Hydrogen and Hydroxide in Brine

Introduction

In the usual chlor-alkali model, sodium ions approach the negative cathode but water accepts the electrons. The result is hydrogen gas and hydroxide ions. Sodium remains dissolved and accompanies hydroxide in the cathode-side liquid, giving aqueous sodium hydroxide rather than sodium metal.

Core explanation

The cathode half-reaction is 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq). Two water molecules supply four hydrogen atoms and two oxygen atoms. The products have two hydrogen atoms in H₂ and two more in the two hydroxide ions; both oxygen atoms remain in hydroxide. Left charge is −2 from electrons, and right charge is −2 from hydroxide. The balanced equation thus conserves atoms and charge.

One mole H₂ requires two moles electrons and forms two moles OH⁻. A current amount corresponding to 0.500 mol electrons would ideally give 0.250 mol H₂ and 0.500 mol OH⁻. Sodium-ion migration helps balance the produced negative charge, so an ideal full brine cell may be described as producing 0.500 mol NaOH formula amount in solution for that electron amount.

The cathode is called negative in this driven cell because the external power source supplies electrons to it. Reduction does not mean sodium ions necessarily become sodium metal. In an aqueous medium, water is a competing reactant and ordinarily produces hydrogen in the brine case. Depositing sodium from water would also be chemically unstable because sodium metal reacts vigorously with water.

Hydroxide buildup raises the pH near the cathode. A separator controls mixing with the anode-side chlorine. If Cl₂ contacts OH⁻, other chlorine-containing species can form, reducing the desired separate chlorine and sodium hydroxide outputs. Therefore cathode chemistry and cell architecture are connected.

Hydrogen collection requires care in real equipment because hydrogen is flammable and mixtures with chlorine or air can be hazardous. A stoichiometric calculation predicts an ideal amount but not actual gas purity, collection efficiency or safety of a cell design.

Step-by-step reasoning

1. Identify aqueous brine and cathode as the reduction electrode. 2. Write water plus electrons on the reactant side. 3. Balance H₂ and OH⁻ products for atoms and charge. 4. Use two electrons per H₂ and one OH⁻ per electron for amounts. 5. Explain sodium's role as a dissolved counter-ion.

Visual explanation

Draw water molecules touching a negative cathode, H₂ bubbles rising and OH⁻ symbols remaining near the surface. Na⁺ symbols move into that region but do not become metal.

Real-world analogy

A spectator arriving at a station can help balance a crowd without becoming the performer on stage. Sodium ions migrate toward the cathode region, while water is the species actually reduced.

Real-world example

In a membrane chlor-alkali cell, the cathode compartment becomes alkaline as hydrogen evolves. Sodium ions cross toward that side so sodium hydroxide solution can be collected separately from anode chlorine.

Why?

Why does NaOH appear when the cathode does not deposit sodium? Na⁺ remains an ion and balances the OH⁻ produced by water reduction in the solution.

Common misconception

“Na⁺ is attracted to the cathode, so sodium metal must be the product.” Attraction and migration do not decide the reduction reaction; water is reduced under ordinary brine conditions.

Worked example

If 0.0800 mol H₂ forms ideally, its cathode reaction consumed 2 × 0.0800 = 0.160 mol electrons and produced 0.160 mol OH⁻. With sufficient Na⁺ in the cathode-side liquid, that corresponds to 0.160 mol NaOH formula amount. It does not correspond to 0.160 mol sodium metal.

Quick check

1. How many moles OH⁻ form ideally per mole H₂ at the brine cathode? Answer: The balanced water-reduction half-reaction forms two moles OH⁻ for every one mole H₂.

Exam focus

Show water as the cathode reactant and balance hydrogen, hydroxide and electrons. State sodium's counter-ion role.

Advanced insight

The pH and sodium-ion concentration near the cathode may differ from the bulk until transport and mixing smooth gradients. Membrane selectivity affects product concentration and efficiency.

Summary

Brine cathode reduction turns water into H₂ and OH⁻, using two electrons per H₂. Sodium stays dissolved and balances hydroxide, producing alkaline sodium hydroxide solution under the ideal cell model.

Practice questions

1. Balance the brine cathode reaction. Answer: 2H₂O + 2e⁻ → H₂ + 2OH⁻ balances atoms and charge. 2. If 0.10 mol electrons is consumed, how much H₂ forms ideally? Answer: Two electrons per H₂ give 0.050 mol H₂. 3. Does sodium metal appear in the usual aqueous brine cathode product list? Answer: No. Sodium remains as Na⁺ in solution while water reduction forms hydrogen and hydroxide.