1. Corrosion
The Chemical Process

2. What is Corrosion? Why are metals such as sodium stored in oil?
What is this process known as?
Does iron undergo this process?
What will make iron corrode more quickly?
They are so reactive – they combine readily with oxygen in the air
This is known as dry corrosion.
Although iron is less reactive it does undergo dry corrosion but it is slow at room temperature
The addition of moisture will make iron corrode more quickly
This is known as wet corrosion.

3. Dry Corrosion Direct combination of a metal with oxygen.
When a piece of iron reacts with air it forms
Particularly if it is heated.
This process is called
Iron has been to iron oxide, this is why there are no deposits of iron on Earth but lots of Iron Oxide, such as haematites (Fe2O3)
Iron oxide
Oxidation
oxidised

4. The iron we use is extracted from this ore.
This extraction process involves
of the iron ore to iron
When oxygen is lost from a substance , that substance has been
The production of iron form haematite can be represented by the equation
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
The iron oxide has lost oxygen
it has been reduced
Reduction cannot occur with oxidation occurring at the same time.
In this reaction CO has gained oxygen to form CO2
it has been oxidised. Oxidation And Reduction always occur simultaneously.
Oxidation and reduction are important chemical processes.
reduction
reduced

5. Dry Corrosion at the Atomic Level
Look at the piece of Mg ribbon in front of you. What do you notice?
This layer is _____________, which results from the corrosion of Mg in air. Mg has been oxidised.
The equation for this reaction is 2Mg(s) + O2  2MgO(s)

6. MgO consists of Mg2+ and O2- ions. So what happens in the reaction:
Mg atoms have no charge so must have lost 2 electrons to form Mg2+ and O2 has no charge so the O atoms in the O2 molecule must have each gained 2 electrons to form O2- ions.
So we can write the reaction with 2 separate equations ( we call half equations) to show what is happening at the atomic level: Mg(s)  Mg2+(s)+ 2e & O2(g) + 4e-  2O2- (s)

7. So we can see that the oxidation of magnesium, which we earlier described in terms of its reaction with oxygen , can be seen to involve the transfer of electrons from Mg to O.
It is this transfer of electrons (ie losing )that we use as broader more widely used definition of OXIDATION & and the gaining of electrons (rather than the losing of Oxygen ) as our definition of REDUCTION.

8. Definition Oxidation is the loss of electrons
Reduction is the gain in electrons
REMEMBER OILRIG- Oxidation involves Loss Reduction involves Gain For an atom to lose electrons there must be another to gain them. Therefore the two are simultaneous.
These reactions are called REDOX reactions.
This definition is more useful as it includes all reactions in which electrons are exchanged and not restricted to those involving oxygen.

9. REDOX Reactions.
Many redox reactions involve reactants other than oxygen.
Eg. When potassium reacts with chlorine, potassium chloride (an ionic substance containing K+ and Cl- ions ) is formed.Write the half equations for these reactions and hence state which reactant has been oxidised and which has been reduced during the reaction.

10. K(s)  K+(s) + e- Cl2(g) + 2 e-  2Cl-(s) oxidised reduced
There are 2 half equations. The first is the formation of K+ ions The second is the formation of Cl- ions. So the potassium has been____________, and the chlorine has been _____________.
K(s)  K+(s) + e-
Cl2(g) + 2 e-  2Cl-(s)
oxidised
reduced

11. Another Example
When a strip of copper is suspended in a solution of silver nitrate, long crystals of silver metal can be seen. The solution changes to a pale blue colour, indicating the presence of Cu2+ ions.
Write the half equations that represent what is happening in the beaker and work out which is oxidised and which is reduced?

12. The first half equation is the copper metal becoming Cu2+ ions in solution.
Cu(s)  Cu2+(aq) + 2e-
The second half equation is the Ag+ions in solution becoming Ag metal
Ag+(aq) +e-  Ag(s)
Therefore Cu is __________.
oxidised
& Ag+ is __________.
reduced

13. Overall Redox Equations.
When we are writing equations that represent a redox reaction we:
Write the half equations first (containing electrons)
And add them together to give us the overall equation for the reaction.
It is important to remember that overall redox equations do not contain any electrons.

14. 2K(s) + Cl2(g)  2K+(s) + 2Cl-(s)
Okay so lets write an overall equation for our examples.
Eg 1
2x( )
K(s)  K+(s) + e-
Cl2(g) + 2e-  2Cl-(s)
Note that the number of electrons are not the same so we have to multiply the first equation by 2 to make them the same.
So the equations are now:
2K(s)  2K+(s) + 2e-
Cl2(g) + 2e-  2Cl-(s)
2K(s) + Cl2(g)  2K+(s) + 2Cl-(s)
Overall Eqn

15. Remember In both half and overall equations for redox reactions the :
The number of atoms of each elements present in the products is equal to the number present in the reactants.
As atoms are conserved in all chemical reactions.
The total charge on the product side of the equation is equal to the total charge on the reactant side of the equation.
Electrons are also conserved in chemical equations.

16. Another Example Li(s)  Li+ (s) + e- O2(g) + 4e-  2O2- (s)
When lithium is oxidised by atmospheric oxygen the reactants may be represented by the following half equations:
Li(s)  Li+ (s) + e-
O2(g) + 4e-  2O2- (s)
Identify the half equation representing the oxidation reaction and write the overall equation.
Step 1 Oxidation is where electrons are lost. The lithium has the electron on the product side so therefore has lost electrons, therefore the lithium has been oxidised.

17. 4Li(s) + O2(g)  4Li+ (s) + 2O2- (s)
To write the overall equation we have to look at the half equations and balance the electrons .
We can see that the first equation has 1 electron on the product side, but the second equation has 4 electrons so the first equation has to be multiplied by 4 i.e.
(Li(s)  Li+ (s) + e-) x 4
4Li(s)  4Li+ (s) + 4e-
Giving
O2(g) + 4e-  2O2- (s)
4Li(s) + O2(g)  4Li+ (s) + 2O2- (s)

18. Oxidants and Reductants.
An oxidant (oxidising agent or oxidiser) is a substance that causes another substance to be oxidised .
A reductant (reducing agent or reducer) is a substance that causes another substance to be reduced.
Lets look at an example and see what this is all about:

19. Example Zn(s)  Zn2+(aq) + 2e- 2H+(aq) + 2e- H2(g)
The reaction between zinc and hydrochloric acid can be represented by the following equations:
Zn(s)  Zn2+(aq) + 2e-
Oxidation (loss of electrons)
2H+(aq) + 2e- H2(g)
Reduction (gain of electrons)
Is oxidised
Is reduced
Thus is the reductant
Bc it caused reduction
Thus is the oxidant
Bc it caused oxidation

20. A diagram to help sort it out
Oxidises the
Electrons
Reductant
Oxidant
Reduces the
Gains electrons and is therefore reduced
Loses electrons and therefore is oxidised
The relationship between oxidants and reductants in oxidation – reduction reactions