IB syllabus > redox(sl)

These notes were written for the old 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

9.2 - Redox equations


9.2.1: Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction


To work out what is going on in a chemical reaction which involves redox we first need to identify the oxidation states of the species on either side of the reaction. Once this is done we can deduce how many electrons have been transferred and produce half-equations.

For example:In the reaction between chlorine and potassium iodide solution the products are iodine and potassium chloride solution. We can write an equation for the reaction:

Cl2 + 2KI --> I2 + 2KCl

Now we must identify the oxidation state of each species.

The chlorine atoms start off in the zero oxidation state (element), but after the reaction they are now in the -I oxidation state (the chloride ion)

This means that each chlorine atom has gained one electron. We can write the half equation:

Cl2 + 2e --> 2Cl-

Now the iodine species start off as iodide ions and end up as iodine (element). The change in oxidation state is from -I to zero.

2I- - 2e --> I2

These, then are the two half equations in the redox reaction.


9.2.2: Deduce redox equations using half-equations. H+ and H2O should be used where necessary to balance half -equations in acid solution. The balancing of equations for reactions in alkaline solution will not be assessed.


Balancing redox equations

Redox equations are constructed from half equations showing the reduction or oxidation of the species involved.

By convention electrode potential half equations are written as reductions - electrons are added to the species being reduced on the left hand side of the equation:

Zn2+(aq) + 2e Zn(s)

Writing down the half equations in the correct form (reduction for copper and oxidation for zinc)

Zn(s) Zn2+(aq) + 2e
Cu2+(aq) + 2e Cu(s)

We check to make sure that the electrons are balanced. In this case they are, and the equations can be added together directly

Zn(s) Zn2+(aq) + 2e
Cu2+(aq) + 2e Cu(s)
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

If the number of electrons on both sides are different then the half equations must be multiplied through by appropriate quantities to balance the number of electrons on both sides.


9.2.3: Define the terms oxidising agent and reducing agent


Oxidising agents

These are the chemicals that cause the oxidation in a redox reaction. We call the reacting compounds in a reaction the reagents (short form of the words reacting agents).

We consider that the removal of electrons from a species is oxidation and these electrons have to be taken away by another compound or species. This species that attracts the electrons is said to be the oxidising agent i.e. the reagent that causes the oxidation.

Reducing agents

Similarly the reagent that causes reduction in a redox reaction is said to be the reducing agent.

The oxidising agent takes the electron and is itself reduced, the reducing agent loses the electrons and is itself oxidised.

2KI + Br2 2KBr + I2
Iodide ions get oxidised Bromine gets reduced
Iodide - reducing agent Bromine - oxidising agent
         
Cr2O72- + 3SO2 + 2H+ 2Cr3+ + 3SO42- + H2O
Chromium(VI) gets reduced Sulphur(IV) gets oxidised      
Chromium(VI) oxidising agent Sulphur(IV) reducing agent      
         

 


9.2.4: Identify the oxidising and reducing agents in redox equations


The first stage in identifying the oxidising and reducing agents in a redox reaction is to assign oxidation numbers (states) to the species in the reactant and the products. The species that loses electrons is oxidised and this is caused by the species that gains the electrons. Hence the species gaining the electrons is the oxidising agent.

By the same logic the species losing the electrons is the reducing agent.

The oxidising agent gets reduced

The reducing agent gets oxidised.


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