1.4 - Oxidation Numbers
Module 5
Lesson 1.4 Oxidation Numbers
Key Concepts
Although half-reactions indicate electrons, chemical equations do not. Therefore, how can it be determined whether a given reaction is a redox reaction? Fortunately, chemists can use "oxidation numbers" (also called oxidation states) to identify reactions in which electron transfer has occurred. Oxidation numbers are not electrical charges, but rather they are positive or negative numbers that are assigned to an atom based on its electronegativity. To distinguish oxidation numbers from actual electrical charges, oxidation numbers are written as positive or negative numbers with the sign preceding the number.
Because oxidation numbers are powerful tools in the analysis of electrochemical systems, you must memorize the information in the following table.
| Atom or Ion | Oxidation number | Examples |
| Atoms existing as free elements | 0 | Cu is 0 Br in Br2 is 0 |
| Hydrogen in most compounds Hydrogen in hydrides (bonded to a metal) |
+1 -1 |
H in H2SO4 is +1 H in MgH2 is -1 |
| Oxygen in almost all compounds, Oxygen in peroxides |
-2 -1 |
O in H2O is -2 O in H2O2 is -1 |
| Alkali metals in compounds | +1 | K in KNO3 is +1 |
| Alkaline-earth metals in compounds | +2 | Ba in BaSO4 is +2 |
| Halogens in most binary ionic compounds | -1 | Cl in CuCl is -1 |
| All monatomic ions | Charge on ion | Sn2+ is +2 P 3- is -3 |
| Rules for determining oxidation numbers for unknown entities | ||
|---|---|---|
| Sum of oxidation numbers of the elements in a neutral compound is zero. |
CaCO3 Ca is +2 O is -2 x 3=-6 To equal 0 C is +4 |
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Sum of oxidation numbers of the atoms in a radical is equal to the charge on that radical. |
NO3- O is -2 x 3 = -6 To equal 1- N is +5 |
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As you can see from the first row of the table, pure elements are assigned an oxidation state of zero.
Watch
Oxidation Numbers
Determining Oxidation Numbers
Watch the following video to learn more about how oxidation numbers are assigned and used.
© chemistNATE
Read page 583 to 585 in the textbook to familiarize yourself further with the concept of oxidation numbers.
Check Your Understanding
Complete "Practice" questions 1 to 5 on page 585 of the textbook. Click on the link below to check your answers.
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S in SO2
x + 2(-2) = 0
x = +4
The oxidation number of S in SO2 is +4
-
Cl in HClO4
+1 + x + 4(-2) = 0
x = 7
The oxidation number of Cl in HClO4 is +7.
-
S in SO4-2
x + 4(-2) = -2
x = +6
The oxidation number of S in SO4-2 is +6
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Cr in Cr2O72-
2(x) + 7(-2) = -2
x = +6
The oxidation number of Cr in Cr2O72- is +6
-
I in MgI2
The compound contains Mg2+ and I-.
I- is a monatomic ion; thus, the oxidation number of I in MgI2 is -1.
-
H in CaH2
The compound contains Ca2+ and H- and is a hydride.
The oxidation number of H in CaH2 is -1
Page 585 Practice Question 2
| a. | N2O(g) | 2x + 1(-2) = 0 | x = +1 |
| b. | NO(g) | x +1(-2) = 0 | x = +2 |
| c. | NO2(g) | x + 2(-2) = 0 | x = +4 |
| d. | NH3(g) | x + 3(+1) = 0 | x = -3 |
| e. | N2H4(g) | 2x + 4(+1) = 0 | x = -2 |
| f. | NaNO3(s) contains Na+ and NO3- | Using NO3-, x + 3(-2) = -1 | x = +5 |
| g. | N2(g) is an element. The oxidation # is 0. | ||
| h. | NH4Cl(s) contains NH4+ and Cl- | Using NH4+, x + 4(+1) = +1 | x = -3 |
Page 585 Practice Question 3
| a. | C(s) (graphite) is an element; therefore, the oxidation number is 0. | ||
| b. | C6H12O6 | 6x + 12(+1) + 6(-2) = 0 | x = 0 |
| c. | Na2CO3 contains Na+ and CO32- | Using CO32-, x + 3(-2) = -2 | x = +4 |
| d. | CO | x + 1(-2) = 0 | x = +2 |
Page 585 Practice Question 4
Remember that coefficients in an equation do not affect the value of oxidation numbers.
O in ClO3-, ClO2, and H2O is -2.
H in H+ and H2O is +1
Cl in ClO3- is +5 (x + 3(-2) = -1).
Cl in Cl- is -1
Cl in ClO2 is +4 (x + 2(-2) = 0).
Cl in Cl2 is 0 (element).
Page 585 Practice Question 5