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Chapter 4: Problem 44

For the complete redox reactions given here, write the half-reactions and identify the oxidizing and reducing agents. (a) \(4 \mathrm{Fe}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}\) (b) \(\mathrm{Cl}_{2}+2 \mathrm{NaBr} \longrightarrow 2 \mathrm{NaCl}+\mathrm{Br}_{2}\) (c) \(\mathrm{Si}+2 \mathrm{~F}_{2} \longrightarrow \mathrm{SiF}_{4}\) (d) \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{HCl}\)

Short Answer

Expert verified
(a) Oxidizing agent: \(\mathrm{O_{2}}\), Reducing agent: \(\mathrm{Fe}\);\n(b) Oxidizing agent: \(\mathrm{Cl_{2}}\), Reducing agent: \(\mathrm{Br^-}\);\n(c) Oxidizing agent: \(\mathrm{F_{2}}\), Reducing agent: \(\mathrm{Si}\);\n(d) Oxidizing agent: \(\mathrm{Cl_{2}}\), Reducing agent: \(\mathrm{H_{2}}\).

Step by step solution

01

Break Down a into oxidation and reduction half-reactions

Starting with (a) \(4\mathrm{Fe}+3\mathrm{O_{2}} \longrightarrow 2\mathrm{Fe_{2}O_{3}}\)\nFor the oxidation half-reaction: \(4\mathrm{Fe} \longrightarrow 8\mathrm{Fe^{+3}} + 24\mathrm{e^-}\).\nFor the reduction half-reaction: \(3\mathrm{O_{2}} + 24\mathrm{e^-} \longrightarrow 6\mathrm{O^{2-}}\).
02

Identifying oxidizing and reducing agents

The reducing agent is the substance that gets oxidized, in this case, it's \(\mathrm{Fe}\), and the oxidizing agent is the substance that gets reduced, which is \(\mathrm{O_{2}}\) in this reaction.
03

Repeat For Remaining Reactions

Repeat steps 1 and 2 for the rest of the given reactions: \n\n(b) \(\mathrm{Cl_{2}}+2\mathrm{NaBr} \longrightarrow 2\mathrm{NaCl}+\mathrm{Br_{2}}\)\nOxidation: \(2\mathrm{Br^-} \longrightarrow \mathrm{Br_{2}} + 2\mathrm{e^-}\) \nReduction: \(\mathrm{Cl_{2}} + 2\mathrm{e^-} \longrightarrow2\mathrm{Cl^-}\)\nOxidizing agent is \(\mathrm{Cl_{2}}\), and the reducing agent is \(\mathrm{Br^-}\).\n\n(c) \(\mathrm{Si}+2\mathrm{F_{2}}\longrightarrow\mathrm{SiF_{4}}\)\nOxidation: \(\mathrm{Si} \longrightarrow \mathrm{Si^{4+}} + 4\mathrm{e^-}\)\nReduction: \(2\mathrm{F_{2}} + 4\mathrm{e^-}\longrightarrow4\mathrm{F}\)\nOxidizing agent is \(\mathrm{F_{2}}\), and the reducing agent is \(\mathrm{Si}\).\n\n(d) \(\mathrm{H_{2}}+\mathrm{Cl_{2}}\longrightarrow2\mathrm{HCl}\)\nOxidation: \(\mathrm{H_{2}} \longrightarrow 2\mathrm{H^+} + 2\mathrm{e^-}\)\nReduction: \(\mathrm{Cl_{2}} + 2\mathrm{e^-} \longrightarrow2\mathrm{Cl^-}\)\nOxidizing agent is \(\mathrm{Cl_{2}}\), and the reducing agent is \(\mathrm{H_{2}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation and Reduction Half-Reactions
Understanding redox reactions begins with recognizing the individual processes of oxidation and reduction that occur. A redox reaction is made up of two half-reactions: one that describes the loss of electrons (oxidation) and another that describes the gain of electrons (reduction).

In the given example (a)
  1. Oxidation Half-Reaction: \(4\mathrm{Fe} \longrightarrow 8\mathrm{Fe^{+3}} + 24\mathrm{e^-}\).

    2. Here, iron (\(\mathrm{Fe}\)) loses electrons, thus undergoing oxidation. Each iron atom loses three electrons, indicating a change in its oxidation state from 0 to +3.
  2. Reduction Half-Reaction: \(3\mathrm{O_{2}} + 24\mathrm{e^-} \longrightarrow 6\mathrm{O^{2-}}\).

    3. Conversely, oxygen (\(\mathrm{O_{2}}\)) gains electrons and is reduced, its oxidation state changes from 0 to -2. The interaction between these half-reactions demonstrates how electrons are transferred in the reaction. By breaking down complex reactions into their respective half-reactions, we make them easier to analyze and understand.
Oxidizing and Reducing Agents
Identifying oxidizing and reducing agents is crucial for fully grasping redox reactions. An oxidizing agent, or oxidant, gains electrons and is reduced during the reaction, while a reducing agent, or reductant, loses electrons and is oxidized.

Referring back to example (a),
  1. The reducing agent is iron (\(\mathrm{Fe}\)), which gives up electrons.
  2. The oxidizing agent is oxygen (\(\mathrm{O_{2}}\)), which accepts electrons.

Recognizing the role of these agents helps us understand the flow of electrons in the reaction. In all the given examples, from (a) to (d), the substance that yields electrons becomes the reducing agent, and the substance that accepts electrons becomes the oxidizing agent, allowing the chemical reaction to proceed.
Chemical Reaction Balancing
Balancing chemical reactions is a fundamental step in the study of chemistry, ensuring that the same number of each type of atom appears on both sides of the equation, in compliance with the law of conservation of mass. In redox reactions, this balancing act also includes ensuring that the same number of electrons is lost and gained.

In our examples, balancing is achieved by 'equalizing' the number of electrons lost in the oxidation half-reaction with those gained in the reduction half-reaction. For instance, in reaction (c) we have:
  1. Oxidation: \(\mathrm{Si} \longrightarrow \mathrm{Si^{4+}} + 4\mathrm{e^-}\).
  2. Reduction: \(2\mathrm{F_{2}} + 4\mathrm{e^-} \longrightarrow 4\mathrm{F^-}\).

A balanced redox equation signals that all atoms and the charge are conserved. Balancing reactions often requires practice, but once mastered, it provides a solid foundation for understanding complex chemical processes.

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Most popular questions from this chapter

A \(0.8870-\mathrm{g}\) sample of a mixture of \(\mathrm{NaCl}\) and \(\mathrm{KCl}\) is dissolved in water, and the solution is then treated with an excess of \(\mathrm{AgNO}_{3}\) to yield \(1.913 \mathrm{~g}\) of \(\mathrm{AgCl}\). Calculate the percent by mass of each compound in the mixture.

Which of the following metals can react with water? (a) \(\mathrm{Au},(\mathrm{b}) \mathrm{Li},(\mathrm{c}) \mathrm{Hg},(\mathrm{d}) \mathrm{Ca},(\mathrm{e}) \mathrm{Pt}\)

A useful application of oxalic acid is the removal of rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) from, say, bathtub rings according to the reaction \(\begin{aligned} \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+& 6 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \longrightarrow \\ & 2 \mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}(a q)+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{H}^{+}(a q) \end{aligned}\) Calculate the number of grams of rust that can be removed by \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) solution of oxalic acid.

Give a chemical explanation for each of the following: (a) When calcium metal is added to a sulfuric acid solution, hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stops even though none of the reactants is used up. (b) In the activity series, aluminum is above hydrogen, yet the metal appears to be unreactive toward steam and hydrochloric acid. (c) Sodium and potassium lie above copper in the activity series. In your explanation, discuss why \(\mathrm{Cu}^{2+}\) ions in a \(\mathrm{CuSO}_{4}\) solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place \(\mathrm{M}\) in the activity series? (e) Before aluminum metal was obtained by electrolysis, it was produced by reducing its chloride \(\left(\mathrm{AlCl}_{3}\right)\) with an active metal. What metals would you use to produce aluminum in that way?

A 22.02-mL solution containing \(1.615 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with a 28.64-mL solution containing \(1.073 \mathrm{~g}\) \(\mathrm{NaOH}\). Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive.
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