1. Lesson #2 Half Reactions
Redox Chemistry
Lesson #2
Half Reactions

2. A Quick Review
Remember that Redox Chemistry deals with the changes in oxidation state that occur during chemical reactions. There are two components to Redox:
Oxidation: an increase in the oxidation state of an atom from reactant to product
Reduction: a decrease in the oxidation state of an atom from reactant to product

3. Now for Half-Reactions
A half-reaction is a chemical equation that only “tells half of the story” in a redox system. There are two possible half-reactions.
An oxidation half-reaction: illustrates ONLY the oxidation in a redox system
A reduction half-reaction: illustrates ONLY the reduction in a redox system

4. Understanding what is actually happening
Oxidation
Reduction
As stated earlier, an increase in the oxidation state.
Need to realize that this is caused by the LOSS of electrons.
That means that in oxidation half-reactions, the electrons will be presented as PRODUCTS.
As stated earlier, a decrease in the oxidation state.
Need to see here that this is caused by the GAINING of electrons.
Therefore, the electrons will be presented as REACTANTS in these half-reactions.

5. Using an example from Lesson #1
Zn S  ZnS
As we saw in Lesson #1, this composition reaction involves redox because of the changes in oxidation state.
Zn S  ZnS
oxidation
reduction

6. Half-Reactions – the Oxidation
Zn S  ZnS
oxidation
reduction
In this reaction, the Zn has been oxidized – which means that it has lost electrons. To have changed from an oxidation state of 0 as a reactant to an oxidation state of +2 in the product, it must have lost 2 electrons. Here is how the oxidation half-reaction is written:
Zn  Zn e-

7. Half-Reaction – the Reduction
Zn S  ZnS
oxidation
reduction
You can also see that the S has changed from an oxidation state of 0 as a reactant to a -2 in the product. That means that the S has gained 2 electrons which will be represented as reactants. Here is the reduction half-reaction:
S e-  S-2

8. What if there is a diatomic element involved in the reaction?
First thing to realize is that you can have a diatomic as either a reactant or a product.
That means that it could either be in an oxidation or a reduction.
The issue here will be in the balancing.
You have account for both the numbers of atoms AND the numbers of electrons.
The following example will illustrate this.

9. Consider the reaction between Mg and HCl.
Mg HCl  MgCl H2
Charting the oxidation states and the changes gives us this.
Mg HCl  MgCl H2
oxidation
reduction

10. Now for the half-reactions
Mg HCl  MgCl H2
oxidation
reduction
The oxidation half-reaction is fairly straight-forward:
Mg  Mg e-
But the reduction half-reaction is a bit more complicated because of the diatomic. Here it is…
H e-  H2
Notice how we need two H ions to each gain 1 electron and yield a diatomic hydrogen molecule.

11. Now try these – write an oxidation and a reduction half-reaction for each given equation.
2 H O2  2 H2O
2 AlBr Cl2  2 AlCl Br2
Mg Ag(NO3)  Mg(NO3) Ag