Question

For a spontaneous reaction \(\mathrm{A}(a q) \rightarrow \mathrm{A}^{-}(a q)+\) \(\mathrm{B}^{+}(a q),\) answer the following questions: (a) If you made a voltaic cell out of this reaction, what half-reaction 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 \(E_{\text { cell }}^{\circ} ?[\) Section 20.3\(]\)

Short answer

For the given spontaneous reaction, the half-reactions occurring in a voltaic cell are: Anode (oxidation): \(B^+ \rightarrow B + e^-\) Cathode (reduction): \(A + e^- \rightarrow A^-\) The half-reaction with higher potential energy is the oxidation half-reaction at the anode: \(B^+ \rightarrow B + e^-\). The sign of \(E_{\text {cell}}^{\circ}\) must be positive for the reaction to proceed spontaneously in a voltaic cell.

Step-by-step solution

Identify the oxidation and reduction half-reactions

For this reaction: \[ A(aq) \rightarrow A^-(aq) + B^+(aq) \] We must find out which species is getting oxidized and which one is getting reduced. When we look at the charges, we can deduce the following: A is neutral and becomes negatively charged (A^-). This means A must have gained one electron (reduction): \[A + e^- \rightarrow A^-\] B^+ is positively charged and combines with A^- to form a neutral species. This means B must have lost one electron (oxidation): \[B^+ \rightarrow B + e^-\]

Identify the half-reactions at the cathode and anode

In a voltaic cell, the oxidation half-reaction occurs at the anode, and the reduction half-reaction takes place at the cathode. For this reaction, Anode (oxidation): \[B^+ \rightarrow B + e^-\] Cathode (reduction): \[A + e^- \rightarrow A^-\]

Determine which half-reaction has higher potential energy

The half-reaction with higher potential energy is the one that occurs at the anode because it is the site of oxidation and electron loss. Higher potential energy: \[B^+ \rightarrow B + e^-\]

Determine the sign of E_cell°

The cell potential, E_cell°, is directly related to the Gibbs free energy change (ΔG°) of the cell reaction. For spontaneous reactions (like the one given), ΔG° is negative, indicating a decrease in Gibbs free energy. Thus, the cell potential, E_cell°, must be positive for the reaction to proceed spontaneously in a voltaic cell. Sign of E_cell°: Positive

Key concepts

Oxidation-Reduction Reactions

In electrochemistry, oxidation-reduction (redox) reactions are processes where electrons are transferred between substances. These reactions are fundamental in generating electricity in voltaic cells or batteries. Oxidation involves the loss of electrons, while reduction involves the gain of electrons.
For example, consider the reaction \( A(aq) \rightarrow A^-(aq) + B^+(aq) \). Here, \( A \) gains an electron and becomes \( A^- \), indicating reduction. Conversely, \( B^+ \) loses an electron and forms \( B \), signifying oxidation.

• Oxidation: \( B^+ \rightarrow B + e^- \)
• Reduction: \( A + e^- \rightarrow A^- \)

Recognizing which component is oxidized or reduced helps in understanding the direction and spontaneity of the reaction.

Voltaic Cell

A voltaic cell, also known as a galvanic cell, converts chemical energy into electrical energy through redox reactions. In such a cell, two half-reactions occur separately at electrodes, called the anode and cathode.

The anode is where oxidation occurs, releasing electrons, while reduction happens at the cathode, where electrons are gained. This separation of half-reactions creates an electric current from electron flow.
In our example:

• Anode: \( B^+ \rightarrow B + e^- \) (oxidation)
• Cathode: \( A + e^- \rightarrow A^- \) (reduction)

The movement of electrons from the anode to the cathode generates electricity, powering devices connected to the cell.

Half-Reactions

Half-reactions are simplified representations of redox reactions, showing either the oxidation or reduction process separately. These reactions help determine which direction electrons will flow and how the cell will function.

Each half-reaction has its potential energy level, measured as electrode potential. Understanding these reactions allows us to predict possible electric potential of electrochemical cells and determine spontaneity. For our exercise, the half-reactions are:

• Anode (oxidation): \( B^+ \rightarrow B + e^- \)
• Cathode (reduction): \( A + e^- \rightarrow A^- \)

This separation is crucial for analyzing electrochemical processes and improving cell efficiency.

Gibbs Free Energy

Gibbs free energy, a thermodynamic property, explains the spontaneity of reactions. In electrochemistry, it connects to the cell's electric potential through the relation: \[ \Delta G^\circ = -nFE^\circ_{cell} \] where \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( E^\circ_{cell} \) is the standard cell potential.

A spontaneous reaction has negative Gibbs free energy (\( \Delta G^\circ < 0 \)). Therefore, for a voltaic cell, the cell potential \( E^\circ_{cell} \) must be positive, as seen in our example. This positive value indicates an ability to do work, illustrating that the cell efficiently converts chemical energy into electrical energy, a rewarding affirmation of its functionality.