Unit 3. Oxidation and Reduction Chemistry 020, R. R. Martin

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1 1 Introduction Unit 3 Oxidation and Reduction Chemistry 020, R. R. Martin Another important type of reaction in aqueous solution involves the transfer of electrons between two species. This is called an oxidationreduction or a redox reaction. What happens when zinc pellets are added to an acid? The zinc dissolves, and a gas is formed. The net ionic equation for this process is Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Zn loses electrons, and H + gains electrons. You can think of this reaction as two separate processes (halfequations or half reactions) Oxidation (loss of electrons): Zn(s) Zn e and 1

2 Reduction (gain of electrons): 2H + (aq) + 2e H 2 (g) OIL RIG: Oxidation is loss, reduction is gain In reality, free electrons are not involved in redox reactions; oxidation and reduction always occur together. All the electrons produced in the oxidation half reaction have to be consumed in the reduction half reaction. (the total number of electrons does not change!!!) When you add the two halfequations together, the electrons have to cancel out : Oxidation (loss of electrons): Zn(s) Zn e Reduction (gain of electrons): 2H + (aq) + 2e H 2 (g) Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) 2

3 The species that accepts electrons is called the oxidizing agent ( oxidant, it oxidizes the other species, but in doing this it is reduced) The species that donates electrons is called reducing agent ( reductant, it reduces the other species, but in doing that it is oxidized). The oxidant oxidizes the reductant and the reductant reduces the oxidant. red ag reduced + ox ag oxidized red ag oxidized + ox ag reduced Here, Zn is the reducing agent, H + agent. is the oxidizing Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) 3

4 2. Oxidation Numbers are required for the proper balancing of redox equations. They simplify the electron bookkeeping. Each atom in a compound can be assigned an oxidation number. Rules for assigning Oxidation Numbers : 1. Any allotrope of any element in the free state has an oxidation number of zero. (i.e. C (Diamond), C (Graphite), C (Gas) for C or O 2(Gas), O 3(Gas) for Oxygen) 2. For a monatomic ionic species the oxidation number is equal to the net charge of the species. (i. e. for O 2, F, Cu +, Fe 2+ and Al 3+ the oxidation numbers are 2, 1, +1, +2 and +3 respectively). 3. The following rules are applied to specific atoms in their chemical compounds. If a conflict occurs, the atom listed first in the list takes preference: a) F = 1 b) Li, Na, K, Rb, Cs = +1 c) Be, Mg, Ca, Sr, Ba = +2 d) H = +1 e) O = 2 f) Cl, Br, I = 1 4. The sum of all the oxidation numbers in a polyatomic chemical species is equal to the net charge on the species. (i.e. carbon tetrachloride, CCl 4 : the net charge is zero, since Cl is 1, the oxidation number of C must be +4.) N.B. The rule that occurs first in the list always takes precedence so that: 4

5 Using these rules the oxidation number of O in OF 2 is +2 and in H 2 O 2 it is 1. In BrF 5 the oxidation number for Br is +5. Example: Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Zn is oxidized (0 +2) H + is reduced (+1 0) An easy way to recognize a redox equation is to note changes in oxidation number: 2Al(s) + 3Cu 2+ (aq) 2Al 3+ (aq) + 3Cu(s) CO 3 2 (aq) + 2H + (aq) CO 2 (g) + H 2 O 5

6 An old Exam Question: Chemistry 020, Nov 2004 The oxidation # of N in NH 4 + is 3 The oxidation # of N in the product, N 2 is zero The N has lost electrons and acts as a reducing agent The oxidation # of O goes from 2 to zero in O 2 in other words O is oxidized. The oxidation # of H does not change. Ans is D 6

7 3. Balancing Redox Equations Acidic solution: Cu + NO 3 Cu 2+ + NO 2 a) Separate the equation into two halfreactions, an oxidation and a reduction and balance in all species except H and O: Cu Cu 2+ NO 3 NO 2 b) Assign oxidation #s and balance electrons: Cu Cu e 1e + (N= 5+) NO 3 NO 2 (N=4+) c) Make the solution Acidic by adding H + to balance charge: (oxidation) Cu Cu e (reduction) 2H + + 1e + NO 3 NO 2 d) Add water to balance oxygen: Cu Cu e 7

8 2H + + 1e + NO 3 NO 2 + H 2 O e) Multiply each halfreaction by an appropriate factor so that the # of electrons in each reaction is the same, add the two reaction together and tidy up the result (Remove any reactants that appear on both sides of the resulting equation): Cu Cu e (2H + + 1e + NO 3 NO 2 + H 2 O) X 2 4H + + Cu + 2NO 3 2NO H 2 O f) Check for balance in atoms and charge. Note: also make sure that the final equation is in terms of the smallest whole numbers. 8

9 Basic solution: Br 2(aq) Br (aq) + BrO 3 (aq) (Note this is a disproportionation reaction: A substance in an intermediate oxidation state goes to both higher and lower states.) a) Separate the equation into two halfreactions, an oxidation and a reduction (By assigning oxidation #s) and balance in all species except H and O: (reduction) Br 2(aq) 2Br (aq) (oxidation) Br 2(aq) 2BrO 3 (aq) b) balance electrons: 2e + Br 2(aq) 2Br (aq) (ox # + 0) Br 2(aq) (ox # = 5+) 2BrO 3 (aq) + 10e c) Make the solution Basic by adding OH to balance charge: 2e + Br 2(aq) 2Br (aq) 12OH + Br 2(aq) 2BrO 3 (aq) + 10e 9

10 d) Add water to balance oxygen: 2e + Br 2(aq) 2Br (aq) 12OH + Br 2(aq) 2BrO 3 (aq) + 10e + 6H 2 O e) Multiply each halfreaction by an appropriate factor so that the # of electrons in each reaction is the same, add the two reaction together and tidy up the result (Remove any reactants that appear on both sides of the resulting equation): (2e + Br 2(aq) 2Br (aq) ) x 5 12OH + Br 2(aq) 2BrO 3 (aq) + 10e + 6H 2 O 10e + 5Br 2(aq) 12OH + Br 2(aq) 10Br (aq) + 2BrO 3 (aq) + 10e + 6H 2 O yields: 5Br 2(aq) 12OH + Br 2(aq) 10Br (aq) + 2BrO 3 (aq) + 6H 2 O 6Br 2(aq) + 12OH 10Br (aq) + 2BrO 3 (aq) + 6H 2 O 10

11 f) Check for balance in atoms and charge. Note: also make sure that the final equation is in terms of the smallest whole numbers. NB: Divide by 2 To get the final answer: 3Br 2(aq) + 6OH 5Br (aq) + BrO 3 (aq) + 3H 2 O 11

12 An old exam question: Separate into an oxidation and a reduction: (oxidation) Zn Zn 2+ (reduction) VO 3 V 2+ Balance electrons: Zn Zn e VO 3 + 3e V 2+ Balance charge with H + : Zn Zn e + H + VO 3 + 3e + 6H + V 2+ Balance oxygen with H 2 O Zn Zn e + H + VO 3 + 3e + 6H + V H 2 O Eliminate electrons: 12

13 (Zn Zn e + H + ) x 3 (VO 3 + 3e + 6H + V H 2 O) x 2 3 Zn 3 Zn e + 3 H + 2 VO 3 + 6e + 12 H + 2 V H 2 O Adding yields: 3 Zn + 2 VO H + 3 Zn V H 2 O 13

14 Separate into an oxidation and a reduction: (oxidation) SO 3 (reduction) MnO 4 SO 4 MnO 2 Balance electrons: SO 3 SO 4 + 2e MnO 4 + 3e MnO 2 Balance charge with OH : SO OH SO 4 + 2e MnO 4 + 3e MnO OH Balance oxygen with H 2 O 14

15 SO OH SO 4 + 2e + H 2 O MnO 4 + 3e + 2 H 2 O MnO OH Eliminate electrons: 3 x (SO OH SO 4 + 2e + H 2 O) 2 x (MnO 4 + 3e + 2 H 2 O MnO OH ) 3 SO OH 3 SO e + 3 H 2 O 2 MnO 4 + 6e + 4 H 2 O 2 MnO OH adding yields: 3 SO MnO 4 + H 2 O 3 SO MnO OH 15

16 4. Redox Titrations Recall what we said about acid/base titrations: Acidbase titrations involves four quantities: 1. Concentration of acid 2. concentration of base 3. volume of acid 4. volume of base ( volume means volume to reach the equivalence point). Three of these are required to calculate the fourth! With redox titrations, it is very similar: 1. Concentration of oxidant, 2. concentration of reductant, 3. volume of oxidant, 4. volume of reductant ( volume means volume to reach the equivalence point). Three of these are required to calculate the fourth! The equivalence point is reached when oxidant and reactant are present in stoichiometric amounts. 16

17 Consider the titration of an acidic Fe 2+ solution with permanganate: MnO 4 + Fe 2+ Fe 3+ + Mn 2+ (not balanced) The balanced equation for this redox reaction is MnO 4 (aq) + 8H + (aq) + 5Fe 2+ (aq) 5Fe 3+ (aq) + Mn 2 +(aq) + 4H 2 O What volume of M KMnO 4 solution is required to react completely with ml of M Fe(NO 3 ) 2? Moles Fe(NO 3 ) 2 = M x V = 0.25 moles/liter x liters = x 10 3 moles Moles KMnO 4 required = Moles Fe(NO 3 ) 2 /5 = x 10 3 moles = M x V = moles/liter x a liters a = volume KMnO 4 required = 2.01 x 10 3 liters = 2.02 ml 17

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