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Chapter 4 4.10 – Balancing Redox Reactions: The Half-Reaction Method.

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Presentation on theme: "Chapter 4 4.10 – Balancing Redox Reactions: The Half-Reaction Method."— Presentation transcript:

1 Chapter 4 4.10 – Balancing Redox Reactions: The Half-Reaction Method

2 Oxidation numbers Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 4/2 1.An atom in its elemental state has an oxidation number of 0. 2.A monatomic ion has an oxidation number identical to its charge 3.An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion 1.Hydrogen can be either +1 or – 2.Oxygen usually has an oxidation number of -2 4.Halogens usually have an oxidation number of -1 5.The sum of the oxidation numbers is 0 for a neutral compound and is equal to the net charge for a polyatomic ion Rules for Assigning Oxidation Numbers Oxidation Number (State): A value which indicates whether an atom is neutral, electron-rich, or electron-poor.

3 Writing the half Oxidation and Reduction Reactions Write the separate half oxidation and reduction reactions for the following equation. 2Cs(s) + F 2 (g) 2CsF(s) 3

4 Balancing Redox Reaction using half equation Assign oxidation numbers to each atom from the given unbalanced equation Split the equation into half-reaction Complete and balance each half reaction Balance all atoms except O and H Balance oxygen atoms by adding H 2 O to one side of the equation Balance hydrogen atoms by adding H + ions to one side of the equation Balance the number of electrons being transferred Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 4/4

5 Balancing Redox Reactions: The Half-Reaction Method Combine the half-reaction to obtained the final balanced equation The electrons must be cancelled Simplify the equation by reducing coefficients and canceling repeated species After you’re done, double check your balanced equation Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 4/5

6 Additional steps for balancing equations in basic solution Add the desired number of – OH react with H + ions in the reaction; add to both side of the reaction. This is a neutralization step Simplify the equation by noting that H + combines with – OH to give H 2 O Cancel any repeating H 2 O and – OH ions and reduce reaction to the lowest coefficients Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 4/6

7 Example Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 4/7 Balance the following net ionic equation in acidic solution: Cr 3+ (aq) + I O 3 1- (aq) I 1- (aq) + Cr 2 O 7 2- (aq)

8 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 4/8 Write the two unbalanced half-reactions. I O 3 1- (aq) I 1- (aq) Cr 3+ (aq)Cr 2 O 7 2- (aq) Reduction Oxidation

9 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 4/9 Balance both half-reactions for all atoms except O and H. I O 3 1- (aq) I 1- (aq) 2Cr 3+ (aq)Cr 2 O 7 2- (aq)

10 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/10 Balance each half-reaction for O by adding H 2 O, and then balance for H by adding H 1+. I O 3 1- (aq) + 6H 1+ (aq) 3H 2 O(l) + I 1- (aq) 2Cr 3+ (aq) + 7H 2 O(l)14H 1+ (aq) + Cr 2 O 7 2- (aq)

11 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/11 Balance each half-reaction for charge by adding electrons to the side with greater positive charge. I O 3 1- (aq) + 6H 1+ (aq) + 6e- + 3H 2 O(l) + I 1- (aq) 2Cr 3+ (aq) + 7H 2 O(l)6e- + 14H 1+ (aq) + Cr 2 O 7 2- (aq) oxidation: reduction: Electrons were place on the wrong side of equation on the previous note. This is the correction equations

12 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/12 Multiply each half-reaction by a factor to make the electron count the same in both half-reactions. I O 3 1- (aq) + 6H 1+ (aq) + 6e- 3H 2 O(l) + I 1- (aq) 2Cr 3+ (aq) + 7H 2 O(l)6e- + 14H 1+ (aq) + Cr 2 O 7 2- (aq) oxidation: reduction: Electrons were place on the wrong side of equation on the previous note. This is the correction equations

13 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/13 Add the two balanced half-reactions together and cancel species that appear on both sides of the equation. I O 3 1- (aq) + 6H 1+ (aq) + 6e- + 3H 2 O(l) + I 1- (aq) 2Cr 3+ (aq) + 7H 2 O(l)6 e- + 14H 1+ (aq) + Cr 2 O 7 2- (aq) oxidation: reduction: I O 3 1- (aq) + 2Cr 3+ (aq) + 4H 2 O(l) 8H 1+ (aq) + I 1- (aq) + Cr 2 O 7 2- (aq) Acidic Condition

14 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/14 MnO 2 (s) + BrO 3 1- (aq)MnO 4 1- (aq) + Br 1- (aq) Balance the following net ionic equation in basic solution:

15 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/15 Write the two unbalanced half-reactions. Br O 3 1- (aq) Br 1- (aq) MnO 2 (s)MnO 4 1- (aq) Oxidation Reduction

16 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/16 Balance both half-reactions for all atoms except O and H. Br O 3 1- (aq) Br 1- (aq) MnO 2 (s)MnO 4 1- (aq) Reduction Oxidation

17 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/17 Br O 3 1- (aq) + 6H 1+ (aq)3H 2 O(l) + Br 1- (aq) MnO 2 (s) + 2H 2 O(l)4H 1+ (aq) + MnO 4 1- (aq) Balance each half-reaction for O by adding H 2 O, and then balance for H by adding H 1+. Oxidation Reduction

18 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/18 Br O 3 1- (aq) + 6H 1+ (aq) + 6e - 3H 2 O(l) + Br 1- (aq) MnO 2 (s) + 2H 2 O(l)3e - + 4H 1+ (aq) + MnO 4 1- (aq) Balance each half-reaction for charge by adding electrons to the side with greater positive charge. oxidation Reduction

19 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/19 Multiply each half-reaction by a factor to make the electron count the same in both half-reactions. 2 Br O 3 1- (aq) + 6H 1+ (aq) + 6e - 3H 2 O(l) + Br 1- (aq) MnO 2 (s) + 2H 2 O(l)3e - + 4H 1+ (aq) + MnO 4 1- (aq)

20 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/20 2MnO 2 (s) + H 2 O(l) + BrO 3 1- (aq) 2H 1+ (aq) + 2MnO 4 1- (aq) + Br 1- (aq) Br O 3 1- (aq) + 6H 1+ (aq) + 6e - 3H 2 O(l) + Br 1- (aq) 2MnO 2 (s) + 4H 2 O(l)6e - + 8H 1+ (aq) + 2MnO 4 1- (aq) Add the two balanced half-reactions together and cancel species that appear on both sides of the equation. Acidic

21 Balancing Redox Reactions: The Half-Reaction Method Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 4/21 Since the reaction occurs in a basic solution, “neutralize” the excess H 1+ by adding OH 1- and cancel any water (if possible) 2MnO 2 (s) + H 2 O(l) + BrO 3 1- (aq) + 2OH 1- (aq) 2OH 1- (aq) + 2H 1+ (aq) + 2MnO 4 1- (aq) + Br 1- (aq) 2H 2 O 2MnO 2 (s) + BrO 3 1- (aq) + 2OH 1- (aq) H 2 O(l) + 2MnO 4 1- (aq) + Br 1- (aq) Basic

22 Example Balance the following net-ionic equation by the half-reaction method. Cu(s) + NO 3 - (aq)  Cu 2+ (aq) + NO 2 (g) Acidic condition I - (aq) + MnO 4 - (aq)  I 2 (aq) + MnO 2 (g) Basic Condition Fe(OH) 2 (s) + O 2 (g)  Fe(OH) 3 (s) Basic condition


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