2.  Oxidation is a chemical change in which electrons are lost by an atom or group of atoms.  Zn(s)  Zn 2+ (aq) + 2 e - (oxidized0  Example: The reaction between magnesium metal and oxygen to form magnesium oxide involves the oxidation of magnesium.  2 Mg(s) + O2(g)  2 MgO(s)  The equation above is also an example of Oxidation.

3.  reduction is a chemical change in which electrons are gained by an atom or group of atoms.  Cu 2+ (aq) + 2 e -  Cu(s) ( reduced)  The reaction between magnesium oxide and carbon at 2000C to form magnesium metal and carbon monoxide is an example of the reduction of magnesium oxide to magnesium metal.  MgO(s) + C(s)  Mg(s) +CO (this is reduction)

4. 1. The oxidation number of an atom in an element is zero. 2. The oxidation number of an atom in a monatomic ion is equal to the charge of the ion. 3. The oxidation number of oxygen is -2 in all oxides, it is -1 in peroxides. 4. The oxidation number of hydrogen is +1 in all compounds except hydrides(binary compounds of a metal and hydrogen) where it is -1. 5. The oxidation number of fluorine is -1 in all of its compounds. The other halogens(Cl, Br, I) have an oxidation number of -1 in binary compounds, except when the other element is oxygen or another halogen above it in the periodic table. 6. The sum of the oxidation numbers of the atoms in a compound is zero. monatomic ion is equal to the charge of the ion. The sum of the oxidation numbers of the atoms in a polyatomic ion equals the charge on the ion.

5.  Oxidation-reduction equations are often too difficult to balance by the inspection method(trial and error) we have used up to this point. There are two systematic methods for balancing redox equations. One is the half-reaction or ion-electron method and the second is the oxidation-state method. In this summary, I am only going to cover the half- reaction method. First we will list the rules, and then we will go through a couple of examples in detail.

6. 1. Write a skeleton equation or half-reaction that includes those reactants and products that contain the elements undergoing a change in oxidation number. 2. Write a half-reaction equation for the oxidizing agent, with the element undergoing a reduction on each side of the equation. The element should not be written as a free element or ion unless it really exists as such. It should be written a part of a real molecular or ionic species. 3. Write another half-reaction equation for the reducing agent, with the element undergoing an increase in oxidation number on each side. 4. Balance each half-reaction as to number of atoms of each element. In neutral or acidic solution, H 2 O and H + may be used for balancing oxygen and hydrogen atoms. The oxygen atoms are balanced first. For each excess oxygen atom on one side of an equation, balance is obtained by adding one H 2 O to the other side. Then H + is used to balance hydrogens. Note that O 2 and H 2 are not used to balance the oxygen and hydrogen atoms unless they are actually reactants or products.If the solution is alkaline, OH - and H 2 O may be used. For each excess oxygen on one side of an equation, balanced is obtained by adding one H 2 O to the same side and 2 OH - to the other side. If hydrogen is still unbalanced after this is done, balance is obtained by adding one OH - for each excess hydrogen on the same side as the excess and one H 2 O to the other side. If both oxygen and hydrogen are in excess on the same side of the half- reaction, an OH - can be written on the other side for each paired excess of H and O.

7. 5. Balance each half-reaction as to the number of charges by adding electrons to the the left or right side of the half-reaction. If the preceding rules have been followed carefully, it will be found that electrons must be added to the left side of the half-reaction for the oxidizing agent and to the right side of the half- reaction for the reducing agent. 6. Multiply each half-reaction by a number chosen so that it makes the number of electrons the same in each half-reaction. 7. Add the two half-reactions resulting from the multiplications. In the resultant equation, cancel any terms common to both sides. All electrons must cancel. 8. Transform the net ionic equation that results form step 8 into a molecular equation. This is done by adding to each side of the equation equal numbers of the ions which do not undergo electron transfer (spectator ions) but which are present along with the reactive components in neutral chemical substances.