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School Chemistry notes: Summary of electrode equations you need to know

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SUMMARY of electrode half-equations and products

ELECTROCHEMISTRY revision notes on electrolysis, cells, experimental methods, apparatus, batteries, fuel cells and industrial applications of electrolysis

(Suitable for AQA, Edexcel and OCR GCSE chemistry students)

8. Summary of electrode reactions and products

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8. Summary of the products of the electrolysis of various electrolytes

What are the products of the electrolysis of molten aluminium oxide, aqueous copper sulphate solution, aqueous sodium chloride solution (brine), hydrochloric acid, sulphuric acid, molten lead(II) bromide, molten calcium chloride? add overall equations?

Electrolyte

solution or melt

 negative cathode product negative electrode

cathode half-equation

 positive anode product positive electrode

anode half-equation

 comments
molten aluminium oxide

Al2O3(l)

 molten aluminium Al3+(l) + 3e– ==> Al(l) oxygen gas

2O2–(l) – 4e– ==> O2(g)

or  2O2–(l)  ==> O2(g) + 4e–

 The industrial method for the extraction of aluminium its ore
aqueous copper(II) sulfate

CuSO4(aq)

 copper deposit any conducting electrode e.g. carbon rod, any metal including copper itself

Cu2+(aq) + 2e– ==> Cu(s)

 oxygen gas inert electrode like carbon (graphite rod) or platinum

(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–

 The blue colour of the copper ion will fade as the copper ions are converted to the copper deposit on the cathode
aqueous copper (II) sulphate

CuSO4(aq)

 copper deposit any conducting electrode e.g. carbon rod, any metal including copper itself

Cu2+(aq) + 2e– ==> Cu(s)

 copper(II) ions – the copper anode dissolves copper anode

Cu(s) – 2e– ==> Cu2+(aq)

or  Cu(s) ==> Cu(s) + 2e–

 This is the basis of the method of electroplating any conducting solid with a layer of copper. When using both copper cathode and anode, the blue colour of the copper ion does not decrease because copper deposited at the (–) cathode = the copper dissolving at the (+) anode.
Copper(II) chloride

CuCl2(aq)

 copper deposit Cu2+(aq) + 2e– ==> Cu(s) chlorine gas

2Cl–(aq) – 2e– ==> Cl2(g)

 or  2Cl–(aq) ==> Cl2(g) + 2e– 
molten sodium chloride

NaCl(l)

 molten sodium Na+(l) + e– ==> Na(l) chlorine gas

2Cl–(l) – 2e– ==> Cl2(g)

 or  2Cl–(l) ==> Cl2(g) + 2e–

 This a method used to manufacture sodium and chlorine.
aqueous sodium chloride solution (brine)

NaCl(aq)

 hydrogen

2H+(aq) + 2e– ==> H2(g)

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

or 2H2O(l) + 2e– ==> H2(g) + 2OH–(aq)

 chlorine gas

2Cl–(aq) – 2e– ==> Cl2(g)

 or  2Cl–(aq) ==> Cl2(g) + 2e–

 This is the process by which hydrogen, chlorine and sodium hydroxide are manufactured
hydrochloric acid

HCl(aq)

 hydrogen gas 2H+(aq) + 2e– ==> H2(g)

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

 chlorine gas

2Cl–(aq) – 2e– ==> Cl2(g)

 or  2Cl–(aq) ==> Cl2(g) + 2e– 

 All acids give hydrogen at the cathode.

Theoretically the gas volume ratio is H2:Cl2 is 1:1, BUT, chlorine is slightly so there seems less chlorine formed than actually was.
sulphuric acid

sulfuric acid

H2SO4(aq)

 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– 

 All acids give hydrogen at the cathode. Whereas hydrochloric acid gives chlorine at the anode, the sulfate ion does nothing and instead oxygen is formed. This is the classic 'electrolysis of water'.

Theoretically the gas volume ratio is H2:O2 is 2:1 which you see with the Hofmann Voltammeter
molten lead(II) bromide

PbBr2(l)

 molten lead Pb2+(l) + 2e– ==> Pb(l) bromine vapour

2Br–(l) – 2e– ==> Br2(g)

 or  2Br–(l) ==> Br2(g) + 2e–

 A good demonstration in the school laboratory – brown vapour and silvery lump provide good evidence of what's happened
molten calcium chloride

CaCl2(l)

 solid or molten calcium Ca2+(l) + 2e– ==> Ca(s) chlorine gas

2Cl–(aq) – 2e– ==> Cl2(g)

 or  2Cl–(aq) ==> Cl2(g) + 2e–

 The basis of the industrial method for the manufacture of calcium metal
Molten anhydrous zinc chloride

ZnCl2(l)

 solid zinc Zn2+(l) + 2e– ==> Zn(s)

 

 chlorine gas

2Cl–(aq) – 2e– ==> Cl2(g)

 

 A good demonstration in the school laboratory - safer than using lead bromide
Silver nitrate

AgNO3(aq)

 solid silver

Ag+(aq) + 2e– ==> Ag(s)

 oxygen gas

4OH–(aq) – 4e– ==> 2H2O(l) + O2(g)

 electroplating experiment
Sodium bromide

 NaBr(aq)

 hydrogen gas 2H+(aq) + 2e– ==> H2(g)  

2Br–(aq) – 2e– ==> Br2(aq)

 School experiment
Potassium iodide

KI(aq)

 hydrogen gas 2H+(aq) + 2e– ==> H2(g)  

2I–(aq) – 2e– ==> I2(aq/s)

 School experiment
Sulfate salts of reactive metals > hydrogen theoretically hydrogen gas 2H+(aq) + 2e– ==> H2(g)  

4OH–(aq) – 4e– ==> 2H2O(l) + O2(g)

 School experiment

Similar results with most nitrate salts of reactive metals
           
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9. A shorter summary of electrode reactions (half-equations) and products

 

 

 

Equation

reference

number

(–) negative cathode electrode where reduction of the attracted positive cations is by electron gain to form metal atoms or hydrogen [from Mn+ or H+, n = numerical positive charge]. The electrons come from the positive anode (see below).

(+) positive anode electrode where the oxidation of the atom or anion is by electron loss. Non–metallic negative anions are attracted and may be oxidised to the free element. Metal atoms of a metal electrode can also be oxidised to form positive metal ions which pass into the liquid electrolyte. The released electrons move round in the external part of the circuit to produce the negative charge on the cathode electrode.

So, before each electrode equation is a (–) for a negative cathode electrode = a reduction reaction equation or a (+) for a positive anode electrode = an oxidation reaction equation

The electrode half-equations are shown on the left with examples of industrial processes where this electrode reaction happens on the right. Unless otherwise stated, the electrodes are inert i.e. they do not chemically change e.g. platinum or carbon–graphite.

 PLEASE NOTE – all electrode equations are a summary–simplification of what happens on an electrode surface in electrolysis. There may be e.g. two equations which are totally equivalent to each other to describe WHAT IS ACTUALLY FORMED e.g. the formation of hydrogen or oxygen and in some cases other products may be formed too.

1

a reduction electrode reaction

(–) Na+(l) + e– ==> Na(l) (sodium metal)

 sodium ion reduced to sodium metal atoms: typical of electrolysis of molten chloride salts to make chlorine and the metal
2

an oxidation electrode reaction

(+) 2Cl–(l/aq) – 2e– ==> Cl2(g)

 or  2Cl–(l/aq) ==> Cl2(g) + 2e–

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

 chloride ion oxidised to chlorine gas molecules: electrolysis of molten chloride salts(l) or their concentrated aqueous solution(aq) or conc. hydrochloric acid(aq) to make chlorine
3

a reduction electrode reaction

(–) 2H+(aq) + 2e– ==> H2(g) (hydrogen gas)

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

or 2H2O(l) + 2e– ==> H2(g) + 2OH–(aq)

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?

 hydrogen ion or water reduced to hydrogen gas molecules: electrolysis of many salt or acid solutions to make hydrogen
4

a reduction electrode reaction

(–) Cu2+(aq) + 2e– ==> Cu(s) (copper deposit)

 copper(II) ion reduced to copper atoms: deposition of copper in its electrolytic purification or electroplating using copper(II) sulphate solution, electrode can be copper or other metal to be plated
5

an oxidation electrode reaction

(+) Cu(s) – 2e– ==> Cu2+(aq) (copper dissolves)

or  Cu(s) ==> Cu(s) + 2e–

 copper atoms oxidised to copper(II) ions: dissolving of copper in its electrolytic purification or electroplating (must have positive copper anode)
6

a reduction electrode reaction

(–) Al3+(l) + 3e– ==> Al(l) (aluminium)

 aluminium ions reduced to aluminium atoms: extraction of aluminium in the electrolysis of its molten oxide ore(l) 
7

an oxidation electrode reaction

(+) 2O2–(l) – 4e– ==> O2(g) (oxygen gas)

or  2O2–(l)  ==> O2(g) + 4e–

 oxide ion oxidised to oxygen gas molecules: electrolysis of molten oxides e.g. anode reaction in the extraction of aluminium from molten bauxite.
8

an oxidation electrode reaction

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

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

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

or 2H2O(l) ==> 4H+(aq) + O2(g) + 4e–

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?

 There are two equations that describe the formation of oxygen in the electrolysis of water.

hydroxide ions or water molecules are oxidised to oxygen gas molecules: electrolysis of many salt solutions such as sulphates, sulphuric acid etc. gives oxygen (chlorides ==> chlorine in concentrated solution, but can also give oxygen in diluted solution)
9

 a reduction electrode reaction

(–) Pb2+(l) + 2e– ==> Pb(l) (lead deposit)

 lead(II) ions reduced to lead atoms: electrolysis of molten lead(II) bromide(l) 
10

an oxidation electrode reaction

(+) 2Br–(l/aq) – 2e– ==> Br2(g/l) (bromine)

or  2Br– ==> Br2 + 2e–

 bromide ions oxidised to gas/liquid bromine molecules: electrolysis of molten bromide salts(l) or their concentrated aqueous solution(aq) or conc. hydrobromic acid(aq) to make bromine
11

a reduction electrode reaction

(–) Zn2+(aq) + 2e– ==> Zn(s) (zinc deposit)

zinc ions reduced to zinc atoms: galvanising steel (the electrode) by electroplating from aqueous zinc sulphate solution, (or from molten zinc chloride?)
12

a reduction electrode reaction

(–) Ag+(aq) + e– ==> Ag(s) (silver deposit)

 silver ions reduced to silver atoms: silver electroplating from silver salt solution(aq), electrode can be other metal
13

a reduction electrode reaction

(–) Ca2+(l) + 2e– ==> Ca(s) (calcium metal)

calcium ions reduced to calcium atoms e.g. in molten calcium chloride or bromide etc.
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Electrolysis Quiz (GCSE 9-1 HT Level (harder)

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