Electrolysis of Molten Salts

Introduction

An ionic compound can be decomposed using electrical energy after it is melted so its ions can move. Molten sodium chloride is the standard example: sodium ions gain electrons at one electrode, and chloride ions lose electrons at the other. The overall reaction produces sodium metal and chlorine gas, while keeping every atom and charge accounted for.

Core explanation

Solid sodium chloride has ions held in a crystal lattice and does not conduct in the same way as a molten sample. When molten, Na⁺ and Cl⁻ ions can move toward electrodes. At the cathode, Na⁺ gains an electron: Na⁺ + e⁻ → Na. At the anode, chloride loses electrons: 2Cl⁻ → Cl₂ + 2e⁻. The anode reaction uses two chloride ions because elemental chlorine is Cl₂.

Double the sodium half-reaction so electrons match: 2Na⁺ + 2e⁻ → 2Na. Add it to 2Cl⁻ → Cl₂ + 2e⁻ and cancel electrons. The overall ionic form is 2Na⁺ + 2Cl⁻ → 2Na + Cl₂. Writing the molten salt as formula units gives 2NaCl(l) → 2Na(l) + Cl₂(g) under an appropriate high-temperature cell. Sodium's product phase depends on cell temperature; molten sodium is a suitable label for the common industrial example.

Count atoms in the overall formula equation: two Na and two Cl before, two Na atoms and one Cl₂ molecule after. The one-reactant, two-product arrangement makes it decomposition. Electron transfer at both electrodes makes it redox. External electrical energy drives the process; the reverse formation of salt from sodium and chlorine is energetically favourable under ordinary conditions.

The word “molten” is crucial. Aqueous NaCl contains water as well as Na⁺ and Cl⁻, and water can react at the electrodes. Under common aqueous-brine electrolysis conditions, hydrogen and hydroxide can form at the cathode rather than sodium metal, while chlorine can form at the anode. Thus 2NaCl(l) → 2Na + Cl₂ is not a valid description of a typical sodium chloride solution in water.

Other molten salts can be decomposed similarly in principle, but their products, electrode conditions and practical processes differ. For molten lead(II) bromide, a simplified overall equation is PbBr₂(l) → Pb(l) + Br₂(g) under conditions in which those phases apply. Pb²⁺ receives two electrons and two Br⁻ lose two electrons to form Br₂. The balanced formula equation has one Pb and two Br on each side.

Both sodium and chlorine products require professional handling, separation and specialised equipment. The lesson is about interpreting the equations, not operating a cell. The overall equation omits equipment design, energy losses and competing reactions that matter in industry.

Step-by-step reasoning

1. Confirm that the salt is molten rather than dissolved in water. 2. Identify mobile cations and anions and their electrode products. 3. Balance each half-reaction for atoms and electrons, then add and cancel electrons. 4. Write the overall formula equation and check atom counts and suitable state labels.

Visual explanation

Draw a molten NaCl pool with Na⁺ arrows toward the negative cathode and Cl⁻ arrows toward the positive anode. At the cathode, two sodium ions become two sodium atoms; at the anode, two chloride ions pair as Cl₂. The electron flow in the external circuit closes the accounting.

Real-world analogy

Imagine two queues of tagged pieces moving to separate counters. One counter gives each piece a token, while the other collects matching tokens from a pair. The total tokens exchanged must balance. In electrolysis, electrons are those counted transfers, and the ions' identities determine the products.

Real-world example

Molten sodium chloride can be electrolysed industrially to make sodium metal and chlorine gas. The balanced equation 2NaCl(l) → 2Na(l) + Cl₂(g) shows the ideal material ratio. Separation of products matters because sodium and chlorine can react back to form salt.

Why?

Why must NaCl be molten for the simple sodium-producing equation? The ions need mobility to reach electrodes, and the absence of water avoids water reduction competing with Na⁺. A dissolved salt has different available reactants and therefore can yield different products.

Common misconception

“Electrolysing any NaCl-containing liquid makes sodium metal.” Water in an aqueous solution can be reduced instead. Read (l) as molten NaCl and (aq) as NaCl dissolved in water; they imply different electrochemistry.

Worked example

Build the overall molten-salt equation from half-reactions. Cathode: 2Na⁺ + 2e⁻ → 2Na. Anode: 2Cl⁻ → Cl₂ + 2e⁻. Add: 2Na⁺ + 2Cl⁻ → 2Na + Cl₂. Package the reactant ions as 2NaCl(l): 2NaCl(l) → 2Na(l) + Cl₂(g) for the stated high-temperature setting. Two Na and two Cl atoms occur on each side.

Quick check

1. Why is Cl₂, rather than Cl, the anode product in molten NaCl electrolysis? Answer: Elemental chlorine forms diatomic Cl₂ molecules, so two chloride ions supply its two chlorine atoms.

Exam focus

Distinguish molten from aqueous salts. State cathode reduction and anode oxidation, balance electrons and use elemental diatomic products correctly. Check state labels against temperature and avoid predicting sodium metal from ordinary brine.

Advanced insight

OpenStax's electrolysis treatment contrasts molten NaCl with aqueous NaCl and shows why water changes the cathode product. In an industrial molten cell, applied voltage must exceed the reversible requirement once resistance and overpotentials are included; the balanced equation alone does not determine energy consumption.

Summary

Molten-salt electrolysis uses mobile ions and electrical energy to decompose an ionic compound. For molten NaCl, sodium forms at the cathode and chlorine at the anode: 2NaCl(l) → 2Na(l) + Cl₂(g). Aqueous salt electrolysis can produce different products because water participates.

Practice questions

1. Balance the electrolysis of molten sodium chloride. Answer: 2NaCl(l) → 2Na(l) + Cl₂(g) under suitable high-temperature conditions. 2. Which ion is reduced at the cathode in the molten NaCl example? Answer: Na⁺ gains an electron to form sodium metal. 3. Why is aqueous sodium chloride not equivalent to molten sodium chloride? Answer: Water is present in the aqueous solution and can react at electrodes, changing the product set.