Problem 23 - Chemistry The Central Science In Si Units

Question

Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (acidic solution) (b) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (d) \(\mathrm{S}(s)(a q) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (acidic solution) (e) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{OH}^{-}(a q)\) (basic solution)

Step-by-Step Solution

Answer

(a) Reduction; (b) Oxidation; (c) Oxidation; (d) Reduction; (e) Reduction; (f) Reduction.

Step 1: Balancing Cr Half-Reaction

For the half-reaction \( ext{Cr}_{2} ext{O}_{7}^{2-} ightarrow ext{Cr}^{3+}\) in acidic solution:1. Balance the Cr atoms: \( ext{Cr}_{2} ext{O}_{7}^{2-} ightarrow 2 ext{Cr}^{3+}\).2. Balance the oxygen atoms with water: \( ext{Cr}_{2} ext{O}_{7}^{2-} + 14 ext{H}^{+} ightarrow 2 ext{Cr}^{3+} + 7 ext{H}_{2} ext{O}\).3. Balance the charges with electrons: For charge balance, add 6 electrons to the right side: \( ext{Cr}_{2} ext{O}_{7}^{2-} + 14 ext{H}^{+} + 6e^{-} ightarrow 2 ext{Cr}^{3+} + 7 ext{H}_{2} ext{O}\). It is a reduction reaction because electrons are gained.

Step 2: Balancing Mn Half-Reaction

For the half-reaction \( ext{Mn}^{2+} ightarrow ext{MnO}_{4}^{-}\) in acidic solution:1. Balance the Mn atoms: \( ext{Mn}^{2+} ightarrow ext{MnO}_{4}^{-}\).2. Balance the oxygen atoms with water: \( ext{Mn}^{2+} + 4 ext{H}_{2} ext{O} ightarrow ext{MnO}_{4}^{-}\).3. Balance the hydrogen atoms with \( ext{H}^{+}\): \( ext{Mn}^{2+} + 4 ext{H}_{2} ext{O} ightarrow ext{MnO}_{4}^{-} + 8 ext{H}^{+}\).4. Balance the charges with electrons: Add 5 electrons to the left: \( ext{Mn}^{2+} + 4 ext{H}_{2} ext{O} ightarrow ext{MnO}_{4}^{-} + 8 ext{H}^{+} + 5e^{-}\). It is an oxidation reaction because electrons are lost.

Step 3: Balancing Iodine Half-Reaction

For the half-reaction \( ext{I}_{2} ightarrow ext{IO}_{3}^{-}\) in acidic solution:1. Balance the iodine atoms: \( ext{I}_{2} ightarrow 2 ext{IO}_{3}^{-}\).2. Balance the oxygen atoms with water: \( ext{I}_{2} + 6 ext{H}_{2} ext{O} ightarrow 2 ext{IO}_{3}^{-}\).3. Balance the hydrogen atoms with \( ext{H}^{+}\): \( ext{I}_{2} + 6 ext{H}_{2} ext{O} ightarrow 2 ext{IO}_{3}^{-} + 12 ext{H}^{+}\).4. Balance the charges with electrons: Add 10 electrons to the right: \( ext{I}_{2} ightarrow 2 ext{IO}_{3}^{-} + 12 ext{H}^{+} + 10e^{-}\). It is an oxidation reaction because electrons are lost.

Step 4: Balancing Sulfur Half-Reaction

For the half-reaction \( ext{S} ightarrow ext{H}_{2} ext{S}\) in acidic solution:1. Balance the sulfur atoms: \( ext{S} ightarrow ext{H}_{2} ext{S}\).2. Balance the hydrogen atoms with \( ext{H}^{+}\): \( ext{S} + 2 ext{H}^{+} ightarrow ext{H}_{2} ext{S}\).3. Balance the charges with electrons: Add 2 electrons to the right: \( ext{S} + 2 ext{H}^{+} + 2e^{-} ightarrow ext{H}_{2} ext{S}\). It is a reduction reaction because electrons are gained.

Step 5: Balancing Nitrogen Half-Reaction

For the half-reaction \( ext{NO}_{3}^{-} ightarrow ext{NO}_{2}^{-}\) in basic solution:1. Balance the nitrogen atoms: Already balanced as is.2. Balance the oxygen atoms with \( ext{H}_{2} ext{O}\): \( ext{NO}_{3}^{-} ightarrow ext{NO}_{2}^{-} + ext{H}_{2} ext{O}\).3. Balance the hydrogen atoms with \( ext{OH}^{-}\): \( ext{NO}_{3}^{-} + ext{OH}^{-} ightarrow ext{NO}_{2}^{-} + ext{H}_{2} ext{O}\).4. Balance the charges with electrons: Add 2 electrons to the left: \( ext{NO}_{3}^{-} + ext{OH}^{-} + 2e^{-} ightarrow ext{NO}_{2}^{-} + ext{H}_{2} ext{O}\). It is a reduction reaction because electrons are gained.

Step 6: Balancing Peroxide Half-Reaction

For the half-reaction \( ext{H}_{2} ext{O}_{2} ightarrow ext{OH}^{-}\) in basic solution:1. Balance the oxygen atoms with \( ext{H}_{2} ext{O}\) if needed: \( ext{H}_{2} ext{O}_{2} ightarrow 2 ext{OH}^{-}\).2. Balance the hydrogen atoms with \( ext{OH}^{-}\): Already balanced.3. Balance the charges with electrons: Add 2 electrons to the right: \( ext{H}_{2} ext{O}_{2} + 2e^{-} ightarrow 2 ext{OH}^{-}\). It is a reduction reaction because electrons are gained.

Key Concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, also known as redox reactions, are processes where electrons are transferred between substances. One element undergoes oxidation, meaning it loses electrons, while another element undergoes reduction, meaning it gains electrons.
To determine whether a reaction involves oxidation or reduction:

• Identify the changes in oxidation state of the elements involved. Oxidation involves an increase in oxidation state, while reduction involves a decrease.
• Recognize oxidizing and reducing agents. The substance that gets reduced (gains electrons) is the oxidizing agent, and the one that gets oxidized (loses electrons) is the reducing agent.

Oxidation-reduction reactions are fundamental in chemistry, governing processes like combustion, metabolism, and corrosion. In the exercise, the balancing process is crucial to ensure the transfer of electrons is accurately represented. For example, balancing \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+} \] requires adjusting oxygen with water and hydrogen ions with charges to reflect electron exchange accurately.

Acidic and Basic Solutions

Balancing redox reactions can occur in either acidic or basic solutions, and the approach slightly differs. In acidic solutions, hydrogen ions (\( \mathrm{H}^{+} \)) are available to help balance the reaction.
Steps for balancing in acidic solutions:

• Balance atoms other than hydrogen and oxygen first.
• Add water (\( \mathrm{H}_2\mathrm{O} \)) to balance oxygen atoms.
• Use hydrogen ions (\( \mathrm{H}^{+} \)) to balance hydrogen atoms.
• Adjust for charge by adding electrons to balance the oxidation and reduction sides.

Basic solutions require hydroxide ions (\( \mathrm{OH}^{-} \)) for balancing. Once the acidic balance is achieved, neutralize added hydrogen ions by adding \( \mathrm{OH}^{-} \), converting \( \mathrm{H}^{+} \) into water.For example, in basic solutions, like \( \mathrm{NO}_{3}^{-} \longrightarrow \mathrm{NO}_{2}^{-} \), hydroxide is used to balance hydrogen, and the charge balance is maintained by electrons.

Electron Transfer

Electron transfer is the cornerstone of redox reactions. It involves the movement of electrons from one element or compound (the reducing agent) to another (the oxidizing agent). This electron flow is what drives the transformation seen in redox reactions.
To understand electron transfer:

• Assign oxidation numbers to determine which species is reduced and which is oxidized.
• Balance electron transfer to ensure the electrons lost in oxidation are gained in reduction.
• Customize the stoichiometry in the reaction so that each electron-space precisely accounts for the electron economy of the reaction.

In practical terms, the Cr half-reaction \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O} \) exemplifies a balanced electron transfer, highlighting how electrons facilitate converting dichromate ions to chromium ions in acidic medium.