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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°

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

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Most popular questions from this chapter

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?

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?

If the equilibrium constant for a one-electron redox reaction at 298 \(\mathrm{K}\) is \(8.7 \times 10^{4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an aluminum strip placed in a solution of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) , and the other has a nickel strip placed in a solution of \(\mathrm{NiSO}_{4}\) . The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \mathrm{Ni}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. $$ \begin{array}{l}{\text { (a) } \mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)} \\\ {\text { (b) } 2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)} \\\ {\text { (c) } 3 \mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow 3 \mathrm{S}(s)+} \\\\{\quad\quad 2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$
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