Oxidation-Reduction Summary and Study Assignment

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1 Oxidation-Reduction Summary and Study Assignment The atoms in elements, ions or molecules involved in electrochemical reactions are characterized by the number of protons the atom has compared to the number of electrons it is assigned; we refer to the difference between the number of protons and the number of assigned electrons as the oxidation state of the atom. For monatomic ions, the oxidation state of the ion is simply the ion s charge. In the case of covalent bonding (electrons are shared) between atoms of differing electronegativity, the bonding electrons are assigned to the atom with the largest electronegativity when determining the oxidation state of the bonded atoms. For polyatomic ions the sum of the oxidation states must equal the total charge of the ion. For molecules, the sum of the oxidation states must equal zero. In electrochemical reactions, also called oxidation-reduction or redox reactions, atoms undergo changes in their oxidation state. In other words, redox reactions involve the transfer of VALENCE electrons from one reactant to another. Definitions Oxidation: The loss or apparent loss of electrons; an increase in oxidation number for the atom. Reduction: The gain or apparent gain of electrons; a decrease in oxidation number for the atom. Oxidation number (O.N.): A bookkeeping number system to determine which atom is oxidized or reduced. Equals the difference between the number of protons and the number of assigned electrons an atom has; sign can be or +. Oxidizing : Oxidizes another substance. The oxidizing agent is reduced. Reducing : Reduces another substance. The reducing agent is oxidized. Oxidation (O.N.) number rules For an atom in its elemental form (K, S 8, N 2, Au, etc.) O.N. = 0 For a monatomic ion O.N. = the charge on the ion. (Ti 3+ O.N. = +3) Fluorine is always 1 in compounds. Cl, Br, and I are 1 as well except when combined with fluorine or oxygen. Hydrogen is +1 and oxygen is usually 2 in most compounds. Except the metal hydrides MH x, H is 1. For the peroxides, O 2 2-, O is 1. The sum of oxidation numbers must equal zero for a neutral compound or equal the ionic charge for a polyatomic ion. Examples FeCl 2 : Ionic compound. The O.N. of Cl is 1. The O.N. of Fe is +2 C 2 H 6 : Covalent compound. The O.N. of H is +1. The O.N. of each C is 3 HSO 4 : Polyatomic ion. The O.N. of H is +1. The O.N. of each O is 2. The O.N. of S is +6. Change in oxidation numbers and redox reactions In a oxidation-reduction reaction one element is oxidized and one element is reduced. By assigning oxidation numbers to the elements one can then see which element is oxidized (has an increase in oxidation number) and which element is reduced (has a decrease in oxidation number). The oxidized element has a loss or an apparent loss of electrons. The change in oxidation number indicates the number of electrons lost by the element. The reduced element has a gain or an apparent gain of electrons. The change in oxidation number indicates the number of electrons gained by the element. Dr. Larson/Daley 1 5/23/08

2 Examples 1. N 2 + 2O 2 > 2NO 2 N 2 : each N has an O.N. of 0. NO 2 : O has an O.N. of -2.; N has an O.N. of +4. O 2 : each O has an O.N. of 0. N has an increase in O.N. from 0 to +4. N is oxidized by losing 4 e. N 2 is the reducing agent. O has a decrease in O.N. from 0 to 2. O is reduced by gaining 2 e. O 2 is the oxidizing agent. 2. ClO 3 (aq) + 6H + (aq) + 6Br (aq) > Cl (aq) + 3H 2 O(l) + 3Br 2 (l) ClO 3 : each O has a O.N. of -2.; Cl has a O.N. of +5. Cl : Cl has a O.N. of -1. H + : H has a O.N. of +1. H 2 O: O has a O.N. of -2.; each H has a O.N. of +1. Br : Br has a O.N. of -1. Br 2 : each Br has a O.N. of 0. Cl has a decrease in O.N. from +5 to 1. Cl is reduced by gaining 6 e. ClO 3 is the oxidizing agent. Br has an increase in O.N. from -1 to 0. Br is oxidized by losing 1 e. Br is the reducing agent. Equivalent Mass Equivalent mass of a reactant in redox chemistry is defined as the mass of the reactant that gains (reduction) or loses (oxidation) 1 mole of electrons. Consider the examples from above: 1. The equivalent mass of O 2 can be determined as follows: (32.00 g O 2 /mol O 2 )x(1 mol O 2 /4 mol electrons) = g O 2 /mol electrons The equivalent mass of N 2 can be determined as follows: (28.02 g N 2 /mol N 2 )x(1mol N 2 /8 mol electrons) = g N 2 /mol electrons 2. The equivalent mass of ClO 3 can be determined as follows: (83.45 g ClO 3 /mol ClO 3 )x(1 mol ClO 3 /6 mol electrons) = g ClO 3 /mol electrons The equivalent mass of Br can be determined as follows: (79.90 g Br /mol Br )x(1mol Br /1 mol electrons) = g Br /mol electrons NOTE: Equivalent mass of a reactant depends upon the reaction of interest. Specifically, equivalent mass depends upon the change of oxidation state undergone by the reactant. Therefore, a given reactant may have more than one possible equivalent mass. In fact, a given reactant may act as a reducing agent in one reaction and an oxidizing agent in a different reaction! For example, consider the following reaction: N 2 + 3Ba > Ba 3 N 2 In this case, N is reduced from an O.N. of 0 to an O.N. of 3 and N 2 is the oxidizing agent. The equivalent mass of N 2 in this case is g N 2 /mol electrons) Dr. Larson/Daley 2 5/23/08

3 Balancing Oxidation-Reduction Reactions Steps for balancing using the half-reaction method 1. Rewrite the equation in NET-IONIC form. 2. Assign oxidation numbers to each element. 3. Identify the oxidized and reduced species from the changes in oxidation number between reactants and products. 4. Divide the reaction into oxidation and reduction half-reactions. Include only the reactants (and their products) that have a change oxidation number. 5. Balance each half-reaction in acid medium by: a) balancing elements other than O and H by inspection b) balancing O atoms by adding H 2 O to the side with fewer O atoms c) balancing H atoms by adding H + ions to the side with fewer H atoms d) balancing charge by adding e to the side with the most positive ionic charge. 6. Find the common multiple of electrons between half-reactions by multiplying each half-reaction by an integer to make e lost equal to e gained. 7. Add the half-reactions together and cancel identical substances (including phases!) appearing on each side. 8. If the reaction occurs in basic medium, add OH ions to both sides and convert each H + to H 2 O. 9. If possible, add the spectator ions back after balancing the overall equation. Example of the half-reaction method Step 1: KMnO 4 (aq) + KI(aq) MnO 2 (s) + KIO 3 (aq) (NOT BALANCED) MnO 4 (aq) + I (aq) MnO 2 (s) + IO 3 (aq) (Skeleton net ionic equation, not balanced, spectator ions omitted!) Step 2 and 3: Mn changes from a +7 oxidation state to a +4 oxidation state. Mn is reduced. I changes from a 1 oxidation state to a +5 oxidation state. I is oxidized. Step 4: Reduction half-reaction MnO 4 MnO 2 Oxidation half-reaction I IO 3 Dr. Larson/Daley 3 5/23/08

4 Step 5: For the reduction half-reaction the steps are: a) Atoms other than O and H MnO 4 MnO 2 b) O atom with H 2 O MnO 4 MnO 2 + 2H 2 O c) H atom with H + 4H + + MnO 4 MnO 2 + 2H 2 O d) Charge with e 3e + 4H + + MnO 4 MnO 2 + 2H 2 O Repeating for the oxidation half-reaction gives: I + 3H 2 O IO 3 + 6H + + 6e Step 6.: Multiply the reduction half-reaction by 2x: 2 [3e + 4H + + MnO 4 MnO 2 + 2H 2 O] = 6e + 8H + + 2MnO 4 2MnO 2 + 4H 2 O Step 7: 6e + 8H + + 2MnO 4 2MnO 2 + 4H 2 O + I + 3H 2 O IO 3 + 6H + + 6e = 6e + 8H + + 2MnO 4 + I + 3H 2 O 2MnO 2 + 4H 2 O + IO 3 + 6H + + 6e 2H + (aq) + 2MnO 4 (aq) + I (aq) 2MnO 2 (s) + H 2 O(l) + IO 3 (aq) (Solution for acidic medium) Step 8: ONLY If in basic medium add OH (aq) to both sides 2OH + 2H + + 2MnO 4 + I 2MnO 2 + H 2 O + IO 3 + 2OH 2H 2 O + 2MnO 4 + I 2MnO 2 + H 2 O + IO 3 + 2OH H 2 O(l) + 2MnO 4 (aq) + I (aq) 2MnO 2 (s) + IO 3 (aq) + 2OH (aq) (Solution for basic medium) Step 9: Add back in spectator ions. Let s do the basic media solution. H 2 O(l) + 2KMnO 4 (aq) + KI(aq) 2MnO 2 (s) + KIO 3 (aq) + 2KOH(aq) Dr. Larson/Daley 4 5/23/08

5 Oxidation-Reduction Exercise Assigning Oxidation Numbers Give the oxidation number of nitrogen in each of the following: NH 2 OH N 2 N 2 H 4 NH 4 + NO Give the oxidation number of manganese in each of the following: MnO 4 - Mn 2 O 3 Mn MnSO 4 MnO 2 2- Types of Reactions Combination, decomposition, single displacement and combustion reactions are all examples of oxidation-reduction reactions. Note that there are reactions that do not fall into these categories that are redox reactions. Precipitation reactions that are categorized as double displacement (Also called exchange reactions in some textbooks.) are NOT oxidation-reduction reactions. All acid-base reactions, whether they are Arrhenius, Bronsted-Lowry or Lewis, are NOT oxidation-reduction reactions. Directions: For each of the following reactions classify the reaction as: (a) redox, (b) precipitation, or (c) acid-base. Reaction Reaction Type 1. 2Al(s) + 3Cl 2 (g) > 2AlCl 3 (s) 2. Ag + (aq) + Cl (aq) > AgCl(s) 3. Mg(s) + 2HCl(aq) > MgCl 2 (aq) + H 2 (g) 4. Cu 2+ (aq) + H 2 S(aq) > CuS(s) + 2H + (aq) 5. Ba(s) + 2H 2 O(l) > Ba 2+ (aq) + 2OH (aq) + H 2 (g) 6. 2H + (aq) + CO 2 3 (aq) > CO 2 (g) + H 2 O(l) 7. HCl(aq) + NH 3 (aq) > NH 4 Cl(aq) 8. 2CH 3 CH 3 (g) + 7O 2 (g) > 4CO 2 (g) + 6H 2 O(g) 9. AgCl(s) + 2NH 3 (aq) > Ag(NH 3 ) 2+ (aq) + Cl (aq) Dr. Larson/Daley 5 5/23/08

6 Oxidizing and Reducing s and Equivalent Mass Complete the table below ONLY for the reactions classified as REDOX REACTIONS from above. If the reaction is not redox, leave blank. Reaction # Reducing Electrons Lost per Formula Unit of Reducing Equivalent Mass of Reducing Oxidizing Electrons Gained per Formula Unit of Oxidizing Equivalent Mass of Oxidizing Dr. Larson/Daley 6 5/23/08

7 Balancing Oxidation-Reduction Reactions For each of the following reactions balance the reaction using the half-reaction method. 1. Fe 2+ (aq) + H 2 O 2 (aq) Fe 3+ (aq) + H 2 O(l) 2. I 2 (s) + S 2 O 3 2 (aq) S4 O 6 2 (aq) + I (aq) 3. PbO(s) + NH 3 (aq) N 2 (g) + Pb(s) Dr. Larson/Daley 7 5/23/08

8 4. MnO 4 (aq) + C 2 O 4 2 (aq) Mn 2+ (aq) + CO 2 (g) (BASIC MEDIUM) 5. NaClO (aq) + NaCrO 2(aq) NaCl (aq) + Na 2 CrO 4(aq) (Try adding back the spectator ions at the end.) Dr. Larson/Daley 8 5/23/08

9 6. H 2 S (aq) + Cr 2 O 7 2 (aq) S (s) + Cr 3+ (aq) 7. PbO 2(s) + NaCl (aq) NaClO (aq) + Na[Pb(OH) 3 ] (aq) (What is the medium? Try adding back the spectator ions at the end.) 8. Br 2(l) Br (aq) + BrO 3 (aq) Note: This reaction is a special type of redox reaction called a disproportination or self redox reaction where a single reactant (Br 2 in this case) is both oxidized and reduced. Dr. Larson/Daley 9 5/23/08