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Electrochemistry Lesson 5 Balancing Redox Reactions.

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Presentation on theme: "Electrochemistry Lesson 5 Balancing Redox Reactions."— Presentation transcript:

1 Electrochemistry Lesson 5 Balancing Redox Reactions

2 Balancing Redox Full Reactions in Acid Solution MnO 4 - + C 2 O 4 2- →MnO 2 + CO 3 2- MnO 4 - → MnO 2 C 2 O 4 2- → CO 3 2- 2MnO 4 - + 3C 2 O 4 2- + 6H 2 O + 8H + → 2MnO 2 + 6CO 3 2- + 12H + + 4H 2 O 2MnO 4 - + 3C 2 O 4 2- + 2H 2 O → 2MnO 2 + 6CO 3 2- + 4H + 1.Separate into half reactions2.Balance each half reaction 3.Add the two half reactions4.Simplify 5.Check + 4H + + 3e - + 2H 2 O2(2( 3(3( + 2e - + 4H + ) 2 ) -8 24

3 Balancing Redox Full Reactions in Basic Solution 2MnO 4 - + 3C 2 O 4 2- + 2H 2 O → 2MnO 2 + 6CO 3 2- + 4H + 1.Balance in acid solution2.Add OH - to neutralize the H + 4OH - 4H 2 O2

4 Balancing Redox Full Reactions in Basic Solution 2MnO 4 - + 3C 2 O 4 2- + → 2MnO 2 + 6CO 3 2- +-12 4OH - 2H 2 O

5 Balancing Redox Full Reactions in Acid Solution Fe+O 2 → H 2 O+Fe(OH) 3 Fe +→+ Fe(OH) 3 O 2 → H 2 0 4Fe + 12H 2 O + 3O 2 +12H + → 12H + + 4Fe(OH) 3 + 6H 2 O 4Fe + 6H 2 O + 3O 2 → 4Fe(OH) 3 3H 2 O3H + + 3e - ) + H 2 O)+4H + 4e - + 3( 4(4( 00 6

6 Balance the redox reaction As→H 2 AsO 4 - +AsH 3 (alkaline) You must separate into two half reactions!This is called a Disproportionation reaction ( same reactant gets reduced and ozidized) 3As + 12H 2 O + 15H + + 5As → 3H 2 AsO 4 - + 18H + + 5AsH 3 8As + 12H 2 O → 3H 2 AsO 4 - + 3H + + 5AsH 3 3OH - 3OH - + 8As + 9H 2 O → 3H 2 AsO 4 - + 5AsH 3-3 As H 2 AsO 4 - AsH 3 + 4H 2 O+6H + + 5e - 3H + ++ 3e - 3(3( ) 5(5( ) → → 3

7 Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions!

8 Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions! IPO 4 - → + PO 4 3-

9 Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions!(another Disproportionation) IPO 4 - → + PO 4 3-

10 Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions! IPO 4 - → IO 3 - + PO 4 3- IPO 4 - → I 2 + PO 4 3-

11 Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions! IPO 4 - → IO 3 - + PO 4 3- 2IPO 4 - → I 2 + 2PO 4 3-

12 Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions! IPO 4 - + 3H 2 O → IO 3 - + PO 4 3- + 6H + 2IPO 4 - → I 2 + 2PO 4 3-

13 Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions! IPO 4 - + 3H 2 O → IO 3 - + PO 4 3- + 6H + + 3e - 2IPO 4 - + 4e - → I 2 + 2PO 4 3-

14 Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions! 4(IPO 4 - + 3H 2 O → IO 3 - + PO 4 3- + 6H + + 3e - ) 3(2IPO 4 - + 4e - → I 2 + 2PO 4 3- ) Continue…

15 Continue: Balance the redox reaction IPO 4 - →IO 3 - +I 2 + PO 4 3- (alkaline) You must separate into two half reactions! 4(IPO 4 - + 3H 2 O → IO 3 - + PO 4 3- + 6H + + 3e - ) 3(2IPO 4 - + 4e - → I 2 + 2PO 4 3- ) 12H 2 O + 10IPO 4 - → 3I 2 + 10PO 4 3- + 24H + + 4IO 3 - 24OH - 24OH - + 10IPO 4 - → 3I 2 + 10PO 4 3- + 4IO 3 - + 12H 2 O-34

16 Homework Page207 Exercise#24 do 5 acidic redox equations and 5 basic redox equations.


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