7. The electrolysis of hydrochloric acid

The products of electrolysing hydrochloric acid are hydrogen gas and chlorine gas

Pure hydrogen chloride gas (HCl) and pure liquid water (H₂O) are covalent compounds.

In advanced level chemistry they would be referred to as polar compounds, but not ionic.

But, because they are essentially covalently bonded, they are very poor conductors of electricity, with very few free ions to carry an electric current (not even liquified hydrogen chloride gas).

However, hydrogen chloride gas dissolves in, and reacts with, water to form hydrochloric acid, with 100% ionisation to give a solution of hydrogen ions H⁺ and chloride ions Cl^- in a 1 : 1 ratio and this solution readily conducts electricity and undergoes electrolysis.

HCl (g) + aq ===>  HCl(aq)  ===>  H⁺(aq)  +  Cl^-(aq)

At advanced level chemistry this equation would be written as ...

HCl(g)  +  H₂O(l)  ===>  H₃O⁺(aq)  +  Cl^-(aq)

... to show the formation of the oxonium ion H₃O⁺ (not required for GCSE level chemistry, where H⁺, the simple symbol for the hydrogen ion, is sufficient)

OK, so now for the details of the electrolysis of hydrochloric acid.

You can demonstrate the electrolysis of hydrochloric acid using the Hoffman voltammeter with platinum electrodes (right diagram).  The Hofmann voltammeter is filled with the electrolyte (hydrochloric acid) by opening the taps at the top of the outer tubes to allow any gas to escape. The gases formed on the electrolysis of the dilute hydrochloric acid can be collected via the same taps.

In the simple electrolysis cell (left diagram), the graphite (carbon) electrodes are, through a large rubber bung, 'upwardly' dipped into an solution of dilute hydrochloric acid. In this cheap and simple apparatus the gaseous products (hydrogen and chlorine) are collected in small test tubes inverted over the carbon electrodes. You have to fill the little test tubes with the electrolyte (dil. hydrochloric acid), hold the liquid in with your finger and carefully invert them over the nearly full electrolysis cell. The electrolysis will only take place when electricity is passed through the dilute hydrochloric acid solution.

In both experiments above the platinum or carbon electrodes are inert. You can also use an even simpler apparatus illustrated on the right, using two inert metal wire electrodes.

The electrolysis will only take place when electricity is passed through the hydrochloric solution

The electrolyte hydrochloric acid, provides a high concentration of hydrogen ions H⁺ and chloride ions Cl^– to carry the current during the electrolysis process.

The electrode reactions and products of the electrolysis of dil. hydrochloric acid are illustrated by the theory diagram above.

Note: The majority of liquid water consists of covalent H₂O molecules, but there are trace quantities of H⁺ and OH^– ions from the reversible self–ionisation of water: H₂O(l) H⁺(aq) + OH^–(aq)

At the positive anode, the chloride ion, Cl^-, is discharged to give chlorine in preference to the hydroxide ion giving oxygen, but only hydrogen can be formed at the negative cathode, because the hydrogen ion is the only positive ion present.

The electrode half-equations for the electrolysis of dilute hydrochloric acid HCl(aq)

(a) The negative cathode electrode reaction for the electrolysis of hydrochloric solution

The positive hydrogen ions H⁺ (from hydrochloric acid) are attracted to the negative cathode electrode, and are reduced by electron gain to form hydrogen gas at the negative electrode.

2H⁺_(aq) + 2e^– ==> H_2(g)

positive ion reduction by electron gain

or 2H₃O⁺_(aq) + 2e^– ==> H_2(g) + 2H₂O_(l)

All acids give hydrogen at the negative cathode.

(b) The positive anode electrode reaction for the electrolysis of hydrochloric solution

The (+) anode attracts both the OH^– (from water) and Cl^– ions (from hydrochloric acid).

Only the chloride ion is discharged in appreciable quantities.

Chloride ions are oxidised by electron loss to form chlorine gas at the positive electrode.

2Cl^–_(aq) – 2e^– ==> Cl_2(g)

 or  2Cl^–_(aq) ==> Cl_2(g) + 2e^– 

negative ion oxidation by electron loss

Theoretically the gas volume ratio is H₂:Cl₂ is 1:1, BUT, chlorine is slightly soluble in water, so there seems to be less chlorine formed than actually was.

Also, if most of the chloride ions have been discharged as chlorine molecules, you then start to get some oxygen gas formed at the anode i.e. like in the electrolysis of water.

2H₂O_(l) – 4e^– ==> 4H⁺_(aq) +  O_2(g)

or  4OH^–_(aq) – 4e^– ==> 2H₂O_(l) +  O_2(g) (oxygen gas)

This is due to the increasingly lower probability of the chloride ion being oxidised and discharged as chlorine gas, versus the probability of water or hydroxide ion being oxidised and discharged as oxygen gas.

 

Overall equation for the electrolysis of hydrochloric acid:

2HCl(aq) ==> H₂(g) + Cl₂(g)

This could be written more accurately at advanced level as an ionic equation:

2H⁺(aq) + 2Cl^–(aq) ==> H₂(g) + Cl₂(g)

Extra COMMENTS on the electrolysis of hydrochloric chloride solution

1. Tests for the gases formed in the electrolysis of sodium chloride solution

The (–) cathode gas - colourless gas gives a squeaky pop with a lit splint – hydrogen

The (+) anode gas - pale green gas turns damp blue litmus red and then bleaches it white – chlorine

Chlorine forms a weakly acid solution in water, which is why it turns blue litmus pink-red, but it is NOT the crucial observation, that's the bleaching action of chlorine.

2. You can collect samples of gases through the taps on the Hofmann voltammeter or from the little test tubes in the simple school electrolysis cell.

3. Theoretically, in the electrolysis of hydrochloric acid solution, the gas volume ratio for H₂ : Cl₂ is 1 : 1, BUT chlorine is slightly soluble in water and  therefore the volume of chlorine gas observed is less than predicted.

Why a 1 : 1 product ratio? It takes two electrons to reduce two hydrogen ions to a hydrogen molecule. It takes the removal of two electrons, one from each chloride ion, to form a chlorine molecule. So, for the same quantity of current passing (electron flow), you should expect to form equal numbers of hydrogen and chlorine molecules.

 Doc Brown's Chemistry 

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