IB syllabus > redox(sl) > 10.2 

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

10.2 - Reactivity

10.2.1: Deduce a reactivity series based upon the chemical behaviour of a group of oxidising and reducing agents. Displacement reactions of metals and halogens (see 3.3.1) provide a good experimental illustration of reactivity. Standard electrode potentials or reduction potentials are not required.

Reactivity series

It is possible to organise a group of similar chemicals that undergo either oxidation or reduction according to their relative reactivity. Oxidation (and reduction) is a competition for electrons. The oxidising species (agents) remove electrons from other species and can force them to become reducing agents (releasers of electrons)

A good example of this competition for electrons is the behaviour of metals. Metals always react by losing electrons (oxidation) they are then reducing agents. However if a metals is in competition with metal ions the more reactive metal can oblige the less reactive metal (in the form of ions) to accept electrons. This is called a displacement reaction.


Zinc reacts with a solution containing copper ions. The zinc metal is more reactive than copper metal and so it can force the copper metal ions to accept electrons and become metal atoms.

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

The zinc metal passes its electrons to the copper ions. We observe that the zinc develops a pink layer of coper on its surface and the blue copper ion solution fades in colour.

We say that the zinc displaces the copper ions from solution.

Experimental observations

If we observe that there is a reaction between a metal and another metal ion in solution this tells us that the solid metal is more reactive than the metal of the dissolved metal ions.

Given this information we can deduce that the most reactive of the three metals is iron, followed by copper, followed by silver. This allows us to arrange the metals into a reactivity series based on these specific reactions

Reduction of metal oxides by metals

metal A + metal B oxide metal A oxide + metal B

When a metal A is heated with a metal B oxide there will be a reaction if the free metal A is more reactive than the metal B of the metal B oxide. This is because the metal B in the metal B oxide is in the form of a metal ion - it has already lost electrons.

There is a competition between the metal ion (in the oxide) and the free metal for the electrons. The more reactive of the two metals will win the competition. Consequently if there is a reaction between a metal and a metal oxide then this tells us that the free metal is more reactive than the metal in the metal oxide.

Experimental observations

We can use this information to arrange the metals in order of reactivity

Sodium most reactive
Copper least reactive

Sodium has the greatest electron releasing power (and conversely the copper ions - Cu2+ - would have the greatest electron attracting power)

Predictions from reactivity series

Once a reactivity series is produced it can be used to predict reactions of pairs of reactant. For example in the table above it should be appreciated that magnesium will react with copper oxide reducing it to copper metal.

Any metal that is more reactive will react with compounds of less reactive metals.

Reactivity series involving non-metals

Metals react by losing electrons - they are reducing agents. Non-metals react by gaining electrons - they are oxidising agents. In the same way that metals can be ordered in terms of reducing strength, the non-metals can be ordered in terms of their oxidising strength. The halogens are a typical example of a non-metal reactivity series.

Reactivity of the halogens

Fluorine most reactive
Iodine least reactive

Fluorine is so reactive that we cannot isolate it in the laboratory very easily, as it reacts with both water and glass. As a result we don't usually deal with fluorine at pre-university level

but compare only the other three (astatine is very rare and radioactive)

Do not confuse this order of reactivity with that of the metals - these are non-metals, their reactivity is in terms of oxidising power - i.e. chlorine is the best oxidising agent out of chlorine, bromine and iodine.

1. Chlorine will displace bromine from solutions containing bromide ions

Cl2 + 2Br- Br2 + 2Cl-

In this reaction the chlorine is oxidising the bromide ions by removing an electron from them. Bromine is liberated from the solution and may be detected by its orange/red colour

2. Bromine will displace iodine from solutions containing iodide ions

Br2 + 2I- I2 + 2Br-

In this reaction the bromine is oxidising the iodide ions by removing an electron from them. Iodine is liberated from the solution and may be detected by its orange/brown colour which turns blue/black in the presence of starch indicator.

It is predictable, then, that chlorine will also displace iodine from a solution containing iodide ions

10.2.2: Deduce the feasibility of a redox reaction from a given reactivity series.

Prediction of feasibility

Once a reactivity series is constructed depending on the reduction or oxidation ability of each species, we can use it to predict the feasibility (probability) of a reaction occurring between any two pairs of reactants.

If one of the substances is a reducing agent - i.e. it reacts by losing electrons then this must react with an oxidising agent - i.e. a species that gains electrons.



K K+ + 1e

best reducing agents (left hand side species)


Mg Mg2+ + 2e



Zn Zn2+ + 2e



Fe Fe2+ + 2e



Cu Cu2+ + 2e



H2 2H+ + 2e



2I- I2 + 2e



2Br- Br2 + 2e



2Cl- Cl2 + 2e

best oxidising agents (right hand side species)

Any species from the right hand side of one of the redox equilibria (the oxidising agent) can be predicted to react with any species above it on the left hand side of the redox equilibria (the reducing agent).

The species on the right hand side of the equilibria will gain electrons to go to the right hand side. They can only gain these electrons form species that are abopve them on the left hand side of the series.

We can therefore predict that chlorine (right hand side) will react with copper (left hand side) to form copper ions nad chloride ions according to the equation:

Cl2 + Cu Cu2+ + 2Cl-

Similarly, we can predict that iodide ions (left hand side) will NOT react with zinc ions (left hand side) as the zinc ions are poor oxidising agents and the iodide ions poor reducing agents.

Note: although a reaction may be predicted as feasible it does not mean that it will happen spontaneously. If the activation energy is high then it may need an extra "push" to get it going. - for example the reaction between chlorine and hydrogen needs a spark or ultraviolet light and then it is explosively fast.

10.2.3: Describe and explain how a redox reaction is used to produce electricity in a voltaic cell. Students should be able to draw a diagram of a simple half-cell, and show how two half-cells can be connected by a salt bridge to form a whole cell. Suitable examples of half-cells are Mg, Zn, Fe and Cu in solutions of their ions.

Electrical cells

As we have seen certain species lose electrons (reducing agents) and other species gain electrons (oxidising agents) when reacting. If these species are not mixed together but connected electrically around an external circuit then these electrons will flow around the external circuit producing an electroic current.

Each of the reacting species is then called a half cell and the whole set up is called an electrochemical cell. It is the basis behind the electrical battery.

An electrochemical cell

In this cell the zinc metal has a tendency to dissolve as ions leaving itselectrons on the electrode. The copper, which is a weaker reducing agent, is forced to accept the electrons and use then to turn the copper ions into copper at the copper electrode. As these electrons flow around the outer (external) circuit they constitute an electric current

The 'salt bridge' is usually a filter paper soaked in potassium nitrate solution (neither of these ions react with any other ions in the experiment). This 'salt bridge' then allows ions to move in both directions equalising any build up of electrical charge in the beakers.

The zinc forces the copper ions to accept electrons and the overall cell equation can be constructed by adding together the two 'half equations' above.

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

This type of cell can be constructed using any pair of reducing and oxidising agents. The greater the difference in the reactivity of one type of species (i.e. the reducing species) the greater the cell potential (voltage)

Consequently a cell constructed from zinc | zinc sulphate in one half cell and silver ! silver nitrate solution in the other half cell will have a greater voltage that the cell above (there is a greater difference in reactivity between zinc and silver than between zinc and copper)

Electrochemical cell - general

General cell construction


Copper - Iron cell

In this cell the iron is the more powerful reducing agent and will preferentially lose electrons. These electrons force the copper redox equation to go in the direction of receiving electrons (reduction) - i.e. the copper Cu2+ ions pick up electrons and are deposited on the electrode as copper metal atoms

The iron | Fe2+ solution beaker is called a half-cell and the copper | copper ions solution is said to be the other half-cell.


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