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Write unbalanced oxidation and reduction half-reactions for the following processes: (a) \(\mathrm{Te}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{TeO}_{2}(s)+\mathrm{NO}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
(a) Oxidation: \( \mathrm{Te} \rightarrow \mathrm{TeO}_{2} \); Reduction: \( \mathrm{NO}_{3}^{-} \rightarrow \mathrm{NO} \). (b) Oxidation: \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} \); Reduction: \( \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} \).

Step by step solution

01

Identifying the Reducing and Oxidizing Agents for Reaction (a)

For the reaction \( \mathrm{Te}(s) + \mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{TeO}_{2}(s) + \mathrm{NO}(g) \), identify the species being oxidized and the one being reduced. \( \mathrm{Te} \) starts as \( \mathrm{Te}(s) \) and transforms into \( \mathrm{TeO}_{2}(s) \), indicating oxidation as it gains oxygen atoms. \( \mathrm{NO}_{3}^{-} \) is reduced to \( \mathrm{NO}(g) \) as it loses oxygen atoms.
02

Writing the Oxidation Half-Reaction for Reaction (a)

In the oxidation half-reaction, \( \mathrm{Te}(s) \) is converted to \( \mathrm{TeO}_{2}(s) \). This process can be initially represented as: \( \mathrm{Te}(s) \rightarrow \mathrm{TeO}_{2}(s) \) without balancing.
03

Writing the Reduction Half-Reaction for Reaction (a)

For the reduction process, \( \mathrm{NO}_{3}^{-}(aq) \) is reduced to \( \mathrm{NO}(g) \). The unbalanced reduction half-reaction is represented as: \( \mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{NO}(g) \).
04

Identifying the Reducing and Oxidizing Agents for Reaction (b)

In the reaction \( \mathrm{H}_{2} \mathrm{O}_{2}(aq) + \mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \), identify that \( \mathrm{Fe}^{2+} \) is oxidized to \( \mathrm{Fe}^{3+} \), and \( \mathrm{H}_{2} \mathrm{O}_{2} \) is reduced to \( \mathrm{H}_{2} \mathrm{O} \).
05

Writing the Oxidation Half-Reaction for Reaction (b)

For the oxidation half-reaction, \( \mathrm{Fe}^{2+}(aq) \) is oxidized to \( \mathrm{Fe}^{3+}(aq) \). The unbalanced reaction is: \( \mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) \).
06

Writing the Reduction Half-Reaction for Reaction (b)

In the reduction half-reaction, \( \mathrm{H}_{2} \mathrm{O}_{2}(aq) \) is reduced to \( \mathrm{H}_{2} \mathrm{O}(l) \). Represent this initially as: \( \mathrm{H}_{2} \mathrm{O}_{2}(aq) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) \).

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Key Concepts

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

Half-Reactions
In redox chemistry, understanding half-reactions is key to deconstruct complex processes. Half-reactions break a redox reaction into two parts, showing either the oxidation or reduction process.

They allow us to focus on individual electron transfer events, simplifying the analysis:
  • **Oxidation Half-Reaction:** Describes the species losing electrons (being oxidized).
  • **Reduction Half-Reaction:** Describes the species gaining electrons (being reduced).
These half-reactions are especially useful because they can be balanced individually before combining to give a balanced redox reaction.

Let's see an example with the reaction \( \mathrm{Te}(s) + \mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{TeO}_{2}(s) + \mathrm{NO}(g) \). Here, we would separate into half-reactions: oxidation where \( \mathrm{Te}(s) \) forms \( \mathrm{TeO}_{2}(s) \), and reduction where \( \mathrm{NO}_{3}^{-}(aq) \) converts into \( \mathrm{NO}(g) \).

By considering these steps, one may more easily manage even complicated reactions.
Redox Chemistry
Redox chemistry involves the study of oxidation-reduction reactions, which are processes where electrons are transferred between chemical species.

Here’s the basic framework of a redox reaction:
  • **Oxidation:** Loss of electrons by a molecule, atom, or ion.
  • **Reduction:** Gain of electrons by a molecule, atom, or ion.
These events occur simultaneously — when one species is oxidized, another is always reduced.

In the given reactions, such as \( \mathrm{H}_{2} \mathrm{O}_{2}(aq) + \mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \), you see this interplay of oxidation and reduction clearly.

Understanding the flow of electrons helps chemists determine the directing force of the reaction and the energy changes involved. This is fundamental in processes like battery operations and various biological systems, including cellular respiration.
Oxidizing Agents
In redox chemistry, an oxidizing agent is the substance that is reduced in a reaction and, in the process, causes another substance to be oxidized.

Key characteristics include:
  • An oxidizing agent gains electrons during the reaction.
  • It is typically reduced itself.
  • Commonly involves elements like oxygen, fluorine, or nitrate.
Taking our first reaction: \(\mathrm{Te}(s) + \mathrm{NO}_{3}^{-}(aq) \rightarrow \mathrm{TeO}_{2}(s) + \mathrm{NO}(g)\), the \(\mathrm{NO}_{3}^{-}\) is reduced to \(\mathrm{NO}\), gaining electrons from \(\mathrm{Te}\). Hence, \(\mathrm{NO}_{3}^{-}\) is the oxidizing agent.

Learning to identify these agents allows us to predict products and understand the driving forces of reactions in various chemical and industrial applications.
Reducing Agents
Contrarily to oxidizing agents, reducing agents are substances that lose electrons in a redox reaction, enabling another substance to be reduced.

Important points about reducing agents include:
  • A reducing agent donates electrons to another species.
  • It is oxidized in the process.
  • Frequently include elements or compounds like hydrogen, metals, or sulfur.
For instance, in the second reaction: \(\mathrm{H}_{2} \mathrm{O}_{2}(aq) + \mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{H}_{2} \mathrm{O}(l)\), \(\mathrm{Fe}^{2+}\) is oxidized to \(\mathrm{Fe}^{3+}\), donating electrons to \(\mathrm{H}_{2} \mathrm{O}_{2}\). Therefore, \(\mathrm{Fe}^{2+}\) acts as the reducing agent.

Identifying reducing agents is crucial for manipulating redox reactions, in both natural processes and chemical syntheses, such as in metallurgical industries.

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Most popular questions from this chapter

Balance the following net ionic equation by the half-reaction method. The reaction takes place in acidic solution. $$ \mathrm{NO}_{3}^{-}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{NO}(g)+\mathrm{Cu}^{2+}(a q) \quad \text { Unbalanced } $$

The following cell reactions occur spontaneously: $$ \begin{aligned} &\mathrm{B}+\mathrm{A}^{+} \longrightarrow \mathrm{B}^{+}+\mathrm{A} \\ &\mathrm{C}+\mathrm{A}^{+} \longrightarrow \mathrm{C}^{+}+\mathrm{A} \\ &\mathrm{B}+\mathrm{C}^{+} \longrightarrow \mathrm{B}^{+}+\mathrm{C} \end{aligned} $$ (a) Arrange the following reduction half-reactions in order of decreasing tendency to occur: \(\mathrm{A}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{A}, \mathrm{B}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{B}\), and \(\mathrm{C}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{C}\) (b) Which of these substances \(\left(\mathrm{A}, \mathrm{A}^{+}, \mathrm{B}, \mathrm{B}^{+}, \mathrm{C}, \mathrm{C}^{+}\right)\) is the strong- est oxidizing agent? Which is the strongest reducing agent? (c) Which of the three cell reactions delivers the highest voltage?

The sodium-sulfur battery has molybdenum electrodes with anode and cathode compartments separated by \(\beta\) -alumina, a ceramic through which sodium ions can pass. Because the battery operates at temperatures above \(300^{\circ} \mathrm{C}\), all the reactants and products are present in a molten solution. The cell voltage is about \(2.0 \mathrm{~V}\). (a) What is the cell reaction if the shorthand notation is \(\mathrm{Mo}(s)\left|\mathrm{Na}(s o l n), \mathrm{Na}^{+}(s o l n) \| \mathrm{S}(s o l n), \mathrm{S}^{2-}(s o l n)\right| \mathrm{Mo}(s) ?\) (b) How many kilograms of sodium are consumed when a \(25 \mathrm{~kW}\) sodium- sulfur battery produces current for \(32 \mathrm{~min} ?\)

Aluminum, titanium, and several other metals can be colored by an electrochemical process called anodizing. Anodizing oxidizes a metal anode to yield a porous metal oxide coating that can incorporate dye molecules to give brilliant colors. In the oxidation of aluminum, for instance, the electrode reactions areCathode (reduction): \(6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow 3 \mathrm{H}_{2}(g)\) Anode (oxidation): \(2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\frac{\mathrm{Al}_{2} \mathrm{O}_{3}(s)+6 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-}}{\text {Overall reaction: }} 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g)\) The thickness of the aluminum oxide coating that forms on the anode can be controlled by varying the current flow during the electrolysis. How many minutes are required to produce a \(0.0100 \mathrm{~mm}\) thick coating of \(\mathrm{Al}_{2} \mathrm{O}_{3}\left(\right.\) density \(\left.3.97 \mathrm{~g} / \mathrm{cm}^{3}\right)\) on a square piece of aluminum metal \(10.0 \mathrm{~cm}\) on an edge if the current passed through the piece is \(0.600 \mathrm{~A} ?\)

A galvanic cell has a silver electrode in contact with \(0.050 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) and a copper electrode in contact with \(1.0 \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) (a) Write a balanced equation for the cell reaction, and calculate the cell potential at \(25^{\circ} \mathrm{C}\). (b) Excess \(\mathrm{NaBr}(a q)\) is added to the \(\mathrm{AgNO}_{3}\) solution to precipitate AgBr. What is the cell potential at \(25^{\circ} \mathrm{C}\) after the precipitation of \(\mathrm{AgBr}\) if the concentration of excess \(\mathrm{Br}^{-}\) is \(1.0 \mathrm{M}\) ? Write a balanced equation for the cell reaction under these conditions. \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgBr}\) at \(25^{\circ} \mathrm{C}\) is \(5.4 \times 10^{-13}\).) (c) Use the result in part (b) to calculate the standard reduction potential \(E^{\circ}\) for the half-reaction $$ \mathrm{AgBr}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q) $$

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