MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 19: OXIDATION AND REDUCTION
WORKSHEET SOLUTIONS
1. __________ reactions always involve the transfer of electrons.
(A) Double displacement (B) Single displacement (C) Decomposition
(D) Oxidation reduction (E) Synthesis
2. An atom that loses electrons is:
(A) oxidized (B) reduced (C) transmuted
(D) promoted
(E) in ground state
3. An atom that gains electrons is:
(A) oxidized (B) reduced (C) transmuted
(D) promoted
(E) in ground state
4. __________ help determine which element is oxidized and which is reduced:
(A) Skeletal equations (B) Balanced equations (C) Oxidation numbers
(D) Chemical formulas (E) Molecular formulas
5. If the oxidation number increases, then the element is:
(A) reduced (B) transmuted (C) promoted (D) in ground state
(E) oxidized
6. If the oxidation number decreases, then the element is:
(A) reduced (B) transmuted (C) promoted (D) in ground state
(E) oxidized
7. __________ is an atom’s ability to attract electrons.
(A) Oxidation (B) Reduction (C) Ionization energy
(D) Atomic size
(E) Electronegativity
8. Halogens are assigned a _____ oxidation number.
(A) - 2 (B) - 1 (C) 0 (D) + 1 (E) + 2
9. Draw a Lewis Dot Structure (LDS) diagram for the following:
9a. Hydrogen
9b. Nitrogen
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VAN NUYS HIGH SCHOOL
9c. Ammonia (NH3)
10. Assign oxidation numbers to hydrogen and nitrogen based on the LDS number for ammonia.
10A.
The nitrogen atom shares a pair of electrons with each of the three hydrogen atoms. Nitrogen is the
more electronegative element because it is farther to the right on the periodic table than hydrogen. This
means nitrogen has 3 “extra” electrons that give it a – 3 oxidation number. The hydrogen atoms each
lost one electron so they have a + 1 oxidation number.
Questions 11 - 13. Assign oxidation numbers to the following common polyatomic ions:
11. Carbonate (CO3-2)
11A.
The three oxygen atoms in CO3-2 are group 16 elements, so they have a – 2 oxidation number. Since
there are three oxygen atoms, they have a combined oxidation number of – 6 (3 x – 2 = - 6). The net
charge on the carbonate ion is – 2, so the carbon must have a + 4 oxidation number.
12. Phosphate (PO4-3)
12A.
The four oxygen atoms in PO4-3 each have a – 2 oxidation number, for a combined – 8 oxidation
number. Since the overall charge on the phosphate ion is – 3, phosphorus must have a + 5 oxidation
number (- 8 + 5 = - 3).
13. Ammonium (NH4+)
13A.
The four hydrogen atoms in NH4+ each have a + 1 oxidation number. Since the overall charge on
ammonium is + 1, the nitrogen atom must have a – 3 oxidation number ( + 4 – 3 = + 1).
14. What is the oxidation number of nitrogen in magnesium nitride (Mg3N2)?
14A.
Since magnesium is an alkaline metal, it has a + 2 oxidation number. This means the Mg3 part of this
species has a + 6 charge. In order to balance out this + 6 charge, both nitrogen atoms must have a – 3
oxidation number ( - 3 x 2 = - 6).
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
15. Find the oxidation numbers for iodine, oxygen, and carbon in the following redox reaction:
I2O5 + 5 CO  I2 + 5 CO2
15A.
For I2O5: Because oxygen is a group 16 element, it has a – 2 oxidation number. Since there are 5
oxygen atoms, the O5 part of this species has a combined oxidation number of – 10. In order for the two
iodine atoms to balance out the five oxygen atoms, each iodine atom must have a + 5 oxidation
number.
For CO: Oxygen has a – 2 oxidation number. Therefore, each carbon must have a + 2 oxidation
number.
For I2: I2 is a pure element, so each iodine atom has a zero oxidation number.
For CO2: Oxygen has a – 2 oxidation number. Since there are two oxygen atoms in CO2, the O2 part
of this species adds up to a – 4 oxidation number. As a result, the carbon atoms in CO2 have a + 4
charge.
Final oxidation numbers
I2(+5)O5(-2) + 5 C(+2)O(-2)  I2(0) + 5 C(+4)O2(-2)
16. For the above reaction, identify the element that is oxidized and the element that is reduced.
16A.
Carbon is oxidized and iodine is reduced.
17. Predict the oxidation numbers for the carbonate ion, CO3-2.
17A.
Each oxygen has a – 2 oxidation number. Since there are 3 oxygen atoms in CO3-2, they must add up
to a - 6 oxidation number. It might be tempting to assign the carbon atom a +6 oxidation number, but
the carbonate ion is not neutral because it has a -2 net charge. Therefore, the carbon must have a + 4
oxidation number (- 6 + 4 = - 2).
18. Zinc metal and chlorine gas react to form zinc chloride:
2 Zn(s) + Cl2(g)  2 ZnCl2(s)
18a. Provide the before and after oxidation numbers for zinc and chloride.
A.
For Zn: Elemental zinc has a + 0 oxidation number.
For Cl2: Elemental chlorine has a + 0 oxidation number.
For ZnCl2: Since chlorine is a halogen, each chlorine atom has a – 1 oxidation number for a total of
– 2 charge. The zinc atom must have a + 2 charge to balance out the chlorine atoms.
Final oxidation numbers
Zn(0) + Cl2(0)  2 Zn(+2)Cl2(-1)
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VAN NUYS HIGH SCHOOL
18b. Identify the element that is reduced.
A. Chlorine
18c. Identify the element that is oxidized.
A. Zinc
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 19: OXIDATION AND REDUCTION
QUIZ SOLUTIONS
1. An atom that loses electrons is:
1A. oxidized
2. An atom that gains electrons is:
2A. reduced
3. __________ help determine which element is oxidized and which is reduced:
3A. Oxidation numbers
4. If the oxidation number increases, then the element is:
4A. oxidized
5. If the oxidation number decreases, then the element is:
5A. reduced
6. __________ is an atom’s ability to attract electrons.
6A. Electronegativity
7. Hydrogen is assigned a _____ oxidation number.
7A. + 1
8. Alkaline metals are assigned a _____ oxidation number.
8A. + 2
9. Draw a Lewis Dot Structure (LDS) diagram for the following:
9a. Carbon
9b. Fluorine
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9c. Carbon tetraflouride (CF4)
10. Assign oxidation numbers to carbon and fluorine based on the LDS diagram for carbon
tetrafluoride.
10A.
The four flourine atoms are halogens and have a -1 oxidation number. Flourine is the more
electronegative element because it is a halogen. This means each of the four fluorine atoms has an
“extra” electron that gives a total – 4 oxidation number. The carbon atom must have a +4 oxidation
number to balance out the fluorine oxidation numbers.
Questions 11 - 13. Assign oxidation numbers to the following common polyatomic ions:
11. Sulfite (SO3-2)
11A.
The three oxygen atoms in SO3-2 are group 16 elements, so they have a – 2 oxidation number. Since
there are three oxygen atoms, they have a combined oxidation number of – 6 (3 x – 2 = - 6). The net
charge on the sulfite ion is – 2, so the sulfur must have a + 4 oxidation number.
12. Permanganate (MnO4-)
12A.
The four oxygen atoms in MnO4- each have a – 2 oxidation number, for a combined – 8 oxidation
number. Since the overall charge on the permanganate ion is – 1, manganese must have a + 7
oxidation number (- 8 + 7 = - 1).
13. Nitrite (NO2-)
13A.
The two oxygen atoms in NO2- each have a - 2 oxidation number. Since the overall charge on nitrite is
- 1, the nitrogen atom must have a + 3 oxidation number ( - 4 + 3 = - 1).
14. What is the oxidation number of phosphorus in phosphoric acid (H3PO4)?
14A.
The three hydrogen atoms have a + 1 oxidation number. They have a combined oxidation number of
+ 3. The four oxygen atoms have a - 2 oxidation number. They have a combined oxidation number of
– 8. This means the phosphorus atom must have a + 5 oxidation number to balance out the molecule
(+ 3 – 8 = - 5).
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
15. Find the oxidation numbers for sulfur, oxygen, and carbon in the following redox reaction:
CO2(g) + SO2(g)  CO(g) + SO3(g)
Reactants:
For CO2: Oxygen has a – 2 oxidation number. Since there are two oxygen atoms in CO2, the O2 part
of this species adds up to a – 4 oxidation number. As a result, the carbon atoms in CO2 have a + 4
oxidation number.
For SO2: Oxygen has a – 2 oxidation number. Since there are two oxygen atoms in SO2, the O2 part
of this species adds up to a – 4 oxidation number. As a result, the sulfur atom must have a + 4
oxidation number.
Products:
For CO: Oxygen has a – 2 oxidation number. Therefore, carbon must have a + 2 oxidation number.
For SO3: Oxygen has a – 2 oxidation number. The three oxygen atoms have a combined – 6 oxidation
number. Sulfur must have a + 6 oxidation number to balance out the molecule.
Final oxidation numbers
C(+4)O2(-2) + S(+4)O2(-2)  C(+2)O(-2) + S(+6)O3(-2)
16. For the above reaction, identify the element that is oxidized and the element that is reduced.
16A. Sulfur is oxidized and carbon is reduced.
17. Predict the oxidation numbers for the chromate ion, CrO4-2.
17A.
Each oxygen has a – 2 oxidation number. Since there are 4 oxygen atoms in CrO4-2, they must add up
to a - 8 oxidation number. It might be tempting to assign the chromium atom a +8 oxidation number,
but the chromate ion is not neutral. It has a -2 net charge. Therefore, the chromium must have a +6
oxidation number (- 8 + 6 = - 2).
18. Sulfur dioxide and oxygen react to form sulfur trioxide:
2 SO2(g) + O2(g)  2 SO3(g)
18a. Provide the before and after oxidation numbers for sulfur and oxygen.
A.
For SO2: Each oxygen atom has a – 2 oxidation number. Since there are two oxygen atoms in SO2,
they have a combined – 4 oxidation number. Sulfur therefore has a + 4 oxidation number.
For O2: Elemental oxygen has a + 0 oxidation number.
For SO3: Each oxygen atom has a – 2 oxidation number. Since there are three oxygen atoms in SO3,
they have a combined – 6 oxidation number. Sulfur therefore has a + 6 oxidation number.
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VAN NUYS HIGH SCHOOL
Final oxidation numbers
2 S(+4)O2(-2)(g) + O2(+0)(g)  2 S(+6)O3(-2)(g)
18b. Identify the element that is reduced.
A. Oxygen
18c. Identify the element that is oxidized.
A. Sulfur
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