F. Electrochemistry · Electrochemistry

Faraday's Law: The Stoichiometry of Electrolysis

F2 F3.5 A1.13About 70 minutes

Learning goals

I can calculate the charge passed in a cell from the current and the time.
I can convert between charge and moles of electrons using the Faraday constant.
I can calculate the mass of metal or volume of gas produced in electrolysis, or the time or current needed.
I can apply Faraday's law to plating, refining and battery capacity.

Words to know

electric current (I)
The rate of flow of charge, in amperes (A). 1 A = 1 coulomb per second.

coulomb (C)
The SI unit of electric charge. Q = I × t, with t in seconds.

Faraday constant (F)
The charge of one mole of electrons: 96 485 C/mol (often rounded to 9.65 × 10⁴ C/mol).

Faraday's law of electrolysis
The amount of substance produced or used at an electrode is directly proportional to the quantity of charge passed through the cell.

ampere-hour (A·h)
A unit of charge used on batteries: 1 A·h = 3600 C; 1 mA·h = 3.6 C.

Counting electrons with an ammeter

A jewellery maker wants exactly 0.50 g of silver on a ring. An aluminium smelter wants to know how much metal a potline makes per day. Both questions have the same answer: count the electrons.

In the 1830s, Michael Faraday showed that the amount of substance made at an electrode is proportional to the charge passed. Double the current, or double the time, and you double the product. This is Faraday's law of electrolysis.

Careful: in Grade 11 Physics (SPH3U) you met Faraday's law of electromagnetic induction. Same scientist, different law. This lesson is about electrolysis.

The three key relationships

Charge from current and time: Q = I × t (Q in coulombs, I in amperes, t in seconds).
Moles of electrons from charge: n(e⁻) = Q ÷ F, where F = 96 485 C/mol is the Faraday constant, the charge on one mole of electrons. (It equals Avogadro's number times the charge of one electron: 6.022 × 10²³ × 1.602 × 10⁻¹⁹ C ≈ 96 485 C.)
Mole ratio from the half-reaction: for example, Cu²⁺ + 2e⁻ → Cu means 2 mol e⁻ make 1 mol Cu. Then use molar mass (or molar volume for gases) as in SCH3U stoichiometry.

The Faraday road map. Current and time give charge; dividing by F gives moles of electrons; the half-reaction gives moles of product; molar mass or molar volume gives the amount you can weigh or measure.

Worked example

Example 1: Copper plating

A current of 2.00 A passes through a copper(II) sulfate solution for 30.0 min. What mass of copper is plated on the cathode? (M of Cu = 63.55 g/mol)

Time in seconds: t = 30.0 min × 60 s/min = 1800 s.
Charge: Q = I × t = 2.00 A × 1800 s = 3600 C.
Moles of electrons: n(e⁻) = 3600 C ÷ 96 485 C/mol = 0.037 31 mol.
Half-reaction: Cu²⁺ + 2e⁻ → Cu, so n(Cu) = 0.037 31 ÷ 2 = 0.018 66 mol.
Mass: m = 0.018 66 mol × 63.55 g/mol = 1.186 g.

1.19 g of copper (three significant figures, matching 2.00 A and 30.0 min).

Worked example

Example 2: How long to plate silver?

How many minutes does it take to plate 5.00 g of silver onto a spoon using a current of 0.750 A? (M of Ag = 107.87 g/mol)

Run the road map backwards. n(Ag) = 5.00 g ÷ 107.87 g/mol = 0.046 35 mol.
Half-reaction: Ag⁺ + e⁻ → Ag, so n(e⁻) = n(Ag) = 0.046 35 mol.
Charge: Q = n(e⁻) × F = 0.046 35 × 96 485 = 4472 C.
Time: t = Q ÷ I = 4472 C ÷ 0.750 A = 5963 s = 99.4 min.

About 99.4 min (about 1 h 39 min).

Worked example

Example 3: A gas product

In a chlor-alkali cell, a current of 10.0 A flows for 1.00 h. What volume of chlorine gas forms at SATP (Vₘ = 24.8 L/mol)?

t = 1.00 h = 3600 s. Q = 10.0 × 3600 = 36 000 C.
n(e⁻) = 36 000 ÷ 96 485 = 0.3731 mol.
Anode: 2Cl⁻ → Cl₂ + 2e⁻, so n(Cl₂) = 0.3731 ÷ 2 = 0.1866 mol.
V = n × Vₘ = 0.1866 × 24.8 = 4.63 L.

4.63 L of Cl₂ at SATP.

At the same current (2.00 A), mass grows in a straight line with time, as Faraday's law says. Silver builds up about 3.4 times as fast as copper by mass: each silver ion needs only one electron, and silver atoms are heavier.

Worked example

Example 4: The current for an aluminium smelter

Aluminium is made by electrolysis (Al³⁺ + 3e⁻ → Al). What constant current would produce 1.00 kg of aluminium per hour? (M of Al = 26.98 g/mol)

n(Al) = 1000 g ÷ 26.98 g/mol = 37.06 mol.
n(e⁻) = 3 × 37.06 = 111.2 mol.
Q = 111.2 × 96 485 = 1.073 × 10⁷ C.
I = Q ÷ t = 1.073 × 10⁷ C ÷ 3600 s = 2.98 × 10³ A.

About 2980 A, for only 1 kg per hour. Real smelter cells run at currents of hundreds of thousands of amperes, which is why aluminium smelters are built where electricity is plentiful, such as Québec with its hydroelectric power.

Cells in series

When several electrolytic cells are connected in series, the same current flows through each one for the same time, so the same number of moles of electrons passes through each. The amounts of product differ only because of the half-reactions.

For example, if a silver cell and a copper cell are in series and 1.50 g of silver is deposited: n(e⁻) = n(Ag) = 1.50 ÷ 107.87 = 0.013 91 mol, so n(Cu) = 0.006 953 mol and m(Cu) = 0.442 g. No current or time is needed.

Battery capacity is a charge

Battery labels often give capacity in milliampere-hours (mA·h). It is a charge: 1 mA·h = 0.001 A × 3600 s = 3.6 C. So Faraday's law tells you how much chemical a battery must use up to deliver its rated charge.

Just for fun

A single mole of electrons carries 96 485 C. If you could pass that charge through a 60 W, 120 V bulb (0.5 A), it would keep the light on for about 53 hours. Electrons are tiny, but there are a lot of them in a mole.

Hands-on activity

How much zinc is in your battery's charge?

You need: a calculator and the labels or online specification sheets of batteries you own (rechargeable AA cells and phones usually list a capacity in mA·h; alkaline cells often don't, so use 2500 mA·h as a typical value for an AA alkaline at low drain). Don't open any battery.

Record the capacity of three batteries in mA·h.
Convert each to coulombs: Q = capacity (mA·h) × 3.6 C per mA·h.
Convert to moles of electrons: n(e⁻) = Q ÷ 96 485.
An alkaline cell's anode reaction uses zinc: Zn → Zn²⁺ + 2e⁻ (in alkaline cells the zinc ends up as zinc oxide). Calculate the minimum mass of zinc an alkaline AA must contain to deliver 2500 mA·h.
For a lithium-ion phone battery, each lithium atom gives one electron. Calculate the mass of lithium that moves from anode to cathode during one full discharge of your phone (M of Li = 6.94 g/mol).

Model answer for step 4: 2500 × 3.6 = 9000 C; 9000 ÷ 96 485 = 0.0933 mol e⁻; n(Zn) = 0.0466 mol; m = 0.0466 × 65.38 = 3.05 g of zinc. For a 4000 mA·h phone battery: 14 400 C; 0.149 mol e⁻ = 0.149 mol Li = 1.04 g of lithium. A real cell contains extra beyond these minimums.

Your turn

A current of 0.500 A flows for 10.0 min. How much charge passes?

How many moles of electrons carry 1930 C?

What mass of silver is deposited by 0.0200 mol of electrons?

A current of 3.00 A is passed through molten NaCl for 20.0 min. What mass of sodium forms? (Na 22.99 g/mol)

How many moles of electrons are needed to make 1.00 mol of aluminium from Al³⁺?

What volume of hydrogen gas at SATP is produced at the cathode when 0.250 mol of electrons pass during water electrolysis?

Two cells in series contain Ag⁺ and Cu²⁺. If 0.020 mol of Cu forms, how many moles of Ag form?

More practice: warm-up

What is the value of the Faraday constant?

Write the equation linking charge, current and time.

Convert 5.00 min to seconds.

How many coulombs are in 1 A·h?

How many moles of electrons are needed to deposit 1 mol of Cu from Cu²⁺?

More practice: on your own

What mass of zinc is plated by a current of 1.50 A in 45.0 min? (Zn 65.38 g/mol)

A part needs 0.250 g of nickel plated from Ni²⁺ in 15.0 min. What current is required? (Ni 58.69 g/mol)

How long does it take a current of 5.00 A to produce 2.00 L of O₂ at SATP from water? (2H₂O → O₂ + 4H⁺ + 4e⁻)

An unknown metal M forms M³⁺ ions. Passing 0.600 A for 30.0 min deposits 0.735 g of M. Find the molar mass and identify the metal.

Cells containing AgNO₃(aq) and CuSO₄(aq) are connected in series. After some time, 2.16 g of Ag has formed. What mass of Cu has formed?

Challenge

A hydrogen electrolyzer runs at 1.80 V and 500 A for 24.0 h. (a) What mass of H₂ does it make? (b) How much electrical energy does it use (E = V × Q)? (c) Express the energy per kilogram of hydrogen in kW·h (1 kW·h = 3.6 × 10⁶ J).

A chrome-plating bath uses Cr³⁺. A bumper needs a layer 2.0 × 10⁻⁴ cm thick over 1.50 × 10³ cm² of surface. Chromium's density is 7.19 g/cm³ (M = 52.00 g/mol). At 25.0 A, how long does plating take, assuming all the current plates chromium?

In an electroplating lab, 0.900 g of copper was expected but only 0.855 g was weighed. Calculate the current efficiency and suggest two causes.

Check your understanding

Quiz

What is the charge passed when 4.00 A flows for 5.00 min?

How many moles of electrons are needed to deposit 0.100 mol of Cu from Cu²⁺(aq)?

The same charge is passed through solutions of Ag⁺ and Al³⁺. Compared with the moles of Ag formed, the moles of Al formed are:

Which change would double the mass of metal plated in an electrolysis?

A current of 0.965 A flows for 1000 s through AgNO₃(aq). About what mass of silver forms?

A phone battery is rated 3000 mA·h. What charge can it deliver?

For parents and teachers

Big idea (F2, quantitative): in electrolysis, electrons are a reactant you can count with an ammeter and a clock. Q = I × t, n(e⁻) = Q ÷ F, then the half-reaction gives the mole ratio.

How to help: Ask your teen to draw the road map (current and time → charge → moles of electrons → moles of product → mass) and then solve one problem forwards and one backwards. Check units at every step.

Common difficulties:

Using minutes or hours instead of seconds.
Forgetting the electron ratio from the half-reaction (2 for Cu²⁺, 3 for Al³⁺).
Confusing this Faraday's law with Faraday's law of induction from physics.
Rounding too early; keep extra digits until the final answer.

Safety: the activity uses only battery labels. Do not open batteries.