Question

Challenge Determine the net change of oxidation number of each of the elements in these redox equations. a. \(C+O_{2} \rightarrow C O_{2}\) b. \(C l_{2}+Z n l_{2} \rightarrow Z n l_{2}+I_{2}\) c. \(C d O+C 0 \rightarrow C d+C O_{2}\)

Short answer

a. Net change for Carbon: +4, Oxygen: -2 b. Net change for Chlorine: -1, Zinc: 0, Iodine: +1 c. Net change for Cadmium: -2, Oxygen: 0, Carbon: +2

Step-by-step solution

Understand oxidation numbers

Oxidation numbers are assigned to elements in compounds or ions to help us determine how the electrons are shared or transferred in a chemical reaction. In general, the sum of the oxidation numbers in a compound must equal the overall charge of the compound. When an element undergoes a redox reaction, its oxidation number changes. The difference between the oxidation number before and after the reaction is the net change of oxidation number.

Assign oxidation numbers to elements in reactions a, b, and c

When assigning oxidation numbers, we should follow these rules: 1. The oxidation number of elements in their elemental state (e.g., O2) is always 0. 2. The oxidation number of monatomic ions equals their charge. 3. The sum of oxidation numbers of elements in a compound equals their overall charge. Now let's assign oxidation numbers for each element in each reaction. a. \(C+O_{2} \rightarrow C O_{2}\) C is in its elemental form, so its oxidation number is 0. In \(O_2\), O has an oxidation number of 0. In \(CO_2\), C has an unknown oxidation number, but O has an oxidation number of -2. Since the overall charge of \(CO_2\) is 0, the oxidation number of C is +4. b. \(Cl_{2}+ZnI_{2} \rightarrow ZnCl_{2}+I_{2}\) In \(Cl_2\), Cl has an oxidation number of 0. In \(ZnI_2\), Zn has an oxidation number of +2 and I has an oxidation number of -1. In \(ZnCl_2\), Zn has an oxidation number of +2 and Cl has an oxidation number of -1. In \(I_2\), I has an oxidation number of 0. c. \(CdO+CO \rightarrow Cd+CO_{2}\) In \(CdO\), Cd has an oxidation number of +2 and O has an oxidation number of -2. In \(CO\), C has an unknown oxidation number, but O has an oxidation number of -2. Since the overall charge of \(CO\) is 0, the oxidation number of C is +2. In \(CO_2\), C has an oxidation number of +4 and O has an oxidation number of -2. Cd is in its elemental form, so its oxidation number is 0.

Calculate the net change of oxidation number for each element

Now we will determine the net change in the oxidation number for each element in the reactions. a. \(C+O_{2} \rightarrow C O_{2}\) Carbon: Initial oxidation number = 0, Final oxidation number = +4, Net change = +4 Oxygen: Initial oxidation number = 0, Final oxidation number = -2, Net change = -2 b. \(Cl_{2}+ZnI_{2} \rightarrow ZnCl_{2}+I_{2}\) Chlorine: Initial oxidation number = 0, Final oxidation number = -1, Net change = -1 Zinc: Initial oxidation number = +2, Final oxidation number = +2, Net change = 0 Iodine: Initial oxidation number = -1, Final oxidation number = 0, Net change = +1 c. \(CdO+CO \rightarrow Cd+CO_{2}\) Cadmium: Initial oxidation number = +2, Final oxidation number = 0, Net change = -2 Oxygen: Initial oxidation number = -2, Final oxidation number = -2, Net change = 0 Carbon: Initial oxidation number = +2, Final oxidation number = +4, Net change = +2

Key concepts

Redox Equations

In chemistry, redox (reduction-oxidation) equations represent chemical reactions where electrons are transferred between atoms. This transfer results in a change in the oxidation state or number of these atoms. Redox reactions are fundamental to understanding energy flow in biological systems, batteries, and corrosion, among other things.

Take, for example, the combustion of carbon in oxygen to form carbon dioxide, represented as \(C+O_{2} \rightarrow CO_{2}\). Carbon gains oxygen, which implies an increase in its oxidation number as it forms carbon dioxide, a process indispensable to life as it releases energy in cellular respiration.

Understanding redox reactions allows you to predict whether a reaction can occur spontaneously and to balance chemical equations. It's a primary tool in the chemist's toolkit, connecting microscopic atomic interactions with macroscopic phenomena we can observe.

Oxidation Number Assignment

To understand redox reactions, one must first be adept at assigning oxidation numbers to elements within compounds. An oxidation number is a bookkeeping tool used to keep track of electrons in oxidation-reduction reactions. It's like an imaginary charge that an atom would have if all bonds were 100% ionic.

For instance, in the reaction \(CdO+CO \rightarrow Cd+CO_{2}\), cadmium (Cd) in \(CdO\) and carbon (C) in \(CO\) both have oxidation numbers that reflect their supposed charges based on the general rules of oxidation numbers. Here’s a brief rundown:

• The oxidation number of any element in its elemental form, like \(O_{2}\) or \(Cl_{2}\), is zero.
• For a monatomic ion, the oxidation number is the same as its charge.
• In a neutral compound, the sum of the oxidation numbers is zero, whereas in a polyatomic ion, it is equal to the ion's charge.

By following these rules, you can deduce which atoms are oxidized or reduced in the reaction.

Net Change of Oxidation Number

The net change in oxidation numbers during a redox reaction indicates how many electrons were gained or lost by each atom involved. It’s kind of like keeping score in a game, but here we’re scoring electrons, not points.

In our example with the chemical reaction \(CdO+CO \rightarrow Cd+CO_{2}\), the net changes in oxidation numbers help us understand the electron transfer. Cadmium starts with an oxidation number of +2 in \(CdO\) and goes to 0 in its elemental form, thus gaining two electrons, a net change of -2. For carbon, the transition from \(CO\) where it has a +2 oxidation number to \(CO_{2}\) with a +4 oxidation number illustrates the loss of two electrons, a net change of +2.

These changes underpin the action of the reaction, revealing the transfer of electrons from one species to another and cementing the concepts of oxidation (loss of electrons) and reduction (gain of electrons) in a concrete manner.