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 (sl)

9.1 - Introduction to oxidation and reduction

Oxidation is the loss of electrons, reduction is the gain of electrons

Oxidation half equation... Mg Mg2+ + 2e-.

Reduction half equation... O + 2e- O2-.

Calculation of oxidation number

There are a few rules to remember

  1. The oxidation number of an element is zero.
  2. The oxidation number of an ion is equal to the charge of the ion.
  3. Hydrogen has an oxidation number of +1 (this doesn't apply to hydrides when hydrogen is -1).
  4. Oxygen has an oxidation number of -2 (except in peroxides when it is -1).

It is then simply a matter of adding together all of the oxidation numbers of the elements in a compound and making sure that the total is = 0

Example: Sulfur in sulfuric acid

The formula is H2SO4 and this must equal 0
therefore (2 x H) + S + (4 x O) = 0
therefore (2 x 1) + (4 x -2) = -S
therefore S = 8 - 2 = + 6

When naming the oxidation number of an atom within a compound it is normal to use a Roman Numeral. This becomes necessary when there is some ambiguity in using just the name.

Example: Phosphoric acid is a term that could apply to acids containing phosphorous with different oxidation states. To specify the acid the oxidation state of the phosphorus must be stated.

H3PO3 - phosphoric (III) acid

H3PO4 - phosphoric (V) acid

Similarly copper oxide could be either CuO or Cu2O, therefore it is important to differentiate by including the oxidation state of the copper in the name

CuO - Copper (II) oxide

Cu2O - Copper (I) oxide

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9.2 - Redox equations

If an element is oxidized, its oxidation number will go up (get more positive). If an element is reduced, its oxidation number will go down. To find out, simply write down the oxidation numbers for each element within the compounds as explained previously.

Examples: Iron (II) hydroxide iron (III) hydroxide

As the oxidation state of the iron increases from II to III then it has been oxidised

sodium dichromate (VI) + sulphuric acid + ethanol chromium (III) sulphate + ethanal + sodium sulphate + Water

The oxidation state of the chromium changes from VI to III and so it has been reduced

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9.3 - Reactivity

An oxidizing agent is an element which causes oxidation (and is reduced as a result) by removing electrons from another species.

A reducing agent is an element which causes reduction (and is oxidized as a result) by giving electrons to another species.

In the previous example the Ethanol is the reducing agent as it has brought about the reduction of Chromium (VI) to Chromium (III).

Why can't the sulphuric acid be the reducing agent? answer

A more reactive metal will displace a less reactive one from a compound and a more reactive halogen will displace a less reactive one from a compound.

This can be generalized to say a stronger reducing agent will displace a weaker one from a compound, and a stronger oxidizing agent will displace a weaker one from a compound.

Thus, if a metal displaces another, we know it must be more reactive and ditto for halogens (which are the given examples).

Feasibility of reaction

If an oxidising agent is placed with a reducing agent then they are likely to react. The feasibility of a reaction can be estimated from the substances positions in a reactivity series.

Example: Zinc will displace Copper from a solution of Copper sulphate and therefore it should be placed above Copper in a reactivity series.

Magnesium is above Zinc in the same series and so by inspection it should be expected to also displace Copper from its salt solution

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9.4 - Voltaic cells

Voltaic cells generate electricity from chemical reactions that occur at the electrodes in the cell. The cell can be described in terms of two 'half-cells'. The general schematic is as follows:

Electricity from chemical reactions

As a redox reaction involves transfer of electrons form one species to another it can be used to produce an electric current if the electrons are made to pass around an external circuit to get from the reducing agent to the oxidising agent. This set-up is known as an electrochemical cell.

In this cell the Zinc anode dissolves and releases electrons which pass around the external wires to the Copper electrode where they are given to the Copper ions (2+) which are then deposited as Copper atoms on the electrode.

At the anode: Zn Zn2+ + 2e

At the cathode: Cu2+ + 2e Cu

The overall cell reaction is found by adding the two equations together:

Zn + Cu2+ Zn2+ + Cu

The salt bridge is there to allow ions to pass through from one half cell to the other so that the half cells do not develop an electrical charge.

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9.5 - Electrolytic cells

This is, fundamentally, the reverse of an electrochemical cell. In this case a greater electromotive force is applied from the external circuit using a battery or power source and this forces the species within the cell to perform the reverse reaction to that which they would normally tend to do.

It is important not to confuse electrochemical cells, which generate electricity by means of a redox reaction and electrolytic cells which use electricity to perform chemical reactions.

Electrolysis of molten ionic liquids

The ionic substance is forced to reverse the ionisation that produced the original ions. Hence electrolysis of molten lead II bromide will give lead and bromine

PbBr2 consists of Pb2+ ions and Br- ions in a giant ionic lattice. When melted (molten) these ions are free to move.

The Pb2+ ions go to the cathode where they pick up electrons and become Lead atoms

Pb2+ + 2e Pb

The Br- ions go to the anode where they lose electrons and become Bromine molecules

2Br- Br2

Conduction of current

Electricity is a flow of electrons around a circuit.

The electrolyte (liquid that conducts electricity) does so by means of ions depositing and picking up electrons from each electrode. Electrons are picked up by positive ions at the cathode and electrons are deposited by negative ions at the anode. Although the electrons have not technically been passed through the conducting liquid, from the point of view of the battery they have gone into the solution via the cathode and are exiting via the anode. As far as the battery is concerned there is a flow of electrons through the cell.


  1. Electron leaves battery and goes to cathode

  2. Electron leaves cathode and goes to positive ion (making it neutral and releasing it from the liquid)

  3. Negative ion drops off electron at the anode (and becomes neutral in the process)

  4. Electron goes up from the anode to the battery

Net result: The battery sends out 1 electron from the negative side and receives 1 electron at the positive side.


The ability of electricity to force chemical change makes the deposition of metals possible in an electrolytic cell.

If copper sulphate solution is electrolysed then the copper ions present in the solution will migrate to the cathode where they pick up electrons are get deposited as copper atoms. A layer of copper builds up on the cathode. Use can be made of this to plate conducting (metal) items by inserting them into the circuit instead of a normal cathode. If they are used to pass the elctrons into solution them they themselves get a deposit of metal. We call this electroplating. It may be carried out for any metal low down in the electrochemical (reactivity) series such as Copper, Nickel, Silver etc.

Electroplating silver

Notice that in the diagram the anode is made of silver. This allows it to dissolve during the process (releasing electrons to the circuit) and replace the silver ions being removed from the solution at the cathode.

The anode reaction:

Ag(s) Ag+(aq) + 1e

The cathode reaction:

Ag+(aq) + 1e Ag(s)

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Redox animation

Useful links


1. Because all of the atoms from sulphuric acid have the same oxidation number in the products as they did in the beginning

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