Chemistry 2e · Electrochemistry

Electrolysis

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On this page 9 sections

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

In 30 seconds

A galvanic cell runs a spontaneous reaction and produces electricity; an does the reverse — it uses electrical energy to force a nonspontaneous redox reaction to run. This process is , and it is how industry makes aluminum, chlorine, sodium hydroxide, and hydrogen; how metals are refined and plated; and how rechargeable batteries are recharged. This topic explains the anatomy of an electrolytic cell (note the flipped / sign convention), the competition between possible half-reactions in aqueous solutions, and Faraday's laws, which predict exactly how much product a given current and time will produce.

Why this matters

Electrolysis turns electricity into chemical change, which makes it central to modern industry: aluminum (Hall–Héroult process), chlorine and sodium hydroxide (), copper refining, and of jewelry, coins, and corrosion-resistant coatings. The same principles describe how rechargeable batteries are recharged, and they define the future of "green hydrogen" produced by splitting water. Faraday's laws give the quantitative link between electrical charge and amount of substance — the basis for electroplating specifications and industrial process design — and understanding explains why real cells need more voltage than the textbook minimum.

The college version

Core Concepts

Anatomy of an electrolytic cell

An electrolytic cell has two electrodes immersed in an electrolyte and connected to an external power source (a battery or DC supply). The anode is the electrode where oxidation occurs and is connected to the positive terminal; the cathode is where reduction occurs and is connected to the negative terminal. Notice the sign flip: in a galvanic cell the anode is negative, but in an electrolytic cell the anode is positive — the external supply pushes electrons into the cathode and pulls them from the anode, driving the reaction uphill against its natural direction.

Choosing what reacts: competition in aqueous solution

When the electrolyte is a molten salt, only its ions can react — electrolysis of molten NaCl gives sodium metal at the cathode and chlorine gas at the anode. In aqueous solution, however, water itself is a candidate for both half-reactions: water can be reduced to H₂ (at sufficiently negative potentials) or oxidized to O₂. The actual products are decided by which reaction needs less applied voltage under the conditions — that is, which is less unfavorable. A practical consequence: electrolyzing aqueous sodium chloride produces H₂ and OH⁻ at the cathode rather than sodium metal, because reducing water is far easier than reducing Na⁺. This competition is the reason chlor-alkali plants make NaOH and Cl₂, not Na metal.

Faraday's laws: charge quantifies product

The amount of product is fixed by the charge passed, not by the identity of the electrode. One mole of electrons carries charge F = 96,485 C. For a current I flowing for time t, the charge is Q = It, and the moles of electrons are Q/F. Since each mole of product requires n moles of electrons (from the balanced half-reaction), the mass of product is:

$m = M I t n F$

where M is the molar mass. This single equation does all electroplating and electrolysis stoichiometry.

Overvoltage and why real cells need more

The theoretical voltage required to drive an electrolysis is the negative of the cell potential of the reverse (spontaneous) reaction — for water splitting, about 1.23 V. Real cells need more: electrode kinetics, slow steps, and gas-bubble formation create an overvoltage (overpotential), so practical water electrolysis runs at roughly 1.5–2 V or more. Overvoltage also changes product choices — for example, it is partly why H₂ rather than sodium forms at the cathode in aqueous NaCl. This is why textbook potentials predict products, but engineering must add the real-world overhead.

Common Confusions

Do Not Confuse	With	Difference
Anode sign in electrolysis	Anode sign in a galvanic cell	Electrolytic anode is positive; galvanic anode is negative
Products of molten NaCl electrolysis	Products of aqueous NaCl electrolysis	Molten gives Na metal and Cl₂; aqueous gives H₂, OH⁻, and Cl₂ (water competes)
n in Faraday's law	n = moles of product	n is moles of electrons per mole of product from the half-reaction
Theoretical voltage	Actual applied voltage	Overvoltage means real cells need extra voltage (kinetics, bubbles, resistance)
Electrolysis of water	Electrolysis of brine	Water gives H₂ + O₂; brine gives H₂ + Cl₂ + NaOH

Eli explains

The same idea, in plain words

Explain it like I’m 10

A battery makes electricity from chemistry; electrolysis is the opposite — you plug in electricity to force chemistry to happen, like using a pump to push water uphill. If you send electricity through salty water, the water splits into bubbles of hydrogen and oxygen gas. The more electricity you send (and the longer you send it), the more gas you get — it's a direct trade: so many electrons in, so many atoms out.

Worked example

Worked Example 1 — How long to plate 10.0 g of silver? Silver plating reduces Ag⁺: Ag⁺ + e^- → Ag(s), so n = 1 and M = 107.87 g/mol. Rearranging Faraday's law for time:

$t = m n F M I = (10.0 g)(1)(96, 485 C/mol) (107.87 g/mol)(2.50 A) = 3.58 \times 103 s \approx 59.6 min$

Unit check: g · C/mol(g/mol) · A = g · C/mol · molg · C/s = s. A 2.50 A current deposits 10 g of silver in about an hour.

Worked Example 2 — Mass of copper deposited. Electrolysis of CuSO₄ solution reduces Cu²⁺ + 2e^- → Cu(s), so n = 2 and M = 63.55 g/mol. With I = 5.00 A for t = 2.00 h = 7.20 × 10³ s:

$m = M I t n F = (63.55 g/mol)(5.00 A)(7.20 \times 103 s) (2)(96, 485 C/mol) = 11.9 g$

This is the same calculation an electroplater or copper refinery uses to plan how long to run a bath for a target coating mass — and why current and time must be controlled precisely in industrial plating.

Key takeaways

Electrolysis drives nonspontaneous reactions with external electrical energy.
In an electrolytic cell the anode is positive and the cathode is negative — opposite of a galvanic cell; oxidation still occurs at the anode.
In aqueous solution, water competes with the solute ions for both half-reactions; compare required voltages to predict products.
Faraday's law: m = MItnF, with n = moles of electrons per mole of product from the balanced half-reaction.
Q = It converts current × time into charge; moles of electrons = Q/F.
Overvoltage means real electrolysis needs more voltage than the theoretical minimum.
Applications: aluminum production, chlor-alkali (Cl₂ + NaOH), copper refining, electroplating, water splitting for H₂.

Check yourself

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

In an electrolytic cell, which electrode is positive, and what happens there?
Show answer
The anode is positive, and oxidation occurs there (e.g., Cl⁻ or water losing electrons).
Why does electrolyzing aqueous NaCl produce H₂ instead of sodium metal at the cathode?
Show answer
Water is reduced to H₂ at a much less negative potential than Na⁺ needs, so water wins the competition in aqueous solution.
State and the meaning of each symbol.
Show answer
m = MIt/(nF): mass deposited equals molar mass × current × time divided by (electrons per mole of product × Faraday constant).
How much charge (in coulombs) is needed to deposit 1 mol of silver (Ag⁺ + e⁻ → Ag)?
Show answer
Q = nF = (1 mol e^-)(96,485 C/mol) = 96,485 C — one faraday per mole of silver.
Why does a commercial water-splitting cell operate above the theoretical 1.23 V?
Show answer
Overvoltage: slow electrode kinetics, gas-bubble buildup, and solution resistance require extra voltage beyond the thermodynamic minimum.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Electrolysis: Using electrical energy to force a nonspontaneous redox reaction
Electrolytic cell: Cell with an external power supply driving the reaction
Anode: Electrode where oxidation occurs
Cathode: Electrode where reduction occurs
Faraday's law: m = MIt/(nF) relating charge to mass of product
Overvoltage: Extra voltage needed beyond the theoretical minimum
Chlor-alkali process: Electrolysis of aqueous NaCl to make Cl₂ and NaOH
Electroplating: Coating an object by reducing metal ions onto it as the cathode

Sources & references

openstax.org — Chemistry 2e

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