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Chapter 20: Problem 5

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

Expert verified
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

01

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): \[B^+ \rightarrow B + e^-\]
02

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): \[A + e^- \rightarrow A^-\]
03

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: \[B^+ \rightarrow B + e^-\]
04

Determine the sign of E_cell掳

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

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Key Concepts

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

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.

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Most popular questions from this chapter

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s)} \\ {\text { (b) } \mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q)} \\ {\text { (c) } \mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q)} \\ {\text { (d) } 2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{NO}(g)+} \\ \quad {4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q)}\end{array} $$

(a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell.(b) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard hydrogen electrode?

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{c}{\mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)} \\ {\mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \quad \longrightarrow 2 \mathrm{I}^{-}(a q)}\end{array} $$ The cell is operated at 298 \(\mathrm{K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 M\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\) (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) were equal to \(0.15 \mathrm{M},\) at what concentration of I \(\mathrm{I}^{-}\) would the cell have zero potential?

Given the following half-reactions and associated standard reduction potentials: $$ \begin{array}{c}{\text { AuBr }_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{V}} \\ {\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

Copper corrodes to cuprous oxide, \(\mathrm{Cu}_{2} \mathrm{O},\) or cupric oxide, \(\mathrm{CuO},\) depending on environmental conditions. (a) What is the oxidation state of copper in cuprous oxide? (b) What is the oxidation state of copper in cupric oxide? (c) Copper peroxide is another oxidation product of elemental copper. Suggest a formula for copper peroxide based on its name. (d) Copper(III) oxide is another unusual oxidation product of elemental copper. Suggest a chemical formula for copper(II) oxide.
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