Problem 10 - Introductory Chemistry Atoms First

Question

Use the shortcut rules to assign oxidation states to all atoms. \(\mathrm{Cu}^{2+}\)

Step-by-Step Solution

Answer

In the given chemical species \(\mathrm{Cu}^{2+}\), there is only one atom, copper (Cu). Since it is a simple ion with a charge of +2, its oxidation state is equal to its charge. Therefore, the oxidation state of copper in \(\mathrm{Cu}^{2+}\) is +2.

Step 1: Identify the ion

We are given a copper (Cu) ion that carries a charge of +2. This means we have a positively charged Cu atom, which is denoted as \(\mathrm{Cu}^{2+}\).

Step 2: Determine the oxidation state of the ion

In a simple ion, the oxidation state is equal to the charge on the ion. The oxidation state of copper in \(\mathrm{Cu}^{2+}\) is, therefore, +2.

Step 3: Summary

Using the shortcut rules, we have determined that the oxidation state of copper in the chemical species \(\mathrm{Cu}^{2+}\) is +2.

Key Concepts

Redox Chemistry

Redox chemistry, also known as oxidation-reduction chemistry, involves the study of chemical processes in which electrons are transferred between atoms, ions, or molecules. This transfer of electrons leads to a change in the oxidation states of the participating entities. In these reactions, one species undergoes oxidation (loses electrons) and another undergoes reduction (gains electrons).

It's vital to understand that oxidation and reduction always occur together; when an atom gives up an electron, there must be another atom to accept it. This complementary change is described by the mnemonic 'LEO the lion says GER', meaning 'Loss of Electrons is Oxidation; Gain of Electrons is Reduction'. Recognizing these changes is crucial because redox reactions are fundamental to many processes, including energy generation in batteries, metabolic pathways in the body, and the rusting of metals.

Oxidation states, a concept linked intimately with redox chemistry, help chemists keep 'score' of the electrons during a chemical reaction, thus, allowing an understanding of which element is oxidized and which is reduced.

Copper Ion

Copper ions are a common participant in redox reactions and have notable importance in both inorganic chemistry and biological systems. Copper typically exhibits two common oxidation states: +1 (cuprous) and +2 (cupric). The \(\mathrm{Cu}^{2+}\) ion, discussed in our original exercise, is an example of the cupric form, which is more prevalent and stable in aqueous solutions than the cuprous form.

Cupric ions are characterized by their distinctive blue color in solution and play a vital role in biological electron transport chains, as seen in mitochondrial processes. They are also fundamental in various enzymatic reactions that mediate important biological functions. In industrial applications, copper ions are key components in materials such as catalysts, pigments, and electrical conductors. Understanding the chemical behavior of copper ions, including their oxidation states, is critical both for academic study and practical applications.

Oxidation Number Assignment

Assigning oxidation numbers is an essential skill when analyzing redox reactions. The oxidation number is a theoretical charge that an atom would have if all bonds to it were ionic. In simpler terms, it's a bookkeeping tool that tracks electron ownership in compounds and ions during chemical reactions.

Several rules help in the determination of these numbers. For a monatomic ion, like \(\mathrm{Cu}^{2+}\) from our exercise, the oxidation number equals the charge of the ion. Therefore, the oxidation state of copper in this ion is +2.

Application in Compounds

In molecular compounds, the rules expand to include assigning an oxidation number of -2 to oxygen (with some exceptions), +1 to hydrogen (when bonded to nonmetals), and using the electronegativity of atoms to determine the electron distribution in covalent bonds.

Understanding how to assign oxidation numbers correctly facilitates the balance of redox equations and predicts the direction of electron flow in reactions, which is integral to the mastery of redox chemistry.