Electrolysis and Faraday’s Law
Convert charge and current-time data through electron stoichiometry to chemical amount, with efficiency assumptions.
AP Chemistry Topic 9.11 · CED printed p. 178 / PDF p. 185. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Half-reaction electron balance
- Current and charge
- Dimensional analysis
Current measures charge flow per time: , and one ampere is one coulomb per second. For constant current q = It; if current varies, use the area under its current–time graph or a justified average. Do not multiply amperes by minutes without converting time to seconds. Faraday’s constant, about 96485 C per mole of electrons, converts charge to electron amount: ne = q/F. Charge in this lesson is q; the capital Q used for a reaction quotient is a different quantity.
Next use the balanced electrode half-reaction. Mz+ + z e− → M deposits one mole of metal per z moles of electrons, so nM = q/(zF) and mass = nM × molar mass. Oxidation can remove material with the analogous electron ratio. Work backward from an observed mass change to charge or current when needed. Equal charge can produce different metal amounts because ionic electron requirements differ, and different masses because molar masses differ too. These are stoichiometric conversions, not memorized metal-specific formulas.
A theoretical result usually assumes 100% current efficiency, a known current/time record, no competing electrode reactions, enough reactant, and material retained on the electrode being measured. A smaller deposited mass may reflect a competing process or lost deposit, not automatically a smaller electron charge. Be explicit about direction: reduction can deposit metal, while oxidation can dissolve it; an anode need not lose mass if an inert support hosts a dissolved-species or gas reaction. Data interpretation should identify what was actually weighed and which half-reaction carried the charge.
Let's work through it
- A model Zn anode loses 0.654 g by Zn → Zn2+ + 2e−. With MZn = 65.4 g mol−1, nZn = 0.654/65.4 = 0.0100 mol.
- The oxidation releases 0.0200 mol electrons. q = 0.0200 × 96485 = 1.93 × 10³ C.
- If this occurs in 600. s, the average current is q/t = 3.22 A. A constant-current assumption is needed only to interpret that as the value throughout the interval.
- For another two-electron metal process, the same electron amount corresponds to 0.0100 mol metal, but its mass needs that metal’s own molar mass.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Why divide electron moles by two for Cu2+ plating?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
Why divide electron moles by two for Cu2+ plating?
Worked answer and reason
The reduction consumes two electrons per copper atom.
The half-reaction gives the mole conversion factor.
Retrieval 2
What changes if the current varies with time?
Worked answer and reason
Use total integrated charge or a justified time average, not an arbitrary single reading.
Chemical amount depends on accumulated electron flow.
Catch the tempting mistake
- Every coulomb contributes to the measured desired metal deposit.
- That is a 100% current-efficiency assumption; competing reactions or deposit loss can lower measured mass.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 9 problem 16 · ap synthesis · data
- Unit 9 problem 17 · ap application · symbolic
Scientific calculator workspace
Use this workspace for arithmetic, logarithms and elementary functions. Keep intermediate precision and report final values with appropriate units and significant figures. Graphing and general equation solving need a suitable approved tool.
Scientific calculator
This calculator requires JavaScript. Use the physics formula sheets and show your working manually.
+ − * / ^ ! · pi · e · sqrt · sin, cos, tan · asin, acos, atan · atan2(y,x) · ln, log · abs · exp · hypot(x,y). Use * between factors. Keep units in your written working.
For the teacher
- Ask for a units-canceling chain from A and s to C, mol electrons, mol metal, and grams. Require efficiency assumptions in the final interpretation.
- The shared electrochem model does not calculate charge, current, time, or plating mass. Use the independent Faraday data and dimensional-analysis tasks here for those requirements.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 9.11.A
- Calculate the amount of charge flow based on changes in the amounts of reactants and products in an electrochemical cell.
- 9.11.A.1
- Faraday’s laws can be used to determine the stoichiometry of the redox reaction occurring in an electrochemical cell with respect to the following: i. Number of electrons transferred ii. Mass of material deposited on or removed from an electrode (as in electroplating) iii. Current iv. Time elapsed v. Charge of ionic species EQN: I = q/t
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 — Required topic-numbered learning objectives and essential knowledge, all continuations and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
- OpenStax Chemistry 2e §17.7, Electrolysis — Faraday charge-to-chemical-amount conversions and electrolytic operation.
Glossary and deeper reading · 7 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
-
Half-reaction
A half-reaction describes either the oxidation or reduction part of a redox process. It includes electrons explicitly so that charge and electron transfer can be balanced. For example, zinc forming a zinc ion and two electrons is an oxidation half-reaction.
-
Mole
A mole is an amount of substance containing exactly 6.02214076 × 10²³ specified entities. The entities can be atoms, molecules, ions, or formula units, so the counted entity must be named. For example, one mole of oxygen molecules contains two moles of oxygen atoms.
-
Reaction quotient
The reaction quotient describes a mixture's current composition using the equilibrium-expression form. It uses species activities, approximated by concentrations or partial pressures in suitable models, with powers from balanced coefficients. For example, a quotient below the equilibrium constant favors net forward reaction toward equilibrium.
-
Molar mass
Molar mass is the mass of one mole of a specified substance. It is commonly expressed in grams per mole and connects a sample's mass to its amount of substance. For example, water has a molar mass of approximately 18.02 grams per mole.
-
Oxidation
Oxidation is electron loss, represented by an increase in oxidation number. It occurs together with reduction so that electrons are conserved in the overall reaction. For example, a zinc atom becoming a zinc ion with charge +2 undergoes oxidation.
-
Current efficiency
Current efficiency is the fraction of transferred charge that produces the specified electrode process. Competing reactions or lost material can make the observed yield lower than the ideal charge-based prediction. For example, an 80 percent efficiency means only 0.80 of the total charge contributes to the selected deposition process.
-
Reduction
Reduction is electron gain, represented by a decrease in oxidation number. It is paired with oxidation in an electron-transfer reaction. For example, a copper ion with charge +2 gaining two electrons to form copper metal undergoes reduction.