Redox Chemistry Lesson 2 Half Reactions A Quick

Redox Chemistry Lesson #2 Half Reactions

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

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

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.

Using an example from Lesson #1 Zn + S Zn. S As we saw in Lesson #1, this composition reaction involves redox because of the changes in oxidation state. oxidation 0 0 +2 -2 Zn + S Zn. S reduction

Half-Reactions – the Oxidation oxidation 0 0 +2 -2 Zn + S Zn. S 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+2 + 2 e-

Half-Reaction – the Reduction oxidation 0 0 +2 -2 Zn + S Zn. S 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 + 2 e- S-2

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

Consider the reaction between Mg and HCl. Mg + 2 HCl Mg. Cl 2 + H 2 Charting the oxidation states and the changes gives us this. oxidation 0 +1 -1 +2 -1 0 Mg + 2 HCl Mg. Cl 2 + H 2 reduction

Now for the half-reactions oxidation 0 +1 -1 +2 -1 0 Mg + 2 HCl Mg. Cl 2 + H 2 reduction The oxidation half-reaction is fairly straight-forward: Mg Mg+2 + 2 e. But the reduction half-reaction is a bit more complicated because of the diatomic. Here it is… 2 H+1 + 2 e- H 2 Notice how we need two H ions to each gain 1 electron and yield a diatomic hydrogen molecule.

Now try these – write an oxidation and a reduction half-reaction for each given equation. 2 H 2 + O 2 2 H 2 O 2 Al. Br 3 + 3 Cl 2 2 Al. Cl 3 + 3 Br 2 Mg + 2 Ag(NO 3) Mg(NO 3)2 + 2 Ag