Oxidation and reduction IB syllabus > redox (hl) > 19.3

19.3 - Electrolysis


19.3.1: List and explain the factors affecting the products formed in the electrolysis of aqueous solutions. Factors to be considered are position in the electrochemical series, nature of the electrode and concentration. Suitable examples for electrolysis include water, aqueous sodium chloride and aqueous copper II sulphate.


Electrolysis of solutions

Solutions in water contain hydrogen and hydroxide ions as well as the ions of the solute. These H+ and OH- ions compete at the electrodes with the solute ions.

The ions that are successfully released at the electrodes depend on three factors

  1. The position of the ion in the electrochemical series
  2. The concentration of the ion in the solution
  3. The nature of the electrode

1. The position of the ion in the electrochemical series

This is probably better expressed as the position of the redox equilibrium in the electrochemical series. All the redox equilibria are expressed as reductions going from left to right, for example:

Cu2+ + 2e Cu

However, going from the right hand side to the left hand side would be an oxidation. Hence it may be called a redox equilibrium.

The Copper ion | Copper equilibrium has an electrode potential of +0.34 V. This is more positive than hydrogens electrode potential of 0.00 V. This means that copper ions are more easily reduced (i.e. have more tendency to go to the right hand side) than hydrogen ions (remember that as we go down the series the redox equilibria species on the left hand side have better oxidising properties, i.e they can remove electrons from other things more easily - including electrodes).

Consequently in a competition between the two ions the copper ions will preferentially pick up the electrons.

As a rule of thumb, if the metal appears below hydrogen in the electrochemical series then it will be preferentially deposited.


2. The concentration of the ions

When two ions with similar reactivity are in competition then the relative concentration of the two ions becomes an important factor.

A good example of this is the electrolysis of sodium chlorides. When the chloride concentration is large the chloride ions lose electrons and chlorine gas is released at the electrode, but when it is in low concentration the hydroxide ions from the water are preferentially released


3. The nature of the electrode

Usually, inert electrodes such as graphite or platinum are used for electrolysis. These electrodes do not interfere with the reactions occuring at the surface of the electrode, they simply act as a point of connection between the electrical circuit and the solution.

However, if metal electrodes are used in metal ion solutions they can get involved in the reactions by dissolving as ions, leaving their electrons behind (this can only happen when the metal takes the place of the anode, the positive electrode) - this is called electrode participation


 

Example: Electrolysis of sodium chloride solution

The ions present in the solution are:

sodium ions chloride ions hydrogen ions hydroxide ions
Na+ Cl- H+ OH-

At the cathode

The positive ions are attracted to the negative cathode. There is competition between the sodium ions and the hydrogen ions. As the hydrogen ion | hydrogen redox equilibrium appears lower in the electrochemical series than the sodium ion | sodium equilibrium, then the hydrogen ions are preferentially reduced and hydrogen gas is produced at the electrode (bubbles are seen)

2H+ + 2e H2

At the anode

There is competition between the negative ions at the positive anode. The chloride ions compete with the hydroxide ions to release their electrons to the anode. Whe the solution is fairly concentrated the chloride ions preferentially lose electrons to become chlorine atoms (and then molecules)

2Cl- - 2e Cl2

Ions remaining in solution

The ions that are removed from the solution, then, are the hydrogen ions and the chloride ions. This means that the sodium ions and the hydroxide ions remain in the solution - i.e sodium hydroxide is also produced.

Note: When the solution of chloride ions is dilute then OH. ions are preferentially released at the anode.

 

Example: Electrolysis of copper II sulphate solution

The ions present in the solution are:

copper ions sulphate ions hydrogen ions hydroxide ions
Cu2+ SO42- H+ OH-

At the cathode

The positive ions are attracted to the negative cathode. There is competition between the copper ions and the hydrogen ions. As the hydrogen ion | hydrogen redox equilibrium appears higher in the electrochemical series than the copper ion | copper equilibrium, then the copper ions are preferentially reduced and copper metal is deposited at the electrode (a pink layer is observed)

Cu2+ + 2e Cu

At the anode

There is competition between the negative ions at the positive anode. The sulphate ions compete with the hydroxide ions to release their electrons to the anode. The hydroxide ions are much better reducing agents and are preferentially released AS OXYGEN GAS and water

4OH- - 4e 2H2O + O2

Ions remaining in solution

The ions that are removed from the solution, then, are the copper ions and the hydroxide ions. This means that the hydrogen ions and the sulphate ions remain in the solution - i.e sulphuric acid is also produced.

 

Example: Electrolysis of copper(II) sulphate solution using copper electrodes (participating electrodes)

The ions present in the solution are:

copper ions chloride ions hydrogen ions hydroxide ions
Cu2+ SO42- H+ OH-

At the cathode

The positive ions are attracted to the negative cathode. There is competition between the copper ions and the hydrogen ions. As the hydrogen ion | hydrogen redox equilibrium appears higher in the electrochemical series than the copper ion | copper equilibnrium, then the copper ions are preferentially reduced and copper metal is deposited at the electrode (a pink layer is observed)

Cu2+ + 2e Cu

At the anode

In this case, the electrode is made of copper and it is easier for the copper to dissolve leaving its electrons behind on the anode than for any other ion to be released.

Cu - 2e Cu2+

Ions remaining in solution

Copper is deposited at the cathode and is dissolved at the anode. Consequently the concentration of copper ions in solution remains constant. This can be used as a method of purification of copper as only pure copper is deposited at the cathode.

In this purification an anode made of impure copper is turned to pure copper at the cathode leaving the impurities behind (the sludge in the diagram).

 


Summary

Reactive metals (more reactive than hydrogen) are never deposited during electrolysis of aqueous solutions. If the metal ion comes from a metal more reactive than hydrogen then hyrogen gas is liberated at the cathode.

Halide ions (chloride, bromide, iodide) are released preferentially and if these are not present, the hydroxide ions from the water are released at the anode.

Unreactive metals, such as copper or nickel, may participate in reactions at the anode.


19.3.2: List the factors affecting the amount of product formed during electrolysis. Factors are charge on the ion, current and duration of electrolysis


Products of electrolysis

The ions involved in elecrolysis are picking up electrons and depositing electrons at the electrodes. It is apparent that the number of ions involved will be directly related to the number of electrons passing around the external circuit. It will also depend on the charges of the ions involved.


Relationship between current and number of electrons

Each electron posseses an electrical charge and the movement of the electrons around the external circuit may be thought of as the movement of electrical charge 'Q'. This is related to the electrical current as follows:

1 Ampere of current (I) is the passage of 1 Coulomb (C) of charge per second around the circuit

To calculate the number of coulombs of charge it is necessary to multiply the current by the time in seconds

Hence:- Q = It  

One mole of electrons has a charge equivalent to 96,500 coulombs - also called a Faraday of charge (F)


Charge on the ion

So, to release one mole of a singly charged ion at an electrode exactly one Faraday of charge (96,500 C) must pass through the electrode.

If the ion has a double charge then two moles of electrons (2 Faradays) are needed to release 1 mole of ions

Example

Calculate the number of moles of hydrogen released when 5 amps of current passes for 3000 seconds through a solution of sulphuric acid

As:- Q = It  
Then:- Q = 5 x 3000 = 15000 C  
Therefore:- Q = 15000/96500 Faradays  
  Q = 0.155 F  

In the electrolysis the hydrogen gas is released at the cathode as follows:

2H+ + 2e H2

2 moles of electrons release 1 mole of gas
2 Faradays of charge are needed to release 1 mole of hydrogen
Therefore 0.155 F releases 0.155/2 moles of hydrogen
 
Therefore moles of hydrogen released = 0.0777 moles

 

 

Example

Calculate the number of coulombs needed to deposit 6.35g of copper at the cathode in an electrolysis

Copper RAM = 63.5  
No of moles Cu = 6.35/63.5  
= 0.1 moles  
     

In electrolysis copper metal is released at the cathode as follows:

Cu2+ + 2e Cu

2 moles of electrons release 1 mole of copper
Therefore 0.2 moles of electrons release 0.1 moles of copper
0.2 moles of electrons = 0.2 Faradays = 0.2 x 96,500 coulombs
= 19,300 coulombs

Calculate the time that a current of 4 amps must pass to deposit this mass of copper

as Q = It

19,300 coulombs = It
19,300 coulombs = 4 x t
Therefore t = 19,300 / 4
t = 4825 seconds

 

 


19.3.3: Determine the relative amount of products formed during the electrolysis of aqueous solutions


Relative quantities of product

As ions have different charges it follows that diffierent ions will require different amounts of charge for release at an electrode.

For example, copper ions have a 2+ charge, whereas silver ions have a single (1+) charge. It follows, then, that twice as many electrons are needed to deposit one mole of copper than 1 mole of silver

Cu2+(aq) + 2e Cu (s)
1 mole of copper ions + 2 moles electrons   1 mole of copper atoms
     
Ag+(aq) + 1e Ag (s)
1 mole of silver ions + 1 mole electrons   1 mole of silver atoms
     

Therefore for the same number of moles of electrons - i.e the same electrical charge - twice as many moles of silver as copper will be deposited.


Example

In the following diagram the number of moles of silver deposited will be twice the number of moles of copper deposited. The number of moles of copper deposited will be equal to the number of moles of nickel deposited.

Cu2+(aq) + 2e Cu(s)
     
Ni2+(aq) + 2e Ni(s)
     
Ag+(aq) + 1e Ag(s)

 


Useful links

Electrolysis from Purdue University

http://members.aol.com/logan20/faraday.html

Electrolysis

Industrial process details

 


 
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