CHEMISTRY 40S: AQUEOUS SOLUTIONS LESSON 4 NOTES Introduction Oxidation and reduction reactions often occur in aqueous environments. We ve often heard of the term oxidation in reference to the rusting of metal but reduction is not a commonly used term. In this lesson you will learn how to identify oxidation and reduction reactions, or what chemists call "redox" reactions. Learning Outcomes When you have completed this lesson, you will be able to: Relate the role of oxygen to the process of rusting and burning. Define oxidation and reduction. Determine oxidation numbers for atoms in simple compounds and ions. Identify the oxidizing agent, the reducing agent, the substance reduced and the substance oxidized, given a balanced chemical equation. Distinguish between redox and non-redox reactions. History of Oxidation and Reduction The term oxidation has been in use since the discovery of oxygen in 1774 by Joseph Priestly. Antoinne Lavoisier, considered to be the Father of Modern Chemistry, learned of Priestly's work and subsequently used this information as a basis for his demonstration that substances burned because they combined with oxygen. Reactions such as burning and corrosion were called oxidation because the substances were combined with oxygen. Lavoisier and other chemists of his time found that substances increased in mass after they were burned, to form an oxide. They also knew that an oxide decreased in mass when heated. They called the reverse process reduction, since the mass of the substance was reduced. Not until the discovery of electrons, did the meaning of oxidation and reduction take on a new meaning and apply to more reactions. Oxidationreduction reactions, or redox reactions, are defined as chemical changes that occur when electrons are transferred from one reactant to another.
Defining Oxidation and Reduction Oxidation occurs when an atom loses one or more electrons and reduction is when an atom gains one or more electrons. There are two mnemonics to help us remember these terms: "OIL RIG" Oxidation Is Losing electrons, Reduction Is Gaining electrons or "LEO says GER": Losing Electrons is Oxidation, Gaining Electrons is Reduction For example, a reaction Lavoisier studied, the burning of magnesium occurs by the reaction 2 Mg(s) + O 2 (g) 2 MgO(s) In this reaction magnesium begins as a neutral atom and loses two electrons to become a Mg 2+ ion in MgO. Magnesium is oxidized. Oxygen begins as a neutral atom and gains the two electrons from magnesium to become an O 2 ion in MgO. Oxygen is reduced. Similarly, the reaction of magnesium combining with chlorine to produce magnesium chloride is also a redox reaction. Mg(s) + Cl 2 (g) MgCl 2 (s) Magnesium loses two electrons to become a Mg 2+ ion and is oxidized. Each chlorine atom gains one electron from magnesium to become Cl ions. Each chlorine atom is reduced.
Oxidation Numbers Not all redox reactions are as simple as the ones on the previous page, so chemists have designed a bookkeeping system to keep track of electrons in a chemical reaction. They have assigned oxidation numbers to all atoms and ions. The oxidation number represents the charge the atom would have if every bond were ionic. Not every bond is ionic, but chemists assume they are for this system. The oxidation number is not always the actual charge, but it is very helpful to follow electrons in redox reactions. The rules for assigning oxidation numbers are as follows. You may wish to print these out and use them as a reference until they become more familiar with practice. 1. Oxidation numbers are always assigned PER ATOM. 2. The oxidation numbers of all uncombined elements is zero. (e. g. O 2, K(s), H 2, etc.) 3. The oxidation number of monatomic ions equals the charge of that ion. 4. In compounds, the oxidation number for alkali metals (for example, Li, Na, K, etc.) is always +1. 5. In compounds, the oxidation number for the alkaline earth metals (e. g. Be, Mg, Ca, etc.) is always +2. 6. In compounds, the oxidation number of aluminum is always +3. 7. In compounds, the oxidation number of fluorine is 1. 8. In compounds, the oxidation number of hydrogen is +1. An exception is in metal hydrides, such as NaH or MgH 2, when hydrogen is 1. 9. In compounds, the oxidation number of oxygen is 2. An exception is in peroxides, such as H 2 O 2 or Na 2 O 2, when its oxidation number is 1. 10. For any neutral compound, the sum of the oxidation numbers for each atom must be zero. 11. For a poly atomic ion, the sum of the oxidation numbers for each atom must be the charge of that ion. Note the difference between ion charge and oxidation numbers: for example, the magnesium ion, Mg 2+ ion charge = 2+ oxidation number = +2
Assigning Oxidation Numbers Some helpful tips when assigning oxidation numbers: Remember, that when we assign oxidation numbers, we assign them to EACH atom. Assign numbers to those that are known first. Then calculate the others. If there are no rules to assign numbers to a certain atom in a compound, imagine the compound is ionic and assign numbers based on ion charge. Example 1. Assign oxidation numbers to each atom in SO 2. Solution Step 1: Start with atoms which are known. We know oxygen is 2 in compounds. Step 2: Solve for other atoms Since oxygen is 2, the total of the oxidation numbers for oxygen is 2 2 = 4. Remember, oxidation numbers are assigned per atom. The sum for SO 2 must be zero, so S = +4 So, S = +4, O = 2 Example 2. Assign oxidation numbers for each atom in K 2 Cr 2 O 7. Solution Step 1: Start with atoms which are known. Each O is 2. Each K is +1. Step 2: Solve for other atoms. The total for all the oxygen atoms is 2 7 = 14. The total for the potassium atoms is +1 2 = +2.
The total for the "molecule" must be zero. Therefore the total for all the Cr atoms is 14 2 = +12. If there are two chromium atoms, then each chromium atom is +12 2 = +6. So, K = +1, Cr = +6, O = 2. Example 3. Assign oxidation numbers for each atom in Fe(NO 3 ) 3. Solution Step 1: Start with atoms which are known. Each O is 2. Step 2: Solve for other atoms. Both Fe and N are unknown, so let's break the compound into its ions: Fe 3+ and NO 3. The oxidation number for Fe is equal to its charge, +3. The total for all oxygen atoms in the nitrate ion is -2 3 = 6. The sum of oxidation numbers for the nitrate ion must be 1. The oxidation number for nitrogen is 6 1 = +5. So, Fe = +3, N = +5, O = 2. Recognizing Redox Reactions Recall, a redox reaction is one that involves the transfer of one or more electrons from one atom to another. In order to recognize a redox reaction, we must follow the electrons. We can follow the electrons by determining if the oxidation numbers for each element change from reactants to products. If the oxidation numbers of none of the elements change, the reaction is NOT a redox reaction. Example 4. Is the reaction SO 2 + H 2 O H 2 SO 3 a redox reaction?
Solution. If we assign oxidation numbers to each atom, we find the oxidation number of S remains at +4 in the reactants and in the products, O remains unchanged in the products at 2 and H remains unchanged at +1. The oxidation numbers do not change, so electrons were not transferred. Therefore, this is not a redox reaction. Recall, if an atom is oxidized, it will lose electrons. If it loses electrons, it becomes more positively charged and its oxidation number becomes more positive from the reactants to the products. If an atom is reduced, it gains electrons. If it gains electrons, it becomes more negatively charged and its oxidation number becomes more negative, from reactants to products. Example 5. Is the following reaction a redox reaction? Cu(s) + 2 AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2 Ag(s) SolutionAssign oxidation numbers. Cu is zero in the reactants and +2 in the products. Ag is +1 in the reactants and zero in the products. N = +5 in both products and reactants. O = 2 in both products and reactants.
This is redox reaction, since the oxidation numbers of Cu and Ag change. Cu is oxidized because its oxidation number becomes more positive, indicating it has lost electrons.. The oxidation number of Ag becomes more negative, indicating a gain of electrons. Ag is reduced. Exercise 1. Assign oxidation numbers for each element in the following compounds. a. Na 2 Cr 2 O 7 b. KNO 3 c. FeCl 2 d. H 2 C 2 O 4 e. HClO 3 f. KMnO 4 g. MnSO 4 h. H 2 SO 4 i. H 2 j. NH 4 NO 3 k. NO 2 l. C 2 H 5 OH m. MgSO 3 n. Na 2 O 2 o. CaH 2 p. Ca(ClO) 2 q. SrCrO 4 r. MnO 2 s. N 2 O 5 t. O 2 u. Ag 2 S v. Zn(NO 2 ) 2 w. Fe 2 (MnO 4 ) 3 x. Cr 2 O 3 2. Identify which of the following are redox reactions. a. 2 NO 2 N 2 O 4 b. 2 Mg + O 2 2 MgO c. Mg + 2 Ag + + 2 NO 3 Mg 2+ + 2 NO 3 + 2 Ag d. 2 SO 2 + O 2 2 SO 3 e. MgO + SO 3 MgSO 4
Answer Key 1. a. Na = +1, Cr = +6, O = 2 b. K = +1, N = +5, O = 2 c. Fe = +2, Cl = 1 d. H = +1, C = +3, O = 2 e. H = +1, Cl = +5, O = 2 f. K = +1, Mn = +7, O = 2 g. Mn = +2, S = +6, O = 2 h. H = +1, S = +6, O = 2 i. H = 0 j. N = 3, H = +1, N = +5, O = 2 k. N = +4, O = 2 l. C = 2, H = +1, O = 2 m. Mg = +2, S = +4, O = 2 n. Na = +1, O = 1 (peroxide) o. Ca = +2, H = 1 (hydride) p. Ca = +2, Cl = +1, O = 2 q. Sr = +2, Cr = +6, O = 2 r. Mn = +4, O = 2 s. N = +5, O = 2 t. O = 0 u. Ag = +1, S = 2 v. Zn = +2, N = +3, O = 2 w. Fe = +3, Mn = +6, O = 2 x. Cr = +3, O = 2 2.