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Oxidation- Reduction Reaction “redox reaction” Part 1.

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1 Oxidation- Reduction Reaction “redox reaction” Part 1

2 Objectives for the Unit Identify elements that change oxidation state from one side of a chemical equation to the other. Identify reducing agents and oxidizing agents in a chemical equation. Use the half-reaction method to balance redox reactions.

3 Oxidation Old definition: –combination of oxygen with other substances. Oxidation as defined today: –The loss of electrons from an atom or ion. e - e - goes bye-bye

4 Reduction Old definition: –“reducing" a substance into its components. Reduction as defined today: –The gaining of electrons by an atom or ion.

5 Reduction-Oxidation Reaction “Redox reaction” redox reactionA reduction and oxidation reaction is commonly called redox reaction for short Oxidation always accompanies reduction –One atom must lose electrons and another must pick up the electrons. –The number of electrons lost (oxidation) must be equal to the number of electrons gained (reduction)

6 Concept check: Concept check: (answer these questions on you paper using Cornell note format. Answer the question first before advancing the slide) 1. Oxidation means gaining/ losing electrons Answer: losing 2. Reduction means gaining/losing electrons Answer: gaining 3.Can there be a oxidation without reduction? Explain. Answer: no, one atom must give up electrons (oxidation) and another must be there to receive or gain (reduction) electrons 4.Compare a reducing agent with an oxidation agent. Answer: reducing agents are atoms that provide the electrons (started out with electrons) and when they lose the electrons, then they are oxidized. Oxidation agents are atoms that will accept the electrons, (started out with less electrons) and when they gain electrons, they will be reduced.

7 Tip LEO LEO The Lion Goes GER Loose Electrons, Oxidize Gain Electrons, Reduce

8 How do you know this is a "Redox" equation? 2 H 2 + O 2  2 H 2 O What is the charge of H 2 here? What is the charge of O 2 here? What is the charge of H 2 and O here? Think back, how does hydrogen bond with oxygen and why 2 hydrogen for every oxygen? This is a redox reaction because atoms are being reduced (oxygen gained two electrons to go from a charge of 0 to -2) and oxidized (hydrogen loses one electron to go from a charge of 0 to +1)

9 How to identify a redox reaction Oxidation and Reduction must both occur in a Redox reaction. If one particle gains electrons in a reaction, some other particle must lose them.So? Look for the change in oxidation number (charges).

10 How to assign oxidation numbers. You have learned to read the oxidation numbers of many elements from the periodic table, ie: group 1 elements is +1, group 2 elements are +2, group 13 elements are +3, group 15 elements are -3, group 16 elements are -2 and group 17 elements are -1. While that information is important, the following rules are to be your guide when working with Redox equations.  Rules for assigning oxidation numbers: The oxidation number of a free element = 0. The oxidation number of a monatomic ion = the charge on the ion. The oxidation number of hydrogen = + 1 and rarely - 1. The oxidation number of oxygen = - 2 and in peroxides (H 2 O 2 )= - 1. The sum of the oxidation numbers in a polyatomic ion (ions made up of two or more elements) = charge on the ion. Elements in group 1, 2, and aluminum are always as indicated on the periodic table.

11 The oxidation number of elements not covered by the rules must be "calculated" using the known oxidation numbers in a compound. Example #1: K 2 CO 3 To calculate C By rule K is +1 and O is -2 The sum of all the oxidation numbers in this formula equal 0 because there is no charge on this polyatomic ion. Multiply the subscript by the oxidation number for each element. K = (2) ( + 1 ) = + 2 O = (3) ( - 2 ) = - 6 therefore, C = (1) ( + 4 ) = + 4 Example #2: HSO 4 - To calculate S by rule, H is + 1 by rule O is - 2 The sum of all the oxidation numbers in this formula equal -1 because this polyatomic ion has a charge of -1. Multiply the subscript by the oxidation number for each element. H = (1) ( + 1 ) = + 1 O = (4) ( - 2 ) = - 8 therefore, S = (1) ( + 6 ) = + 6

12 Practice Problems: Use the rules above to determine the oxidation number of the element indicated in each formula. 1.Sb in Sb 2 O 5 2.N in Al(NO 3 ) 3 3.P in Mg 3 (PO 4 ) 2 4.S in (NH 4 ) 2 SO 4 5.Cr in CrO Cl in ClO 4- 7.B in NaBO 3 8.Si in MgSiF 6 9.I in IO N in (NH 4 ) 2 S 11.Mn in MnO Br in BrO Cl in ClO – 14.Cr in Cr 2 O Se in H 2 SeO 3 answers

13 Oxidation- Reduction Reaction “redox reaction” Part 2 Reducing Agents and Oxidizing Agents

14 Reducing Agents the reactant that gives up electrons. The reducing agent contains the element that is oxidized (looses electrons). If a substance gives up electrons easily, it is said to be a strong reducing agent.

15 Oxidizing and Reducing Agents Reducing agent: atoms that provide electrons to reduce the other atom

16 Oxidizing agents the reactant that gains electrons. The oxidizing agent contains the element that is reduced (gains electrons). If a substance gains electrons easily, it is said to be a strong oxidizing agent.

17 Oxidizing and Reducing Agents Oxidizing agent: atoms that will gain electrons thus oxidizing the other atom.

18 Example: Fe 2 O 3 (s) + 3CO (g)  2Fe (s) + 3CO 2 (g) Notice that the oxidation number of C goes from +2 on the left to +4 on the right. The reducing agent is CO, because it contains C, which loses e -. Notice that the oxidation number of Fe goes from +3 on the left to 0 on the right. The oxidizing agent is Fe 2 O 3, because it contains the Fe, which gains e

19 Charting Reducing Agents and Oxidizing Agents Practice Problems: In any Redox equation, at least one particle will gain electrons and at least one particle will lose electrons. This is indicated by a change in the particle's oxidation number from one side of the equation to the other. For each reaction below, draw arrows and show electron numbers as in the example here. The top arrow indicates the element that gains electrons, reduction, and the bottom arrow indicates the element that looses electrons, oxidation. An arrow shows what one atom of each of these elements gains or looses. This is the first thing that must be done in balancing a Redox reaction. Learn to do it well.

20 Charting Reducing Agents and Oxidizing Agents Practice Problems 1.Mg + O 2  MgO 2.Cl 2 + I -  Cl - + I 2 3.MnO C 2 O 4 -2  Mn +2 + CO 2 4.Cr + NO 2 -  CrO N 2 O BrO MnO 2  Br - + MnO Fe +2 + MnO 4 -  Mn +2 + Fe +3 7.Cr + Sn +4  Cr +3 + Sn +2 8.NO S  NO 2 + H 2 SO 4 9.IO I -  I 2 10.NO 2 + ClO - NO Cl -

21 Oxidation- Reduction Reaction “redox reaction” Part 3 Balancing Redox Equations by the Half-reaction Method

22 1.Decide what is reduced (oxidizing agent) and what is oxidized (reducing agent). Do this by drawing arrows as in the practice problems. 2. Write the reduction half-reaction. –The top arrow in step #1 indicates the reduction half-reaction. Show the electrons gained on the reactant side. –Balance with respect to atoms / ions. To balance oxygen, add H 2 O to the side with the least amount of oxygen. THEN: add H + to the other side to balance hydrogen. Remember that the arrow in step #1 indicates the number of electrons gained by one atom. 3. Write the oxidation half-reaction. –The bottom arrow in step #1 indicates the oxidation half-reaction. –Show the electrons lost on the product side. –Balance with respect to atoms / ions. To balance oxygen, add H 2 O to the side with the least amount of oxygen. THEN: add H + to the other side to balance hydrogen. Remember that the arrow in step #1 indicates the number of electrons lost by one atom.

23 Balancing Redox Equations by the Half-reaction Method Continue 4. The number of electrons gained must equal the number of electrons lost. –Find the least common multiple of the electrons gained and lost. –In each half-reaction, multiply the electron coefficient by a number to reach the common multiple. –Multiply all of the coefficients in the half-reaction by this same number. 5. Add the two half-reactions. –Write one equation with all the reactants from the half-reactions on the left and all the products on the right. –The order in which you write the particles in the combined equation does not matter. 6. Simplify the equation. –A) Cancel things that are found on both sides of the equation as you did in net ionic equations. –B) Rewrite the final balanced equation. Check to see that electrons, elements, and total charge are balanced. –There should be no electrons in the equation at this time. –The number of each element should be the same on both sides. –It doesn't matter what the charge is as long as it is the same on both sides. If any of these are not balanced, the equation is incorrect. The only thing to do is go back to step #1 and begin looking for your mistake.

24 Practice Problems: Identify the oxidizing agent and reducing agent in each equation: AnswersAnswers 1.H 2 SO 4 + 8HI  H 2 S + 4I 2 + 4H 2 O 2.Au 2 S 3 + 3H 2  2Au + 3H 2 S 3.Zn + 2HCl  H 2 + ZnCl 2 To make working with redox equations easier, we will omit all physical state symbols. However, remember that they should be there. An unbalanced redox equation looks like this: MnO H 2 SO 3 + H +  Mn +2 + HSO H 2 O Study how this equation is balanced using the half-reaction method on the next slide. It is important that you understand what happens in each step.

25 MnO H 2 SO 3 + H +  Mn +2 + HSO H 2 O Oxidizing agent Reducing agent +7 to +3 gain 5 e - Reducing half reaction: Step 2 MnO e -  Mn to +7 lose 3 e - Oxidation half reaction: Step 3 H 2 SO 3  HSO e - Reducing half reaction: Step 5a 3MnO H 2 SO e -  3Mn HSO e - Reducing half reaction: Step 4 3 (MnO e -  Mn +2 ) = 3MnO e-  3Mn +2 Oxidation half reaction: Step 4 5 (H 2 SO 3  HSO e - ) = 5H 2 SO 3  5HSO e- Reducing half reaction: Step 5 3MnO H 2 SO e -  3Mn HSO e - Reducing half reaction: Step 5b 3MnO H 2 SO 3 -  3Mn HSO 4 -

26 3MnO H 2 SO 3 + H +  3Mn HSO H 2 O Notice, some of the above is balance but the H and O are not. Balance the equation the way that you know, by counting the number and kind of atoms. 3MnO H 2 SO 3 + 9H +  3Mn HSO H 2 O

27 Practice Problems: Balance these Redox equations using the half-reaction method. The balanced equation and the detailed half-reaction solution are provided for each equation. Use these helps only after doing your best to balance the equation without help. It is to your advantage to work all these problems to gain experience with different situations that arise when working with redox equations. 1.HNO 3 + H 3 PO 3  NO + H 3 PO 4 + H 2 O 2.Cr 2 O H + + I -  Cr +3 + I 2 + H 2 O 3.As 2 O 3 + H + + NO H 2 O  H 3 AsO 4 + NO 4.CuS + NO 3 -  Cu +2 + NO 2 + S 5.H 2 SeO 3 + Br -  Se + Br 2 6.Fe +2 + Cr 2 O 7 -2  Fe +3 + Cr +3 7.HS - + IO 3 -  I - + S 8.CrO I -  Cr +3 + I 2 9.IO 4- + I -  I 2 10.MnO H 2 O 2  Mn +2 + O 2 11.H 3 AsO 4 + Zn  AsH 3 + Zn +2 Click below to see answer

28 oxidation number answers: 1.Sb N P S Cr Cl B Si I N – 3 11.Mn Br Cl Cr Se + 4 Back

29 Balanced Redox Equations Answers: 1.2HNO 3 + 3H 3 PO 3  2NO + 3H 3 PO 4 + H 2 O 2.14H + + Cr 2 O I -  2Cr I 2 + 7H 2 O 3.3As 2 O 3 + 4H + + 4NO H 2 O  6H 3 AsO 4 + 4NO 4.CuS + 2NO H +  Cu NO 2 + S + 2H 2 O 5.H 2 SeO 3 + 4Br - + 4H +  Se + 2Br 2 + 3H 2 O 6.6Fe +2 + Cr 2 O H +  6Fe Cr H 2 O 7.3HS - + IO H +  I - + 3S + 3H 2 O 8.16H + + 2CrO I -  2Cr I 2 + 8H 2 O 9.8H + + IO I -  4I 2 + 4H 2 O 10.6H + + 2MnO H 2 O 2  2Mn O 2 + 8H 2 O 11.8H + + H 3 AsO 4 + 4Zn  AsH 3 + 4Zn H 2 O Click below to go back to problems.

30 Charting Reducing Agents and Oxidizing Agents Practice Problems 1.Mg + O 2  MgO 2.Cl 2 + I -  Cl - + I 2 3.MnO C 2 O 4 -2  Mn +2 + CO to -2, lose 2 e - 0 to -2, gain 2 e to -1, gain 1 e - -1 to 0, lose 1 e to +2, gain 5 e - +3 to +4, lose 1 e -

31 Charting Reducing Agents and Oxidizing Agents Practice Problems 1.Mg + O 2  MgO 2.Cl 2 + I -  Cl - + I 2 3.MnO C 2 O 4 -2  Mn +2 + CO 2 4.Cr + NO 2 -  CrO N 2 O BrO MnO 2  Br - + MnO Fe +2 + MnO 4 -  Mn +2 + Fe +3 7.Cr + Sn +4  Cr +3 + Sn +2 8.NO S  NO 2 + H 2 SO 4 9.IO I -  I 2 10.NO 2 + ClO - NO Cl -

32 Charting Reducing Agents and Oxidizing Agents Practice Problems 1.Mg + O 2  MgO 2.Cl 2 + I -  Cl - + I 2 3.MnO C 2 O 4 -2  Mn +2 + CO 2 4.Cr + NO 2 -  CrO N 2 O BrO MnO 2  Br - + MnO Fe +2 + MnO 4 -  Mn +2 + Fe +3 7.Cr + Sn +4  Cr +3 + Sn +2 8.NO S  NO 2 + H 2 SO 4 9.IO I -  I 2 10.NO 2 + ClO - NO Cl -

33 Charting Reducing Agents and Oxidizing Agents Practice Problems 1.Mg + O 2  MgO 2.Cl 2 + I -  Cl - + I 2 3.MnO C 2 O 4 -2  Mn +2 + CO 2 4.Cr + NO 2 -  CrO N 2 O BrO MnO 2  Br - + MnO Fe +2 + MnO 4 -  Mn +2 + Fe +3 7.Cr + Sn +4  Cr +3 + Sn +2 8.NO S  NO 2 + H 2 SO 4 9.IO I -  I 2 10.NO 2 + ClO - NO Cl -

34 Charting Reducing Agents and Oxidizing Agents Practice Problems 1.Mg + O 2  MgO 2.Cl 2 + I -  Cl - + I 2 3.MnO C 2 O 4 -2  Mn +2 + CO 2 4.Cr + NO 2 -  CrO N 2 O BrO MnO 2  Br - + MnO Fe +2 + MnO 4 -  Mn +2 + Fe +3 7.Cr + Sn +4  Cr +3 + Sn +2 8.NO S  NO 2 + H 2 SO 4 9.IO I -  I 2 10.NO 2 + ClO - NO Cl -

35 Part 3 Redox Agents Answers 1)ox: H 2 SO 4 red: HI 2)ox: Au 2 S 3 red: H 2 3)ox: HCl red: Zn


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