2.1 Oxidation numbers

  Balancing Reaction

Oxidation number: The charge an atom has or appears to have when electrons are distributed.

To keep track of electrons in redox reactions, we use oxidation numbers.

Rules for assigning oxidation numbers (O.N.)

1. A free elements (or uncombined element) is assigned oxidation number (O.N.) of zero.

Example: Oxidation of $O_2$, $N_2$ etc is zero.

2. The oxidation number of halogens is usually -1.

Exception: When the halogen are bonded to a more electronegative element, then its O.N. changes.

3. The oxidation number of hydrogen is +1.

Exception: When bonded to a less electronegative element, i.e. a metal, then the oxidation number of hydrogen is -1.

4. The oxidation number of oxygen in most compounds is -2.

Exception:

The oxidation number of oxygen in $H_2O_2$ and peroxide ion $O_2^{2 - }$ is -1.

The oxidation number of oxygen in $OF_2$ is +1.

5. The sum of oxidation number of atoms in a neutral compound must equal zero.

6. The charge of a monoatomic ion is same as its oxidation number.

Example:

$Li^+$ has O.N. of $+1$
$Fe^{+3}$ has O.N. of $+3$

7. In the compound, group 1 elements have an oxidation number of $+1$, group 2 elements have an oxidation number of $+2$.

8. The sum of oxidation number of atoms in a polyatomic ion equals the charge on the ion.

Question: Assign oxidation numbers to all the elements in the following.

(a). $Li_2O$
(b). $HNO_3$
(c). $C{r_2}O_7^{2 - }$

Solution:

(a). Lithium $(Li)$ is a group 1 element. So its oxidation number is $+1$.

Also, the oxidation state of oxygen is $-2$.

(b). As we know oxidation number of hydrogen is $+1$ and oxidation number of oxygen is $-2$.

Let the oxidation number of nitrogen be $x$.
$$\mathop H\limits^{ + 1} \mathop N\limits^x \mathop {{O_3}}\limits^{ - 2} $$
As we know sum of all oxidation number of atoms in a neutral compound must equal to zero. Therefore,
$$\left( { + 1} \right) + \left( x \right) + \left( { - 2} \right) \times 3 = 0$$$$1 + x - 6 = 0$$$$x = 5$$
So, the oxidation number of nitrogen in $HNO_3$ is $5$.

(c). As we know oxidation number of oxygen is $-2$.

Charge on the compound is $-2$

Let the oxidation number of $Cr$ be $x$.

$${\left[ {\mathop {C{r_2}}\limits^x \mathop {{O_7}}\limits^{ - 2} } \right]^{2 - }}$$
As we know sum of all oxidation number of atoms in ion must equal to the charge on that ion. Therefore,
$$\left( x \right) \times 2 + \left( { - 2} \right) \times 7 = - 2$$$$2x - 14 = - 2$$$$2x = 12$$$$x = + 6$$

Note: When an element is oxidised, its oxidation number is increased. When an element is reduced, its oxidation number is decreased. Change in oxidation number can be used to decide whether an oxidation or reduction has taken place.

Question: Find the element which undergoes oxidation and reduction in the element shown below.
$$PbO + C \to Pb + CO$$
Solution: The reaction with oxidation number of each element is given by,
$$\mathop {Pb}\limits^{ + 2} \mathop O\limits^{ - 2} + \mathop C\limits^0 \to \mathop {Pb}\limits^0 + \mathop C\limits^{ + 2} \mathop O\limits^{ - 2} $$
Oxidation state of $Pb$ has changed from $+2$ to $0$. So, we can say $PbO$ undergoes reduction.

Similarly, the oxidation state of $C$ has changed from $0$ to $2$. So, we can say $C$ undergoes oxidation.