The ELECTROLYSIS OF WATER

Doc Brown's Chemistry KS4 science–chemistry GCSE/IGCSE/O level/A Level Revision Notes - ELECTROCHEMISTRY revision notes on electrolysis, cells, experimental methods, apparatus, batteries, fuel cells, industrial applications of electrolysis

2. Electrolysis of acidified water (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.

Reminders: Electrolysis (of acidified water) is a way of splitting up (decomposition) of the compound (water) using electrical energy. The electrical energy comes from a d.c. (direct current) battery or power pack supply. A conducting liquid, containing ions, called the electrolyte (dilute sulfuric acid), must contain the compound (water) that is being broken down. The electricity must flow through electrodes dipped into the electrolyte to complete the electrical circuit with the battery. Electrolysis can only happen when the circuit is complete, and an electrical current (electricity) is flowing, then the products of electrolysing dil. sulfuric acid are released on the electrode surfaces where they can be collected. Electrolysis always involves a flow of electrons in the external wires and electrodes and a flow of ions in the electrolyte and there is always a reduction at the negative cathode electrode (which attracts positive ions, cations) and an oxidation at the positive anode electrode (which attracts negative ions, anions) and it is the ions which are discharged to give the products. These revision notes on the electrolysis of acidified water should prove useful for the new AQA chemistry, Edexcel chemistry & OCR chemistry GCSE (9–1, 9-5 & 5-1) science courses.

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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 voltameter 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 voltameter is filled with the electrolyte (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, 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.

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) from the acid, makes water a much better electrical conductor (a much better electrolyte solution). These small extra ion concentrations of hydrogen ions (H+) and hydroxide ions (OH–) are from the self-ionisation of water itself.

Diagram of the electrolysis of water.

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 ion or water molecules are reduced to hydrogen gas molecules by electron gain at the negative electrode

2H+(aq) + 2e– ==> H2(g) (hydrogen gas, bubbles seen on the negative electrode)

positive ion reduction by electron gain

or 2H3O+(aq) + 2e– ==> H2(g) + 2H2O(l)

All three 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.

 

(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 (SO42-) or the traces of hydroxide ions (OH–) are attracted to the positive electrode. But the sulfate ion is too stable and nothing happens (its 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

(i) (+) 2H2O(l) – 4e– ==> 4H+(aq) + O2(g) (oxygen gas)

or 2H2O(l) ==> 4H+(aq) + O2(g) + 4e–

molecule oxidation by electron loss

(ii) 4OH–(aq) – 4e– ==> 2H2O(l) + O2(g) (oxygen gas)

negative ion oxidation by electron loss

or  4OH–(aq) ==> 2H2O(l) + O2(g) + 4e–

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. 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?

 

Overall equation for the electrolysis of water: 2H2O(l) ==> 2H2(g) + O2(g)

 

Extra COMMENTS on the electrolysis of water

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.

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.

The electrolysis of sodium hydroxide gives exactly the same products, hydrogen and oxygen in exactly the same proportions 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.

(c) doc bGenerally speaking, the less reactive a metal, the more easily its ion is reduced to the metal on the electrode surface e.g. in a mixture of positive ions the preference order is

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

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

 

 

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

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, for the 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.

 

3. Chemical Tests for the gases formed from electrolysis of water experiment

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

hydrogen – colourless gas that gives a squeaky pop when ignited with a lit splint.

oxygen – colourless gas that relights a glowing splint.

 

4. What is left?

As the electrolysis proceeds the dilute sulfuric acid gets slightly more concentrated as only water is removed from the electrolyte as hydrogen and oxygen.

 

SUMMARY OF PRODUCTS FROM THE ELECTROLYSIS OF acidified WATER

with inert carbon (graphite) electrodes or inert platinum electrodes

Electrolyte negative cathode product negative electrode

cathode half-equation

positive anode product positive electrode

anode half-equation

dilute solution of sulfuric acid

sulfuric acid

H2SO4(aq)

but can be sulfate salts of reactive metals whose ions are not discharged, and alkaline hydroxides of reactive metals like sodium hydroxide

hydrogen gas 2H+(aq) + 2e– ==> H2(g)

or 2H3O+(aq) + 2e– ==> H2(g) + 2H2O(l)

oxygen gas

(i) 4OH–(aq) – 4e– ==> 2H2O(l) + O2(g)

or  4OH–(aq) ==> 2H2O(l) + O2(g) + 4e–

(ii) (+) 2H2O(l) – 4e– ==> 4H+(aq) + O2(g)

or 2H2O(l) ==> 4H+(aq) + O2(g) + 4e– 

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ELECTROCHEMISTRY INDEX:  1. INTRODUCTION to electrolysis - electrolytes, non-electrolytes, electrode equations, apparatus 2. Electrolysis of acidified water (dilute sulfuric acid) and some sulfate salts and alkalis 3. Electrolysis of sodium chloride solution (brine) and bromides and iodides 4. Electrolysis of copper(II) sulfate solution and electroplating with other metals e.g. silver 5. Electrolysis of molten lead(II) bromide (and other molten ionic compounds) 6. Electrolysis of copper(II) chloride solution 7. Electrolysis of hydrochloric acid 8. Summary of electrode equations and products 9. Summary of electrolysis products from various electrolytes 10. Simple cells (batteries) 11. Fuel Cells e.g. the hydrogen - oxygen fuel cell 12. The electrolysis of molten aluminium oxide - extraction of aluminium from bauxite ore & anodising aluminium to thicken and strengthen the protective oxide layer 13. The extraction of sodium from molten sodium chloride using the 'Down's Cell' 14. The purification of copper by electrolysis 15. The purification of zinc by electrolysis 16. Electroplating coating conducting surfaces with a metal layer 17. Electrolysis of brine (NaCl) for the production of chlorine, hydrogen & sodium hydroxide AND 18. Electrolysis calculations


Electrolysis Quiz (GCSE 9-1 HT Level (harder)

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


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