Electrolytic cells

An electrolytic cell is an electrochemical cell that uses electrical energy from an external source to drive a non-spontaneous redox reaction. This process is called electrolysis.

The key components of an electrolytic cell are:

• Electrolyte - A solution containing free-moving ions, either dissolved (aqueous solution) or molten ionic compound.
• Cathode - The negative electrode where reduction occurs and electrons enter the cell.
• Anode - The positive electrode where oxidation occurs and electrons leave the cell.
• DC power source - Provides the electrical energy to drive the non-spontaneous redox reaction.

In the closed circuit, electrons flow from the negative terminal of the power source to the cathode, where cations are reduced. The electrons then flow through the electrolyte to the anode, where anions are oxidised. Finally, the electrons return to the positive terminal of the power source.

Electrolysis of molten salts

When a molten ionic compound undergoes electrolysis, the cations (usually metals) are reduced at the cathode to form the pure metal, while the anions are oxidised at the anode. The identity of the products depends on the composition of the molten salt.

For example, in the electrolysis of molten sodium chloride (NaCl):

Sodium ions are reduced at the cathode to form liquid sodium: Na^+^ + e^- ➔ Na

Chloride ions are oxidised at the anode to form chlorine gas: 2Cl^-^ ➔ Cl_2_ + 2e^-^

The overall equation is: 2NaCl_(l)_ ➔ 2Na_(l)_ + Cl_2(g)_

This process is used industrially to produce reactive metals like sodium, magnesium, and aluminium from their molten salts. An inert atmosphere is used to prevent the oxidation of the metal product.

Worked example 1 - Electrolysis of molten Al_2_O_3_

Deduce the products of the electrolysis of molten aluminium oxide, Al_2_O_3(l)_, with inert electrodes. Write the overall cell equation and the half-equations occurring at each electrode.

Step 1: Split the electrolyte into ions

Al₂O_3(l) ➔ 2Al³⁺ + 3O^2-

Step 2: Predict direction of ion migration

Al³⁺ ions will migrate towards the cathode.

O^2- ions will migrate towards the anode.

Step 3: Reaction at cathode

Al^3+^ ions will be reduced to liquid aluminium at the cathode.

Al³⁺ + 3e^- ➔ Al

Step 4: Reaction at anode

O^2-^ ions will be oxidised to oxygen gas at the anode.

O^2- ➔ 1⁄2O₂ + 2e^-

Step 5: Overall equation

Ionic: 2Al³⁺ + 3O^2-➔ 2Al + 3⁄2O₂

Full: Al₂O_3(l) ➔ 2Al_(l)_ + 3⁄2O_2(g)

Electrolysis of aqueous solutions

In aqueous solutions, water molecules can also be oxidised or reduced at the electrodes, in addition to the solute ions. The reduction and oxidation of water occur at specific reduction potentials:

Reduction of water: H₂O_(l) + e^- ➔ 1⁄2H_2(g) + OH^-_(aq) E^⦵ = $-$0.83 V

Oxidation of water: H₂O_(l) ➔ 1⁄2O_2(g) + 2H⁺_(aq) + 2e^- E^⦵ = $-$1.23 V

To deduce the products of electrolysis in aqueous solutions, compare the standard electrode potentials of the solute ions and water:

• At the cathode, the species with the most positive reduction potential will be reduced.
• At the anode, the species with the least negative oxidation potential will be oxidised.

For example, in the electrolysis of aqueous concentrated sodium chloride (NaCl):

Cathode: H₂O is reduced instead of Na⁺ because its reduction potential (E^⦵ = $-$0.83 V) is less negative than that of Na^+^ (E^⦵ = $-$2.71 V).

Anode: Cl^-^ is preferentially oxidised because its oxidation potential (E^⦵ = $-$0.83 V) is more negative than that of H_2_O (E^⦵ = −1.23 V).

Worked example 2 - Electrolysis of aqueous CuSO_4_

The standard reduction potentials for Cu²⁺ and H₂O are as follows;

Cu²⁺_(aq) + 2e^- ➔ Cu_(s) E^⦵ = +0.34 V

H₂O_(l) + e^- ➔ 1⁄2H_2(g) + OH^-_(aq) E^⦵ = $-$0.83 V

Deduce the products of the electrolysis of aqueous copper sulfate, CuSO_4(aq), with inert electrodes. Write the overall cell equation and the half-equations occurring at each electrode.

Step 1: Split the electrolyte into ions

CuSO_4(aq) ➔ Cu²⁺_(aq) + SO₄^2-_(aq)

Step 2: Predict direction of ion migration

Cu²⁺ ions will migrate towards the cathode.

SO₄^2- ions will migrate towards the anode.

Step 3: Reaction at cathode

Cu²⁺ has a more positive reduction potential than H₂O, so it is reduced to copper metal at the cathode.

Cu²⁺_(aq) + 2e^- ➔ Cu_(s)

Step 4: Reaction at anode

SO₄^2- ions cannot be oxidised further as sulfur is already in its highest oxidation state of +6.

Therefore, H₂O will be oxidised to oxygen gas at the anode.

H₂O_(l) ➔ 1⁄2O_2(g) + 2H⁺_(aq) + 2e^-

Step 5: Overall equation

Ionic: Cu²⁺_(aq) + SO₄^2-_(aq) + H₂O_(l) ➔ Cu_(s) + 1⁄2O_2(g) + 2H⁺_(aq) + SO₄^2-_(aq)

Full: CuSO_4(aq) + H₂O_(l) ➔ Cu_(s) + 1⁄2O_2(g) + H₂ SO_4(aq)

Effect of electrolyte concentration

The concentration of the electrolyte can affect the products of electrolysis, particularly at the anode. This is because the concentration of ions in the solution determines their availability for oxidation or reduction at the electrodes.

For example, in the electrolysis of NaCl_(aq)_:

• In concentrated NaCl_(aq) - Cl^- is preferentially oxidised to Cl_2_ at the anode because its oxidation potential, E^⦵ = −1.36 V, is more negative than that of H_2_O, E^⦵ = −1.23 V.
• In dilute NaCl_(aq) - H_2_O is more likely to be oxidised to O_2_, either along with Cl^- or as the sole product, because the concentration of Cl^- is lower.

In the case of CuSO_4(aq) electrolysis, the concentration does not affect the products because Cu²⁺ is always preferentially reduced at the cathode, and water is always oxidised at the anode

Effect of electrode material

The nature of the electrode material can affect the products of electrolysis, especially at the anode. Inert electrodes, such as platinum or graphite, are often used to prevent the electrodes from participating in the reactions.

For example, in the electrolysis of CuSO_4(aq):

• With inert electrodes (e.g., platinum or graphite), Cu²⁺ is reduced to Cu_(s) at the cathode, and H₂O is oxidised to O₂ at the anode.
• With a copper anode, the anode is oxidised to Cu²⁺, which then migrates to the cathode and is reduced to Cu_(s). This process is used in electroplating to deposit a layer of copper on the cathode.

Electroplating

Electroplating is the process of using electrolysis to coat an object with a thin layer of metal. It has various practical applications:

• Creating a protective, corrosion-resistant layer on objects.
• Improving the appearance of objects.
• Purifying metals by selectively dissolving impurities.

Electroplating involves the following steps:

1. The metal to be plated (e.g. copper) is used as the anode and is oxidised to form cations in solution.
2. The cations move through the electrolyte to the cathode.
3. At the cathode, the cations are reduced and deposited as a thin layer of metal on the object being plated.

For example, consider the process of electroplating an iron nail with copper using a copper(II) sulfate electrolyte.

The half-reaction equations for the anode and cathode processes are as follows:

Anode (copper oxidation): Cu_(s) ➔ Cu²⁺_(aq) + 2e^-

Cathode (copper(II) reduction): Cu²⁺_(aq) + 2e^- ➔ Cu_(s)

During the electroplating process, copper atoms from the anode dissolve into the solution as Cu²⁺_(aq) ions. These ions then migrate through the electrolyte to the cathode (iron nail), where they are reduced and plated onto the iron surface as a thin layer of Cu_(s).

Worked example 3 - Deducing the half-equations for silver-nickel electroplating

A nickel fork is plated with a thin layer of silver in an electroplating experiment using a silver nitrate (AgNO₃) electrolyte, as shown below.

Give the half-equations for the reactions occurring at each electrode.

Step 1: Reaction at anode

At the anode, silver is oxidised:

Ag_(s) ➔ Ag⁺_(aq) + e^-

Step 2: Reaction at cathode

At the cathode, silver(I) ions are reduced:

Ag⁺_(aq) + e^- ➔ Ag_(s)