AP Chem Score Calculators

AP Chemistry Electrolysis

Faraday's law calculations for mass, time, and gas volume, in four worked examples.

An electrolysis problem looks different each time, plating a metal, timing a deposit, collecting a gas, but it is one chain of conversions that always starts from current and time. Once the chain is memorized, the question just decides where you enter it and where you leave. The Unit 9 review shows one copper example; this page works the other kinds you are likely to see.

The Electrolysis Conversion Chain From Current and Time to Mass

  1. Write the half-reaction and find n, the moles of electrons per mole of product.
  2. Find the charge: q = I × t, with current in amperes and time in seconds.
  3. Convert charge to moles of electrons: divide by F = 96,485 C/mol e−.
  4. Convert to moles of product: divide by n.
  5. Convert to mass with molar mass, or to gas volume with the ideal gas law.
The Electrolysis Roadmap: From Current and Time to Mass current × time I (A), t (s) q = It charge q (C) ÷ F electrons mol e⁻ ÷ n product mol × M mass grams F = 96,485 C/mol e⁻; n = mol e⁻ per mol product, from the half-reaction (Cu²⁺ + 2e⁻ → Cu: n = 2); M = molar mass. Working backward (time or current from a mass): run the same steps in reverse, dividing by molar mass first and multiplying by n and F. Time must be in seconds, because 1 ampere = 1 coulomb per second.

In an electrolytic cell an outside current drives a reaction that would not otherwise occur; reduction happens at the cathode and oxidation at the anode, as in galvanic cells. The conversions above are mole-ratio stoichiometry with electrons as one more species, the same method as in stoichiometry.

Worked Example 1: Mass of Copper Plated by a Steady Current

A current of 2.50 A passes through a solution of Cu2+ for 1.00 hour. What mass of copper is deposited?

Half-reaction: Cu2+ + 2 e− → Cu, so n = 2.
Charge: q = (2.50 A)(3600 s) = 9000 C.
Moles of electrons: 9000 C ÷ 96,485 C/mol = 0.0933 mol.
Moles of copper: 0.0933 ÷ 2 = 0.0466 mol.
Mass: 0.0466 mol × 63.55 g/mol = 2.96 g.

Worked Example 2: Comparing Aluminum With Copper at the Same Charge

What mass of aluminum is deposited by 5.00 A flowing for 30.0 minutes through a solution of Al3+, and how does that compare with the copper in Example 1?

Half-reaction: Al3+ + 3 e− → Al, so n = 3.
Charge: q = (5.00 A)(1800 s) = 9000 C, the same as before.
Moles of electrons: 0.0933 mol, the same.
Moles of aluminum: 0.0933 ÷ 3 = 0.0311 mol.
Mass: 0.0311 mol × 26.98 g/mol = 0.839 g.

The same charge gives only 0.839 g of aluminum against 2.96 g of copper. Aluminum needs 3 electrons per atom instead of 2, so fewer atoms are produced, and each aluminum atom is much lighter than a copper atom. Always take n from the half-reaction rather than guessing.

Worked Example 3: Finding How Long to Plate a Given Mass of Silver

How long does a 0.500 A current take to deposit 1.00 g of silver from Ag+?

Work the chain backward.
Moles of silver: 1.00 g ÷ 107.87 g/mol = 0.00927 mol.
Moles of electrons: Ag+ + e− → Ag has n = 1, so 0.00927 mol e−.
Charge: 0.00927 mol × 96,485 C/mol = 894.5 C.
Time: t = q ÷ I = 894.5 C ÷ 0.500 A = 1789 s = 29.8 minutes.

Worked Example 4: Gas Volumes From the Electrolysis of Water

A 2.00 A current flows for 965 s through water. What volume of O2 forms at the anode at 298 K and 1.00 atm, and what volume of H2 forms at the cathode?

Charge: (2.00 A)(965 s) = 1930 C, so 1930 ÷ 96,485 = 0.0200 mol e−.
Anode: 2 H2O → O2 + 4 H+ + 4 e−, so n = 4 per O2: 0.0200 ÷ 4 = 0.00500 mol O2.
Volume of O2: V = nRT/P = (0.00500)(0.08206)(298) ÷ 1.00 = 0.122 L.
Cathode: 2 H2O + 2 e− → H2 + 2 OH−, so n = 2 per H2: 0.0200 ÷ 2 = 0.0100 mol H2, a volume of 0.245 L.

Twice as much H2 as O2 forms, matching the 2 : 1 ratio in the balanced equation 2 H2O → 2 H2 + O2.

Five Electrolysis Practice Questions With Answers

  1. How many moles of electrons are in 3860 C of charge?
    Show answer3860 ÷ 96,485 = 0.0400 mol e−.
  2. What mass of silver is deposited by 1.00 A flowing for 965 s?
    Show answerq = 965 C = 0.0100 mol e−; n = 1, so 0.0100 mol Ag = 1.08 g.
  3. How long does 0.500 A take to deposit 0.0100 mol of copper?
    Show answer0.0100 mol Cu needs 0.0200 mol e− = 1930 C; t = 1930 ÷ 0.500 = 3860 s, about 64.3 minutes.
  4. Why does a mole of aluminum need more charge than a mole of sodium?
    Show answerAl3+ needs 3 mol e− per mole of metal and Na+ needs 1, so aluminum needs three times the charge per mole.
  5. Why must time be converted to seconds before calculating charge?
    Show answerAn ampere is one coulomb per second, so q = I × t gives coulombs only when t is in seconds.

Common Electrolysis Mistakes

Electrolysis connects directly to the free energy of a reaction, since driving a nonfavored process requires an external energy source; the link is explored in Gibbs free energy. To estimate how a strong Unit 9 would affect your overall result, the AP Chem Score Calculator converts section scores into a 1–5 range.

Frequently Asked Questions

What is the formula for electrolysis calculations?

There is one chain rather than one formula: charge q = I x t (coulombs, with time in seconds), moles of electrons = q / 96,485, moles of product = moles of electrons / n, and mass = moles of product x molar mass. Combined, m = (I x t x M) / (n x F).

What is the Faraday constant?

F = 96,485 coulombs per mole of electrons. It is the charge carried by one mole of electrons and converts between coulombs and moles of electrons.

How do you find n in an electrolysis problem?

Write the half-reaction for the product and read off how many electrons are transferred per mole of product. For Cu2+ + 2e- -> Cu, n = 2; for Ag+ + e- -> Ag, n = 1; for Al3+ + 3e- -> Al, n = 3.

How do you find the time needed to deposit a given mass?

Work backward: convert the mass to moles of product, multiply by n to get moles of electrons, multiply by 96,485 to get charge, and divide by the current to get time in seconds.

Why does the same charge deposit different masses of different metals?

The same charge delivers the same moles of electrons, but different metals need different numbers of electrons per atom (n) and have different molar masses. The mass deposited depends on both.

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