Problem 53 - Chemistry: Principles and Reactions

Question

III Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)

Step-by-Step Solution

Answer

Question: Classify each of the following half-reactions as either oxidation or reduction: (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\) Answer: (a) Reduction (b) Reduction (c) Reduction (d) Oxidation

Step 1: Identify the oxidation states of the elements in each half-reaction

To classify each half-reaction, we will first determine the oxidation states of the elements involved. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) In this half-reaction, the oxidation state of O in \(\mathrm{O}_{2}\) is 0, and the oxidation state in \(\mathrm{O}^{2-}\) is -2. (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) In this half-reaction, the oxidation state of Mn in \(\mathrm{MnO}_{4}^{-}\) is +7 (since O is -2) and the oxidation state in \(\mathrm{MnO}_{2}\) is +4. (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) In this half-reaction, the oxidation state of Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is +6 and the oxidation state in \(\mathrm{Cr}^{3+}\) is +3. (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\) In this half-reaction, the oxidation state of Cl in \(\mathrm{Cl}^{-}\) is -1 and the oxidation state in \(\mathrm{Cl}_{2}\) is 0.

Step 2: Determine the electron transfer and classify each half-reaction

Now we will examine the transfer of electrons in each half-reaction to classify each one as oxidation or reduction. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) The oxidation state of O changes from 0 to -2 and gains 2 electrons. This is a reduction process. (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) The oxidation state of Mn changes from +7 to +4 and gains 3 electrons. This is a reduction process. (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) The oxidation state of Cr changes from +6 to +3 and gains 3 electrons. This is a reduction process. (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\) The oxidation state of Cl changes from -1 to 0 and loses 1 electron. This is an oxidation process.

Step 3: Answer

After analyzing all the half-reactions given, we can classify them as follows: (a) Reduction (b) Reduction (c) Reduction (d) Oxidation

Key Concepts

Oxidation State

The oxidation state, sometimes referred to as oxidation number, is a fundamental concept in chemistry that indicates the hypothetical charge an atom would have if all bonds to atoms of different elements were completely ionic. This theoretical charge is a tool for understanding the changes that occur in a redox (reduction-oxidation) reaction.

When tackling problems that involve assigning oxidation states to different elements, a systematic approach is useful. Generally, we follow certain rules, such as assigning an oxidation state of 0 to diatomic molecules (like \(O_2\)), -2 to oxygen (except in peroxides or when bonded to fluorine), and +1 to hydrogen (when bonded to nonmetals). By using the oxidation states, we can track how electrons are redistributed throughout a chemical reaction, allowing us to identify the substance being oxidized and the one being reduced.

Electron Transfer

Electron transfer is the movement of electrons from one atom or molecule to another and is the essence of redox reactions. The process involves two complementary changes: reduction, where an entity gains electrons, and oxidation, where it loses electrons. Recognizing this transfer is key in chemistry to determine the nature of a reaction.

During the transfer, the oxidation state of one substance will increase (oxidation), while that of the other will decrease (reduction). For example, in the reaction \(\mathrm{Cl}^-\mathrm{(aq)} \longrightarrow \mathrm{Cl}_2\mathrm{(g)}\), chlorine atoms each lose an electron, undergoing oxidation. Monitoring these changes provides insight into the direction and magnitude of the reaction, allowing us to predict the behavior of chemicals under different conditions.

Reduction Process

Reduction is one of the two key components of a redox reaction. It describes the gain of electrons by a molecule, atom, or ion. When a substance undergoes reduction, its oxidation state decreases, which can be tracked through a meticulous observation of the corresponding half-reaction.

In the given textbook solutions, for instance, the reduction process is identified when the manganese ion in \(\mathrm{MnO}_{4}^{-}\) gains electrons to form \(\mathrm{MnO}_{2}\) with a lower oxidation state. The importance of recognizing the reduction process lies in its ubiquitous role in chemical reactions, including metabolism in living organisms and industrial processes like metal refining.

Oxidation Process

In contrast to reduction, oxidation is a chemical process whereby a substance loses electrons. This loss results in an increase in the oxidation state of the substance. Oxidation is not simply the addition of oxygen to a compound, which is a common misconception; it involves any reaction where electrons are transferred away from an atom.

Examining the half-reaction \(\mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{g})\), we can see that the chloride ion loses one electron to become gaseous chlorine, reflecting an oxidation process. This illustrates the fundamental nature of oxidation in electrochemistry and its applications, from generating electricity in batteries to the formation of corrosion.