Question

For a spontaneous reaction \(\mathrm{A}(a q)+\mathrm{B}(a q) \longrightarrow \mathrm{A}^{-}(a q)+\) \(\mathrm{B}^{+}(\mathrm{at})\), 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 halfreaction would be occurring at the anode? (b) Which half-reaction from (a) is higher in potential energy? (c) What is the sign of \(E_{\text {cell? }}^{\text {? }}\) [Section 20.3]

Short answer

In a voltaic cell based on the spontaneous reaction \(A(aq) + B(aq) \rightarrow A^-(aq) + B^{+}(aq)\), the half-reaction occurring at the anode (oxidation) is \(A \rightarrow A^- + e^-\), and the half-reaction at the cathode (reduction) is \(B + e^- \rightarrow B^+\). The half-reaction with a higher standard electrode potential (E°) will have higher potential energy, but specific values are not provided. However, since the reaction is spontaneous, the sign of E_cell is positive.

Step-by-step solution

Identify the oxidation and reduction half-reactions

To determine the half-reactions happening at the anode and cathode, we first need to recognize the oxidation and reduction half-reactions. In the given reaction: A(aq) + B(aq) -> A^-(aq) + B^+(aq) Species A is losing an electron and becoming negatively charged (A -> A^- + e^-). This is the oxidation half-reaction. On the other hand, species B is gaining an electron and becoming positively charged (B + e^- -> B^+). This is the reduction half-reaction.

Determine half-reactions in a voltaic cell

In a voltaic cell, the anode is where the oxidation occurs, and the cathode is where the reduction happens. From Step 1, we have: Oxidation: A -> A^- + e^- Reduction: B + e^- -> B^+ Thus, in the voltaic cell made from this reaction, the half-reaction occurring at the anode is: A -> A^- + e^- And the half-reaction occurring at the cathode is: B + e^- -> B^+

Determine the higher potential energy half-reaction

To find which half-reaction has higher potential energy, we will compare the relative standard electrode potentials (E°) of the two half-reactions. As standard electrode potentials are not given in the problem, we cannot provide a definite answer in this case. However, we can say that the half-reaction with a higher E° value will have higher potential energy.

Find the sign of E_cell

The sign of the cell potential (E_cell) tells us whether the reaction is spontaneous or non-spontaneous. In this case, we are given that the overall reaction is spontaneous. A spontaneous reaction in a voltaic cell has a positive cell potential, which means the following: E_cell = E_cathode - E_anode > 0 In this exercise, although specific values of E° cannot be determined, we can conclude that the sign of E_cell is positive since the reaction is spontaneous.

Key concepts

Oxidation-Reduction Reaction

An oxidation-reduction, or redox reaction, is a fundamental chemical process where electrons are transferred between species. This reaction involves two essential half-reactions: oxidation and reduction. The oxidation half-reaction refers to when a species loses electrons, while the reduction half-reaction is when a species gains electrons. In the given example, the species A loses an electron to become \(A^{-}\), demonstrating oxidation. Conversely, species B gains an electron to transform into \(B^{+}\), representing the reduction process.

• Oxidation involves electron loss: \(A \rightarrow A^{-} + e^{-}\)
• Reduction involves electron gain: \(B + e^{-} \rightarrow B^{+}\)

Understanding these half-reactions is crucial because they determine where each process occurs in a voltaic cell. Oxidation will occur at the anode, and reduction will happen at the cathode. Mastering these concepts allows one to predict how a redox reaction will proceed in an electrochemical context.

Spontaneous Reaction

A spontaneous reaction is a natural process that occurs without external input of energy. Such reactions are accompanied by a change in Gibbs free energy, typically resulting in a negative value, indicating that the process will proceed on its own. In the context of electrochemistry, a spontaneous reaction within a voltaic cell yields a positive cell potential (\(E_{\text{cell}} > 0\)), which means that the reaction is favored to proceed as described.

The overall spontaneity of a reaction in a voltaic cell can be established by examining the standard electrode potentials of the half-reactions involved. The equation for cell potential is expressed as:

• \(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\)
• For a spontaneous reaction: \(E_{\text{cell}} > 0\)

A positive value here confirms the spontaneous nature, and no energy input is required to drive the reaction forward. Such understanding is key in predicting the direction and feasibility of electrochemical processes.

Cell Potential

Cell potential, denoted as \(E_{\text{cell}}\), is a measure of the voltage supplied by a voltaic cell within a reaction. It tells us the energetic difference between the two electrodes: the cathode and the anode. This voltage originates from the potential energy stored during the electron transfer process. The cathode, where reduction occurs, typically has a higher potential than the anode, where oxidation occurs.

To calculate \(E_{\text{cell}}\), the formula used combines the standard electrode potentials of both half-reactions:

• \(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\)

The ability to predict the cell potential allows chemists to determine the feasibility and efficiency of a reaction. A positive potential indicates a spontaneous reaction, applicable in batteries and other energy conversion devices. Understanding and applying these concepts can support the design of more effective and sustainable energy systems.