Question

In each of the following reactions, relate starting materials and products by the processes of reduction, oxidation, disproportionation or no redox change. In some reactions, more than one process is taking place. \((\mathrm{a})\left[\mathrm{HCO}_{3}\right]^{-}+[\mathrm{OH}]^{-}-\left[\mathrm{CO}_{3}\right]^{2-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Au}+\mathrm{HNO}_{3}+4 \mathrm{HCl}-\mathrm{HAuCl}_{4}+\mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}\) (c) \(2 \mathrm{VOCl}_{2}-\mathrm{VOCl}_{3}+\mathrm{VOCl}\) (d) \(\mathrm{SO}_{2}+4 \mathrm{H}^{+}+4 \mathrm{Fe}^{2+}-\mathrm{S}+4 \mathrm{Fe}^{3+}+2 \mathrm{H}_{2} \mathrm{O}\) (e) \(2 \mathrm{CrO}_{2} \mathrm{Cl}_{2}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\left[\mathrm{Cr}_{2} \mathrm{O}_{7}\right]^{2-}+4 \mathrm{Cl}^{-}+6 \mathrm{H}^{+}\) \((\mathrm{f})\left[\mathrm{IO}_{4}\right]^{-}+2 \mathrm{I}^{-}+\mathrm{H}_{2} \mathrm{O} \rightarrow\left[\mathrm{IO}_{3}\right]^{-}+\mathrm{I}_{2}+2[\mathrm{OH}]^{-}\) \((\mathrm{g}) 2 \mathrm{KCl}+\mathrm{SnCl}_{4} \rightarrow \mathrm{K}_{2}\left[\mathrm{SnCl}_{6}\right]\) (h) \(2 \mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HNO}_{2}+\mathrm{HNO}_{3}\)

Short answer

a) No redox. b) Redox. c) Disproportionation. d) Redox. e) No net redox. f) Redox. g) No redox. h) Disproportionation.

Step-by-step solution

Recognize Redox Reaction

Identify if the given reaction involves oxidation or reduction processes. Oxidation is the loss of electrons, while reduction is the gain of electrons.

Balance Charges

For each chemical equation, ensure that the number of electrons lost and gained is balanced. This helps in identifying the oxidizing and reducing agents.

Analyze Each Reaction Separately

Go through each given reaction from (a) to (h) and determine if it's an oxidation, reduction, disproportionation, or no redox change based on the balanced equations and the processes involved.

Analysis of Reaction (a)

The bicarbonate ion \([ ext{HCO}_3^-]\) transforms into carbonate ion \([ ext{CO}_3^{2-}]\), where there is no change in oxidation state. The reaction represents no redox change.

Analysis of Reaction (b)

In \( ext{Au} + ext{HNO}_3 + 4 ext{HCl} ightarrow ext{HAuCl}_4 + ext{NO} + 2 ext{H}_2 ext{O}\), \( ext{Au}\) is oxidized from 0 to +3 (in \( ext{AuCl}_4^- \)), and \( ext{N} ext{O}_3^-\) is reduced to \( ext{NO}\). This is a redox reaction involving oxidation and reduction.

Analysis of Reaction (c)

Here, \([2 ext{VOCl}_2 o ext{VOCl}_3 + ext{VOCl} ]\), vanadium exists in different oxidation states (IV to V and III). This is a disproportionation reaction.

Analysis of Reaction (d)

Analyzing \( ext{SO}_2 + 4 ext{H}^+ + 4 ext{Fe}^{2+} ightarrow ext{S} + 4 ext{Fe}^{3+} + 2 ext{H}_2 ext{O}\), \( ext{SO}_2\) is reduced to sulfur, and \( ext{Fe}^{2+}\) is oxidized to \( ext{Fe}^{3+}\). This reaction encompasses both oxidation and reduction.

Analysis of Reaction (e)

For the reaction \[2 ext{CrO}_2 ext{Cl}_2 + 3 ext{H}_2 ext{O} ightarrow ext{Cr}_2 ext{O}_7^{2-} + 4 ext{Cl}^- + 6 ext{H}^+ \], the conversion of chromium from \( ext{Cr}^{+6}\) to \( ext{Cr}^{+6}\) doesn't imply redox. However, Cl is reduced from \( ext{Cl}_{2}\) to \( ext{Cl}^{-}\). This indicates no net redox change, just a substitution.

Analysis of Reaction (f)

In \([ ext{IO}_4^- + 2 ext{I}^- + ext{H}_2 ext{O} ightarrow ext{IO}_3^- + ext{I}_2 + 2 ext{OH}^-]\), iodide ion \( ext{I}^{-}\) is oxidized to \( ext{I}_2\), and \( ext{IO}_4^-\) is reduced to \( ext{IO}_3^-\). This is a redox reaction involving both oxidation and reduction.

Analysis of Reaction (g)

In \(2 ext{KCl} + ext{SnCl}_{4} ightarrow ext{K}_2[ ext{SnCl}_{6}]\), potassium cation \( ext{K}^+\) gains and loses no electrons; hence, it undergoes no redox change.

Analysis of Reaction (h)

In \(2 ext{NO}_{2} + ext{H}_2 ext{O} ightarrow ext{HNO}_2 + ext{HNO}_3\), nitrogen in \( ext{NO}_{2}\) is both reduced (to \( ext{HNO}_2\)) and oxidized (to \( ext{HNO}_3\)), indicating a disproportionation reaction.

Key concepts

Reduction

Reduction is one part of a redox reaction, often occurring simultaneously with oxidation. It involves the gain of electrons by an atom, ion, or molecule. In a chemical context, when an element decreases its oxidation state, it is considered to be undergoing reduction.
To easily remember this, consider the mnemonic "OIL RIG," which stands for 'Oxidation Is Loss' and 'Reduction Is Gain,' referring to the electrons.

• An important example of reduction, as seen in the original exercise, is reaction (b): nitric acid (HNO₃) is reduced to nitric oxide (NO), indicating the gain of electrons by nitrogen.
• In reaction (f), the iodide ion (I⁻) is reduced to iodine triiodide (IO₃⁻).

Recognizing reduction is crucial because it helps identify which substances act as oxidizing agents (those that gain electrons). In typical reactions, identifying the gain of electrons requires examining the changes in oxidation states of the substances involved.

Oxidation

Oxidation refers to the loss of electrons during a chemical reaction. This results in an increase in the oxidation state of the substance. In redox reactions, oxidation occurs alongside reduction, where one species loses electrons and another gains them.
To keep it simple, think of oxidation as the opposite of reduction. Remember the "OIL" in "OIL RIG," which stands for 'Oxidation Is Loss' of electrons.

• An example from the original task is reaction (b), where gold (Au) is oxidized from its elemental state to a +3 oxidation state in HAuCl₄.
• In reaction (d), the conversion of Fe²⁺ to Fe³⁺ is a classic example of oxidation, as iron loses an electron.

Understanding oxidation is essential for spotting the reducing agents (those that lose electrons) in reactions. These reactions are vital in processes ranging from industrial applications to biological systems.

Disproportionation

Disproportionation is a fascinating type of redox reaction where a single substance undergoes both oxidation and reduction. Often, this leads to the same element forming two different substances with distinct oxidation states.
This type of reaction can initially seem confusing because it involves self-competition in a sense. One part of the molecule is oxidized, while another is simultaneously reduced.

• Looking at the original exercise, reaction (c) with vanadium compounds is a great example, where vanadium shows a shift from an oxidation state of IV to both V and III.
• In reaction (h), nitrogen in NO₂ is disproportionated into nitrous acid (HNO₂) and nitric acid (HNO₃).

Grasping the concept of disproportionation can be highly beneficial, especially for understanding certain chemical reactions in the environment and industrial chemistry, where such reactions play critical roles.

Chemical Equations Analysis

The analysis of chemical equations involves understanding the processes occurring in a reaction, which can include oxidation, reduction, or non-redox changes. A systematic approach usually helps, by examining the reactants and products to determine which substances are oxidized or reduced.
This process is crucial for correctly balancing equations and predicting the substances formed.

• Each reaction, such as in step-by-step solutions of the original exercise, focuses on determining if it is a redox, no change, or a disproportionation reaction. For example, reaction (a) signifies no redox change, as there is no alteration in the oxidation states.
• In reactions where no net redox change occurs, it is important to recognize that other types of chemical processes, such as substitutions, may be taking place.

Analyzing chemical equations enables one to identify and balance redox reactions, a fundamental aspect of chemistry for understanding the interactions during chemical transformations.