Oxidation Numbers

The oxidation number of an element in a compound or radical  is the valency or combining power of the element in that compound or radical. It is also defined as the electrical charge an element appears to have in a given molecule or ion. The importance of oxidation number in chemistry cannot be overemphasized, as it can be used to predict the nature of a reaction. A change in oxidation number is used to reflect an electron gain or loss in a reaction. If the oxidation number increases, it means there is an electron loss (oxidation), while a decrease in oxidation number shows an electron gain (reduction).

Determining the oxidation numbers of element follows a set of rules as discussed below:

Rules For Determining Oxidation Numbers

1) The oxidation number of all elements in the free state is zero. By free state, we mean their uncombined state with any other element. e.g., F2, Mg, O3, Cl2, S, Ca, K, Na, Zn etc.

2) The oxidation number of a simple ion is equal to the charge of the ion. A simple ion is one consisting of a single element. e.g., Na+, Mg2+, Al3+, Zn2+, K+, Ca2+, Cl-, F-, S2-, I- etc with oxidation numbers of +1, +2, +3, +2, +1, +2, -1, -1, -2 and -1 respectively.

3) The oxidation number of a radical is the algebraic sum of the oxidation numbers of all the elements in the make up the radical. A radical is a group of atoms of elements with a common charge. e.g., NH4+ (ammonium ion), OH- (hydroxyl ion), NO3- (trioxonitrate (V) ion), SO4-- (tetraoxosulphate (VI) ion), MnO4- (tetraoxomanganate (VII) ion), Cr2O7-- (heptaoxodichromate (VI) ion), ClO3- (trioxochlorate (VII) ion) with oxidation numbers +1, -1, -1, -2, -1, -2,  and -1 respectively.

4) The algebraic sum of the oxidation numbers of all the elements in a compound is zero. Example, in the compound CaF2 (calcium fluoride), its oxidation number is equal to zero, as shown:

            (+2 x 1) + (-1 x 2) = +2 + (-2) = +2 - 2 = 0

Also, the oxidation number of H2SO4 equals to zero as calculated below:

            (H x 2) + (S x 1) + (O x 4)
         = (+1 x 2) + (+6 x 1) + (-2 x 4)
         = +2 + 6 + (-8)
         = 8-8 = 0

5) The oxidation number of oxygen is -2, except in peroxides, where it has an oxidation number of -1. Example of peroxides include: H2O2 (hydrogen peroxide), Na2O2 (sodium peroxide), K2O2 (potassium peroxide), BaO2 (barium peroxide), CaO2 (calcium peroxide) etc.

6) The oxygen number of hydrogen is +1, except in metallic hydrides, where its oxidation number is -1. Metallic hydrides are compounds which hydrogen forms with metals. Examples include: NaH (sodium hydride), CaH2 (calcium hydride), LiH (lithium hydride), KH (potassium hydride) etc.

7) In most simple compounds, the oxidation number of the elements is equivalent to their valencies. Example in NaCl, the oxidation numbers of Na and Cl are +1 and -1 respectively, and these are the same as their valencies.

Using these arbitrary rules, we can calculate the oxidation number of any element in a compound or radical as long as the oxidation numbers of the other elements are known.

Examples:

Question 1:
Calculate the oxidation number of Mn in the following compounds:
(i) KMnO4 (ii) MnO (iii) MnO2

Answer:
(i) KMnO4
Since KMnO4 is a compound, it means that the algebraic sum of all the oxidation numbers of K, Mn, and O will be equal to zero. Remember also that the valency of K is +1, and the oxidation number of O is -2. Thus,

           (K x 1) + (Mn x 1) + (O x 4) = 0

We multiply the oxidation number of O by 4, because there are four atoms of O in the compound.

          (+1 x 1) + Mn + (-2 x 4) = 0   
                        +1 + Mn + (-8) = 0 
                             +1 + Mn - 8 = 0 
                                           Mn = + 8 - 1
                                                  = +7

Therefore, the oxidation number of Mn in KMnO4 is +7. Hence, the name potassium tetraoxomanganate (VII)

(ii) MnO
MnO is a compound, therefore  the algebraic sum of the oxidation numbers of Mn and O will be equal to zero. Thus,

                     (Mn x 1) + (O x 1) = 0
                             Mn + (-2 x 1) = 0   
                                   Mn + (-2) = 0 
                                        Mn - 2 = 0 
                                             Mn = 0 + 2
                                                   = +2

Therefore, the oxidation number of Mn in MnO is +2. Hence, the name manganese (II) oxide.

(iii) MnO2
MnO2 is aso a compound, therefore the algebraic sum of the oxidation numbers of Mn and O will be equal to zero. So,

                      (Mn x 1)+(O x 2) = 0
                            Mn + (-2 x 2) = 0   
                                  Mn + (-4) = 0 
                                       Mn - 4 = 0 
                                            Mn = 0 + 4
                                                  = +4

Therefore, the oxidation number of Mn in MnO2 is +4, and the name of the compound is manganese (IV) oxide.

Question 2:
What is the oxidation number of Cr in:
(i) Cr2O7-- (ii) Cr2O3 (iii) NaCrO4

Answer:
(i) Cr2O7--
Cr2O7-- is not a compound, but a radical with an overall charge of -2. (Note: the negative signs attached to Cr2O7 represent -2) Therefore,  the algebraic sum of the oxidation numbers of Cr and O should be equal to -2. Hence,

                      (Cr x 2) + (O x 7) = -2
                                2Cr+(-2*7) = -2   
                                2Cr + (-14) = -2 
                                     2Cr - 14 = -2
                                            2Cr = +14 - 2
                                            2Cr = +12
Therefore,
                                              Cr = +12/2
                                                   = +6

Thus, the oxidation number of Cr in Cr2O7-- is +6, and that is why the radical is named heptaoxodichromate (VI) ion.

(ii) Cr2O3
Cr2O3 is a compound. So, the algebraic sum of the oxidation numbers of Cr and O will be equal to zero. Thus,
                          (Cr x 2) + (O x 3) = 0
                                  2Cr + (-2*3) = 0   
                                      2Cr + (-6) = 0 
                                           2Cr - 6 = 0 
                                                 2Cr = 0 + 6
                                                 2Cr = +6
Therefore,
                                                   Cr = +6/2
                                                        = +3

Hence, the oxidation number of Cr in Cr2O3 is +3, and the name is chromium (III) oxide.

(iii) NaCrO4
NaCrO4 is a compound, and the algebraic sum of the oxidation numbers of Na, Cr and O is equal to zero. Thus,

         (Na x 1) + (Cr x 1) + (O x 4) = 0
                  (+1 x 1) + Cr + (-2 x 4) = 0   
                                   1 + Cr + (-8) = 0 
                                        1 + Cr - 8 = 0 
                                                    Cr = + 8 - 1
                                                    Cr = +7

Therefore, the oxidation number of Cr in NaCrO4 is +7, and the name of the compound is sodium tetraoxochromate (VII).

Question 3:
Calculate the oxidation number of S in the following compounds:
(i) SO2 (ii) SO3 (iii) SO4--

Answer:
(i) SO2
SO2 is a molecule. So, the algebraic sum of the oxidation numbers of S and O will be equal to zero. Thus,
                            (S x 1) + (O x 2) = 0
                                    S + (-2 x 2) = 0   
                                          S + (-4) = 0 
                                               S - 4 = 0 
                                                    S = 0 + 4
                                                    S = +4

Hence, the oxidation number of S in SO2 is +4, which explains why the IUPAC name of the gas is sulphur (IV) oxide.

(ii) SO3
SO3 is a molecule too. Hence,

                          (S x 1) + (O x 3) = 0
                                  S + (-2 x 3) = 0   
                                        S + (-6) = 0 
                                             S - 6 = 0 
                                                  S = 0 + 6
                                                  S = +6

So, the oxidation number of S in SO3 is +6 and the name of the gas is sulphur (VI) oxide.

iii) SO4-- 
SO4--  is a radical with an overall charge of -2. Therefore,  the algebraic sum of the oxidation numbers of S and O should be equal to -2. Hence,

                         (S x 1) + (O x 4) = -2
                                 S + (-2 x 4) = -2   
                                       S + (-8) = -2 
                                            S - 8 = -2
                                                 S = + 8 - 2
                                                 S = +6

Thus, the oxidation number of S in SO4-- is +6, and that is why it is called  tetraoxosulphate (VI) ion.

Question 4:
Calculate the oxidation number of Fe in [Fe(CN)6]3-

Answer:
[Fe(CN)6]3- is a complex ion with a net charge of -3. Expectedly, the algebraic sum of the oxidation numbers of its elements and radicals, namely Fe and CN equals -3. Therefore,

                    (Fe x 1) + (CN x 6) = -3

but the oxidation number of cyanide ion (CN-) is -1.

Therefore, the above equation becomes:

                      (Fe x 1) + (-1 x 6) = -3
                                     Fe + (-6) = -3
                                          Fe - 6 = -3
                                               Fe = + 6 - 3
                                                    = +3

Hence, the oxidation number of Fe in [Fe(CN)6]3- is +3, and the ion is hexacyanoferrate (III) ion.

Do These:

Question 1:
What is the change in oxidation number of Mn in the reaction represented by the equation?
MnO4-  +  8H+  +  5e-  ---> Mn2+  +  4H2O
A. +3 to +2   B. +4 to +2   C. +5 to +2   D. +7 to +2

Question 2:
The ionic equation of the reaction between potassium tetraoxomanganate (VII) and ethanedioic acid in acidic medium is as follows:
MnO4-  +  C2O4--  +  H+  ---> Mn2+  +  CO2  +  H2O
(i) Balance the equation
(ii) Name the:
I. oxidizing agent
II. reducing agent
(iii) How many moles of electrons are transferred in the balanced equation?
(iv) How would you determine the endpoint of a titration involving the above reaction?
(To completely answer this question, you may have to make reference to the posts on Balancing Redox Equations and Redox Reactions: Overview)

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