Chemistry: Atoms First 2e · Electrochemistry

Electrolysis

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

is the use of electrical energy to drive a nonspontaneous redox reaction — the exact opposite of a galvanic cell. A galvanic cell converts chemical energy into electricity on its own; an takes electricity from an external power source and forces chemistry to run backward: Na⁺ becomes sodium metal, Al³⁺ becomes aluminum metal, and water splits into H₂ and O₂. The key quantitative idea is Faraday's laws of electrolysis: the mass of substance produced is proportional to the charge passed, so a simple calculation connects current, time, and moles of electrons to grams of product. This topic covers how electrolytic cells are wired (which electrode is the , which is the ), what happens when water competes with dissolved ions, and the industrial and everyday uses of electrolysis — from aluminum smelting and chlorine production to and rechargeable batteries.

Why this matters

Industrial chemistry: Aluminum is produced almost entirely by the Hall–Héroult electrolysis of Al₂O₃; chlorine and sodium hydroxide come from electrolyzing brine (the ).
Everyday objects: Electroplating deposits the silver on cutlery, the gold on jewelry, and the chromium on car bumpers; electrolysis also refines copper for electrical wiring.
Energy technology: Electrolysis of water produces hydrogen fuel; rechargeable batteries are electrolytic cells while charging (the discharge is galvanic).
Body and safety context: Electrolyte balance and ion movement in the body are governed by similar charge-flow principles, and understanding electrolysis helps make sense of how batteries in medical devices are recharged.
Exams: Faraday's-law stoichiometry problems (m = MIt/nF) are among the most reliable calculation points in the electrochemistry chapter.

The college version

Core Concepts

Electrolytic vs. galvanic cells

Both cell types have an anode (oxidation) and a cathode (reduction), and electrons always flow from anode to cathode through the external circuit. The difference is the sign of E^°_cell and the source of energy:

Galvanic (voltaic) cell: spontaneous, E^°_cell > 0, produces electrical energy; anode is negative, cathode is positive.
Electrolytic cell: nonspontaneous, E^°_cell < 0, consumes electrical energy; the anode is connected to the positive terminal of the power source (so anode is positive, cathode negative).

The applied voltage must exceed the cell's inherent reverse potential; the extra amount needed beyond E^°_cell is called .

Electrolysis of molten ionic compounds

Molten NaCl contains mobile Na⁺ and Cl^- ions with no water present. Applying a voltage gives:

$Cathode: Na + + e - \to Na (l) Anode: 2 Cl - \to Cl 2 (g) + 2e -$

The compound must be molten (or dissolved) so ions can move; a solid salt crystal conducts almost no current because its ions are locked in place.

Aqueous solutions: the competition with water

In aqueous electrolysis, water itself can be oxidized or reduced, so it competes with the dissolved ions. For aqueous NaCl, the possible cathode reactions are Na⁺ + e^- → Na (-2.71 V) and 2H₂O + 2e^- → H₂ + 2OH^- (-0.83 V); the less negative (easier) reduction is water, so the products are H₂ and Cl₂, not sodium metal — the basis of the chlor-alkali industry. Predicting aqueous electrolysis products therefore requires comparing the potentials of the ion reactions against the potentials of water's own oxidation and reduction (with overpotential caveats).

Faraday's laws and quantitative electrolysis

One mole of electrons carries a fixed charge, the Faraday constant:

$F = 96,485 C mol -1$

The charge passed by a steady current is q = It, so the moles of electrons are:

$n(e -) = It F$

Each mole of product requires a stoichiometric number of moles of electrons, giving the mass produced:

$m = MIt nF$

where M is the molar mass of the product, I the current in amperes (C s^-1), t the time in seconds, n the moles of electrons per mole of product, and F the Faraday constant. Two moles of product require twice the charge — this proportionality is the essence of Faraday's first law.

Applications of electrolysis

: molten Al₂O₃ in cryolite; Al³⁺ + 3e^- → Al.
Chlor-alkali process: electrolysis of brine gives Cl₂, H₂, and NaOH.
Electroplating: the object to be plated is the cathode; metal ions in solution (e.g., Ag⁺, Cu²⁺, Cr³⁺) are reduced onto it.
Electrorefining: impure copper is the anode; pure copper deposits at the cathode while impurities fall away.
Water electrolysis: 2H₂O → 2H₂ + O₂, a route to hydrogen fuel.
Rechargeable batteries: charging forces the discharge reaction backward — the battery is an electrolytic cell while plugged in.

Common Confusions

Do Not Confuse	With	Difference
Electrolytic cell	Galvanic cell	Electrolytic: nonspontaneous, consumes electricity, anode is positive; galvanic: spontaneous, produces electricity, anode is negative
Anode sign convention	Cathode sign convention	In electrolysis the anode is the positive terminal (connected to + of the supply); in galvanic cells it is negative. Oxidation happens at the anode in both
n(e^-) = It/F	m = MIt/(nF)	The first gives moles of electrons; the second converts to grams of product using molar mass and stoichiometry
Electrolysis of molten NaCl	Electrolysis of aqueous NaCl	Molten gives Na metal and Cl₂; aqueous gives H₂ and Cl₂ because water is reduced more easily than Na⁺
Faraday constant F	Faraday's laws	F = 96,485 C mol^-1 is the charge per mole of electrons; Faraday's laws are the proportional relationships between charge and amount of product
Current I	Charge q	Current is charge per time (C s^-1); charge is current times time. Use q = It before applying Faraday's constant
Time in hours	Time in seconds	Faraday calculations need seconds; convert hours to seconds (3600 s per hour) or the answer is off by 3600×

Eli explains

The same idea, in plain words

Explain it like I’m 10

A galvanic battery makes electricity from chemistry; electrolysis is the reverse — you plug electricity in and force chemistry to happen backward. Imagine pushing a ball up a hill with a machine instead of letting it roll down: the machine (the power supply) does the work. The amount of stuff you make depends on how much electric charge you push through, so bigger current or longer time means more product.

Worked example

Example 1: How much aluminum from a given current?

A Hall–Héroult cell runs at 10.0 A for 2.00 hours to reduce molten Al₂O₃. How many grams of aluminum metal are deposited? (Al³⁺ + 3e^- → Al; M_Al = 26.98 g mol^-1.)

Step 1 — Write the mass formula:

$m = MIt nF$

Step 2 — Convert time to seconds and identify n:

$t = 2.00 h \times 3600 s 1 h = 7200 s, n = 3 mol e - per mol Al$

Step 3 — Substitute with units:

$m = (26.98 g mol -1)(10.0 A)(7200 s) (3)(96,485 C mol -1) = 26.98 \times 10.0 \times 7200 289,455 g = 6.71 g$

Dimensional check: A · s = C, so g mol^-1 · C / (C mol^-1) = g. About 6.7 g of aluminum requires roughly 20,000 C of charge — showing why aluminum smelters run at enormous currents.

Example 2: Volume of hydrogen from water electrolysis

Water is electrolyzed at 1.00 A for exactly 1.00 hour. What volume of H₂ gas is produced at STP? (2H₂O + 2e^- → H₂ + 2OH^-, so 2 mol e⁻ per mol H₂.)

Step 1 — Moles of electrons from charge:

$n(e -) = It F = (1.00 A)(3600 s) 96,485 C mol -1 = 0.03731 mol e -$

Step 2 — Convert to moles of H₂:

$n(H 2) = 0.03731 mol e - 2 mol e - / mol H 2 = 0.01866 mol$

Step 3 — Convert to volume at STP (22.4 L/mol):

$V = n \times 22.4 L mol -1 = 0.01866 \times 22.4 = 0.418 L$

Dimensional check: mol × L/mol = L. One amp-hour of charge yields about 0.42 L of hydrogen gas at STP — the scale of a small laboratory electrolysis cell.

Example 3: Time to electroplate a silver layer

How long must a 2.00 A current run to deposit 0.500 g of silver from a AgNO₃ solution? (Ag⁺ + e^- → Ag; M_Ag = 107.87 g mol^-1, n = 1.)

Step 1 — Solve the mass formula for t:

$m = MIt nF \Rightarrow t = mnF MI$

Step 2 — Substitute:

$t = (0.500 g)(1)(96,485 C mol -1) (107.87 g mol -1)(2.00 A) = 48,242.5 215.74 s = 223.6 s$

Dimensional check: g · C mol^-1 / (g mol^-1 · A) = C/A = s. About 224 seconds (3.7 minutes) — a reminder that electroplating is slow, which is exactly what makes it controllable.

Key takeaways

Electrolytic cells drive nonspontaneous reactions (E^°_cell < 0) using an external power source.
In both cell types: oxidation at anode, reduction at cathode; in electrolytic cells the anode is the positive terminal.
Faraday constant: F = 96,485 C mol^-1 of electrons; charge q = It.
Moles of electrons: n(e^-) = It/F; mass of product: m = MIt/(nF).
Use seconds for time, amperes for current, and the correct n (electrons per mole of product: 1 for Ag, 2 for Cu or H₂, 3 for Al).
In aqueous solution, water competes with ions: electrolyzing brine gives H₂ and Cl₂, not sodium metal.
Electrolysis applications: aluminum smelting, chlor-alkali, electroplating, electrorefining, water splitting, battery charging.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

In an electrolytic cell, is the anode connected to the positive or negative terminal of the power supply? Does oxidation or reduction occur there?
Show answer
Positive terminal; oxidation occurs at the anode.
Write the two half-reactions for electrolysis of molten NaCl.
Show answer
Cathode: Na⁺ + e^- → Na; anode: 2Cl^- → Cl₂ + 2e^-.
Why does electrolyzing an aqueous NaCl solution produce H₂ at the cathode instead of sodium metal?
Show answer
Water is reduced at a less negative potential (-0.83 V) than Na⁺ (-2.71 V), so 2H₂O + 2e^- → H₂ + 2OH^- wins the competition.
How many moles of electrons pass through a cell carrying 2.00 A for 30.0 minutes?
Show answer
n(e^-) = It/F = (2.00 × 1800)/96,485 = 0.0373 mol e^-.
How many grams of copper (M = 63.55 g mol^-1) deposit at a cathode when 1.00 A flows for 1.00 hour and Cu²⁺ + 2e^- → Cu?
Show answer
m = MIt/(nF) = (63.55)(1.00)(3600)/(2 × 96,485) = 1.19 g Cu.
Give two industrial processes that rely on electrolysis.
Show answer
Hall–Héroult aluminum production and the chlor-alkali process (also electroplating, electrorefining, or water electrolysis).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Electrolysis: Using electrical energy to drive a nonspontaneous redox reaction
Electrolytic cell: A cell that consumes electrical energy to force chemistry to happen
Anode: Electrode where oxidation occurs; in electrolysis, the positive terminal
Cathode: Electrode where reduction occurs; in electrolysis, the negative terminal
Faraday constant F: Charge carried by one mole of electrons: 96,485 C mol^-1
Overpotential: Extra voltage above E^°_cell needed to drive the reaction at a practical rate
Electroplating: Depositing a thin metal layer by reducing its ions onto a cathode
Chlor-alkali process: Electrolysis of brine producing Cl₂, H₂, and NaOH
Hall–Héroult process: Electrolytic reduction of molten Al₂O₃ to aluminum
Molten: Melted, so ions are mobile and can conduct current

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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