ELECTROLYSIS 3.

Using an electrolysis cell - investigating the electrolysis of sodium chloride solution (brine) and molten sodium chloride and other halide salts

Aqueous solutions with inert electrodes (carbon or platinum)

The products of electrolysing aqueous sodium chloride solution are hydrogen gas, chlorine gas and sodium hydroxide solution

The electrolysis of aqueous sodium chloride (often referred to as 'brine' solution) is described in terms of apparatus and products formed. What are the products of the electrolysis of aqueous sodium chloride solution (brine)?

The simple apparatus illustrated on the right can be used in simple school or college experiments for the electrolysis of sodium chloride solution (often referred to as 'brine' in the chemical industry). The graphite (carbon) electrodes are, through a large rubber bung, 'upwardly' dipped into an solution of the sodium chloride solution (the electrolyte).

The cell can be made from plastic pipe and a big rubber bung with two holes in it.  In the simple apparatus the gaseous products (hydrogen and chlorine) are collected in small test tubes inverted over the carbon electrodes and chemical tests performed on them. You have to fill the little test tubes with the electrolyte (sodium chloride solution), hold the liquid in with your finger and carefully invert them over the nearly full electrolysis cell.

A more elaborate format is to use a Hoffman Voltammeter (above left diagram) using platinum electrodes and accurately calibrated collecting tubes like burettes. The Hofmann voltammeter is filled with the electrolyte (aqueous sodium chloride solution) 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 'brine' solution can be collected via the same taps. The students should note that nothing happens until you switch on the electricity supply (see simple animation above!). The platinum or carbon electrodes are inert.

The industrial electrodes must be made of an inert material like platinum/titanium which is not attacked by chlorine or alkali, but in the school /college laboratory, the Hofmann voltammeter is a good demonstration (platinum electrodes) and the 'simple cell' for students uses carbon/graphite electrodes which are reasonably inert.

However a simple cell using carbon electrodes can be used by students/pupils to demonstrate the industrial process in the laboratory and the simple apparatus (above right) can also be used in schools using two inert wire electrodes.

The electrolysis will only take place when electricity is passed through the sodium chloride solution.

The electrode reactions and products of the electrolysis of sodium chloride solution (brine) are illustrated by the theory diagram above

The electrolyte sodium chloride solution (brine), provides a high concentration of sodium ions Na⁺ and chloride ions Cl^– to carry the current during the electrolysis process.

Initially there are only traces of hydrogen ions H⁺ and hydroxide ions OH^– from the self-ionisation of water.

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)

Brine is moderately concentrated sodium chloride solution (brine) with carbon (graphite) gives equal volumes of hydrogen gas (hydrogen ions H⁺ discharged at the –ve cathode) and green chlorine gas (chloride ions Cl^– discharged at the +ve anode) with sodium hydroxide left in solution. The electrolysis will only take place when electricity is passed through the sodium chloride solution.

The electrode equations and the theory of what happens in the electrolysis of aqueous sodium chloride

The half-equations for the electrolysis of sodium chloride solution (the electrolyte brine).

(a) The negative cathode electrode reaction for the electrolysis of brine (sodium chloride solution)

The negative (–) cathode attracts the Na⁺ (from sodium chloride) and H⁺ ions (from water). Only the hydrogen ions are discharged at the cathode. The more reactive a metal, the less readily its ion is reduced on the electrode surface.

The hydrogen ions are reduced by electron (e^–) gain to form hydrogen molecules at the negative electrode which attracts positive ions.

2H⁺_(aq) + 2e^– ==> H_2(g)

positive ion reduction by electron gain

other equations

2H₂O_(l) + 2e^–  ==> H_2(g) + 2OH^-_(aq)

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

Nothing happens to the sodium ion, but it is still important (see after the anode reaction has been described).

In fact, if sodium was released (which it isn't), it would immediately react with water to give hydrogen, the same product you get from the reduction of the hydrogen ion.

Test for the cathode gas - colourless gas gives a squeaky pop with a lit splint – hydrogen

(b) The positive anode electrode reaction for the electrolysis of brine (sodium chloride solution)

The positive anode attracts the negative hydroxide OH^– ions (from water) and chloride Cl^– ions (from sodium chloride). Only the chloride ion is discharged in appreciable quantities i.e. it is preferentially oxidised to chlorine.

The chloride ions are oxidised by electron loss to give chlorine molecules at the positive electrode which attracts negative ions.

an oxidation electrode reaction

2Cl^–_(aq) – 2e^– ==> Cl_2(g)

 or  2Cl^– ==> Cl_2(g) + 2e^–

negative ion oxidation by electron loss

Note that you can write these anode oxidation reactions either way round

The chloride ion is oxidised to chlorine gas molecules in any chloride salt solution electrolysed, hydrochloric acid and in any electrolysis of a molten chloride salt.

Test for the anode gas - pale green gas turns damp blue litmus red and then bleaches it white – chlorine (test 2 gas 2)

Usually nothing happens to the hydroxide ion BUT it is important, because, the hydroxide ion, with the unchanged sodium ion, means the residual solution contains sodium hydroxide. In fact this is how sodium hydroxide is manufactured in the chemical industry.

Residual Na⁺ + OH^– = NaOH, a familiar formula! The presence of the alkali sodium hydroxide, can be shown by adding universal indicator/red litmus to the residual brine solution (aqueous sodium chloride) at the end of the experiment.

The indicator will turn from green to purple because of the formation of alkaline sodium hydroxide.

 

Note that, if most of the chloride ions have been discharged as chlorine molecules, you can then get some oxygen gas formed at the anode i.e. like in the electrolysis of water, and chloride ions are being replaced by hydroxide ions which can be oxidised to oxygen at the anode.

2H₂O_(l) – 4e^– ==> 4H⁺_(aq) + O_2(g)

or

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

For more, see Extra COMMENTS 2.

Summary of the possible products from the electrolysis of aqueous sodium chloride solution

The three products from the electrolysis of sodium chloride solution are all of industrial significance:

hydrogen, chlorine and sodium hydroxide.

Overall equation for the electrolysis of brine:

2NaCl(aq) + 2H₂O(l) ==> H₂(g) + Cl₂(g) + 2NaOH(aq)

and the ionic equation is ...

2H₂O_(l) + 2Cl^-_(aq) + 2Na⁺_(aq) ==> 2Na⁺_(aq) + 2OH^-_(aq) + H_2(g) + Cl_2(g)

or more correctly   2H₂O_(l)  +  2Cl^-_(aq)  ===>  2OH^-_(aq)  +  H_2(g)  +  Cl_2(g)

(since the sodium ions are spectator ions)

by treating the sodium ion as a spectator ion, though it is an important end product, in combination with the other residual ion, the hydroxide ion, they constitute sodium hydroxide, the third major product important for the chlor-alkali chemical industry.

Another complication in the electrolysis of sodium chloride solution, is that the chlorine will react with sodium hydroxide to form sodium chlorate(I) NaOCl, which is how a bleach is made - but this situation is usually studied at a more advanced level of chemistry.

For the industrial electrolysis of brine and the uses of the products see The Halogens and Salt page.

Five Extra COMMENTS on the electrolysis of sodium chloride solution and other related electrolysis reactions

Some comments make reference to the diagram of the electrolysis of brine above.

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 (test 1 gas 1)

The (+) anode gas - pale green gas turns damp blue litmus red and then bleaches it white – chlorine (test 2 gas 2)

For the industrial electrolysis of brine and the uses of the products see The Halogens page.

You can collect samples of gases through the taps on the Hofmann voltammeter or from the little test tubes in the simple school electrolyse cell. The universal indicator changes from green (~ pH 7 for the salt solution) to blue-purple (Ph > 7) as the alkali sodium hydroxide is formed.

2. In very dilute sodium chloride solution, oxidation of hydroxide ions or water molecules can produce oxygen gas as well as chlorine gas.

Advanced Level Student Note on the ratio of chlorine to oxygen production:

At low concentrations of chloride ion a competing oxidation of water or hydroxide ion can occur, particularly as the concentration of hydroxide ion is increasing as the electrolysis proceeds.

The increase in oxygen to hydrogen ratio through the electrolysis is essentially a concentration effect.

If you consider the electrode potentials:

O₂/OH^- E^θ = +0.40 V   and for   Cl₂/Cl^- E^θ = +1.36 V,

then, logically, the hydroxide ion OH^- is more easily oxidised than the chloride Cl^- ion.

So, initially the concentration-kinetic factor wins out, the much higher concentration of chloride ions over hydroxide ions leads to the much more probable oxidation of the chloride ion to form chlorine. BUT, as the brine (NaCl_(aq)) becomes depleted in chloride ions, and the hydroxide ion is increasing (a product of the electrolysis), the probability of OH^- ion oxidation to give oxygen is more likely, so you begin to get an increase in the O₂/Cl₂ ratio in the product gases at the positive anode electrode the longer the electrolysis continues - at least until no chloride ions are left to be discharged.

3. Theoretically, in the electrolysis of sodium chloride solution, the gas volume ratio for H₂ : Cl₂ is 1 : 1

BUT chlorine is slightly soluble in water and also reacts with the sodium hydroxide formed (the residual solution).

Therefore the volume of chlorine gas observed is seems to be less than predicted.

Why 1 : 1 gas volume 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.

4. Electrolysis of molten sodium chloride gives silvery sodium metal and pale green chlorine gas.

This is a simpler electrolysis situation where the ionic compound sodium chloride on melting provides a highly concentrated mixture of positive sodium ions and negative chloride ions.

It also illustrates the difference sometimes, between electrolysing the pure molten salt and its aqueous solution in water. Here there is no possibility of hydrogen being formed.

The electrode reactions and products of the electrolysis of the molten ionic compound sodium chloride are illustrated by the theory diagram above

molten sodium chloride electrolyte NaCl(l)

(i) molten sodium formed at the negative cathode electrode which attracts the positive sodium ions

Na⁺_(l) + e^– ==> Na_(l)   a reduction electrode reaction (electron gain)

positive ion reduction by electron gain

sodium ion reduced to sodium metal atoms:

typical of electrolysis of molten chloride salts to make chlorine and the metal

(ii) chlorine gas formed at the positive anode electrode which attracts the negative chloride ions

2Cl^–_(l) – 2e^– ==> Cl_2(g)

 or  2Cl^–_(l) ==> Cl_2(g) + 2e^–   an oxidation electrode reaction (electron loss)

negative oxidation by electron loss

See The extraction of sodium from molten sodium chloride using the 'Down's Cell'

APPENDIX 1.

5. The electrolysis of aqueous solutions of sodium bromide and potassium iodide

The concept diagrams for aqueous sodium chloride are equally valid, just substitute in your head Br or I for Cl.

Because sodium and potassium are reactive metals, so you will get hydrogen ions preferentially discharged at the negative cathode giving hydrogen gas.

Sodium bromide and potassium iodide give colourless solutions when dissolved in water, therefore it is quite easy to spot if the coloured halogen elements are formed on the positive anode electrode.

You can use any simple electrolysis apparatus to these experiments.

Sodium Bromide, NaBr(aq)

Sodium bromide gives hydrogen at the cathode and the element bromine at the anode - you would see a orange-brown colouration appearing around the positive electrode.

cathode (-):   2H⁺_(aq) + 2e^–  ==> H_2(g)

anode (+):   2Br^–_(aq) – 2e^–  ==> Br_2(aq)

Potassium Iodide, KI(aq)

Potassium iodide gives hydrogen at the cathode and the element iodine at the anode - you would see a brown colouration appearing around the positive electrode and may be a dark solid precipitate if sufficient iodine is formed.

cathode (-):   2H⁺_(aq) + 2e^– ==> H_2(g)

anode (+):   2I^–_(aq) – 2e^– ==> I_2(aq/s)

Electrolysis Quiz (GCSE 9-1 FT Level (easier)

 Doc Brown's Chemistry 

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