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Oxidation and reduction (Oxidation-Reduction Reactions, or redox reactions) take place in the world at every moment. In fact, they are directly related to the origin of life. For instance, oxidation of nutrients forms energy and enables human beings, animals, and plants to thrive. If elements or compounds were exposed to oxygen, after a series of reactions the oxygen will be converted into carbon dioxide or water (Combustion Reaction). In order to fully understand redox and combustion reactions, we must first learn about oxidation states.
An oxidation state is a number that is assigned to an element in a chemical combination. This number represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element. The terms “oxidation state” and “oxidation number” are often used interchangeably. The transfer of electrons is described by the oxidation state of the molecule. One might mistaken formal charge for oxidation state but they are different. Oxidation state is commonly used to determine the changes in redox reactions. Oxidation state is numerically similar to valence electrons but different from formal charge. Formal charge determines the arrangement of atoms and the likelihood of the molecule existing.
The Rules: Oxidation State
To effectively assign oxidation states to a compound, the seven basic rules must be followed in order. Remember to use the rule that comes first if two rules conflict with each other. These rules hold true for most compounds.
Abbreviation "O.S" = Oxidation states.
Any individual atom uncombined with other elements has the oxidation state of 0 (zero).
Ex.) The O.S for Ag is 0. The oxidation state for O (oxygen) or O2 is 0 as long as it is uncombined with any other element.
a.) In neutral species, the total sum of the oxidation state of all atoms is 0.
Ex.) The sum of O.S for NaCl is 0 since the O.S of Na = +1 and the O.S of Cl = -1, therefore NaCl total O.S = 0
b.) In ions, the total sum of the oxidation state is the charge of the ion.
Ex.) The O.S of Ca2+ (Calcium) is = +2. The total sum of the O.S of all atoms in CrO42− (Chromate ion) is -2. The total sum of the oxidation states of all atoms in CH3COO- (Acetate ion) is -1.
The total sum of the oxidation state of all atoms in any given species is equal to the net charge on that species.
In a compound, the Oxidation state for Group 1(1A) metal is +1 and for Group 2(2A) metal, the oxidation state is +
Ex.) In NaCl, Na has the oxidation state of +1 since it is a Group 1 Alkali metal. Cl would have an oxidation number of -1 to make the sum of the oxidation states 0 (Rule 2). In MgCl2, Mg has the oxidation state of +2, since it is a Group 2 Alkaline Earth metal. Cl would have an oxidation state of -1, and since there are 2 Cl atoms, the overall charge of the species would again be 0 (Rule 2).
The oxidation state of FLOURINE is -1 in a compound.
Ex.) O.S of F is -1 in HF, SF6
The oxidation state of HYDROGEN is +1 in a compound.
Ex.) O.S of H is +1 in HI, CH4 , NH4+
The oxidtation state of OXYGEN is -2 in a compound.
Ex.) O.S of Oxyen is -2 in OH-, H2O, CO32-
In two-element compounds with metals, Group 15(3A) elements have the oxidation state of -3, Group 16(6A) elements have the oxidation state of -2, and Group 17(7A) elements have the oxidation state of -1.
Ex.) In MgBr2, Br has the oxidation state of -1, since it is a Group 17 element. In Li2S, S has the oxidation state of -2, since it is a Group 16.
Find the oxdidation state of Cr in CrO4-2
1. From Rule #2, we know that the overal sum of O.S for this compound is -2.
2. From Rule #6, we know that Oxygen in a compound is -2. Since we have four oxygen, -2 x 4 = -8.
3. With those information, set up an short equation Cr + (-8) = -2, Cr = +6
Find the oxidation state of C in C2H3O2-
1. From Rule #2, we know that the overal sum of O.S for this compound is -1
2. From Rule #5, we know that Hydrogen in a compound is +1. Since we have three Hydrogen, +1 x 3 = +3
3. From Rule #6, we know that Oxygen in a compound is -2. Since we have two oxygen, -2 x 2 = -4
4. With those information, set up an equation 2C + (+3) + (-4) = -1
Rewrite as 2C + 3 - 4 = -1 -----> 2C = 0 -----> C = 0/2 = 0
The oxidation state for C is 0.
Find the oxidation state of S in S2O32-
1. From Rule #2, we know that the overal sum of O.S for this compound is -2
3. From Rule #6, we know that Oxygen in a compound is -2. Since we have three oxygen, -2 x 3 = -6
4. With those information, set up an equation 2S + (-6) = -2
Rewrite as 2S - 6= -2 -----> 2S = 4 -----> S = 4/2 = 2
The oxidation state for S is 2.
Fractional Values for Oxidation States
The oxidation states are usually in whole numbers, but in some cases, they are in fractional numbers. Consider Fe3O4.
In most cases, atoms of the same element in a given compound have the same oxidation states, but each atom can have a different state than the other. For Fe3O4, two Fe atoms have an oxidation state of +3 and one of +2, which makes the total oxidation state of Fe= 8/3.
Instead of averaging the oxidation states, we must keep them separated. This is called fragmenting, which occurs if there is an ionic compound, and the ions can be separated. The oxidation state of the individual atoms in the different ions can be determined.
Still need help? Watch Video Tutorials!
From the Khan Academy (khanacademy)
Dr. Enderle (UCD) explains the oxidation state for transition metal complex (for more difficult problems)
Hint: try to memorize the rules and follow them in order.
Determine the oxidation states of the underlined elements in
1) Cl2; 2) NaH; 3) H2CO; 4) S2O3 2-; 5) KMnO4.; 6)FeCl3; 7)N2; 8)H2SO4; 9)HClO2; 10)CuSO4;
Transition metal complex
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