Question

Diiodinepentaoxide oxidizes carbon monoxide to carbon dioxide under room conditions, yielding iodine as the second product:

${\mathbf{\text{I}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{5}}}\mathbf{\text{(s) + 5CO(g)}}\mathbf{\to }{\mathbf{\text{I}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s) + 5CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

This can be used in an analytical method to measure the amount of carbon monoxide in a sample of air. Determine the oxidation numbers of the atoms in this equation. Which species is oxidized and which is reduced?

Short answer

Iodine is reduced and carbon is oxidized.

Step-by-step solution

Understanding oxidation and reduction

When a reactant loses electrons during a reaction, it is called oxidation. When a reactant gains electrons during a reaction is called reduction. In a redox reaction both oxidation and reduction takes place simultaneously.

Putting the oxidation number of the elements

We have to find out the oxidation number of the respective atoms to understand the oxidation and reduction process. An increase in oxidation number shows oxidation while an decrease in oxidation number represents reduction.

Oxidation number of an element in its natural form is 0.

The oxidation number of iodine in${\text{I}}_{\text{2}}{\text{O}}_{\text{5}}$ is calculated below.

Let the oxidation number of iodine be x. the molecule is a neutral molecule and hence the total sum of oxidation number should be equal to zero.

$\begin{array}{rcl}2x+5\times \left(-2\right)& =& 0\\ 2x& =& 10\\ x& =& 5\end{array}$

The oxidation number of $\text{CO} \text{and} {\text{CO}}_{\text{2}}$is calculated in the following way.

For CO;

$\begin{array}{rcl}& & \text{y - 2 = 0}\\ & & \text{y = 2}\end{array}$

For ${\text{CO}}_{\text{2}}$,

role="math" localid="1663830338743" $\begin{array}{rcl}z+2\times \left(-2\right)& =& 0\\ z& =& 4\end{array}$

Finding out the oxidized and reduced element

The oxidation state of iodine is decreased from +5 to 0. Thus, it is reduced.

The oxidation state of carbon has increased from +2 to +4. Thus, it is oxidized.