2. The final exam will cover material from all the lectures and all the homework's
final exam
First exam & Second exam
Quiz
HomeWorks & Attendance
Poster
exams and HomeWorks
Assignment
Percent of Final
Grade
First exam & Second
exam
(15 +15)/2
Exam ====
Quiz (2-3)
7
HomeWorks &
Attendance
3
Poster --------
Final Exam 25
17. Free radicals are molecules produced by your body when
it breaks down food, or when exposed to environmental
toxins, tobacco smoke, radiation, unhealthy food and
stress to name a few.
This is due to a chemical process called oxidation, in
which the molecules in the apple give up electrons to the
oxygen in the air.
18. The truth is that there's almost no common disease that is
not associated with oxidative stress. This include all of the
major degenerative diseases we have today:
• cancer
• heart disease and stroke
• diabetes
• Parkinson's
• Alzheimer's
• macular degeneration
• cataracts
• rheumatoid arthritis
• arthritis
• asthma and emphysema
• kidney and liver disorders
• psoriasis
• gum disease
• and many more
29. Exercises - Give the oxidation number for the following atoms:
O2 O = 0 F2 F = 0
S8 S = 0 Cl2 Cl = 0
N2 N = 0 Al Al = 0
Co2+ Co = +2 Mn2+ Mn = +2
Cl ‾ Cl = -1 Cr3+ Cr = +3
Oxidation Number Exercise - answers
32. Exercises - Give the oxidation number for the
following atoms:
NaF Na = +1 IF3 I = +3
SF4 S = +4
PF3 P = +3 SF6 S = +6
PF5 P = +5
OF2 O = +2 NF3 N = +3
Oxidation Number Exercise – answers
33. Exercises –
Give the oxidation number for the following atoms:
Na2O Na = ____ Na2O2 O = ____
KO2 K = _____
NaOH Na =_____ LiH H = _____
CaMgO2 O = ____
MgH2 H = ____ MgF2 Mg = _______
34. Balance all atoms other than O and H in each of the half-reactions.
• والهيدروجين االكسجين غير الذرات عدد اوزن
MnO−4 → Mn2+
2I− → I2
Balance O by adding H2O molecules to the appropriate side.
المناسب الجانب الى الماء جزيئات باضافة االكسجين ذرات عدد اوزن
MnO−4 → Mn2+ + 4H2O
2I− → I2
Balance H by adding H+ ions to the appropriate side.
المناسب للجانب الهيدروجين ايونات باضافة الهيدروجين ذرات عدد اوزن
MnO−4 + 8H+ → Mn2+ + 4H2O
2I− → I2
•
Balance charge by adding electrons to the appropriate side.
•
The first half-reaction has 7+ on the left and 2+ on the right.
باضافة المناسب الجانب الى الكترونات باضافة الشحنة معادلة
5
و االيسر الجانب الى الكترونات
2
االيمن للجانب الكترون
We add 5e− to the left to balance charge.
•
MnO−4 + 8H+ + 5e− → Mn2+ + 4H2O
•
The second half-reaction has 2- on the left and zero on the right.
We add 2e− to the right.
•
2I− → I2 + 2e⁻
35. Multiply the two half-reactions by a number that gives the lowest common
multiple of the electrons transferred in each half-reaction.
In this case, the lowest common multiple of 2 and 5 is 10, so we multiply the
top equation by 2 and the bottom equation by 5.
• 2× [ MnO−4+ 8H+ +5e− → Mn2+ + 4H2O ]
5× [ 2I⁻→ I2 + 2e− ]
•
Add the two half-reactions, cancelling species that appear on each side
of the overall equation.
•
2MnO−4 + 10I− + 16H+ → 2Mn2+ + 5I2 + 8H2O
36.
37.
38.
39. Balance all atoms other than O and H in each of the half-reactions.
• والهيدروجين االكسجين غير الذرات عدد اوزن
MnO−4 → Mn2+
2I− → I2
Balance O by adding H2O molecules to the appropriate side.
المناسب الجانب الى الماء جزيئات باضافة االكسجين ذرات عدد اوزن
MnO−4 → Mn2+ + 4H2O
2I− → I2
Balance H by adding H+ ions to the appropriate side.
المناسب للجانب الهيدروجين ايونات باضافة الهيدروجين ذرات عدد اوزن
MnO−4 + 8H+ → Mn2+ + 4H2O
2I− → I2
•
Balance charge by adding electrons to the appropriate side.
•
The first half-reaction has 7+ on the left and 2+ on the right.
باضافة المناسب الجانب الى الكترونات باضافة الشحنة معادلة
5
و االيسر الجانب الى الكترونات
2
االيمن للجانب الكترون
We add 5e− to the left to balance charge.
•
MnO−4 + 8H+ + 5e− → Mn2+ + 4H2O
•
The second half-reaction has 2- on the left and zero on the right.
We add 2e− to the right.
•
2I− → I2 + 2e⁻
40. Multiply the two half-reactions by a number that gives the lowest common
multiple of the electrons transferred in each half-reaction.
In this case, the lowest common multiple of 2 and 5 is 10, so we multiply the
top equation by 2 and the bottom equation by 5.
• 2× [ MnO−4+ 8H+ +5e− → Mn2+ + 4H2O ]
5× [ 2I⁻→ I2 + 2e− ]
•
Add the two half-reactions, cancelling species that appear on each side
of the overall equation.
•
2MnO−4 + 10I− + 16H+ → 2Mn2+ + 5I2 + 8H2O
41.
42.
43.
44. The species at the top left have the greatest "potential" to be reduced, so they are the strongest oxidizing agents.
The strongest oxidizing agent in the list is F2, followed by H2O2, and so on down to the weakest oxidizing agent, Li+