Presentation on theme: "Oxidation and Reduction (Redox) Lance S. Lund April 19, 2011."— Presentation transcript:
Oxidation and Reduction (Redox) Lance S. Lund April 19, 2011
Assigning Oxidation Numbers 1.The oxidation number of an element is zero. 2.The oxidation number of an atom in a monoatomic ion equals the charge on the ion. 3.The oxidation number of oxygen is -2 in most of its compounds. Exception: In peroxides, the oxidation number of oxygen is -1.
4.The oxidation number of hydrogen is +1 in most of its compounds. Exception: In hydrides, the oxidation number of hydrogen is The oxidation number of fluorine is -1 in all of its compounds. Each of the halogens has an oxidation number of -1 in binary compounds, except when the other element is a halogen above it in the Periodic Table or the other element is oxygen. Assigning Oxidation Numbers
6.The sum of oxidation numbers of the atoms in a compound is zero. The sum of the oxidation numbers of the atoms in a polyatomic ion equals the charge on the ion. Assigning Oxidation Numbers
Determine the Oxidation Numbers Determine the oxidation numbers of each element in the following species: –K 2 SO 4 H 3 PO 4 –Cr 2 O 7 2- Br 2
Redox Reactions The Loss of Electrons is Oxidation –An element that loses electrons is said to be oxidized. The species in which that element is present in a reaction is called the reducing agent. The Gain of Electrons is Reduction –An element that gains electrons is said to be reduced. The species in which that element is present in a reaction is called the oxidizing agent.
Balancing Redox Equations 1.Assign oxidation numbers to each atom. 2.Determine the elements that get oxidized and reduced. 3.Split the equation into half-reactions. 4.Balance all atoms in each half- reaction, except H and O. 5.Balance O atoms using H 2 O. 6.Balance H atoms using H +.
Balancing Redox Equations 7.Balance charge using electrons. 8.Sum together the two half-reactions, so that: e - lost = e - gained 9.If the solution is basic, add a number of OH - ions to each side of the equation equal to the number of H + ions shown in the overall equation. Note that H + + OH - ⟶ H 2 O