What you'll be able to do: Relate current and time to the mass of substance produced or consumed at an electrode using Faraday's law.
Best after: Electrolytic Cells and Driven Reactions
Electrolysis stoichiometry is ordinary stoichiometry with one extra conversion at the front: electric charge becomes moles of electrons. Learn that single bridge and every plating problem becomes a road map you already know how to follow.
These are recommended, not required. You can start this lesson at any time.
In an electrolysis, the balanced half-reaction says exactly how many electrons are consumed per ion converted. Ag⁺ + e⁻ → Ag needs one, Cu²⁺ + 2 e⁻ → Cu needs two, and Al³⁺ + 3 e⁻ → Al needs three.
Once electrons are treated as a reactant with a coefficient, standard stoichiometry takes over.
An ampere is a coulomb per second, so charge is simply q = It. Time must be in seconds, which means converting minutes or hours before multiplying. This unit conversion accounts for a large share of wrong answers on this topic.
For constant current, Q = It and 1 A = 1 C/s. If current varies, total charge is the area under its current-time graph, Q = ∫I dt; an average current can be used only when it represents that interval. Use Q for charge here to distinguish it from calorimetric heat q and the reaction quotient Q used elsewhere. Units and context identify which quantity is meant.
q = I t
One mole of electrons carries 96485 C, a value equal to the elementary charge times Avogadro number. Dividing total charge by F gives moles of electrons.
mol e⁻ = q / F = I t / 96485
Amperes and seconds give coulombs; coulombs divided by F give moles of electrons; the half-reaction ratio gives moles of substance; molar mass gives grams, or PV = nRT gives a gas volume. Any single unknown along the chain can be solved by running the same road map backward.
For copper deposition, m = It × (1 mol e⁻/F) × (1 mol Cu/2 mol e⁻) × M(Cu). A 2.00 A current for 1800 s passes 3600 C and produces 3600(63.55)/[2(96485)] = 1.19 g Cu at 100% current efficiency. Writing the conversion factors shows why the electron coefficient belongs in the denominator.
For hydrogen, 2 H₂O + 2 e⁻ → H₂ + 2 OH⁻ also requires two electrons per mole gas. Passing 1930 C produces approximately 0.0100 mol H₂. At 298 K and dry H₂ pressure 1.00 atm, V = nRT/P ≈ 0.245 L. If collected saturated over water, subtract the water vapor pressure from the corrected total pressure before finding dry H₂ volume or moles. There is no universal molar gas volume independent of T and P.
I, t → q → mol e⁻ → mol substance → mass or volume
For a fixed quantity of charge, the moles deposited are inversely proportional to the ionic charge. The same current for the same time deposits three times as many moles of silver as of aluminum. In cells wired in series the current and time are shared, so this comparison is exact.
Equal charge in series cells means equal moles of electrons, not equal moles or masses of metal. Their mass ratio is (M₁/z₁)/(M₂/z₂), assuming the stated efficiencies match. Two electrons deposit one Cu atom but two Ag atoms; molar mass must still be applied after the electron ratio. The identity and oxidation state of the actual reacting ion determine z, not a metal's most familiar charge from memory.
mol substance = (mol e⁻) / (electrons per formula unit)
Industrial cells rarely convert every electron into the desired product, since side reactions such as hydrogen evolution consume some charge. Current efficiency is the fraction that does useful work, so the actual mass equals the theoretical mass times that efficiency.
actual mass = theoretical mass x current efficiency
Faraday's law counts total charge Q=It, but only the fraction used by the desired electrode reaction produces the target substance. Multiply total charge by current efficiency before converting to moles of electrons and product. Side reactions can lower yield without changing the measured current.
To find time for a target mass at efficiency η, invert the chain: t = mzF/(ηIM), where z is electrons per product unit. For 10.0 g Cu, I = 5.00 A and η = 0.800, t ≈ (10.0)(2)(96485)/[(0.800)(5.00)(63.55)] ≈ 7590 s, or 2.11 h. Lower efficiency increases the required time; multiplying the ideal time by 0.800 would predict less time despite wasting charge. Efficiency is not a replacement for stoichiometry and is distinct from a later product-recovery loss.
q = It, with I in amperes and t in seconds.
F = 96485 C per mole of electrons.
mol e⁻ = It/F.
Taken from the balanced half-reaction, for example 2 e⁻ per Cu.
Cells in series pass the same charge, so their electron counts are equal.
q = I t
mol e⁻ = I t / F
mass = (I t / F) × (1 / z) x M
A current of 2.00 A is passed through a AgNO₃ solution for 30.0 minutes. What mass of silver plates onto the cathode? M(Ag) = 107.87 g/mol.
Try it first: Write the cathode half-reaction and note the electron coefficient.
0 of 4 steps revealed.
How long must a 5.00 A current run to deposit 10.0 g of copper from Cu²⁺? M(Cu) = 63.55 g/mol.
Try it first: Run the road map backward from mass to time.
0 of 4 steps revealed.
Molten alumina is electrolyzed at 1.00 × 10⁴ A for 1.00 hour. What mass of aluminum is produced at 100 percent efficiency? M(Al) = 26.98 g/mol.
Try it first: Aluminum is 3+, which changes the electron ratio.
0 of 4 steps revealed.
A 2.00 A current runs 3600 s through Cu²⁺ solution at 80.0% current efficiency. How much Cu plates? Use F=96485 C/mol and M(Cu)=63.55 g/mol.
Try it first: Only 80.0% of charge reduces Cu²⁺.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Using minutes or hours directly in q = It. | An ampere is a coulomb per second, so only seconds give the correct charge. | Convert time to seconds first. |
| Skipping the electron-to-substance ratio. | Moles of electrons equal moles of product only when the half-reaction needs one electron. | Divide by the electron coefficient from the half-reaction. |
| Multiplying instead of dividing by F. | F converts coulombs to moles, so it belongs in the denominator. | Check that the coulombs cancel in the unit analysis. |
| Assuming equal masses deposit for equal charge. | Different ionic charges and molar masses give different masses per mole of electrons. | Compare moles of electrons first, then convert each metal separately. |
| Ignoring current efficiency in an industrial problem. | Side reactions consume charge, so the real yield is lower than the calculation. | Multiply the theoretical mass by the stated efficiency. |
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After the lesson · Question bank
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Faraday's law makes electrons a stoichiometric reagent. Charge in coulombs is current times time, q = It, with current in amperes and time in seconds. Dividing by the Faraday constant, 96485 C per mole of electrons, converts charge into moles of electrons. From there the balanced half-reaction supplies the mole ratio between electrons and the substance being deposited or evolved, and molar mass converts to grams or the ideal gas law converts to a gas volume. The complete chain is current and time, then coulombs, then moles of electrons, then moles of substance, then mass or volume, and every step can be run in reverse when the unknown is the current or the plating time. Because the electron-to-substance ratio depends on ionic charge, the same quantity of charge deposits half as many moles of Cu²⁺ as of Ag⁺, and one third as many of Al³⁺. When electrolytic cells are connected in series they pass identical charge, which is the basis of the classic comparison problem.
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