SCIENCE · INVESTIGATE AND EXPLAIN

Infer deposited material from charge and stoichiometry

Convert current and time to electron amount, deposited mass and conditional efficiency.

Learning outcome: Convert current and time to electron amount, deposited mass and conditional efficiency.

Starting recap

Current measures charge per second; electron stoichiometry connects charge to chemical amount.

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Try

A 1.0 A current runs for 600 s through a stipulated Cu²⁺ reduction with 90% current efficiency.

Open the experiment

Predict

How do mass and side-reaction charge change if current or time doubles?

Your prediction and reason

Experiment

Compare current 1 and 2 A at 600 s, then time 600 and 1200 s at 1 A. Inspect zero-current and zero-time limits.

Close charge and deposited-mass balances

Worked teaching model · synthetic values · Assigned constant current; Cu²⁺+2e−→Cu; M=63.546 g/mol, F=96485.33212 C/mol e−. A stipulated 90% current efficiency sends the rest to unmodeled side reactions. This is accounting, not a physical electrolysis recipe or measured yield.

Current: 1 A Elapsed time: 600 s

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Conditional calculations, not measured observations
Quantity	Value	Unit
Charge passed	600	C
Electron amount	0.006218562	mol e−
100% current-efficiency copper mass	0.1975824	g
Assigned 90% copper mass	0.1778241	g
Charge in side reactions	60	C

Mass follows charge, electron stoichiometry and current efficiency.

Current in A is C/s, so charge is I×t. Divide by F, then by two electrons per copper atom, then multiply by molar mass. Electrode identity, competing reactions and collection losses require additional evidence in an actual experiment.

Predict before changing a control. Save two contrasting captures and explain what is held fixed. Displayed digits are calculation precision, not experimental uncertainty.

Mass follows charge, electron stoichiometry and current efficiency.

Observation to attach to a capture

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Understand

Build the conversion chain

Charge q = I t for a constant current. One ampere is one coulomb per second. Divide charge by the Faraday constant to obtain moles of electrons, divide by the balanced electron requirement per metal atom, and multiply by molar mass. For Cu²⁺+2e−→Cu, the ideal mass is m = ItM/(2F).

With I=1.0 A and t=600 s, q=600 C. The electron amount is approximately 0.006219 mol, the ideal copper amount is about 0.003109 mol, and the ideal mass is about 0.1976 g using M=63.546 g/mol. The lab also stipulates 90% current efficiency, giving approximately 0.1778 g assigned copper and 60 C sent to unspecified side reactions.

Current efficiency is the fraction of charge associated with the desired reaction. It is not the same as collection efficiency, purity, electrical energy efficiency or a statistical confidence level. A real mass discrepancy requires investigation, not automatic attribution to one of those causes.

Use an inverse calculation

If a fictional run deposits 0.100 g copper in 600 s at 100% current efficiency, q=2F×0.100/63.546≈303.7 C and I≈0.506 A. At 90% efficiency the required total charge is larger by 1/0.9, and the current is about 0.562 A. State the efficiency assumption before reporting an inferred current.

If a different cation requires three electrons per deposited atom, replace the electron coefficient by three. Do not carry copper’s coefficient into another reaction because both products happen to be metals. Balance charge in the half-reaction first.

Separate amount, voltage and energy

Faraday accounting predicts chemical amount from charge under the stated yield assumptions. It does not determine the applied voltage or electrical energy. Electrical input is the time integral of voltage times current, and a real cell may require an applied potential above its reversible opposing value. Gas formation and competing chemistry can consume part of the current.

No physical electrolysis procedure is prescribed here. Actual chemical handling, power supplies, gases, exposure controls and waste disposal require qualified supervision and an approved laboratory protocol. This browser ledger is not evidence of laboratory technique or safety competence.

Transfer and uncertainty

A measured mass gain exceeds the 100%-efficient Faraday prediction. Do not report a current efficiency above one as a successful plating mechanism without qualification. Possible explanations include retained liquid, contamination, another deposited material, an incorrect current/time record or a wrong reaction model. Choose an additional measurement that distinguishes explanations. The numerical model has no such measurement and cannot select a unique cause.

Source scope and continuation

Original teaching explanation, calculations and practice. Framework alignment: AP Chemistry topic 9.11 in the College Board course framework, effective Fall 2024, checked 30 October 2026. These pages do not reproduce restricted examination questions or establish AP authorization. The investigation is a bounded computational model, not a conducted laboratory.

Return to the preceding Unit 9 lesson when its reasoning is unfamiliar.

Complete the Unit 9 mixed review, keeping numerical checks separate from independent reasoning review.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Show your calculation or representation, explain the changed case, and separate the stated model from measured evidence. Convert current and time to electron amount, deposited mass and conditional efficiency.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Explain potential changes with the reaction quotient Next: AP Chemistry preparation and next units

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