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Chapter 8: Problem 83

Given that $$ \begin{aligned} &\mathrm{I}_{2}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{I}^{-} ; E^{\circ} 0.54 \mathrm{~V} \\ &\mathrm{Br}_{2}+2 \mathrm{e} \longrightarrow 2 \mathrm{Br} ; E^{\circ}=1.09 \mathrm{~V} \end{aligned} $$ Which of the following is a true statement? (1) lodide will reduce bromine to bromide ion. (2) lodine is a stronger oxidising agent than bromine. (3) Bromine will reducc iodine to iodide ion. (4) \(\Lambda\) ll arc truc.

Short Answer

Expert verified
Statement 1 is true.

Step by step solution

01

Understand the given reactions

We are given two half-reactions with their standard electrode potentials: 1. \ \( \mathrm{I}_2 + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-} \); \(E^{\circ} = 0.54 \mathrm{~V}\) 2. \ \( \mathrm{Br}_2 + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \); \(E^{\circ} = 1.09 \mathrm{~V}\)
02

Determine oxidizing and reducing agents

The standard electrode potential \(E^{\circ}\) indicates how easily a species gains electrons. The more positive the \(E^{\circ}\), the better an oxidizing agent it is. Since \(E^{\circ}\) for \(\mathrm{Br}_2\) is 1.09 V and for \(\mathrm{I}_2\) is 0.54 V, \(\mathrm{Br}_2\) is a stronger oxidizing agent than \(\mathrm{I}_2\).
03

Analyze the given statements

1. \(\mathrm{I}^-\) can reduce \(\mathrm{Br}_2\) to \(\mathrm{Br}^-\) because \(\mathrm{Br}_2\) is a stronger oxidizing agent.2. \(\mathrm{I}_2\) being a weaker oxidizing agent than \(\mathrm{Br}_2\) means this statement is false.3. \(\mathrm{Br}_2\) cannot reduce \(\mathrm{I}_2\) to \(\mathrm{I}^-\) because it is a stronger oxidizing agent. 4. Since not all statements are true, this is false.
04

Conclude the answer

Based on the analysis, the correct statement is: \(\mathrm{I}^-\) can reduce \(\mathrm{Br}_2\) to \(\mathrm{Br}^-\).

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Key Concepts

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

Standard Electrode Potential
Standard electrode potential, denoted as E^{\text{掳}}, indicates the tendency of a chemical species to gain or lose electrons, and thus, to be reduced or oxidized. The more positive the E^{\text{掳}}, the greater the species' ability to gain electrons and act as an oxidizing agent.
For example, in the reactions given:

    <\begin{aligned}\( &\text{I}_2 + 2 \text{e}^{-} \rightarrow 2 \text{I}^{-} ;E^{\text{掳}} = 0.54 \text{~V} \)\( &\text{Br}_2 + 2 \text{e}^{-} \rightarrow 2 \text{Br}^{-} ;E^{\text{掳}} = 1.09 \text{~V} \)\text{Br}_2, with a higher E^{\text{掳}}, is a better oxidizing agent than \text{I}_2.<
Oxidizing Agents
Oxidizing agents are substances that gain electrons during a chemical reaction and are reduced in the process. They effectively cause the oxidation of another substance. The strength of an oxidizing agent is directly related to its standard electrode potential (E^{\text{掳}}).

    For instance:
    \begin{aligned}\( &\text{Br}_2 + 2 \text{e}^{-} \rightarrow 2 \text{Br}^{-};E^{\text{掳}} = 1.09 \text{~V} \)\( &\text{I}_2 + 2 \text{e}^{-} \rightarrow 2 \text{I}^{-};E^{\text{掳}} = 0.54 \text{~V} \)Since \text{Br}_2 has a higher E^{\text{掳}} value than \text{I}_2, \text{Br}_2 is a stronger oxidizing agent than \text{I}_2.
Reducing Agents
Reducing agents are substances that lose electrons during a chemical reaction and are oxidized in the process. They cause the reduction of another substance. The strength of a reducing agent is inversely related to its standard electrode potential (E^{\text{掳}}).

    In the example reactions:
    \begin{aligned}\( &\text{I}_2 + 2 \text{e}^{-} \rightarrow 2 \text{I}^{-};E^{\text{掳}} = 0.54 \text{~V} \)\( &\text{Br}_2 + 2 \text{e}^{-} \rightarrow 2 \text{Br}^{-};E^{\text{掳}} = 1.09 \text{~V} \)Since \text{I}^{-} has a lower E^{\text{掳}} value compared to \text{Br}^{-}, \text{I}^{-} is a stronger reducing agent than \text{Br}^{-}.
Redox Reactions
Redox reactions involve the transfer of electrons between two species. One species is reduced by gaining electrons, while the other is oxidized by losing electrons. The standard electrode potentials (E^{\text{掳}}) can predict the direction of electron flow in redox reactions.

    Consider the following half-reactions:
    \begin{aligned}\( &\text{I}_2 + 2 \text{e}^{-} \rightarrow 2 \text{I}^{-};E^{\text{掳}} = 0.54 \text{~V} \)\( &\text{Br}_2 + 2 \text{e}^{-} \rightarrow 2 \text{Br}^{-};E^{\text{掳}} = 1.09 \text{~V} \)Given these half-reactions, \text{Br}_2 is a better oxidizing agent than \text{I}_2 due to its higher E^{\text{掳}} value.

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