1. Introduction to the electrolysis of water

Using an electrolysis cell - investigating the electrolysis of acidified water (actually dilute sulfuric acid) AND the aqueous solutions of certain sulfate salts of reactive metals e.g. sodium sulfate, magnesium sulfate and strong alkalis like sodium hydroxide.

The electrolysis of dilute sulfuric acid is described and explained. This is the classic 'electrolysis of water' experiment and good introduction to electrolysis experiment. The electrode products and electrode equations for the electrolysis of water are quoted.  What are the products of the electrolysis of water acidified with dil. sulfuric acid? All the electrode equations for the electrolysis of water are explained and diagrams of the apparatus.

Know the electrolysis of water is an important process to produce the green fuel hydrogen from surplus electricity produced from renewable energy sources such as hydroelectric power or wind turbines.

Electrolysis of dilute sulfuric acid - the products of electrolysing water acidified with sulfuric acid are hydrogen gas and oxygen gas

Two experimental setups are described, the Hofmann voltammeter demonstration (left diagram) and a simple cell (right diagram) for use in schools and colleges for pupils to use. Dilute sulfuric acid is used as the electrolyte in this investigation. The Hofmann voltammeter is filled with the electrolyte (dilute sulfuric 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 sulfuric acid can be collected via the same taps. The platinum or carbon electrodes are inert.

You need inert (non–reactive) electrodes like platinum (left) and much cheaper carbon (graphite electrodes, right). In the simple electrolysis cell, the graphite (carbon) electrodes are inserted through a large rubber bung, 'upwardly' dipped into an solution of acidified water. In this cheap and simple apparatus the gaseous products (hydrogen and oxygen) are collected in small test tubes inverted over the carbon electrodes. You have to fill the little test tubes with the electrolyte (dil. sulfuric acid), hold the liquid in with your finger and carefully invert them over the nearly full electrolysis cell. The gases can be collected and tested.

On the right is another simple apparatus for doing experiments like the electrolysis of water. The electrodes must be made of an inert wire.

Students should note that the electrolysis will only take place when electricity is passed through the dilute sulfuric acid solution.

The electrolyte is dilute sulfuric acid (= acidified water) which, during electrolysis is split into hydrogen and oxygen gases. Theoretically into a 2 : 1 ratio by gas volume - which you can measure with the Hofmann Voltammeter.

This is one experimental method of showing water is a compound composed of the elements hydrogen and oxygen atoms i.e. by splitting liquid water into two gaseous element molecules of hydrogen (H₂) and oxygen (O₂).

Water only ionises to a tiny extent giving minute concentrations of hydrogen ions and hydroxide ions, so the presence of high concentrations of hydrogen ions (H⁺ or H₃O⁺) and sulfate ions (SO₄^2–) from the sulfuric acid, makes water a much better electrical conductor (a much better electrolyte solution) than pure water.

These small extra ion concentrations of hydrogen ions (H⁺) and hydroxide ions (OH^–) are from the self-ionisation of water itself.

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)

2. Description and technical details of the electrolysis of water (dilute sulfuric acid)

Diagram explaining the electrolysis of water showing the ion movement to the electrodes..

The electrode reactions and products of the electrolysis of acidified water are illustrated by the theory diagram above

The half-equations for the electrolysis of water electrode equations (electrolyte of acidified with dilute sulfuric acid) are described and explained below.

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

The negative cathode electrode reaction is a reduction (electron gain).

The hydrogen ions (H⁺) are attracted to the negative cathode and are discharged as hydrogen gas.

The hydrogen ions are discharged and reduced to hydrogen gas molecules by electron gain at the negative electrode

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

Hydrogen gas bubbles seen on the negative cathode electrode)

The positive hydrogen ion is reduced by electron gain to form hydrogen molecules.

You can also, at a higher level, write the reduction half reaction as ...

2H₃O⁺_(aq) + 2e^– ==> H_2(g) + 2H₂O_(l) (cathode half-reaction)

The two equations amount to the same overall change i.e. the formation of hydrogen gas molecules and as far as I know any is acceptable in an exam?

The electrolysis of many salts (of reactive metals) or acid solutions produce hydrogen at the negative cathode electrode.

All acids give hydrogen at the cathode.

Hydrogen ions are reduced to hydrogen in preference to the ions of reactive metals (not so for less reactive metals e.g. electrolysis of copper salt solutions).

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

The positive anode reaction is an oxidation electrode reaction (electron loss).

The negative sulfate ions (SO₄^2-) or the traces of hydroxide ions (OH^–) are attracted to the positive electrode, but the sulfate ion is too stable and nothing happens (it is not discharged). Instead either hydroxide ions or water molecules are discharged and oxidised to form oxygen.

The hydroxide ions or water molecules are oxidised to oxygen gas molecules by electron loss at the positive electrode

The half reaction at the anode is the discharging and oxidation of the hydroxide ion leading to the formation of oxygen molecules.

4OH^–_(aq) – 4e^– ==> 2H₂O_(l) + O_2(g) (anode half-reaction)

negative ion oxidation by electron loss

or  4OH^–_(aq) ==> 2H₂O_(l) + O_2(g) + 4e^– (anode half-reaction)

There are two possible electrode equations that describe the formation of oxygen in the electrolysis of water, but in strongly acid solution equation (i) is probably more appropriate, whereas in alkali, (ii) is the more likely electrode equation.

Both equations amount to the same overall change i.e. the formation of oxygen gas molecules and as far as I know either is acceptable in an exam?

The overall equation for the electrolysis of water is:

2H₂O(l) ==> 2H₂(g) + O₂(g)

Note the 2:1 molar ratio of H₂:O₂ gases.

3. Extra COMMENTS on the electrolysis of water and 'inert' salt solutions

1. The electrolysis of many salts (e.g. sulfates of reactive metals) or sulfuric acid produces hydrogen at the negative cathode electrode and oxygen at the positive anode electrode.

Whereas hydrochloric acid gives chlorine at the anode (as will all chloride salts), the sulfate ion does nothing and instead oxygen is formed.

See Part 7. Electrolysis of hydrochloric acid

So for most sulfate salts of reactive metals e.g. sodium sulfate, magnesium sulfate and potassium sulfate, the electrolysis products of the aqueous salt solution are hydrogen at the negative (–) cathode electrode and oxygen at the positive (+) anode with inert electrodes such as carbon or platinum.

So the electrode equations are the same as above e.g.

(i) 2H⁺_(aq) + 2e^– ==> H_2(g) (at the -ve cathode, half-reaction)

 (ii) 4OH^–_(aq) – 4e^– ==> 2H₂O_(l) + O_2(g)  (at the +ve anode, half-reaction)

The electrolysis of sodium hydroxide gives exactly the same products, hydrogen and oxygen in exactly the same proportions (2:1 by volume) so, as with the salt examples quoted above, the electrode equations are the same as above.

Aqueous solutions containing the cations of reactive metals like potassium, sodium or magnesium are NOT discharged at the negative cathode electrode, but hydrogen ions are, so you get hydrogen instead of a metal formed on the electrode surface.

Generally speaking, the less reactive a metal, the more easily its ion is reduced to the metal on the electrode surface.

A general rule with reference to the reactivity series of metals:

If the metal in the salt solution is more reactive than hydrogen, then the hydrogen ion is most likely to be discharged at the negative cathode giving off hydrogen.

If the metal in the salt solution is less reactive than hydrogen it is the metal ion that is likely to be discharged forming a deposit of the metal on the electrode surface.

These rules apply if a mixture of ions is present in the electrolyte.

4. The theoretical gas volume ratio and tests for the gases

Theoretically the gas volume ratio for H₂:O₂ is 2:1 which you see with the Hofmann Voltammeter in section 1.

This isn't just because the ratio of hydrogen atoms to oxygen atoms is 2 : 1 in the molecule

It is equally important that you realise for every four electrons that flow through the circuit four hydrogen ions are reduced to two molecules of hydrogen AND two molecules of water (or four hydroxide ions) are oxidised to give one molecule of oxygen.

Therefore, passage of the same quantity of electrical current (electron flow), you would expect two molecules of hydrogen to one of oxygen in the electrolysis of water.

Since both gases are only very slightly soluble in water, you should observe an approximate 2:1 ratio for H₂:O₂ gas volumes,

 

Chemical Tests for the gases formed from electrolysis of water experiment

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.

At the negative cathode electrode a colourless gas that gives a squeaky pop when ignited with a lit splint - hydrogen.

At the positive anode a colourless gas that relights a glowing splint - oxygen.

What is left in the electrolyte solution after prolonged electrolysis?

As the electrolysis proceeds the dilute sulfuric acid gets slightly more concentrated as only water is removed from the electrolyte, albeit as hydrogen and oxygen in a 2:1 ratio.