Electrolytic Cells and Electrolysis Mechanisms

An electrolytic cell uses electrical energy from an external source to drive a non-spontaneous redox reaction: oxidation takes place at the anode (the positive electrode) and reduction at the cathode (the negative electrode), the reverse of the signs in a galvanic cell. In brief: the electrolysis of molten NaCl gives sodium at the cathode (Na⁺ + e⁻ → Na) and chlorine at the anode (2Cl⁻ → Cl₂ + 2e⁻), and it needs an applied voltage of more than 4.07 V.

Electrolysis is used to make aluminum, chlorine, sodium hydroxide and hydrogen, to purify copper and to plate objects with metals. In every case a direct current is used to push electrons through a reaction that would not go on its own. This article explains how an electrolytic cell works, how to predict the products in a molten salt and in an aqueous solution, gives a comparison with galvanic cells and describes the main industrial processes.

How does an electrolytic cell work?

An electrolytic cell has two electrodes in an electrolyte (a molten salt or a solution containing ions), connected to a DC power source. The source acts as an “electron pump”: it pushes electrons to one electrode (the cathode, which becomes negative) and removes them from the other (the anode, which becomes positive). Positive ions (cations) are attracted to the cathode, where they are reduced, and negative ions (anions) to the anode, where they are oxidized. For a reaction with a negative E°_cell (positive ΔG°), the applied voltage must exceed |E°_cell|.

Galvanic and electrolytic cells compared

In both types of cell, oxidation occurs at the anode and reduction at the cathode. Only the signs of the electrodes are different.

What happens in a molten salt?

In a molten salt, there is only one cation and one anion, so the products are easy to predict. For molten sodium chloride (at about 801 °C and higher):

• Cathode: Na⁺ + e⁻ → Na(l).
• Anode: 2Cl⁻ → Cl₂(g) + 2e⁻.
• Overall: 2NaCl(l) → 2Na(l) + Cl₂(g).

The standard potential is E° = −2.71 − 1.36 = −4.07 V, so at least 4.07 V is needed. In the industrial Downs cell, CaCl₂ is added to lower the melting point to about 600 °C.

How do you predict products in aqueous solution?

In an aqueous solution, water can also be oxidized or reduced, so there are competing reactions:

• Reduction of water: 2H₂O + 2e⁻ → H₂ + 2OH⁻ (E = −0.83 V at pH 14, −0.41 V at pH 7).
• Oxidation of water: 2H₂O → O₂ + 4H⁺ + 4e⁻ (E = +1.23 V at pH 0, +0.82 V at pH 7).

The rules are:

1. At the cathode, the species with the more positive reduction potential is reduced first. Cations of the very active metals (Li⁺, K⁺, Ca²⁺, Na⁺, Mg²⁺, Al³⁺) are not reduced in water, and hydrogen forms instead. Cu²⁺, Ag⁺ and Ni²⁺ are reduced to the metal.
2. At the anode, the species that is easiest to oxidize reacts. Halide ions (Cl⁻, Br⁻, I⁻) are usually oxidized in preference to water. Sulfate, nitrate and fluoride ions are not oxidized, and oxygen forms instead.
3. Overpotential: the actual voltage needed is higher than the thermodynamic value, because some electrode reactions are slow. The formation of O₂ has a large overpotential, which is why Cl₂ (E° = 1.36 V) is produced in concentrated NaCl solution although the oxidation of water is thermodynamically easier (E° = 1.23 V).
4. The material of the electrode matters: a reactive metal anode (such as copper) may itself be oxidized.

Products of common electrolyses

What are the industrial applications?

• Aluminum (Hall–Héroult process): Al₂O₃ dissolved in molten cryolite (Na₃AlF₆) at about 960 °C is electrolyzed with carbon electrodes: Al³⁺ + 3e⁻ → Al at the cathode, and the carbon anode is consumed to give CO₂. It needs roughly 13–15 kWh per kg of aluminum.
• Chlor-alkali industry: electrolysis of brine gives chlorine, hydrogen and sodium hydroxide, on a scale of tens of millions of tonnes per year.
• Copper refining: impure copper is the anode, and pure copper is deposited on the cathode in an acidic CuSO₄ solution; impurities such as silver and gold fall as “anode mud”.
• Electroplating: the object to be plated is the cathode, and metal ions are reduced onto it.
• Hydrogen production: electrolysis of water with a renewable electricity supply gives “green” hydrogen.

Worked examples

Example 1: the minimum voltage for electrolysis of water

For 2H₂O → 2H₂ + O₂ in acid, the cathode reaction is 2H⁺ + 2e⁻ → H₂ (E° = 0.00 V) and the anode reaction is the reverse of O₂ + 4H⁺ + 4e⁻ → 2H₂O (E° = 1.23 V). E°_cell = 0.00 − 1.23 = −1.23 V. A voltage of at least 1.23 V is needed in theory, and about 1.8–2.0 V in practice because of the overpotentials and resistance.

Example 2: predicting the products for aqueous KI

The ions are K⁺, I⁻, and water. At the cathode, K⁺ (−2.93 V) is much harder to reduce than water (about −0.8 V), so H₂ forms. At the anode, I⁻ is oxidized (E° = 0.54 V for I₂/I⁻) more easily than water (1.23 V), so I₂ forms. A little phenolphthalein turns pink at the cathode, because of OH⁻.

Example 3: copper refining

At the impure copper anode: Cu → Cu²⁺ + 2e⁻. At the pure copper cathode: Cu²⁺ + 2e⁻ → Cu. The overall change is a transfer of copper from the anode to the cathode, so only a small voltage of about 0.2–0.3 V is needed, since E°_cell is zero in theory and the rest is for overcoming resistance. Silver and gold, which are less active than copper, do not dissolve and fall to the bottom, while zinc and iron dissolve but stay in solution.

Who developed electrolysis?

William Nicholson and Anthony Carlisle decomposed water with a voltaic pile in 1800. Humphry Davy used electrolysis in 1807 to isolate potassium and sodium from molten potash and soda, and in the following years calcium, magnesium, strontium and barium. Michael Faraday coined the terms electrolysis, electrode, anode, cathode, ion and electrolyte, with the help of William Whewell, in 1834. Charles Martin Hall and Paul Héroult independently invented the aluminum process in 1886, when both were 22 years old, and it turned aluminum from a precious metal into a cheap structural material within a few decades.

Which mistakes are common?

• Assigning wrong signs to the electrodes: in an electrolytic cell the anode is positive.
• Forgetting water in aqueous electrolysis: Na⁺ is not reduced in water.
• Ignoring overpotential: oxygen formation needs more than 1.23 V.
• Assuming that all electrodes are inert: a copper anode dissolves.
• Mixing up oxidation and the anode: oxidation is always at the anode, in both types of cell.