Chapter 11. Electrochemistry Oxidation and Reduction Reactions. Oxidation-Reduction Reactions. Oxidation-Reduction Reactions

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1 Oxidation-Reduction Reactions Chapter 11 Electrochemistry Oxidation and Reduction Reactions An oxidation and reduction reaction occurs in both aqueous solutions and in reactions where substances are burned in the presence of oxygen gas, involves a transfer of electrons as the driving force of the chemical reaction. The reactants in these reactions will lose or gain electrons and change their charge as they form the products. The driving force of these chemical changes is electron transfer. 1 2 Oxidation-Reduction Reactions Oxidation- reduction reactions are a very important class of chemical reactions. They occur all around us and even within us. The bulk of the energy needed for the functioning of all living organisms, including humans, is obtained from food through oxidation-reduction processes. Such diverse phenomena as the electricity obtained from a battery to start a car, the use of natural gas to heat a home, iron rusting, and the functioning of antiseptic agents to kill or prevent the growth of bacteria all involve oxidationreduction reactions. The knowledge of this type of reaction is fundamental to understanding many biological and technological processes. 3 Oxidation-Reduction Reactions Historically, the word oxidation was first used to describe the reaction of a substance with oxygen. According to this historical definition, each of the following reactions involves oxidation: 4 Fe (s) + 3 O 2 (g) 2 Fe 2 O 3 (s) S (s) + O 2 (g) SO 2 (g) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) The reactant on the far left (in bold) in each of these chemical equations is said to have been oxidized. 4 Oxidation-Reduction Reactions Oxidation-Reduction Reactions Originally, the term reduction referred to processes where oxygen was removed from a compound. A particularly common type of reduction reaction, according to the original definition, is the removal of oxygen from a metal oxide to produce free metal. CuO (s) + H 2 (g) Cu (s) + H 2 O (g) 2 Fe 2 O 3 (s) + 3 C (s) 4 Fe (s) + 3 CO 2 (g) The word reduction comes from the reduction in mass of the metal oxide; the free metal has less mass than the metal oxide. 5 Today the words oxidation and reduction are used in a much broader sense. Current definitions include the previous examples but also include reactions with numerous non-oxygen containing substances. Reactions that involve the transfer of electrons from one reactant to another reactant, regardless of the substances involve, are collectively called oxidation-reduction reactions (redox). 6

2 Oxidation-Reduction Reactions For all redox reactions, the reactants must lose electrons and gain electrons during the chemical process. Redox Reactions: OIL RIG OIL oxidation is loss of electrons (e.g., metals in salts) RIG reduction is gain of electrons (e.g., nonmetals in salts) The name given to the charge on an element, ion or element in a covalent bond is called the oxidation state or oxidation numbers. For example: Na + has an oxidation state of +1. F has an oxidation state of To determine what substance is oxidized and which is reduced, we must determine the original and final oxidation states of each substance during the chemical reaction. To determine the oxidation states of all elements in molecules or ionic compounds, it is necessary to follow a few rules. To assign oxidation numbers to elements in compounds, there are a few simple rules: 1. The oxidation number of any free, uncombined element is zero. Na (s), Mg (s), Cu (s), Fe (s) and Zn (s) 9 10 All naturally occurring diatomic molecules have zero oxidation states. Br 2 I 2 N 2 Cl 2 H 2 O 2 F 2 This makes sense there is no dipole (difference in Electronegativity) between the shared electrons in the diatomic molecules. Nonmetals, in their natural uncombined state, have zero oxidation states. E.g., P 4 (s) and S 8 (s) 2. For simple Group A binary ionic compounds (salts), the oxidation state is the charge of the element: Metals (Group I A, IIA, and IIIA) are assigned a positive oxidation state determined by the number of electrons the element has lost. Nonmetals (Group VA, VIA, VIIA) are assigned a negative oxidation state determined by the number of electrons the element has gained

3 Use the periodic table to help with assigning oxidation numbers to elements. a. IA metals have oxidation numbers of +1. b. IIA metals have oxidation numbers of +2. c. IIIA metals have oxidation numbers of +3. There are a few rare exceptions. d. VA elements have oxidation numbers of 3 in binary compounds with H, metals or NH 4+. e. VIA elements below O have oxidation numbers of 2 in binary compounds with H, metals or NH 4+. f. VIIA elements have oxidation numbers of 1 in binary compounds with H, metals or NH Groups IA metals IIA metals IIIA metals V nonmetals VI nonmetals Oxidation number Monoatomic Ions Li +, Na +, K +, Rb +, Cs + Mg 2+, Ca 2+, Sr 2+, Ba 2+ Al 3+, Ga 3+, In 3+ N 3, P 3, As 3, Sb 3, Bi 3 O, S, Se, Te 3. In the formula for any compound, the sum of the oxidation numbers of all elements in the compound is zero. Na 2 SO 4 = zero charge on formula unit In a polyatomic ion, the sum of the oxidation numbers of the constituent elements is equal to the charge on the ion. SO 4 2- =- 2on the ion. VII nonmetals 1 F, Cl, Br, I Fluorine has an oxidation number of 1 in its compounds. 5. Hydrogen, H, has an oxidation number of +1 unless it is combined with metals, where it has the oxidation number - 1. Examples LiH, BaH 2 6. Oxygen usually has the oxidation number 2. Exceptions: In peroxides O has oxidation number of 1. Examples: H 2 O 2, CaO 2, Na 2 O 2 In OF 2, O has oxidation number of +2. (Fluorine is the most electronegative element and has the greater share of electrons.) 17 18

4 For simple binary ionic compounds that are comprised of transition metals (with more than one possible oxidation state), the oxidation state of the metal is determined by balancing the charge: For our purpose, all transition metals in salts (except zinc and silver) have oxidation states that are determined by the balancing the charge of the anion. e.g. FeO vs. Fe 2 O 3-2 FeO Since the oxidation state of oxygen is 2 in the compound, the iron atom must have a charge of + 2 to give the formula unit an overall net charge of zero. x + (- 2) = 0 x = Fe 2 O 3 Since the oxidation state of oxygen is 2 in the compound and there are three oxygen atoms, the iron must have a charge of +3 to give the formula unit an overall net charge of zero. 2x + 3(- 2) = 0 x = + 3 Zinc always has an oxidation state of +2 when combined in a salt. ZnCl 2, ZnS, ZnSO 4, Zn 3 (PO 4 ) 2 Silver always has an oxidation state of +1 when combined in a salt. AgCl, Ag 2 O, Ag 2 SO 4, Ag 3 PO Example: Assign oxidation numbers to each element in the following compounds: NaNO 3 Na = +1 O = 2 N =? Calculate using rule (-2) + x = 0 x = +5 N = +5 K 2 Sn(OH) 6 K = +1 O = - 2 H = +1 Sn =? 2(+1) + 6(-2) + 6(+1) + x = 0 x = +4 Sn =

5 H 3 PO 4 H = +1 O = - 2 P =? P = +5 3(+1) + 4( 2) + x = 0 = + 5 SO 3 2- O = - 2 S =? S = +4 3(-2) + x = -2 x = HCO 3 Cr 2 O 7 2- O = - 2 H = +1 C =? O = 2 Cr =? C = ( 2) + x = 1 x = +4 Cr = +6 7( 2) + 2(x) = 2 x = Determine the oxidation number of the underlined element in FeSO 4. The oxidation state of iron must be +2 in order to balance the 2 of the sulfate ion. Fe = + 2 O = x + 4( 2) = 0 x = + 6 OR The oxidation state of sulfur can be determined by the sulfate ion: SO 4 x + 4( 2) = 2 x =

6 Determine the oxidation number of the underlined element in Mg(ClO 4 ) 2. OR Mg = + 2 O = x + 8( 2) = 0 x = + 7 ClO 4 x + 4( 2) = 1 Determine the oxidation number of the underlined element in CH 4. H = +1 (least electronegative element) C = 4 Determine the oxidation number of the underlined element in CO 2. O = 2 (most electronegative element) C = + 4 x = Determine the oxidation number of the underlined element in H 2 O. H = + 1 (the least electronegative element) O = 2 Determine the oxidation number of the underlined element in H 2 O 2. H = + 1 (the least electronegative element) O = 1 Oxidation Reduction Reactions: REDOX Oxidizing agents are those elements that cause loss of electrons (cause oxidation) oxidizing agents are reduced by gaining electrons. Reducing agents are those elements that give up electrons (cause reduction) reducing agents are oxidized by giving up electrons In redox reactions, the reactants transfer electrons oxidation and reduction occurs between the reactants Na (s) + Cl 2 (g) 2 NaCl (s) To determine the oxidizing agent and reducing agent, we can divide the reaction of the reactants into halfreactions: 0 +1 oxidation reaction: : 2 Na 2 Na+ + 2 e reducing agent 0 1 reduction reaction: : Cl e 2 Cl oxidizing agent 35 Since we cannot either create or destroy electrons, the numbers of electrons transferred in the reaction must be the same on both sides of the chemical equation: Fe (s) + 3 O 2 (g) 2 Fe 2 O 3 (s) 0 +3 oxidation reaction 4 Fe 4 Fe e 0 2 reduction reaction 3 O e 6 O reducing agent oxidizing agent 36

7 NaBr (aq) + Cl 2(l) 2 NaCl (aq) + Br 2(s) We can remove the spectator ion (Na + ) before we set up the half- reactions. oxidation reaction Br Br 2 + 2e reducing agent A more complicated redox reaction requires that we must identify what is oxidized and what is reduced by assigning oxidation states to all the elements in the reactants and products H 2 SO 4(aq) + 2 HI (g) SO 2(g) + I 2(s) + 2 H 2 O reduction reaction 0 1 Cl e 2 Cl oxidizing agent H 2 SO 4(aq) + 2 HI (g) SO 2(g) + I 2(s) + 2 H 2 O We must identify in the reaction which elements changed oxidation states. The oxidation state of hydrogen is the same on both sides of the equation, so it is not involved in transfer of electrons. The oxidation state of oxygen is the same on both sides of the equation, so it is not involved in transfer of electrons. The oxidation state of sulfur changed and the oxidation state of iodine changed. 1 0 oxidation reaction 2 I I e red agent reduction reaction SO e SO 2 ox agent It is not always obvious how to balance a redox equation. For these more complex oxidation- reduction reactions, we have a system by which to balance the equation by mass and by charge. The half- reaction method is useful in that it is the method that is used in electrolysis. Balance the equation as much as possible without the spectator ions. Set up half- reactions (oxidation and reduction) as we did in the previous reactions

8 Balance the half- reactions by the following rules: First, determine if the reaction has taken place in acidic (H + ) conditions or basic (OH ) conditions. The rules for balancing oxygen and hydrogen are different depending on whether the reaction takes place in acidic or basic solutions. The Half-Reaction Method Half reaction method rules: 1. Write the unbalanced reaction. 2. Break the reaction into 2 half reactions: One oxidation half-reaction and One reduction half-reaction Each reaction must have complete formulas for molecules and ions. 3. Mass balance each half reaction by adding appropriate stoichiometric coefficients The Half-Reaction Method 4. Charge balance the half reactions by adding appropriate numbers of electrons. Electrons will be products in the oxidation half-reaction. Electrons will be reactants in the reduction half-reaction. 5. Multiply each half reaction by a number to make the number of electrons in the oxidation half-reaction equal to the number of electrons reduction half-reaction. 6. Add the two half reactions. 7. Eliminate any common terms and reduce coefficients to smallest whole numbers. In acidic conditions: Balance the oxygen by adding H 2 O molecules to the opposite side of the equation. Balance the hydrogen by adding H+ to the opposite side of the equation Example: C (s) + HNO 3 (aq) CO 2 (g) + NO 2 (g) + H 2 O Set up the half- reactions: C CO e oxidation NO 3 + 1e NO 2 reduction Balance the oxygen by adding H 2 O to the opposite side. (One H 2 O for each oxygen) 2 H 2 O + C CO e NO 3 + 1e NO 2 + H 2 O Balance the hydrogen by adding H + to the opposite side. 2 H 2 O + C CO e + 4 H + 2 H + + NO 3 + 1e NO 2 + H 2 O 47 48

9 The number of e must be the same on both sides of the equation. Balance the charge by multiplying the half- reaction with the appropriate number to give the same number of electrons on both sides of the equation. 2 H 2 O + C CO e + 4 H + 4 (2 H + + NO 3 + 1e NO 2 + H 2 O) 2 H 2 O + C CO e + 4 H + 8 H NO e 4 NO H 2 O Cancel e, water, and H + from opposite sides of the equations, and sum the remaining reactants and products. 2 H 2 O + C CO e + 4 H H NO e 4 NO H 2 O. 4 H NO 3 + C 4 NO 2 + CO H 2 O Add the H + back to the NO 3 ions to complete the equation: 4 HNO 3 + C 4 NO 2 + CO H 2 O In basic conditions: Balance the oxygen by adding 2 OH molecules to the side that needs oxygen and adding one H 2 O to the other side. Balance the hydrogen by adding H 2 O to the side that needs hydrogen and adding one OH to the other side Example: In basic solution, hypochlorite ions (ClO ) oxidize chromite ions (CrO ) to chromate ions (CrO 4 ) and are reduced to chloride ions. Write a balanced net ionic equation for this reaction CrO + ClO CrO 4 + Cl Set up the half-reactions: CrO 2 CrO e oxidation ClO + 2 e Cl reduction Balance the oxygen by adding 2 OH to the opposite side. (2 OH for each oxygen) CrO + 4 OH CrO e ClO + 2 e Cl + 2 OH Balance the hydrogen by adding one H 2 O for every two OH - on the opposite side. CrO + 4 OH CrO H 2 O + 3 e H 2 O + ClO + 2 e Cl + 2 OH 53 54

10 Balance the charge by multiplying the half- reaction with the appropriate number to give the same number of electrons on both sides of the equation. 2 (CrO + 4 OH CrO H 2 O + 3 e ) 3 (H 2 O + ClO + 2 e Cl + 2 OH ) 2 CrO + 8 OH 2 CrO H 2 O + 6 e 3 H 2 O + 3 ClO + 6 e 3 Cl + 6 OH Add the two half reactions. Eliminate any common terms and reduce coefficients to smallest whole numbers CrO + 8 OH 2 CrO H 2 O + 6 e 3 H 2 O + 3 ClO + 6 e 3 Cl + 6 OH 2 CrO + 2 OH + 3 ClO - 2 CrO 4 + H 2 O + 3 Cl Stoichiometry of Redox Reactions It is possible in a few reactions that an element in the reactant is oxidized and it is also the element that is reduced. This is called a disproportionation oxidation-reducion reaction. Redox titrations: Titrations can be preformed to determine the presence of substances that are oxidized or reduced. In this procedure, the standard is a solution of a known oxidizing or reducing agent. Slide 25 p.148 Note that chlorine in the reactant is oxidized and also reduced during the reaction. We will do the half reactions in class Stoichiometry of Redox Reactions Stoichiometry of Redox Reactions Example. What volume of M KMnO 4 is required to oxidize 35.0 ml of M HCl? The balanced reaction is: 2 KMnO HCl 2 KCl+2 MnCl 2+ 5 Cl2+ 8 H2O ( )( ) 35 ml HCl M HCl = 5.25 mmol HCl ( ) mmol HCl mmol KMnO4 ( ) 2 mmol KMnO = 16 mmol HCl 1 ml mmol KMnO4 = 3.28 ml mmol KMnO

11 Stoichiometry of Redox Reactions End of Chapter 11 Example. A volume of 40.0 ml of iron (II) sulfate is oxidized to iron (III) by 20.0 ml of M potassium dichromate solution. What is the concentration of the iron (II) sulfate solution? We will solve this problem and others during class. 61

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