Lesson #2 Half Reactions 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  ZnS As we saw in Lesson #1, this composition reaction involves redox because of the changes in oxidation state. Zn + S  ZnS 0 0 +2 -2 oxidation reduction

Half-Reactions – the Oxidation Zn + S  ZnS 0 0 +2 -2 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+2 + 2 e-

Half-Reaction – the Reduction Zn + S  ZnS 0 0 +2 -2 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 + 2 e-  S-2

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.

Consider the reaction between Mg and HCl. Mg + 2 HCl  MgCl2 + H2 Charting the oxidation states and the changes gives us this. Mg + 2 HCl  MgCl2 + H2 0 +1 -1 +2 -1 0 oxidation reduction

Now for the half-reactions Mg + 2 HCl  MgCl2 + H2 0 +1 -1 +2 -1 0 oxidation 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… H+1 + 2 e-  H2 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 H2 + O2  2 H2O 2 AlBr3 + 3 Cl2  2 AlCl3 + 3 Br2 Mg + 2 Ag(NO3)  Mg(NO3)2 + 2 Ag