Homework

21-1 Do REVIEW & PRACTICE, page
621 HEATH text, questions #1-4.
 Optional: Read HEBDEN WORKBOOK
Section 5-1 "Introduction". Do Exercises p.192
#1-2.

21-2. Do REVIEW & PRACTICE, page
627 HEATH text, questions #1-4.

3. Optional: Read HEBDEN WORKBOOK
Section 5-2 "Oxidation Numbers". Do Exercise
p.194 #3-5.
4. Do HANDOUT "Section 21•1/21•2
Provincial Exam Questions".

21-1 Prov. Exam Answers
1.C
6. C
 2. D
7. C
 3. B
8. C
 4. A
9. B
 5. A
10. A
 Written: 1) oxidation = loss of electrons
reduction =gain of lost electrons
Occurs simultaneously

21-2 Prov. Exam Answers
1. B
 2. B
 3. B
 4. D
 5. C
 6. D
 7. B
 8. D

9. C
10. A
11. C
12. A
13. D
14. A
15. C
16. A
17. A
18. B
19. A
20. C
21. B
22. B
23. D
24. C
25. B
26. C
27. C
28. B
29. A
30. A
21-2 Prov. Exam Answers
25. B
 26. C
 27. C
 28. B
 29. A
 30. A

Any Questions?
21-1
 21-2

8,5
2,5, 7,9,20
21-3 Balancing Redox Reactions –
Using Half Reactions


a) Half Reactions
Sample 1: Co+2 ssd Co
“unbalanced”
Oxidation or reduction?


Co+2 + 2e– ssd Co “balanced”
Reduction!!

Sample 2
NO ssd N2O
“unbalanced”
Oxidation or reduction?


2NO + 2H+ + 2e– ssd N2O + H2O “balanced”
Reduction!!


i) Half reactions must be balanced for
charge and mass

ii) MAJOR OH– (memory aid)
1st balance major atoms (other than O or H),
 2nd balance oxygen using water,
 3rd balance hydrogen using H+
 4th balance charge (-) using e–


iii) Balance NO ssd N2O
(acidic)


1st Balance for atoms other than O and H

2NO ssd N2O


2nd Add water to balance O

2NO ssd N2O + H2O


3rd Add H+ to balance H

2NO + 2H+ ssd N2O + H2O

4th Add e– to balance charge



2NO + 2H+ + 2e– ssd N2O + H2O
Check: 2N’s
balances
2”O”’s 2H’s and charge
iv) Balance MnO4– ssd Mn+2 (acidic)

1st Balance for atoms other than O and H


ssd Mn+2 (Mn already balanced)
2nd Add water to balance O


MnO4–
MnO4–
ssd Mn+2 + 4H2O
3rd Add H+ to balance H

MnO4– + 8H+ ssd Mn+2 + 4H2O

4th Add e– to balance charge

MnO4– + 8H+ + 5e – ssd Mn+2 + 4H2O

Check: 1 Mn , 4 “O” ‘s, 8 H’s, and charge of +2
on each side.
v) Acidic vs. Basic solutions




Acidic – proceed as above
Basic – Has some Bonus steps:
3a) AFTER balancing H+, for each H+ present,
add an equal number of OH– ions to BOTH
sides;
3b) combine H+ and OH– present to form H2O;



3c) cancel out water on both sides (and then
proceed to step 4– electrons)
vi) Balance Cl2 sd ClO3–
(basic)
 1st Balance for atoms other than O and H
Cl2

ssd
ClO
23–
2nd Add water to balance O
Cl2 + 6H2O ssd
2ClO3–


3rd Add H+ to balance H
Cl2 + 6H2O ssd
2ClO3–
+ 12H+

3 A) Bonus: Add OH–, equal to H+ on both sides


ssd 2ClO3– + 12H+
B) Combine H+ and OH–


Cl2 + 6H2O + 12 OH–
Cl2 + 6H2O + 12OH– ssd 2ClO3–
6
+ 12H2O
C) Cancel out H2O on both sides

Cl2 + 12OH– ssd 2ClO3–
+ 6H2O


4th step Add e– to balance charge

Cl2 + 12OH– ssd 2ClO3– + 6H2O
+ 10e–
+ 12 OH–
HW 21-3 half reactions
 Do written first #2,3,8 to practice
 MC #1,2,4,7,11(tough)

b) Full Redox Reactions
i) Procedure:

 Separate redox into its two half
reactions

 Balance each half reaction as above

 Make e– the same for both half
reactions (electrons gained = electrons lost)

 Add the two half reactions together
(cancel out species common to both sides)

i)

Balance in acidic soln
Cu+2 + Al(s) ssd Cu(s) + Al+3
1st Separate half reactions
 Cu+2
 Al(s)



ssd
ssd
Cu(s)
Al+3
2nd Balance half reactions
Cu+2 + 2e– ssd Cu (s)
Al(s)
ssd Al+3 + 3e–
3rd Make e– the same




3( Cu+2 + 2e– ssd Cu (s) )
3 Cu+2 + 6e– ssd 3Cu (s)
2(Al(s) ssd Al+3 + 3e– )
2Al(s) ssd 2Al+3 + 6e–
4th Add half reactions together
3Cu+2 + 6e– ssd 3Cu (s)

2Al(s) ssd 2Al+3 + 6e–
 3Cu+2 + 2Al(s) ssd 3Cu(s) + 2Al+3

iii)
Cr2O7–2
ssd Cr+3

Cr2O7–2
ssd 2 Cr+3

Cr2O7–2
ssd 2 Cr+3 + 7H2O
 Cr2O7–2 + 14H+
ssd 2 Cr+3 + 7H2O
 Cr2O7–2 + 14H+ +6e- ssd 2 Cr+3 + 7H2O

Fe+2 ssd Fe+3
 Fe+2 ssd Fe+3 + 1e
Combine
 Cr2O7–2 + 14H+ +6e- ssd 2 Cr+3 + 7H2O
 6Fe+2 ssd 6Fe+3 + 6e Cr2O7–2 + 14H++ 6Fe+2 ssd 2 Cr+3 + 7H2O

+ 6Fe+3
Check charge: -2 + 14 + 12 = +24 LHS
+6+18= +24 RHS yay

MnO4–
 MnO4–
 MnO4– +8H+
 MnO4– +8H+ +5e-
ssd Mn+2
ssd Mn+2 +4H2O
ssd Mn+2 +4H2O
ssd Mn+2 +4H2O
SO2
 SO2 +2H2O
 SO2 +2H2O
 SO2 +2H2O
ssd
ssd
ssd
ssd

SO4–2
SO4–2
SO4–2 + 4H+
SO4–2 + 4H+ + 2e-
2(MnO4–
+8H+ +5e- ssd Mn+2 +4H2O)
5(SO2 +2H2O
sd SO4–2 + 4H+ + 2e- )
2MnO4– +16H+ +10e- ssd 2Mn+2 +8H2O)
 5SO2 +10H2O ssd 5SO4–2 + 20 H+ + 10e-)

2(MnO4–
+8H+ +5e- ssd Mn+2 +4H2O)
5(SO2 +2H2O
sd SO4–2 + 4H+ + 2e- )
2MnO4– +16H+ +10e- ssd 2Mn+2 +8H2O)
2
4
 5SO2 +10H2O ssd 5SO4–2 + 20H+ + 10e-)


2MnO4–+5SO2 +2H2Osd2Mn+2+ 5SO4–2+ 4H+
 Check Charge: -2 + 0 = -2 LHS
+4 -10 +4 = -2 RHS YAY!
What if the last reaction was in
Base?

JUST FOLLOW THIS, DON’T WRITE
IT!@
MnO4–
ssd Mn+2
 MnO4–
ssd Mn+2 +4H2O
 MnO4– +8H+
ssd Mn+2 +4H2O
 MnO4– +8H+ + 8OH- ssd Mn+2 +4H2O

+ 8OH MnO4– +8H2O ssd Mn+2 +4H2O+ 8OH MnO4– +4H2O ssd Mn+2 +8OH MnO4– +4H2O + 5e- ssd Mn+2 +8OH-







SO2
SO2
SO2
SO2
SO2
SO2
SO2
ssd SO4–2
+2H2O
ssd SO4–2
+2H2O
ssd SO4–2 + 4H+
+2H2O+ 4OH- sd SO4–2 + 4H+ + 4OH+2H2O+ 4OH- sd SO4–2 + 4H2O
+ 4OH- sd SO4–2 + 2H2O
+ 4OH- sd SO4–2 + 2H2O+ 2e-

NOW LCM of 10
MnO4– +4H2O + 5e- ssd Mn+2 +8OH SO2 + 4OHsd SO4–2 + 2H2O+ 2e
And multiply and cancel and add up etc…
LOTS OF WORK… or you could be smart
and…
Do the equation in acid as
before……
2MnO4– +16H+ +10e- ssd 2Mn+2 +8H2O
4+
2
–2
 5SO2 +10H2O ssd 5SO4 + 20H + 10e 2MnO4–+5SO2 +2H2Osd2Mn+2+ 5SO4–2+ 4H+

NOW change to base...!
 2MnO4–+5SO2 +2H2Osd2Mn+2+ 5SO4–2+ 4H+

+ 4OH
+ 4OH-
2MnO4– +5SO2 +2H2Od2Mn+2+ 5SO4–2+ 4H2O
+ 4OH-

2MnO4– +5SO2 +4OH- d2Mn+2+ 5SO4–2+ 2H2O
Check Charge: -2 -4
= -6 LHS
+4 -10 = -6 RHS YAY!

FOR FULL REDOX REACTIONS IN
BASE: Change to base AFTER balancing
everything in acid!