Electrochemistry ⚡
Throughout this unit, we centered most of our discussion around the topics of entropy and Gibbs Free Energy. However, another aspect of energy we have yet to discuss is electricity!
Electrochemistry is the study of redox reactions and how we can use redox reactions to produce electrical energy. To understand electrochemistry and its applications, we will examine cell potentials, which are a measure of the voltage released during redox reactions. Then, we can connect that voltage to spontaneity and equilibrium. We will also look at how electrical power can make nonspontaneous redox reactions occur.
Review of Redox Reactions
Before jumping into the thermodynamics of electrochemistry, we need to review redox reactions from Unit 4. A redox reaction, also known as an oxidation-reduction reaction, is a reaction in which electrons transfer from a reducing agent to an oxidizing agent.
The reducing agent is oxidized (loses electrons), and the oxidizing agent is reduced (gains electrons). A good acronym to remember this concept is OIL RIG (oxidation is loss; reduction is gain). The language of reducing and oxidizing agents is not specifically referenced on the exam, but it is helpful to know in developing a conceptual understanding of the chemistry occurring.
An example of a redox reaction is 2AgNO3 + Cu → Cu(NO3)2 + 2Ag. We see that copper begins at an oxidation number of 0 and ends with an oxidation number of +2. Silver starts with an oxidation number of +1 and ends with an oxidation number of 0. Therefore, copper was oxidized, and silver was reduced. Electrons were transferred from copper to silver.
We can write the above reaction in terms of an oxidation and reduction half-reaction that add together to form the overall reaction:
Copper is oxidized to form Cu2+: Cu → Cu2+ + 2e-
Ag+ is reduced to form Ag: 2Ag+ + 2e- → 2Ag
(Note that we multiplied the second reaction by 2 to balance the electrons!)
Reduction Potentials
When redox reactions occur, the electrons experience an electromotive force, a force pushing electrons from a reducing agent to an oxidizing agent. The stronger the electromotive force, the more spontaneous the reaction. Electromotive force measurements are in volts. We will be using these measurements to calculate cell potential.
We can look at an example of how to calculate the voltage of a redox reaction by using standard reduction potentials. Standard reduction potentials are the voltage of a reduction. Note that voltages can be negative, meaning it takes energy to make the reaction happen. Reduction potentials are a set of calculated constants provided to you. The table of standard reduction potential values we will be referencing is here.
Consider the redox reaction: Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s).
First, we can split the reaction into two half-reactions:
Zn → Zn2+ + 2e- (oxidation)
Pb2+ + 2e- → Pb (reduction)
Looking at our chart, we see that the reduction potential for Pb2+ is -0.13 V, and for Zn2+, the reduction potential is -0.76 V. However, our half-reaction for Zn2+ is an oxidation reaction. Flipping a reduction potential reaction negates the potential, meaning the potential for Zn → Zn2+ + 2e- is +0.76 V. Finally, we can add these two numbers together to find that the potential for the overall reaction is -0.13 + 0.76 = 0.63 V.
NOTE: Multiplying a half-reaction does not change the potential!
To find the E value for any given redox reaction, you can use the equation E = Ered - Eox, meaning we could have also done: (-0.13 V) - (-0.76 V) = +0.63 V in our previous example. Later in this unit, we will look at calculating cellular energy, Ecell, using cathodes and anodes of galvanic cells.
Galvanic/Voltaic Cells
Now that we refreshed redox reactions and learned a bit about reduction potentials, we can better understand what happens when the oxidizing agent and reducing agent are separated and connected through a wire. Remember that the oxidizing agent gains electrons and the reducing agent loses electrons. Therefore, electrons will travel through the wire from the reducing agent to the oxidizing agent. The force at which these electrons travel is the cell potential notated as Ecell.
The part of the cell where oxidation occurs is known as the anode, and reduction occurs at the cathode. Electrons travel from the anode to the cathode. A good way of remembering this is the phrase “An Ox, Red Cat,” which tells you that the anode is where oxidation occurs, and the cathode is where reduction occurs.
These cells are known as galvanic or voltaic cells. Take the reaction Cu + 2Ag+ → 2Ag + Cu2+:
Image From BCCampus
In this image, we see two half-reactions occurring. One of these is the reduction of Ag+ into silver metal, and the other is the oxidation of Cu into Cu2+. Therefore, we have a copper anode and a silver cathode.
When a wire connects these two electrodes in solutions of their respective ions (Cu2+ and Ag+), electrons can travel from the copper to the Ag+ and create Ag. As a result of this chemical process, we see the copper anode shrink and the silver cathode grow because we are losing Cu and forming Cu2+ and losing Ag+ to form Ag. We can add a voltmeter to the wire to measure the electromotive force of this reaction, which we find to be +0.46 V.
The two solutions connect via a salt bridge, which ensures continuous ion flow. A salt bridge is typically a highly soluble salt whose purpose is to act as a source of spectator ions that can migrate into each half-cell to preserve neutrality. They are full of inert ions that will not interact with the reaction. In the example case, we used NaNO3.
As a practice exercise, see if you can write out the half-reactions and find the cell potential using the equation we looked at earlier. You should get the same number (Ecell = 0.46 V). As another exercise, try to logically understand why we can write Ecell = Ecathode - Eanode using E = Ered - Eox.
Electrolytic Cells 🔋
For a galvanic cell, we assume that the Ecell is greater than 0, which means that the reaction occurs spontaneously. However, if we want to drive a nonspontaneous redox reaction, we use voltage to make it happen. Cells where we drive nonspontaneous redox reactions, and Ecell is less than 0, are electrolytic cells.
Suppose we want to rip apart a solution of Na+ and Cl- ions to form the original Na and Cl2 that created the solution: 2NaCl → 2Na + Cl2. This is a nonspontaneous redox reaction, so we can use a battery to create an electromotive force to pull the electron off Cl- and add it to Na+.
First, we can mathematically demonstrate that this reaction is nonspontaneous. Using half-reactions, we can calculate Ecell for this redox reaction:
2Na+ + 2e- → 2Na (E = -2.71 V)
2Cl- → Cl2 + 2e- (E = -1.36 V)
The anode is where oxidation occurs, so Eanode = -1.36 V. The cathode is where reduction occurs, so Ecathode = -2.71 V. We know that Ecell = Ecathode - Eanode = -2.71 V - (-1.36 V) = -1.35 V. Because Ecell for this redox reaction is negative, it will not occur spontaneously. Therefore, we have to use a battery (or some other source of electricity) to actively push electrons from the anode to the cathode.
Because we have molten NaCl (the ions themselves), we use inert electrodes to collect the products, which in this case are Cl2 gas and Na metal. As the electrons are propelled by the voltage of the battery, the Cl- in solution is oxidized into Cl2, and Na+ ions are reduced into Na, meaning gas bubbles will accumulate on the anode and sodium metal will develop on the cathode. Note that the voltage of the battery must be greater than or equal to the voltage of the overall redox reaction. In this example, we must use, at a minimum, a 1.35 V battery.
Learning Summary
We learned that electrochemistry with redox reactions is applicable to measure cell potential in terms of voltage. Oxidation, the gain of electrons, occurs at the anode of an electrochemical cells. Reduction, the loss of electrons, takes place at the cathode.
Galvanic or voltaic cells have spontaneous, thermodynamically favored reactions. Conversely, electrolytic cells involve nonspontaneous, thermodynamically unfavored reactions. Electrolytic cells, due to their thermodynamically unfavorable reactions, require a battery (or other energy source) to produce electromotive force for electron transfer.
Through writing half-reactions and interpreting visual representations of cells, we can analyze how half-reactions occur and in what direction current flows. In combination, these learning objectives support an overall understanding of electrochemistry!
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| ATP to ADP conversion | The hydrolysis of adenosine triphosphate to adenosine diphosphate, a thermodynamically favorable reaction that releases energy to drive unfavorable biological processes. |
| common intermediates | Shared chemical species or compounds that participate in multiple reactions within a coupled reaction system. |
| coupled reactions | Two or more reactions that share common intermediates, where a thermodynamically favorable reaction is linked to drive a thermodynamically unfavorable reaction forward. |
| electrolytic cell | An electrochemical cell in which electrical energy is used to drive a non-spontaneous redox reaction. |
| external source of energy | Energy supplied from outside a system to drive a process that would not occur spontaneously, such as electrical energy or light. |
| photosynthesis | The process by which light energy is used to drive the thermodynamically unfavorable conversion of carbon dioxide and water into glucose. |
| standard Gibbs free energy change | The change in free energy under standard conditions; negative values indicate thermodynamically favored processes that favor products. |
| thermodynamically unfavored | A reaction that does not proceed spontaneously under standard conditions, resulting in a negative cell potential and positive Gibbs free energy change. |
Frequently Asked Questions
What's the difference between galvanic cells and electrolytic cells?
Galvanic (voltaic) and electrolytic cells are the same parts used differently. In both you have two half-cells, electrodes, and ion flow (often via a salt bridge), and oxidation always occurs at the anode while reduction occurs at the cathode (CED 9.8.A.1 & 9.8.A.3). Key differences: - Galvanic cells run spontaneously (ΔG < 0). The redox reaction is thermodynamically favored, E°cell > 0, and the cell produces electrical current as electrons flow from anode → cathode. Electrode masses change according to the half-reactions (CED 9.8.A.2). - Electrolytic cells force a nonspontaneous reaction (ΔG > 0) by applying an external power supply. Electrons are driven by that supply so the half-reactions occur opposite the spontaneous directions and Ecell is effectively made negative without the power source. For AP review, see the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do I remember which electrode is the anode and which is the cathode?
Short rules to remember: oxidation happens at the anode, reduction at the cathode (CED 9.8.A.3). Mnemonics that work in practice: - “An Ox” → Anode = Oxidation. - “Red Cat” → Reduction at the Cathode. - Electrons always flow from anode → cathode (they leave the anode). So if a species is losing electrons, it’s at the anode. Quick note on signs: in a galvanic (spontaneous) cell the anode is the negative electrode and the cathode is positive; in an electrolytic (nonspontaneous) cell an external power source makes the anode positive and the cathode negative. You don’t need to memorize “+”/“–” for the AP (labeling charge isn’t assessed), but knowing electron flow and which half-reaction (ox vs red) occurs where is essential (CED 9.8.A.1–A.3). For extra practice and diagrams, check the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and more problems at (https://library.fiveable.me/practice/ap-chemistry).
Why does oxidation always happen at the anode in both types of cells?
Because “anode” is defined as the electrode where oxidation happens, oxidation always occurs there in both galvanic and electrolytic cells. Mechanistically: oxidation is loss of electrons, so the anode is the site electrons are produced and leave the half-cell. In a galvanic (spontaneous) cell those electrons flow from the anode through the external circuit to the cathode; in an electrolytic (nonspontaneous) cell an external power source pulls electrons away from the anode and forces the reverse overall reaction—but the chemical process at the anode is still oxidation. Which metal/ion is oxidized is predicted by comparing standard reduction potentials; the half-reaction with the lower reduction potential will be reversed (oxidized) at the anode (CED 9.8.A.2–9.8.A.3). For AP review, see the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and more unit resources (https://library.fiveable.me/ap-chemistry/unit-9). For practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).
Can someone explain what a salt bridge actually does in an electrochemical cell?
A salt bridge keeps the two half-cells electrically neutral so the redox reaction can keep running. In a galvanic cell, oxidation at the anode produces excess positive charge (more cations) and reduction at the cathode removes positive charge (or produces anions). If those charge imbalances aren’t fixed, the cell potential stops. The salt bridge (or porous membrane) lets inert ions flow: anions migrate toward the anode compartment and cations toward the cathode compartment to balance charge. It completes the internal circuit without letting the half-cell solutions mix and react directly (so spectator ions, not reactants, move). At the particulate level, this is ion migration that maintains charge neutrality while electrons travel through the external wire from anode → cathode (oxidation at anode, reduction at cathode—CED 9.8.A.1 & 9.8.A.3). In electrolytic cells the direction of ion flow is the same idea but driven by an external power source. For a clear diagram and AP-aligned notes, see the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK). More unit review and practice problems are at (https://library.fiveable.me/ap-chemistry/unit-9) and (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about electron flow direction - do electrons always flow the same way in galvanic vs electrolytic cells?
Short answer: electrons always flow from the anode to the cathode—in every electrochemical cell oxidation happens at the anode (electrons produced) and reduction at the cathode (electrons consumed). The difference between galvanic and electrolytic cells is why that flow happens, not the direction. - Galvanic (voltaic): a spontaneous redox causes electrons to flow through the external wire from anode → cathode, producing a positive cell potential (E°cell > 0). - Electrolytic: an external power supply forces a nonspontaneous redox; you still have oxidation at the anode and reduction at the cathode, so electrons are pushed from anode → cathode by the external source. Remember: ions move through the salt bridge or electrolyte to balance charge, and the AP CED stresses “oxidation at anode, reduction at cathode” for all cells (Topic 9.8.A.3). For a clear diagram and practice problems, check the Fiveable study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and more unit practice (https://library.fiveable.me/practice/ap-chemistry).
What happens at each electrode during electrolysis of water?
In electrolysis of water the key is: oxidation happens at the anode, reduction at the cathode (CED 9.8.A.3). That means gas forms at both electrodes: - Cathode (reduction): H+ (or H3O+) is reduced to H2 gas. Half-reaction in acidic solution: 2 H+ + 2 e– → H2(g). In basic solution water/ OH– is reduced: 2 H2O + 2 e– → H2(g) + 2 OH–. - Anode (oxidation): water (or OH–) is oxidized to O2 gas. Acidic half-reaction: 2 H2O → O2(g) + 4 H+ + 4 e–. In basic solution: 4 OH– → O2(g) + 2 H2O + 4 e–. Electrons are pulled off at the anode by the external power supply and travel through the circuit to the cathode; ions in solution migrate to keep charge balance. You’ll see bubble evolution (H2 at cathode, O2 at anode) rather than metal deposition unless metal ions are present. For AP review, connect this to galvanic vs electrolytic distinctions in 9.8 (see the Topic 9.8 study guide for examples) (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK). For extra practice try the unit problems (https://library.fiveable.me/practice/ap-chemistry).
How do I know if a reaction is thermodynamically favored or unfavored in electrochemical cells?
Check the cell’s standard cell potential and free energy. For any electrochemical cell: - Use E°cell (from standard reduction potentials): E°cell = E°cathode − E°anode. - If E°cell > 0, the reaction is thermodynamically favored (spontaneous) and the cell is galvanic/voltaic. - If E°cell < 0, it’s thermodynamically unfavored (nonspontaneous) and you need an external power supply—an electrolytic cell. Connect to Gibbs: ΔG° = −nFE°cell (n = electrons transferred, F ≈ 96,485 C·mol−1). So positive E°cell → negative ΔG° → favored. For nonstandard conditions use the Nernst equation to find Ecell and decide spontaneity. Remember: oxidation at the anode, reduction at the cathode (CED 9.8.A.2–3). This is exactly what AP will test—calculate E°cell or ΔG° and state spontaneity. For a quick review see the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK), the Unit 9 overview (https://library.fiveable.me/ap-chemistry/unit-9), and practice problems (https://library.fiveable.me/practice/ap-chemistry).
What's the point of having two separate half-cells instead of just mixing everything together?
If you mixed everything, the redox reaction would just run to completion in one place—no separated flow of electrons, no useful electrical work, and no measurable cell voltage. Two half-cells force oxidation to happen at the anode and reduction at the cathode in separate locations so electrons must travel through an external wire (doing electrical work) rather than through solution. The salt bridge (or porous membrane) keeps charge balanced by letting ions migrate, preventing charge buildup that would stop the reaction. Keeping half-cells separate also lets you control and predict direction using standard reduction potentials and draw clear diagrams of electron flow for AP tasks (see CED 9.8.A.1–3). For practice drawing cells, predicting E°cell, and salt-bridge roles, check the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why do some electrodes gain mass while others lose mass during the reaction?
If you watch an electrochemical cell, electrode mass changes because oxidation at the anode removes metal atoms as ions, and reduction at the cathode deposits ions as solid metal. Oxidation = loss of electrons (anode), so a metal anode (like Zn → Zn2+ + 2e−) loses atoms to solution and its mass drops. Reduction = gain of electrons (cathode), so metal ions in solution (Ag+ + e− → Ag(s)) plate onto the cathode and its mass increases. In electrolytic cells the same chemistry (anode oxidized, cathode reduced) happens but the direction is forced by an external power supply, so which electrode gains/loses mass depends on the imposed reaction. Gas-evolving reactions (H2, O2) or concentration changes can mask mass change. This is all covered in Topic 9.8 (CED 9.8.A.1–3). For a concise review, see the Fiveable Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
I don't understand how current flows through the salt bridge - aren't salts solid?
Good question—salts used in salt bridges aren’t left as solid crystals; they’re dissolved or held in a gel so ions can move. The salt bridge contains a concentrated, mobile electrolyte (e.g., KCl or KNO3) in a gel or solution. In a galvanic cell electrons flow through the external wire from anode to cathode, so the salt bridge lets ions move through the internal circuit to keep each half-cell electrically neutral (anions migrate toward the anode half-cell, cations toward the cathode half-cell). Importantly, actual electrons do NOT move through the salt bridge—only ions. That ion migration prevents charge buildup that would stop the spontaneous reaction (CED EK 9.8.A.1: “ion flow through the salt bridge”). For a quick review and diagrams, check the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and more practice at the Unit 9 page (https://library.fiveable.me/ap-chemistry/unit-9) or the AP practice problems (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between a voltaic cell and a galvanic cell or are they the same thing?
They’re the same thing: “galvanic” and “voltaic” both mean a spontaneous electrochemical cell that produces electrical energy from a thermodynamically favored redox reaction. Key points to remember for AP Chem (CED 9.8.A): - Oxidation always occurs at the anode and reduction at the cathode (9.8.A.3). - Electrons flow through the external circuit from anode → cathode; ions move through the salt bridge to maintain charge. - Galvanic/voltaic cells have E°cell > 0 (spontaneous). By contrast, electrolytic cells drive nonspontaneous reactions with an external power supply (E°cell < 0 under standard conditions) (9.8.A.2). - Don’t worry about labeling electrodes as “positive/negative” for the AP exam—CED excludes that as an assessed skill. For a clear visual and practice, check the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and more unit review (https://library.fiveable.me/ap-chemistry/unit-9). For extra practice problems, use (https://library.fiveable.me/practice/ap-chemistry).
How do I draw and label an electrochemical cell diagram for the AP exam?
Draw a cell diagram using the standard line notation and label the physical parts. For a galvanic cell (spontaneous): - Left = anode (oxidation): e.g. Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) = right = cathode (reduction). Use single vertical bars for phase boundaries and a double bar for the salt bridge. - Show a salt bridge, voltmeter, and indicate electron flow (e−) from anode → cathode through the external wire. Note macroscopic/particulate changes: anode mass decreases (metal → cations), cathode mass increases (cations → metal), ions move through the salt bridge to maintain charge. For an electrolytic cell (nonspontaneous): - Same diagram form but include an external power supply forcing e− from the anode to cathode (direction may be opposite of spontaneous). Label which half-reaction is occurring at each electrode (oxidation at anode, reduction at cathode). On the AP exam: always state oxidation at anode and reduction at cathode; don’t worry about labeling electrodes as “+” or “−” (that won’t be assessed). For examples and practice, see the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and Unit 9 resources (https://library.fiveable.me/ap-chemistry/unit-9); practice questions available at (https://library.fiveable.me/practice/ap-chemistry).
Why do we need a voltage measuring device and what does it actually measure?
You need a voltage-measuring device (voltmeter) because it tells you the cell potential—the electrical potential difference (EMF) between the two half-cells. Practically, it measures how strongly electrons want to flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). For an open-circuit cell a good voltmeter (high internal resistance) reads Ecell: a positive Ecell means the overall redox is spontaneous (galvanic/voltaic); a negative reading means the reaction as written is nonspontaneous (you’d need an external power supply to make it happen, as in an electrolytic cell). What it actually measures: - The difference in electric potential (volts) set up by the two half-reactions (can be E°cell for standard conditions or Ecell from the Nernst equation when concentrations differ). - It doesn’t measure current flow (amps) unless you use an ammeter or change the setup; and a proper voltmeter’s high resistance ensures it doesn’t appreciably alter the cell’s reactions. For AP-style work you’ll connect Ecell to thermodynamics: ΔG° = −nFE°cell and use standard potentials or the Nernst equation to calculate values (see the Topic 9.8 study guide for examples: https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK). For more practice problems, check (https://library.fiveable.me/practice/ap-chemistry).
Can you explain at the particle level what's happening when gas bubbles form at an electrode?
At the particle level, gas bubbles form because electrochemical reactions at an electrode produce neutral gas molecules (e.g., H2 from 2H+ + 2e− → H2 or O2 from 2 H2O → O2 + 4 H+ + 4 e−). Ions in the solution are reduced or oxidized at the electrode surface and become neutral molecules. Those neutral molecules start as single molecules adsorbed on the electrode, then collide and cluster (nucleation). Once a tiny cluster reaches a critical size it’s energetically favorable to grow into a bubble. As more gas molecules form, the bubble grows, detaches when buoyancy overcomes surface adhesion, and rises away. Factors that change bubble behavior: current density (faster production → faster growth), electrode material/roughness (provides nucleation sites), and overpotential (affects reaction rate). This explanation ties to the CED: evolution of a gas at an electrode and particulate descriptions in 9.8.A.1. For practice problems and review on galvanic vs electrolytic cells, see the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK) and more practice questions (https://library.fiveable.me/practice/ap-chemistry).
I missed the electrochemistry lab - how do you actually set up a galvanic cell in real life?
You can set one up pretty simply—think two half-cells connected so electrons flow through a wire and ions through a salt bridge. Example (Daniell cell): place a strip of Zn(s) in 1.0 M ZnSO4(aq) and a Cu(s) strip in 1.0 M CuSO4(aq). Put each half-cell in its own beaker. Connect the metal strips with a voltmeter or wire with an alligator clip; the Zn electrode is the anode (oxidation: Zn → Zn2+ + 2e–) and electrons flow from Zn to Cu. Complete the circuit with a salt bridge (KNO3 or KCl in agar) between beakers to allow ion flow and keep charge neutrality. Measure Ecell with the voltmeter (should be ~1.10 V for standard Zn/Cu). Watch for electrode mass change (anode loses mass, cathode gains metal) and possible gas evolution if relevant. Wear goggles, gloves, and work in a fume hood if gases form. This directly maps to AP CED Topic 9.8 (identify electrodes, salt bridge, electron/ion flow, spontaneous vs. nonspontaneous). For a compact review, see the Topic 9.8 study guide (https://library.fiveable.me/ap-chemistry/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK). For more practice problems, check the AP Chem practice bank (https://library.fiveable.me/practice/ap-chemistry).