Question

Iodine pentoxide \(\left(\mathrm{I}_{2} \mathrm{O}_{5}\right)\) is sometimes used to remove carbon monoxide from the air by forming carbon dioxide and iodine. Write a balanced equation for this reaction, and identify species that are oxidized and reduced.

Short answer

Balanced equation: \(\text{I}_2\text{O}_5 + 5\text{CO} \rightarrow 5\text{CO}_2 + \text{I}_2\). CO is oxidized; Iodine in \(\text{I}_2\text{O}_5\) is reduced.

Step-by-step solution

Write the Unbalanced Equation

First, we identify the chemical reaction involving iodine pentoxide and carbon monoxide. The reactants are \(\text{I}_2\text{O}_5\) and \(\text{CO}\) and the products are \(\text{CO}_2\) and \(\text{I}_2\). Therefore, the unbalanced chemical equation is: \[ \text{I}_2\text{O}_5 + \text{CO} \rightarrow \text{CO}_2 + \text{I}_2 \]

Balance the Equation

To balance the equation, we ensure that the number of each type of atom is the same on both sides. Begin by balancing iodine atoms. There are 2 iodine atoms in \(\text{I}_2\) on the right, so use 1 \(\text{I}_2\text{O}_5\) on the left. This gives us: \[ \text{I}_2\text{O}_5 + \text{CO} \rightarrow \text{CO}_2 + \text{I}_2 \] Next, balance the carbon atoms. There is 1 carbon in \(\text{CO}_2\), so we add a coefficient of 5 in front of \(\text{CO}\): \[ \text{I}_2\text{O}_5 + 5\text{CO} \rightarrow 5\text{CO}_2 + \text{I}_2 \] Now, balance the oxygen atoms: \(\text{I}_2\text{O}_5\) provides 5 oxygens and \(5 \times \text{CO}\) gives 5 oxygens, resulting in \(5 \times \text{CO}_2\). The equation is now balanced: \[ \text{I}_2\text{O}_5 + 5\text{CO} \rightarrow 5\text{CO}_2 + \text{I}_2 \]

Identify Oxidized and Reduced Species

Determine which species are oxidized and reduced based on oxidation states. In \(\text{CO}\), carbon has an oxidation state of +2, and in \(\text{CO}_2\), it is +4. Carbon is oxidized as its oxidation state increases. In \(\text{I}_2\text{O}_5\), iodine has an oxidation state of +5, and in \(\text{I}_2\), it is 0. Iodine is reduced as its oxidation state decreases.

Key concepts

Iodine Pentoxide

Iodine pentoxide, expressed as \( ext{I}_2 ext{O}_5\), is a chemical compound comprised of iodine and oxygen. It plays a significant role in various chemical reactions, particularly as a strong oxidizing agent. This means it has the ability to accept electrons from other species during chemical reactions, thereby influencing the reaction pathway.

In the reaction involving the removal of carbon monoxide (CO) from the air to form carbon dioxide (CO\(_2\)) and iodine (I\(_2\)), iodine pentoxide acts as the oxidizer. This facilitates the conversion of CO to CO\(_2\), simultaneously reducing itself to iodine gas. Such reactions underscore the importance of iodine pentoxide in purifying air systems by effectively removing harmful gases like CO.

Its application is particularly valued in closed environments, such as submarines or space habitats, where air quality control is vital.

Oxidation-Reduction

Oxidation-reduction reactions, often abbreviated as redox reactions, are chemical processes in which oxidation and reduction occur simultaneously. In these reactions, one species loses electrons (oxidation) while another gains them (reduction). This electron transfer is what drives the chemical change.

In the provided exercise, the reaction between iodine pentoxide and carbon monoxide features both oxidation and reduction steps:

• The carbon in carbon monoxide (CO) is oxidized because it goes from an oxidation state of +2 to +4 in carbon dioxide (CO\(_2\)).
• The iodine in iodine pentoxide (\( ext{I}_2 ext{O}_5\)) is reduced from an oxidation state of +5 to 0 in elemental iodine (I\(_2\)).

Such transformations are central to understanding how substances interact and change in chemical reactions, illustrating the dynamic nature of matter.

Chemical Reactions

Chemical reactions are processes where reactants undergo transformation to form products. These transformations involve breaking and forming of chemical bonds, and balancing these reactions is crucial to ensuring that the law of conservation of mass is followed.

The exercise provides an excellent example of a chemical reaction:

• Reactants: Iodine pentoxide (\( ext{I}_2 ext{O}_5\)) and carbon monoxide (CO).
• Products: Carbon dioxide (CO\(_2\)) and iodine (I\(_2\)).

To balance the chemical equation for this reaction, we ensure that the number of each type of atom on the reactant side is equal to the number on the product side. This involves a meticulous process of adjusting coefficients to match all elements across the equation. Such balanced equations are critical for accurately representing the stoichiometry of the reaction.

Oxidation States

Oxidation states, or oxidation numbers, help chemists track how electrons are transferred in chemical reactions. These numbers indicate the degree of oxidation of an atom within a compound, and calculating them helps determine which atoms are oxidized or reduced during the reaction.

In the given exercise:

• Carbon in CO has an oxidation state of +2, while in CO\(_2\), it is +4, reflecting oxidation as the oxidation state increases.
• Iodine in \( ext{I}_2 ext{O}_5\) has an oxidation state of +5, which decreases to 0 in I\(_2\), indicating reduction.

This change in oxidation states not only shows which species are oxidized or reduced but also drives the redox reaction. Understanding how to assign and interpret oxidation states is essential for identifying redox reactions in chemistry.