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Chapter 20: Problem 66

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \mathrm{AgCl}(s)+\mathrm{e}^{-\longrightarrow} \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two cell compartments have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=2.55 M\), respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cl \(^{-}\) ] will increase, decrease, or stay the same as the cell operates.

Short Answer

Expert verified
(a) The cathode is the electrode with a higher concentration of Cl鈦�, which is the one with [Cl鈦籡 = 2.55 M. (b) The standard emf of the cell is 0 V, as both electrodes have the same half-reaction. (c) Applying the Nernst equation, the cell emf for the given concentrations is approximately 0.236 V. (d) As the cell operates, [Cl鈦籡 will decrease at the cathode and increase at the anode.

Step by step solution

01

Identify the Cathode and Anode

The cathode is the electrode where reduction occurs, while the anode is where oxidation occurs. Given the half-reaction: \(\mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)\) This is a reduction half-reaction because electrons are on the reactant side and Ag is reduced. Thus, the electrode with a higher concentration of Cl^- will be the cathode, as it will capture more electrons and facilitate the reaction. In this case, \(2.55\,\mathrm{M} > 0.0150\,\mathrm{M}\), so the electrode with \([\mathrm{Cl}^{-}] = 2.55\,\mathrm{M}\) is the cathode.
02

Calculate the Standard Emf

The standard emf of the cell (\(E^{0}_{cell}\)) is the difference between the standard reduction potentials of cathode (\(E^{0}_{cathode}\)) and anode (\(E^{0}_{anode}\)). In this case, as both the cathode and the anode are based on the same half-reaction, their standard reduction potentials will be the same, so the standard emf of the cell will be zero. \(E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = 0\)
03

Calculate the Cell Emf for the Given Concentrations

To calculate the cell emf for the given concentrations, we will apply the Nernst equation: \(E_{cell} = E^{0}_{cell} - \frac{RT}{nF}\ln{Q}\) where: \(E^{0}_{cell}\) = standard emf of the cell (calculated in Step 2) \(R\) = ideal gas constant = \(8.314\,\mathrm{J\,K^{-1}mol^{-1}}\) \(T\) = temperature (assuming \(298\,\mathrm{K}\)) \(n\) = number of electrons transferred in the reaction (in this case, \(n = 1\)) \(F\) = Faraday's constant = \(96485\,\mathrm{C/mol}\) \(Q\) = reaction quotient The reaction quotient, \(Q\), can be calculated using the Cl^- concentrations in the cathode and anode: \(Q = \frac{[\mathrm{Cl}^{-}]_{anode}}{[\mathrm{Cl}^{-}]_{cathode}} = \frac{0.0150\,\mathrm{M}}{2.55\,\mathrm{M}}\) Now, we can plug all these values into the Nernst equation: \(E_{cell} = 0 - \frac{(8.314\,\mathrm{J\,K^{-1}mol^{-1}})(298\,\mathrm{K})}{(1)(96485\,\mathrm{C/mol})}\ln{\left(\frac{0.0150\,\mathrm{M}}{2.55\,\mathrm{M}}\right)}\) Solving this equation to find the cell emf: \(E_{cell} \approx 0.236\,\mathrm{V}\)
04

Predict the Behavior of [Cl^-] as the Cell Operates

As the cell operates, the cathode will reduce AgCl into Ag and Cl^-, while the anode will oxidize Ag and Cl^- into AgCl. The reaction progress will tend to lower the [Cl^-] concentration at the cathode and increase the [Cl^-] concentration at the anode, maintaining a constant overall cell potential. Thus, for each electrode: - Cathode: [Cl^-] will decrease. - Anode: [Cl^-] will increase.

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Key Concepts

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

Voltaic Cell
A voltaic cell, also known as a galvanic cell, is a type of electrochemical cell that transforms chemical energy into electrical energy through spontaneous redox reactions. The key components include two electrodes: the anode and the cathode, each residing in separate compartments called half-cells. A salt bridge or porous disc connects these compartments, facilitating ion exchange while preventing the solutions from mixing.

In a voltaic cell, the anode is where oxidation occurs and electrons are released, traveling through an external circuit to the cathode, where reduction happens. This movement of electrons generates an electric current, which is harnessed to perform work, like powering a device.

The silver-silver chloride voltaic cell we've been discussing operates using the silver chloride reduction half-reaction \( \text{AgCl}(s) + \text{e}^- \rightarrow \text{Ag}(s) + \text{Cl}^- (aq) \).In this specific example, the differing \([\text{Cl}^-]\) concentrations between compartments determine which acts as the cathode and anode.
Nernst Equation
The Nernst equation is a fundamental relationship in electrochemistry for determining the electromotive force (emf) of an electrochemical cell under non-standard conditions. It allows us to calculate the cell potential when the concentrations of the reactants and products are not at standard conditions (1 M concentration, 1 atm pressure, 298 K temperature).

The general form of the Nernst equation is:\[E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\]where:
  • \(E_{cell}\) is the cell potential under specific conditions.
  • \(E^0_{cell}\) is the standard cell potential, based on the standard reduction potentials of the cathode and anode.
  • \(R\) is the universal gas constant (8.314 J K-1 mol-1).
  • \(T\) is the temperature in Kelvin.
  • \(n\) is the number of moles of electrons exchanged in the reaction.
  • \(F\) is Faraday's constant (96485 C mol-1).
  • \(Q\) is the reaction quotient, which reflects the ratio of concentrations of products to reactants.
    • In our specific voltaic cell scenario, the Nernst equation aids in computing the non-standard cell emf using the given \([\text{Cl}^-]\) concentrations for the cathode and anode, resulting in a calculated cell potential.
Cell Emf
Electromotive force, or emf, is the potential difference between the two electrodes in a voltaic cell when no current flows. It represents the cell's ability to drive electrons through a circuit, and is a crucial measure of the cell's energy-producing capacity.

The standard emf (\(E^0_{cell}\)) depends on the intrinsic reduction potential of the electrodes involved. In the discussed voltaic cell with silver-silver chloride electrodes reacting via the same half-reaction, \(\text{AgCl}(s) + \text{e}^- \rightarrow \text{Ag}(s) + \text{Cl}^- (aq)\),this results in both electrodes having identical standard reduction potentials, yielding a standard emf of zero.

However, under actual working conditions, the cell emf will differ due to varying ion concentrations. Here, the Nernst equation comes into play, showing how even with an initial standard emf of zero, the actual emf arises from the concentration gradient of \([\text{Cl}^-]\).This results in a calculated non-zero cell potential, in line with our previous calculations.

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Most popular questions from this chapter

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=-0.858 \mathrm{~V} \\ & E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(\mathrm{aq})+2 \mathrm{OH}^{-}(a q) \\ \mathrm{En}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}(s) & E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \\ E_{\mathrm{red}}^{\circ}=-0.14 \mathrm{~V} \end{aligned} $$ (a) Write the cell reaction for the combination of these half-cell reactions that leads to the largest positive cell emf, and calculate the value. (b) Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf, and calculate that value.

Gold metal dissolves in aqua regia, a mixture of concentrated hydrochloric acid and concentrated nitric acid. The standard reduction potentials \(\begin{aligned} \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-}-\mathrm{Au}(s) & E_{\mathrm{red}}^{\circ} &=+1.498 \mathrm{~V} \\ \mathrm{AuCl}_{4}^{-}(a q)+3 \mathrm{e}^{-}--\rightarrow \mathrm{Au}(\mathrm{s})+4 \mathrm{Cl}^{-}(a q) & \\\ E_{\mathrm{red}}^{\circ} &=+1.002 \mathrm{~V} \end{aligned}\) are important in gold chemistry. (a) Use half-reactions to write a balanced equation for the reaction of Au and nitric acid to produce \(\mathrm{Au}^{3+}\) and \(\mathrm{NO}(\mathrm{g})\), and calculate the standard emf of this reaction. Is this reaction spontaneous? (b) Use half-reactions to write a balanced equation for the reaction of \(\mathrm{Au}\) and hydrochloric acid to produce \(\mathrm{AuCl}_{4}^{-}(a q)\) and \(\mathrm{H}_{2}(g)\), and calculate the standard emf of this reaction. Is this reaction spontaneous? (c) Use half-reactions to write a balanced equation for the reaction of Au and aqua regia to produce \(\mathrm{AuCl}_{4}^{-}(a q)\) and \(\mathrm{NO}(\mathrm{g})\), and calculate the standard emf of this reaction. Is this reaction spontaneous under standard conditions? (d) Use the Nernst equation to explain why aqua regia made from concentrated hydrochloric and nitric acids is able to dissolve gold.

This oxidation-reduction reaction in acidic solution is spontaneous: $$ \begin{array}{r} 5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)-\rightarrow \\ 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ A solution containing \(\mathrm{KMnO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is poured into one beaker, and a solution of \(\mathrm{FeSO}_{4}\) is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at \(298 \mathrm{~K}\) when the concentrations are the following: \(\mathrm{pH}=0.0, \quad\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M}, \quad\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

A common shorthand way to represent a voltaic cell is to list its components as follows: anode|anode solution || cathode solution|cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}\); sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{H}^{+}\right| \mathrm{H}_{2} ;\) sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+& 6 \mathrm{H}^{+}(a q)--\rightarrow \\ & \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ \(\mathrm{Pt}\) is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{Cl}^{-}\). Sketch the cell.

If you were going to apply a small potential to a steel ship resting in the water as a means of inhibiting corrosion, would you apply a negative or a positive charge? Explain.
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