The following notes were written for the previous IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

Oxidation and reduction and (hl)

19.1 - Standard electrode potentials

Every oxidation or reduction process can be expressed by a half equation showing only the species being oxidised (or reduced) and any water or hydrogen ions needed in the process.

Constructing half equations

The atom or species being oxidised (or reduced) is written in its two different species on each side of the equation. The appropriate nuber of electrons is placed to balance the change in oxidation number of the particular atom being oxidised (or reduced). If necessary use water or hydrogen ions to balance the equation.

Example 1: The oxalate ion C2O42- changes to carbon dioxide when oxidised. Construct the half equation.

Step 1: place the two species on either side

C2O42- 2CO2

Step 2: Examine the oxidation number of the carbon on each side and add the necessary electrons. In this case carbon is changing from the III oxidation state to the IV oxidation state and therefore each carbon atom needs to lose 1 electron. There are two carbon atoms and so 2 electrons must be lost.

C2O42- 2CO2 + 2e

Step 3: The equation is now checked for balancing and is balanced in terms of both atoms and charges

Example 2: Sulphur dioxide is a good reducing agent and turns to sulphate ions in the process. Construct the half equation.

Step 1: Write out the two species on each side

SO2 SO42-

Step 2: Sulphur is in the form of sulphur (IV) at first and becomes Sulphur (VI), losing two electorns in the process

SO2 SO42- + 2e

Step 3: Here we can see that the atoms are not balanced and we must provide oxygen on the left hand side by adding a suitable number of water molecules. This will leave hydrogen ions on the right hand side

SO2 + 2H2O SO42- + 2e + 4H+

On inspection the equation is now balanced in terms of both electrical charges and atoms.

Combination of half equations to give balanced redox equations

Each half equation represents one half of the redox process - either the reduction or the oxidation. A reduction process can be combined with an oxidation process to give a redox equation by simply ensuring that the number of electrons is the same for each equation and then adding them and cancelling out any common factors.

Example: Construct the equation for the reaction between manganate (VII) ions and iron (II) ions in aqueous acidic solution

MnO4- + 8H+ + 5e Mn2+ + 4H2O

In this example the Manganate (VII) ion is being reduced to Manganese (II) ions. This process cannot happen in isolation- it needs something to provide the 5 electrons. A reducing agent.

A typical reducing agent is iron (II) sulphate as it contains the iron (II) ions that can release an electron to give iron (III) ions

Fe2+ Fe3+ + 1e

These two equations can be combines by simple addition to give the ionic redox reaction that occurs when manganate (VII) ions react with Iron (II) ions. But before they can be added the electrons must be equalised by multiplying the second equation by 5

5Fe2+ 5Fe3+ + 5e

Now add the reduction and the oxidation half equations

MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

This represents the redox process

In general the procedure is as follows:

  1. Write down the two half equations, one reduction and the other oxidation.
  2. Equalise the number of electrons by suitable multiplication of one or both half equations
  3. Add the two equations together and cancel out the electrons on both sides
More information about redox equations  

Standard electrode potentials

Standard electrode potential is the potential difference between a given half cell (at 1 mol dm-3 conc) and the standard hydrogen electrode

The standard hydrogen electrode consists of a solution of H3O+ ions at 1 mol dm-3 in a beaker. Placed into this is a platinum electrode surrounded by a gas tube submerged in the solution, with hydrogen at 1 atm inside. The circuit to the other half cell is then attached to the platinum electrode, and a salt bridge saturated in potassium chloride. The entire process should take place at 298K and 1 atm pressure.

Measuring cell potentials

Cell potential is the potential difference between two half cells. They can be calculated by comparing the standard electrode potential of each half cell and taking the difference.

Example: What is the emf of a cell made from a zinc|zinc sulphate half cell connected to a copper|copper sulphate half cell?

zinc|zinc sulphate half cell Eº = -0,76 V

copper|copper sulphate half cell Eº = +0,34 V

E cell = the difference between the two values = 1,10 V

Direction of electron flow will be from the better reducing agent i.e. in this case from the zinc (the half cell with the greater negative value) to the copper half cell

Table of Standard electrode potentials

Spontaneity of reaction

Electrode potentials can be used to predict the feasibility of a reaction. If the electrode potentials are compared for two processes and the Eº for a hypothetical cell determined from the relationship

Eº = E(red) - E(ox)

  • If the value of Eº calculated is positive and greater than +0,3V (approximately) then the reaction is likely to occur.
  • If the value calculated is between 0 and +0,3V then its likely to be an equilibrium.
  • If the value is negative then it is unlikely to occur.

Example: Will hydrogen peroxide react with iron (II) ions in acidic solution?

Fe3+(aq) + e- Fe2+(aq) ...................Eº = 0,77V

H2O2(aq) + 2H+(aq) + 2e- 2H2O(l)... Eº = 1,78V

If the two were to react the equation would be (by combining the half equations above)

H2O2(aq) + 2H+(aq) + 2Fe2+(aq) 2H2O(l) + 2Fe3+(aq)

In other words the H2O2 gets reduced and the Fe2+ oxidised therefore as:

Eº = E(red) - E(ox)

= 1,78V - 0,77V = +1,01V

The reaction is feasible and spontaneous

More information about standard electrode potentials  

19.2 - Electrolysis

Electrolysis is the situation when redox cells are forced to run in reverse by attaching an electricity source to overcome the potential difference. In aqueous solutions, the ions present in water compete at the electrodes, and will sometimes be oxidised/reduced in preference to the dissolved salts.

It is possible to use the standard electrode potentials to predict this, in that species above water (when it is on the left) will not be oxidised, and species below water (on the right) will not be reduced in an aqueous solution.


At the cathode:

Species more reducing than the H+|H2 (half cell = 0V) will not be released at the cathode in aqueous solution. Instead the reaction is: 2H+(aq) + 2e H2(g)

At the anode:

Species with a lower oxidising power than the O2|H2O couple will not be released from the anode. The usual reaction is 4OH- - 4e 2H2O

Highly concentrated solutions may overcome this to some degree. It is possible for Cl2 to be released from a strong solution of chloride ions.

Example: Predict the products of electrolysis of strong calcium chloride solution.

At the cathode

Species present Ca2+ and H+. Ca2+ is higher in the reactivity seriers than hydrogen and therefore cannot be released. The reaction is therefore: 2H+(aq) + 2e H2(g)

At the anode

Species present OH- and Cl- . The chloride concentration is strong and so it is preferentially oxidised and the reaction is: 2Cl-(aq) Cl2(g) + 2e

Species remaining in solution: Calcium ions and hydroxide ions

Quantities produced by electrolysis

Faraday's law states that the mass of product produced will be proportional to the charge passed.

The charge (Coulombs)= current (amps) x time (seconds)


Faraday's law may also be restated as...the number of faradays required to discharge 1 mol of an ion at an electrode equals the number of charges on that ion.

1 Faraday = 96500 Coulombs

Example: Calculate the mass of copper released by a current of 10A passing for 200 seconds through a Copper II sulphate solution.

Charge = Amps x seconds = 10 x 200 = 2000 coulombs

Number of Faradays = 2000/96500 = 0,0207 Faradays

The reaction occurring at the cathode is:

Cu2+(aq) + 2e- Cu(s)

Therefore: 2 Faradays will produce 1 mole of copper

Hence 0,0207 Faradays will produce 0,0207/2 moles of copper = 0,0104 moles

Therefore the mass of copper produced = 0,0104 x 63,5 = 0,658g


Copper ions leave the impure copper anode and copper metal is deposited pure on the metal cathode.

More information about electrolysis  


Table of electrode potentials (Word)

electrode potentials

Standard electrode potentials

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