells run in reverse, with an external power source pushing a nonfavored reaction. Faraday's Law lets you calculate how many grams of metal plate out from the current and time, using the Faraday constant to convert charge into moles of electrons.
Multiple-choice questions love handing you two standard reduction potentials and asking you to identify the anode and cathode, compute E°cell, and decide if the cell is thermodynamically favored. Practice questions in this style give you cell notation like Al(s)|Al³⁺(aq)||Fe²⁺(aq)|Fe(s) and ask for the sign relationship between ΔG° and E°cell, or hand you a ΔG° value (say +48.2 kJ/mol) and ask what that implies about the cell's voltage. The pattern to lock in is simple. E° > 0 means ΔG° < 0 means favored; E° < 0 means ΔG° > 0 means an external potential is required. FRQs typically go further, asking you to label a cell diagram, write the half-reactions, predict the direction of electron flow and ion flow through the salt bridge, explain a mass change at an electrode, or reason about how a concentration change shifts the cell potential using Q versus K (not Le Châtelier).
Both are electrochemical cells, and in both, oxidation happens at the anode and reduction at the cathode. The difference is the direction of energy conversion. A galvanic cell runs a thermodynamically favored reaction (E° > 0, ΔG° < 0) and produces electrical energy, like a battery. An electrolytic cell uses an external power source to drive a nonfavored reaction (E° < 0, ΔG° > 0), like electroplating a metal. If a question says a power supply or applied voltage is involved, you're looking at electrolysis.
An electrochemical cell separates a redox reaction into two half-cells, with oxidation at the anode and reduction at the cathode in every cell type.
Galvanic (voltaic) cells run thermodynamically favored reactions and produce a positive E°cell; electrolytic cells need an external voltage to drive unfavored reactions.
E°cell and ΔG° always have opposite signs, so a positive cell potential means a negative free energy change and a thermodynamically favored reaction.
The salt bridge maintains charge neutrality by letting ions flow between half-cells; without it, electron flow stops almost immediately.
Under nonstandard conditions, the cell potential depends on Q. The farther the reaction is from equilibrium, the larger the voltage, and it drops to zero as the cell reaches equilibrium.
Use Q versus K reasoning, not Le Châtelier's principle, to explain voltage changes, because an operating cell is not at equilibrium.
It's a device that splits a redox reaction into two half-cells connected by electrodes, a wire, and a salt bridge, so chemical and electrical energy can be interconverted. It's the core concept of Topics 9.8-9.10 in Unit 9.
No. The anode is negative in a galvanic cell but positive in an electrolytic cell. What never changes is the chemistry, since oxidation always occurs at the anode and reduction at the cathode.
Galvanic cells are one type of electrochemical cell. 'Electrochemical cell' is the umbrella term covering both galvanic cells (favored reaction, E° > 0, produce voltage) and electrolytic cells (unfavored reaction, require an external power source).
Calculate E°cell from the standard reduction potentials of the two half-reactions. If E°cell is positive, ΔG° is negative and the reaction is favored. For example, a Zn/Cu cell gives E°cell = +0.34 V - (-0.76 V) = +1.10 V, so it's favored.
No, and the CED calls this out directly (9.10.A.2). An operating cell is not at equilibrium, so explain voltage changes by comparing Q to K. The farther Q is from K, the larger the cell potential.