Problem 5 - Chemistry: The Central Science in SI Units, Global Edition

Question

For a spontaneous reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow \mathrm{A}^{-}(a q)+\) \(\mathrm{B}^{+}(a q),\) answer the following questions: (a) If you made a voltaic cell out of this reaction, what halfreaction would be occurring at the cathode, and what half reaction would be occurring at the anode? (b) Which half-reaction from (a) is higher in potential energy? (c) What is the sign of \(E_{\text {cell }}^{\circ}\) ?

Step-by-Step Solution

Answer

The cathode half-reaction is \(\mathrm{B} + e^{-} \longrightarrow \mathrm{B}^{+}\), and the anode half-reaction is \(\mathrm{A} \longrightarrow \mathrm{A}^{-} + e^{-}\). We cannot determine which half-reaction has higher potential energy without the standard electrode potentials for A and B. However, since the reaction is spontaneous, the sign of \(E_{\text {cell}}^{\circ}\) is positive.

Step 1: Identify the redox process

First, let's write the given reaction: \(\mathrm{A}(aq)+\mathrm{B}(aq) \longrightarrow \mathrm{A}^{-}(aq)+\mathrm{B}^{+}(aq)\) In this reaction, A is losing an electron to become A⁻, and B is gaining an electron to become B⁺. So, A is being oxidized (loss of electron) and B is being reduced (gain of electron).

Step 2: Identify the cathode and anode half-reactions

In a voltaic cell, the cathode is where the reduction occurs, and the anode is where oxidation occurs. So, we can identify the half-reactions: Cathode (reduction): \(\mathrm{B} + e^{-} \longrightarrow \mathrm{B}^{+}\) Anode (oxidation): \(\mathrm{A} \longrightarrow \mathrm{A}^{-} + e^{-}\)

Step 3: Determine the half-reaction with higher potential energy

To determine which half-reaction has higher potential energy, we need to compare their standard electrode potentials, denoted as \(E^{\circ}\). However, we are not given the values for \(E_{\text {A}}^{\circ}\) and \(E_{\text {B}}^{\circ}\) in the problem. Since we don't have this information, we cannot definitively answer this part of the question.

Step 4: Determine the sign of E_cell

The cell potential, \(E_{\text {cell}}^{\circ}\), is given by the difference between the cathode's potential and the anode's potential: \[E_{\text {cell}}^{\circ} = E_{\text {cathode}}^{\circ} - E_{\text {anode}}^{\circ}\] As we don't have the standard electrode potentials for A and B, we can't directly calculate the value of \(E_{\text {cell}}^{\circ}\). However, we are told that the reaction is spontaneous. For a spontaneous reaction, the cell potential must be positive: \[E_{\text {cell}}^{\circ} > 0\] So, without knowing the actual values of the electrode potentials, we can conclude that the sign of \(E_{\text {cell}}^{\circ}\) is positive.

Key Concepts

Redox Reactions

Redox reactions are vital to understanding electrochemistry and involve the transfer of electrons between substances. In the reaction \( \mathrm{A}(aq)+\mathrm{B}(aq) \rightarrow \mathrm{A}^{-}(aq)+\mathrm{B}^{+}(aq) \), A loses an electron to become \( \mathrm{A}^{-} \), and B gains an electron to become \( \mathrm{B}^{+} \). This means A is oxidized (loss of electrons), and B is reduced (gain of electrons). Hence, the acronym "OIL RIG" (Oxidation Is Loss, Reduction Is Gain) can help remember this process.

Identifying which substances undergo oxidation or reduction is crucial because it tells us which direction the electrons flow. This flow of electrons is what drives electrochemical processes.

Voltaic Cells

Voltaic or galvanic cells convert chemical energy into electrical energy through redox reactions. In our given reaction, B, which is reduced, acts at the cathode because cathodes host reduction reactions. Meanwhile, A, which is oxidized, acts at the anode, because anodes host oxidation reactions.

To set up a voltaic cell:

• The half-reaction at the cathode is: \( \mathrm{B} + e^{-} \rightarrow \mathrm{B}^{+} \)
• The half-reaction at the anode is: \( \mathrm{A} \rightarrow \mathrm{A}^{-} + e^{-} \)

This separation of half-reactions in different compartments allows for the movement of electrons through an external circuit, generating electricity that can power devices.

Standard Electrode Potentials

Standard electrode potentials \( (E^{\circ}) \) provide a measure of a substance's tendency to gain or lose electrons. Unfortunately, the exercise didn't provide specific values for \( E_{\text{A}}^{\circ} \) or \( E_{\text{B}}^{\circ} \), so we couldn't calculate which half-reaction has higher potential energy.

However, understanding these potentials lets us predict the direction of electron flow in a cell. A higher \( E^{\circ} \) value means a stronger tendency to gain electrons (reduction), while a lower \( E^{\circ} \) value indicates a strong tendency to lose electrons (oxidation). This aids in determining which side of a cell acts as the cathode or anode.

Spontaneity of Reactions

The spontaneity of electrochemical reactions is linked to the cell potential \( (E_{\text{cell}}^{\circ}) \). Calculated as the difference between the cathode and anode potentials, if \( E_{\text{cell}}^{\circ} > 0 \), the reaction is spontaneous.

In the exercise, it was noted that the reaction is spontaneous, indicating a positive overall cell potential. This spontaneous nature means the reaction can proceed without external energy, suitable for generating electricity.

Understanding the concept of spontaneity helps predict whether a voltaic cell will work efficiently. It's key for designing cells and batteries for practical applications, ensuring that they deliver sufficient energy.