/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 102 A voltaic cell is constructed th... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 102

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{aligned} \mathrm{Cu}^{*}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{Cu}(s) \\ \mathrm{l}_{2}(s)+2 \mathrm{c}^{-} & \longrightarrow 2 \mathrm{I}^{-}(a q) \end{aligned} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 \mathrm{M}\) and \(\left[1^{-}\right]=3.5 \mathrm{M}\). (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 \({ }^{-}\)would the cell have zero potential?

Short Answer

Expert verified
#Step 2: Calculate the standard cell potential, E掳# #tag_title# (Calculate the standard cell potential) #tag_content# To calculate E掳, we first need to find the standard reduction potential, E掳, for each half-reaction: Cu鈦�(aq) + e鈦� 鈫� Cu(s) E掳 = +0.522 V I鈧�(s) + 2e鈦� 鈫� 2I鈦�(aq) E掳 = +0.536 V Now we can write the net reaction and calculate the standard cell potential: Cu鈦�(aq) + I鈧�(s) 鈫� Cu(s) + 2I鈦�(aq) E掳 = (+0.522 V) - (+0.536 V) = -0.014 V #Step 3: Calculate the cell potential, E, for given concentrations# #tag_title# (Calculate the cell potential) #tag_content# Using the Nernst equation, we can calculate the cell potential, E: E = -0.014 V - (8.314 J/mol K 脳 298 K * ln(0.25 M/3.5虏 M))/(2 mol 脳 96485 C/mol) E = -0.014 V - 0.05916 V 脳 (ln(0.0214)) E = -0.014 V + 0.0701 V E = 0.0561 V #Step 4: Determine the anode and cathode of the cell# #tag_title# (Determine the anode and cathode) #tag_content# Since the cell potential, E, calculated in step 3, is positive, the reaction is spontaneous as written. Within the net reaction, Cu鈦� is reduced to Cu(s) and I鈧�(s) is oxidized to I鈦�(aq). Therefore, Cu鈦�(aq) + e鈦� 鈫� Cu(s) is the cathode half-reaction, and I鈧�(s) + 2 e鈦� 鈫� 2I鈦�(aq) is the anode half-reaction. #Step 5: Analyze the anode under standard conditions# #tag_title# (Analyze the anode under standard conditions) #tag_content# Since E掳 is negative (-0.014 V), the net reaction is non-spontaneous under standard conditions. However, our initial conditions make the reaction spontaneous. If we switch back to standard conditions, E掳 would remain negative, so the answer to part (b) would not be the same under standard conditions. #Step 6: Determine the concentration of I鈦� for zero cell potential# #tag_title# (Determine the concentration of I鈦�) #tag_content# To find the concentration of I鈦� when the cell potential is zero and [Cu鈦篯 = 0.15 M, we can use the Nernst equation and set E = 0 V: 0 V = -0.014 V - (8.314 J/mol K 脳 298 K * ln(0.15 M/[I鈦籡虏))/(2 mol 脳 96485 C/mol) Solving for [I鈦籡, we find that the concentration of I鈦� needed for zero cell potential is approximately [I鈦籡 = 2.45 M.

Step by step solution

01

(Write the Nernst equation for the cell potential)

The Nernst equation helps us calculate the cell potential under non-standard conditions. It is given by: \(E = E^0 - (\frac{RT}{nF})\ln Q\) where \(E\) is the cell potential, \(E^0\) represents the standard cell potential, \(R\) is the gas constant (8.314 J/mol K), \(T\) is the temperature (298 K), \(n\) is the number of electrons transferred, \(F\) is the Faraday constant (96,485 C/mol), and \(Q\) is the reaction quotient.

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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 device that transforms chemical energy into electrical energy. This transformation occurs through spontaneous oxidation-reduction reactions. In a voltaic cell, there's a flow of electrons from the anode to the cathode through an external circuit, which produces electric current. The cell itself is made up of two half-cells, each containing a different redox reaction.

In the context of the given exercise, one half-cell involves the reduction of Cu ions to solid copper, while the other involves the reduction of iodine to iodide ions. These reactions occur in their respective half-cells, and by setting them up in a voltaic cell configuration, we can harness electrical energy from the reactions.

Key components to remember in a voltaic cell are:
  • Anode: The electrode where oxidation occurs. Electrons leave the cell here.
  • Cathode: The electrode where reduction happens. Electrons enter the cell here.
  • Salt Bridge: Allows ions to flow between the two half-cells to maintain charge balance.
Nernst Equation
The Nernst Equation is a vital formula in electrochemistry, helping us calculate the cell potential (\( E \)) of a voltaic cell under non-standard conditions. The equation adjusts the standard cell potential (\( E^0 \)) by considering the concentration of the reacting species, the temperature, and the number of electrons transferred in the reaction.

The equation is:
\[ E = E^0 - \left( \frac{RT}{nF} \right) \ln Q \]

Where each variable stands for the following:
  • \( E^0 \): Standard cell potential in volts.
  • \( R \): Universal gas constant, 8.314 J/mol K.
  • \( T \): Temperature, here given as 298 K.
  • \( n \): Number of moles of electrons transferred in the reaction.
  • \( F \): Faraday's constant, 96,485 C/mol.
  • \( Q \): Reaction quotient, representing the ratio of product and reactant activities.
This equation is crucial when concentrations deviate from 1 M, life deviates from standard conditions.
Standard Cell Potential
The standard cell potential (\( E^0 \)) is a measure of how much voltage the cell can produce under standard conditions: 1 M concentration for each ion, 1 atm pressure for any gases, and a temperature of 298 K. It is determined by the relative strengths of the two half-reactions making up the cell.

To calculate the \( E^0 \) for the entire cell, we take the difference between the standard reduction potentials of the cathode and anode reactions. This is defined as:
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} \]

This value tells us if a reaction is spontaneous. A positive \( E^0_{cell} \) indicates that the reaction can proceed spontaneously in the forward direction. In our exercise, determining \( E^0_{cell} \) requires looking up standard reduction potentials for each half-reaction involved.
Reaction Quotient
The reaction quotient (\( Q \)) is a key player in determining the direction the reaction will proceed, as well as in calculating the cell potential using the Nernst equation. It is similar to the equilibrium constant, but \( Q \) is applicable at any point in the reaction, not just at equilibrium.

The formula for the reaction quotient is:
\[ Q = \frac{[products]}{[reactants]} \]
Where each concentration is raised to the power equivalent to its coefficient in the balanced equation. In a voltaic cell, concentrations that deviate significantly from standard conditions will affect \( Q \), and thus the cell potential.

In the exercise, \( Q \) would be calculated based on the concentrations of copper ions and iodide ions. By altering these concentrations, the reaction will either move towards equilibrium or away from it, affecting the overall cell potential.

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Most popular questions from this chapter

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) \(\mathrm{Ni}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)\) (acidic solution)

Given the following half-reactions and associated standard reduction potentials: \(\mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)\) \(\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) \quad E_{\text {red }}^{\mathrm{o}}=-0.858 \mathrm{~V}\) \(\mathrm{HO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{2}(a q)+\mathrm{e}^{\circ} \longrightarrow \mathrm{Eu}^{2}(a q)=-0.43 \mathrm{~V}\) \(1 \mathrm{O}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)\) (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate

A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s) \longrightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode half-cell is \(1.00 \mathrm{M}\) and the cell generates an emf of \(+0.22 \mathrm{~V}\), what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? (b) If the anode half-cell contains \(\left[\mathrm{SO}_{4}{ }^{2-}\right]=1.00 \mathrm{M}\) in equilibrium with \(\mathrm{PbSO}_{4}(s)\), what is the \(K_{4 p}\) of \(\mathrm{PbSO}_{4}\) ? Batteries and Fuel Cells (Section 20.7)

From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger reducing agent: (a) Fe(s) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Ca}(s)\) or \(\mathrm{Al}(s)\) (c) \(\mathrm{H}_{2}\) (g, acidic solution) or \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) (d) \(\mathrm{BrO}_{3}^{-}(a q)\) or \(\mathrm{lO}_{3}^{-}(a q)\) 20.44 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\operatorname{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{2}(\mathrm{~g})\) 20.45 By using the data in Appendix E, determine whether each of the following substances is likely to serve as an exidant or a reductant: (a) \(\mathrm{Cl}_{2}\) (g), (b) \(\mathrm{MnO}_{4}^{-}\)(aq, acidic solution), (c) \(\mathrm{Ba}\) (s), (d) \(\mathrm{Zn}(\) s). 20.46 Is each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(\mathrm{aq})\), (b) \(\mathrm{Ca}(\mathrm{s})\), (c) \(\mathrm{CO}_{3}^{-}(\mathrm{aq})\), (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) ? \(20.47\) (a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{C}_{2}, \mathrm{O}_{2}\) (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: \(\mathrm{Zn}, \mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{AL}\). 20.48 Based on the data in Appendix E, (a) which of the following is the strongest oxidizing agent and which is the weakest in acidic solution: \(\mathrm{Br}_{2}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Zn}, \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-} ?\) (b) Which of the following is the strongest reducing agent, and which is the weaket in acidic solution: \(\mathrm{F}^{-}, \mathrm{Zn}, \mathrm{N}_{1}{ }^{+}\), \(\mathrm{I}_{\mathrm{n}} \mathrm{NO}\) ?

A voltaic cell that uses the reaction $$ \mathrm{PdCl}_{4}{ }^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{CT}(a q)+\mathrm{Cd}^{2+}(a q) $$ has a measured standard cell potential of \(+1.03 \mathrm{~V}\). (a) Write the two half-cell reactions. (b) By using data from Appendix \(E\), determine \(E_{\text {rel }}^{0}\) for the reaction involving \(\mathrm{Pd}\). (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.
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