Oxidation and reduction reactions v Found often in aqueous environments v Examples include, rusting of metals v Cracking a glow stick, where

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2 Oxidation and reduction reactions v Found often in aqueous environments v Examples include, rusting of metals v Cracking a glow stick, where electrons are transferred to convert the chemical energy into light energy.

Antoine Lavoisier Father of Modern Chemistry Joseph Priestley Antoine

3 The term Oxidation was introduced since the discovery of oxygen in 1774 by Joseph Priestley. Antoine Lavoisier Father of Modern Chemistry Joseph Priestley Antoine Lavoisier the Father of modern chemistry used this information to believe that when substances burned, they combine with oxygen to form an oxide.

4 Ø Reactions such as burning and corrosion are called oxidation because they combined with oxygen. Ø Lavoisier and other chemists found that substances increased in mass after they were burned to form an oxide. Ø But also decreased in mass when heated. Ø In modern times, oxidation reduction is the transfer of electrons from one reactant to another.

5 Oxidation Numbers Represents the number of electrons the element has lost or gained. IF there is a change in oxidation state (number), then the element has undergone oxidation or reduction If an element has an oxidation state of +2, then it is said to have lost 2 electrons If an element has an oxidation state of -2, then it has gained 2 electrons. Oxidation numbers are similar but not the same as their charges.

6 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. Assume a compound is neutral unless otherwise specified (HNO 3 vs NO 3 - )

7 The rules for assigning oxidation numbers are as follows. 1. The oxidation numbers of all uncombined elements is zero. (e. g. O2, K(s), H2, etc.) 2. The oxidation number of monatomic ions equals the charge of that ion. 3. In compounds, the oxidation number for alkali metals (for example, Li, Na, K, etc.) is always In compounds, the oxidation number for the alkaline earth metals (e. g. Be, Mg, Ca, etc.) is always In compounds, the oxidation number of aluminum is always In compounds, the oxidation number of fluorine is In compounds, the oxidation number of hydrogen is +1. An exception is in metal hydrides, such as NaH or MgH 2, when hydrogen is 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 For any neutral compound, the sum of the oxidation numbers for each atom must be zero. 10. For a polyatomic 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+ vs oxidation number = +2

8 TOP DOWN METHOD Write the individual oxidation number below the element, and the total at the top of the element. SO 2

9 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

10 Example 2. Assign oxidation numbers for each atom in K2Cr2O7. 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 = +6. So, K = +1, Cr = +6, O = 2.

11 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.

13 1. Assign oxidation numbers for each element in the following compounds. a. Na 2 Cr 2 O b 7. d. H 2 C 2 O e 4. g. MnSO h 4. j. NH 4 NO k 3. m. MgSO n 3. p. Ca(ClO) q 2. KNO 3 HClO 3 H 2 SO 4 NO 2 Na 2 O 2 SrCrO 4 c. FeCl 2 f. KMnO 4 i. H 2 l. C 2 H 5 OH o. CaH 2 r. MnO 2 s. N 2 O 5 t. O u 2. Ag 2 S v. Zn(NO 2 ) w 2. Fe 2 (MnO 4 ) x. Cr 2 O 3

14 Ø The reaction Lavoisier studied, the burning of magnesium occurred by the reaction. 2 Mg (s) + O 2(g) 2 MgO (s) Ø The Net reaction is: Dissociates! Ø 2 Mg (s) + O 2(g) 2Mg O 2- Ø The Mg(s) lost 2 electrons while Ø O 2 gained 2 electrons in the process

16 "OIL RIG" Oxidation Is Losing electrons, Reduction Is Gaining electrons "LEO says GER": Losing Electrons is Oxidation, Gaining Electrons is Reduction

18 Identify each of the following changes as either oxidation or reduction. I 2 + 2e- à 2I - Reduction Kà K + + e- Fe 2+ à Fe 3+ + e- Ag + +e- à Ag Oxidation Oxidation Reduction

19 1. Find the oxidation state for the same element on both sides of the reaction 2. Determine if electrons are lost or gained 3. Remember electrons are negative, so becoming more negative means gaining electrons! 4. 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?

20 Example 4. Is the reaction SO 2 + H 2 O H 2 SO 3 a redox reaction? Solution. 1. Assign oxidation numbers to each atom, SO 2 + H 2 O H 2 SO 3 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.

21 Is the following reaction a redox reaction? Cu(s) + 2 AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2 Ag(s) Solution 1. Assign oxidation numbers. 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.

22 Identify what is oxidized and what is reduced in the following processes. 2Br- + Cl 2 à Br 2 + 2Cl - 2Ce + 3Cu 2+ à 3Cu + 2Ce 3+ 2Zn + O 2 à 2ZnO 2Na + 2H + à 2Na + + H 2 Determine the oxidation number for the Bolded NaClO 4 AlPO 4 HNO 2 NH 4 + AsO 4 3- CrO 4 2-

23 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 NO 3 Mg NO Ag d. 2 SO 2 + O 2 2 SO 3 e. MgO + SO 3 MgSO 4

CHEMISTRY 40S: AQUEOUS SOLUTIONS LESSON 4 NOTES. When you have completed this lesson, you will be able to:

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