F3.5 Explain industrial electrochemistry applications | Ontario Grade 12 Chemistry (SCH4U)

Ontario Grade 12 Chemistry

Electrochemistry

How electrical energy helps produce and purify useful materials

A metal coating can protect an object from corrosion, and purified metals are important in electrical equipment. Making or purifying these materials often involves an observable change: metal appears on a surface, or a liquid metal collects during production. At the particle level, these changes involve electrons moving through an external circuit and ions gaining or losing electrons at electrodes. This lesson connects those changes to three industrial applications of electrochemistry.

What you will learn

Explain how an electrolytic cell uses electrical energy to cause a chemical change.
Describe the electrode reactions in aluminum production, the chlor-alkali process, and copper refining.
Relate electron transfer and ion movement to the products made in these industrial processes.

1. From redox reactions to electrolytic cells

In a redox reaction, electrons move from one substance to another. Oxidation is the loss of electrons. Reduction is the gain of electrons. These definitions help explain what happens at electrodes in industrial processes.

An electrode is a conducting surface where oxidation or reduction occurs. In an electrolytic cell, a power supply pushes electrons through the external circuit and drives a chemical change that requires electrical energy. This differs from a galvanic cell, which produces electrical energy from a chemical reaction. In both cell types, oxidation occurs at the anode and reduction occurs at the cathode.

The power supply removes electrons from the anode and delivers electrons to the cathode. Positive ions in the electrolyte move toward the cathode, where they can gain electrons. Negative ions move toward the anode, where they can lose electrons. The electrolyte is a liquid or solution that contains mobile ions and carries charge within the cell.

Industrial cells are designed so that the desired products can be collected and, when needed, kept apart. Their size and operating conditions vary, but the same basic ideas apply: identify the ions, track electrons, and determine what forms at each electrode.

\text{anode: oxidation; cathode: reduction}

Oxidation is electron loss; reduction is electron gain.
Oxidation takes place at the anode, and reduction takes place at the cathode.
An electrolytic cell uses electrical energy to drive a chemical change.

2. Producing aluminum and industrial chemicals

Aluminum is used in products such as transportation equipment and packaging. It is not obtained by simply heating aluminum-containing ore with carbon. Industrial production instead uses electrolysis, which supplies the energy needed to reduce aluminum ions to aluminum metal.

In the production cell, aluminum oxide is dissolved in a hot liquid mixture that includes cryolite. This allows the ions to move through the cell. Aluminum ions gain electrons at the cathode and form liquid aluminum, which can be collected. Oxygen-containing ions lose electrons at the anode. The carbon anodes are consumed as they react with oxygen, so they must be replaced.

The half-reactions show electron transfer at each electrode. The overall equation includes the reaction of carbon with oxygen. Together, these equations explain why aluminum production needs both electrical energy and carbon anodes.

Electrolysis also makes important industrial chemicals. In the chlor-alkali process, concentrated aqueous sodium chloride solution, called brine, is electrolyzed. Chloride ions lose electrons at the anode and form chlorine gas. Water gains electrons at the cathode, producing hydrogen gas and hydroxide ions. Sodium ions remain in solution with hydroxide ions, giving sodium hydroxide solution. The products are useful in chemical manufacturing and other applications.

The products must be collected so that they do not mix in ways that interfere with their use. Industrial equipment therefore keeps the electrode regions separated while allowing ions to move as needed. The specific design is not required to understand the key chemistry: the electrode reactions determine the products.

\mathrm{Al^{3+}(l) + 3e^- \rightarrow Al(l)}

Electrolysis reduces aluminum ions to liquid aluminum at the cathode.
Carbon anodes react with oxygen-containing products and are gradually used up.
The chlor-alkali process produces chlorine, hydrogen, and sodium hydroxide from brine and water.

3. Coating and purifying metals

Electroplating puts a thin metal layer onto another object. A metal object to be coated is connected as the cathode. The electrolyte contains ions of the coating metal. At the cathode, these ions gain electrons and become metal atoms that build up on the object. For example, when copper ions are reduced, copper metal forms on the cathode.

A copper coating can give an object a useful surface. The metal layer is produced by reduction, not by the object attracting already formed copper particles. The copper ions in solution gain electrons at the cathode and become part of the solid coating.

Electrorefining is used to purify copper. An impure copper slab is the anode, and a thin sheet of pure copper is the cathode. The electrolyte contains copper ions. Copper atoms from the impure anode lose electrons and enter solution as copper ions. Copper ions gain electrons at the cathode and form solid copper. As the anode dissolves and the cathode grows, copper is transferred and purified.

The half-reactions for copper refining make the transfer clear. The anode supplies copper ions to the solution, while the cathode removes copper ions from it. Impurities do not all behave in the same way: some remain in solution, and some collect below the anode. The intended product is the pure copper deposited at the cathode.

\mathrm{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-}

In electroplating, the object receiving the coating is the cathode.
In copper electrorefining, impure copper is the anode and pure copper is deposited at the cathode.
Metal atoms become ions by losing electrons; metal ions become atoms by gaining electrons.

4. Reading industrial cell equations

Half-reactions show what happens at one electrode. They include electrons so that the loss or gain of charge is visible. To describe a complete process, the electrons lost in oxidation must equal the electrons gained in reduction. Adding the balanced half-reactions gives an overall equation when the process has been specified.

State symbols add useful information:

(s)

means solid,

(l)

means liquid,

(g)

means gas, and

(aq)

means dissolved in water. A state symbol matters in industrial electrochemistry because a product may be collected as a metal, a gas, or a solution.

The equations in this lesson are qualitative descriptions of the reactions. They do not by themselves specify how much material a plant produces. Production amounts depend on operating conditions and the amount of electrical charge supplied; this lesson focuses on identifying industrial applications and explaining their electrode reactions.

When interpreting a process, first identify the cell type and products. Then locate the anode and cathode, decide which particles reach each electrode, and write the oxidation and reduction reactions. Finally, check that atoms and total charge are balanced.

\text{electrons lost} = \text{electrons gained}

Electrons lost in oxidation must be accounted for in reduction.
State symbols show whether a substance is solid, liquid, gas, or dissolved in water.
Check both atoms and total charge in each balanced reaction.

Worked example

Explaining the products of the chlor-alkali process

A factory electrolyzes concentrated aqueous sodium chloride. Use the electrode reactions to explain which products form and where they form.

Identify the cell and particles
Electrolysis is being used, so this is an electrolytic cell. The brine contains sodium ions and chloride ions, and the water can also take part in the electrode reactions.
Find the anode product
Oxidation occurs at the anode. Chloride ions lose electrons, and chlorine gas forms. Two chloride ions release the two electrons needed for one chlorine molecule.
$\mathrm{2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-}$
Find the cathode products
Reduction occurs at the cathode. Water gains the two electrons released at the anode. This produces hydrogen gas and hydroxide ions.
$\mathrm{2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)}$
Combine the products
The sodium ions remain in solution and balance the hydroxide ions, so sodium hydroxide is present in the solution. Adding the electrode reactions and accounting for the sodium ions gives the balanced overall reaction.
$\mathrm{2NaCl(aq) + 2H_2O(l) \rightarrow Cl_2(g) + H_2(g) + 2NaOH(aq)}$

Answer: Chlorine gas forms at the anode. Hydrogen gas forms at the cathode. Sodium hydroxide remains in solution. The two half-reactions each involve two electrons, so electron transfer is balanced.

Check: The overall equation conserves sodium, chlorine, hydrogen, and oxygen atoms. Its net charge is zero on both sides.

Common mistakes and how to avoid them

Calling the anode the reduction electrode because it is connected to the power supply.

Correction: Oxidation occurs at the anode in both galvanic and electrolytic cells. Identify the reaction at the electrode rather than relying on a guessed sign.

Saying that sodium metal is produced in the chlor-alkali process described here.

Correction: In aqueous brine, water is reduced at the cathode to form hydrogen gas and hydroxide ions. Sodium ions remain in solution.

Treating copper electroplating and copper electrorefining as the same setup.

Correction: Both deposit copper at the cathode, but electrorefining uses an impure copper anode as the copper source, while electroplating coats an object.

Lesson summary

Industrial electrolysis uses electrical energy to drive chemical changes.
Aluminum production reduces aluminum ions at the cathode and uses carbon anodes.
The chlor-alkali process makes chlorine, hydrogen, and sodium hydroxide.
Electroplating coats an object; electrorefining transfers copper from an impure anode to a pure cathode.

Check your understanding

Question 1

In copper electrorefining, where does copper metal deposit?

At the cathode, where copper ions gain electrons
At the anode, where copper atoms gain electrons
In the electrolyte, where copper ions lose electrons without an electrode
At the cathode, where copper ions lose electrons

Show answer and explanation

At the cathode, where copper ions gain electrons

Reduction occurs at the cathode. Copper ions gain electrons there and form solid copper.

Question 2

Which set of products forms in the chlor-alkali process described in this lesson?

Chlorine gas, hydrogen gas, and sodium hydroxide
Sodium metal, oxygen gas, and hydrochloric acid
Chlorine gas, oxygen gas, and sodium metal
Hydrogen gas and copper metal

Show answer and explanation

Chlorine gas, hydrogen gas, and sodium hydroxide

Chloride ions are oxidized to chlorine, water is reduced to hydrogen and hydroxide ions, and sodium ions remain in solution with hydroxide ions.

Key terms

Electrolysis: The use of electrical energy to drive a chemical change.
Electrode: A conducting surface where oxidation or reduction occurs.
Electrolyte: A liquid or solution containing mobile ions that carries charge in a cell.
Electroplating: Using electrolysis to deposit a metal coating on an object.
Electrorefining: Using electrolysis to purify a metal by transferring it from an impure anode to a cathode.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Chemistry (SCH4U), expectation F3.5. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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F3.5 · Explain industrial electrochemistry applications