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Chapter 18: Problem 140

Consider a cell based on the following half-reactions: $$\begin{aligned}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{C}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{C}^{\circ}=0.77 \mathrm{~V} \end{aligned}$$ a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$\mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q)$$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The cell under standard conditions can be represented as: Anode: Fe, where the oxidation reaction is \(Fe^{3+} + e^- \to Fe^{2+}\), and the concentration of [Fe鲁鈦篯 = 1 M and [Fe虏鈦篯 = 1 M. Cathode: Au, where the reduction reaction is \(Au^{3+} + 3e^- \to Au\), and the concentration of [Au鲁鈦篯 = 1 M. Using the Nernst equation with the given cell potential of 0.31 V and the reaction \(Au^{3+}(aq) + 4\mathrm{Cl}^-(aq) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(aq)\), the equilibrium constant K is calculated to be approximately \(6.23 \times 10^{-11}\).

Step by step solution

01

Draw the cell under standard conditions

To draw the cell under standard conditions, we first need to identify the half-reactions that occur at the anode and the cathode based on their reduction potentials. $$\begin{aligned}\mathrm{Au}^{3+} + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}\ & \ \mathscr{E}^{\circ} = 1.50 V \\\ \mathrm{Fe}^{3+} + \mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \ \mathscr{E}^{\circ} = 0.77 V \end{aligned}$$ Since the reduction potential of Au鲁鈦� is greater than that of Fe鲁鈦�, Au鲁鈦� will be reduced at the cathode, and Fe鲁鈦� will be oxidized at the anode. The cell representation is as follows: Anode | Cathode -|- \(Fe^{3+} + e^- \to Fe^{2+}\) | \(Au^{3+} + 3e^- \to Au\) Fe | Au Now, label the direction of electron flow, and the concentrations: - Anode: Fe, oxidation occurs, electrons flow from the anode to the cathode - Cathode: Au, reduction occurs, electrons flow to the cathode from the anode Concentrations: - Anode compartment: [Fe鲁鈦篯 = 1 M, [Fe虏鈦篯 = 1 M (standard conditions) - Cathode compartment: [Au鲁鈦篯 = 1 M (standard conditions)
02

Calculate the value of K for the given reaction

Use the Nernst equation and the given cell potential (0.31 V) to find the equilibrium constant K for the reaction: $$\mathrm{Au}^{3+}(aq) + 4\mathrm{Cl}^-(aq) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(aq)$$ The Nernst equation is: $$\mathscr{E} = \mathscr{E}^\circ - \frac{RT}{nF} \ln{Q}$$ Where: - \(\mathscr{E}\) is the cell potential under non-standard conditions - \(\mathscr{E}^{\circ}\) is the cell potential under standard conditions - R is the gas constant (8.314 J/mol路K) - T is the temperature in Kelvin (298 K for 25掳C) - n is the number of electrons transferred (3 in this case) - F is the Faraday constant (96485 C/mol) - Q is the reaction quotient Since \(\mathrm{Au}^{3+}\) is being reduced in the cell, the overall reaction in the cell can be expressed as: $$\mathrm{Au}^{3+} + 3\mathrm{e}^{-} + 4\mathrm{Cl}^{-} \rightleftharpoons \mathrm{Au} + \mathrm{AuCl}_{4}^{-}$$ Let the cell potential under non-standard conditions be 0.31 V. Notice that only 0.10 M of Cl鈦� ions are added to the Au鲁鈦� compartment for the non-standard cell potential. Now, we can use the Nernst equation to find the reaction quotient, Q: $$0.31 = 1.50 - \frac{8.314 \times 298}{3 \times 96485} \ln{Q}$$ Solve for Q: $$Q = \exp{\left(\frac{3(0.31-1.50) \times 96485}{8.314 \times 298}\right)} \approx 6.23 \times 10^{-11}$$ For the given reaction, at equilibrium, \(Q = K\), and so the equilibrium constant K is: $$K \approx 6.23 \times 10^{-11}$$

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Key Concepts

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

Half-Reactions
In electrochemical cells, the reactions occurring at the electrodes are often split into two separate processes called half-reactions. Each half-reaction represents either an oxidation or a reduction process.
For the provided cell, we have:
  • The cathode reaction: \(\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Au}\), with a standard reduction potential of 1.50 V.
  • The anode reaction: \(\mathrm{Fe}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\), with a standard reduction potential of 0.77 V.
Understanding half-reactions helps us determine which elements are oxidized or reduced.
In this exercise, gold (Au) is reduced, while iron (Fe) is oxidized. The electrons flow from the anode, where the oxidation occurs, to the cathode, where the reduction takes place. This flow is crucial for generating electric current in the cell.
Nernst Equation
The Nernst equation allows us to calculate the cell potential under non-standard conditions.
The general form is:\[\mathscr{E} = \mathscr{E}^\circ - \frac{RT}{nF}\ln{Q}\]Here,
  • \(\mathscr{E}\): cell potential at non-standard conditions.
  • \(\mathscr{E}^\circ\): standard cell potential.
  • \(R\): universal gas constant (8.314 J/mol路K).
  • \(T\): temperature in Kelvin (298 K in this case).
  • \(n\): number of moles of electrons exchanged (3 for this reaction).
  • \(F\): Faraday's constant (96485 C/mol).
  • \(Q\): reaction quotient.
In this problem, the Nernst equation helps find \(Q\) when the non-standard cell potential \(\mathscr{E}\) is given as 0.31 V. This equation captures how changes in concentration and reaction conditions affect the cell potential.
Cell Potential
Cell potential (or electromotive force, EMF) is the measure of the energy available from a redox reaction occurring within an electrochemical cell.
It's determined by the difference in reduction potentials between the cathode and the anode.
Under standard conditions, it can be calculated as:\[\mathscr{E}^\circ_{\text{cell}} = \mathscr{E}^\circ_{\text{cathode}} - \mathscr{E}^\circ_{\text{anode}}\]For the cell in the exercise:
  • Cathode (Au鲁鈦� to Au): \(1.50\, \text{V}\).
  • Anode (Fe鲁鈦� to Fe虏鈦�): \(0.77\, \text{V}\).
Thus, \(\mathscr{E}^\circ_{\text{cell}} = 1.50\, \text{V} - 0.77\, \text{V} = 0.73\, \text{V}\).
These calculations allow us to predict how much voltage a cell can produce, aiding in designing batteries or understanding their efficiency under varying conditions.
Equilibrium Constant
The equilibrium constant \(K\) gives us the ratio of concentrations of products to reactants at equilibrium for a reversible reaction.
It's a crucial parameter in determining reaction extents in chemical systems.
In our case, we found the \(K\) value using the reaction where \(\mathrm{Au}^{3+}(aq) + 4\mathrm{Cl}^{-}(aq) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(aq)\)
By applying the Nernst equation and substituting in the given cell potential (0.31 V), we calculated \(Q\), and subsequently \(K\), reflecting the system's equilibrium status.
This value \(K \approx 6.23 \times 10^{-11}\) indicates how the formation of \(\mathrm{AuCl}_{4}^{-}\) is influenced by the added \(\mathrm{Cl}^-\) ions. A small \(K\) value means that at equilibrium, very few products form, emphasizing the dominance of reactants, which can profoundly impact reaction predictability in practical applications.

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Most popular questions from this chapter

Nerve impulses are electrical "signals" that pass through neurons in the body. The electrical potential is created by the differences in the concentration of \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions across the nerve cell membrane. We can think about this potential as being caused by a concentration gradient, similar to what we see in a concentration cell (keep in mind that this is a very simple explanation of how nerves work; there is much more involved in the true biologic process). A typical nerve cell has a resting potential of about \(-70 \mathrm{mV}\). Let's assume that this resting potential is due only to the \(\mathrm{K}^{+}\) ion concentration difference. In nerve cells, the \(\mathrm{K}^{+}\) concentration inside the cell is larger than the \(\mathrm{K}^{+}\) concentration outside the cell. Calculate the \(\mathrm{K}^{+}\) ion concentration ratio necessary to produce a resting potential of \(-70 . \mathrm{mV}\). $$\frac{\left[\mathrm{K}^{+}\right]_{\text {inside }}}{\left[\mathrm{K}^{+}\right]_{\text {outside }}}=?$$

Sketch a galvanic cell, and explain how it works. Look at Figs. \(18.1\) and \(18.2 .\) Explain what is occurring in each container and why the cell in Fig. \(18.2\) "works" but the one in Fig. \(18.1\) does not.

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{MAg}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\). a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\) \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \quad K=?\)

A solution at \(25^{\circ} \mathrm{C}\) contains \(1.0 \mathrm{M} \mathrm{Cd}^{2+}, 1.0 \mathrm{MAg}^{+}, 1.0 \mathrm{M} \mathrm{Au}^{3+}\), and \(1.0 \mathrm{M} \mathrm{Ni}^{2+}\) in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 \mathrm{M}) \| \mathrm{Cu}^{2+}(2.50 \mathrm{M})\right| \mathrm{Cu}$$ The mass of each electrode is \(200 . \mathrm{g}\). a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after \(10.0 \mathrm{~A}\) of current has flowed for \(10.0 \mathrm{~h}\). (Assume each half-cell contains \(1.00 \mathrm{~L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{~h}\). d. How long can this battery deliver a current of \(10.0 \mathrm{~A}\) before it goes dead?
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