/*! 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 34 A voltaic cell that uses the rea... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 34

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

Short Answer

Expert verified
The two half-cell reactions for the given redox reaction in the voltaic cell are: Oxidation reaction (anode): \(Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-\) Reduction reaction (cathode): \(PdCl_{4}^{2-}(aq) + 2e^- \rightarrow Pd(s) + 4Cl^-(aq)\) Using the given standard cell potential and the standard reduction potential of the Cd鈦郝�/Cd half-cell, we can calculate the standard reduction potential of the Pd reaction: \(E^{\circ}_{PdCl_{4}^{2-}/Pd} = 1.433 V\). The voltaic cell consists of a Cd anode in a solution with Cd虏鈦� ions and a Pd cathode in a solution with PdCl4虏鈦� ions. Electrons flow from the anode (Cd) to the cathode (Pd) through a wire, and a salt bridge connects the two beakers to maintain electrical neutrality.

Step by step solution

01

Write the two half-cell reactions for the given redox reaction

First, let's identify the two half-cell reactions for the given redox reaction. We can see that Pd is being reduced (from PdCl4虏鈦� to Pd) and Cd is being oxidized (from Cd to Cd虏鈦�). The two half-cell reactions are: Oxidation half-cell reaction: \(Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-\) Reduction half-cell reaction: \(PdCl_{4}^{2-}(aq) + 2e^- \rightarrow Pd(s) + 4Cl^-(aq)\)
02

Determine the reduction potential E for the Pd reaction

It's essential to use the data from Appendix E (standard electrode potentials) to find the standard reduction potential of the Pd reaction. We first need to find the standard reduction potential for the Cd鈦郝�/Cd half-cell from the data. According to Appendix E, \(E^{\circ}_{Cd^{2+}/Cd} = -0.403 V\) Recall that the overall standard cell potential (\(E^{\circ}_{cell}\)) is equal to the difference between the standard reduction potential of the cathode half-cell and the standard reduction potential of the anode half-cell: \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\) We have \(E^{\circ}_{cell} = +1.03 V\) and the Cd reaction is the anode, so its reduction potential (\(E^{\circ}_{Cd^{2+}/Cd}\)) is -0.403 V. Now we can solve for the standard reduction potential of the Pd reaction (cathode): \(1.03 V = E^{\circ}_{PdCl_{4}^{2-}/Pd} - (-0.403 V)\)
03

Calculate E for the Pd reaction

Solve for the Pd reduction potential: \(E^{\circ}_{PdCl_{4}^{2-}/Pd} = 1.03 V + 0.403 V = 1.433 V\)
04

Sketch the voltaic cell and label its components

Now, let's sketch the voltaic cell: - Draw two separate beakers; one will contain the anode and its half-cell solution, and the other will contain the cathode and its half-cell solution. - In the left beaker, place a piece of solid Cd (labeled "Cd") as the anode and a solution containing Cd虏鈦� ions; in the right beaker, place a piece of solid Pd (labeled "Pd") as the cathode and a solution containing PdCl4虏鈦� ions. - Connect the two electrodes with a wire, and place a voltmeter in the middle of the wire to measure the cell potential. - Place a salt bridge between the two beakers to allow the flow of ions and maintain electrical neutrality in both solutions. Now label the overall direction of electron flow: - Electrons flow from the anode (Cd) to the cathode (Pd) through the wire. - Negative ions from the salt bridge (such as the counterion from the PdCl4虏鈦�) move towards the anode half-cell, and positive ions (such as K鈦�) move towards the cathode half-cell. By following these steps, we successfully broke down the exercise and provided a step-by-step solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Redox Reactions
Redox reactions involve the transfer of electrons between two substances. One substance gets oxidized, losing electrons, while the other gets reduced, gaining electrons. In a voltaic cell, redox reactions drive the flow of electricity.
Here鈥檚 how this works in a practical reaction:
  • In the oxidation half-reaction, a substance loses electrons. For example, in our given reaction, Cadmium (Cd) is oxidized to Cadmium ions (Cd虏鈦�) with the release of electrons:
    \( Cd(s) \rightarrow Cd^{2+}(aq) + 2e^- \).
  • In the reduction half-reaction, a different substance gains electrons. Here, Palladium tetrachloride \( (PdCl_4^{2-}) \) gets reduced to Palladium (Pd), as shown in the half-cell reaction:
    \( PdCl_4^{2-}(aq) + 2e^- \rightarrow Pd(s) + 4Cl^-(aq) \).
This exchange of electrons is crucial for generating electrical energy in voltaic cells.
Cell Potential
The cell potential, also known as the electromotive force (EMF), measures the voltage or potential difference between two half-cells in a voltaic cell. This figure is key to understanding how much electrical energy can be generated.
  • In our sample voltaic cell, the overall cell potential is \(+1.03 \, V\). This indicates that the reaction is spontaneous, allowing the voltaic cell to generate energy.
  • The cell potential is calculated by finding the difference between the reduction potential of the cathode and the reduction potential of the anode:
    \( E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \).
  • Since the anode reaction is less positive, its potential is subtracted from the cathode's, resulting in a positive overall voltage.
This overall positive voltage is what powers devices connected to the voltaic cell.
Half-Cell Reactions
A voltic cell operates on two separate half-cell reactions, each occurring in different compartments. Each half-cell is associated with a specific electrode and involves either oxidation or reduction.
  • In the oxidation half-cell, Cadmium metal undergoes a reaction to form Cadmium ions and free electrons. This takes place at the anode.
  • The reduction half-cell reaction involves Palladium tetrachloride being reduced to Palladium metal as it gains electrons. This happens at the cathode.
Each of these reactions is vital as they independently contribute to the cell's electrical output by facilitating electron flow through the external circuit.
Electrode Potentials
Electrode potentials reflect a half-cell's ability to gain or lose electrons. They are measured in volts and are crucial for determining a cell's overall potential.
  • Every half-reaction has a standard electrode potential, denoted as \( E^{\circ} \), which indicates its tendency to be reduced.
  • In our example, the standard reduction potential for the cadmium half-cell (\( Cd^{2+}/Cd \)) is \(-0.403 \, V\), making it an anode due to its lower potential.
  • For the Palladium reaction \( (PdCl_4^{2-}/Pd) \), the standard reduction potential is calculated to be \( +1.433 \, V \), making it the cathode.
  • The difference between these potentials (cathode minus anode) gives the overall cell potential.
Understanding these values is essential for predicting how efficiently a voltaic cell will operate.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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: \(\mathrm{pH}=0.0, \quad\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M}, \quad\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ., Vol. \(50,1973,61\) ). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage ( \(2 \mathrm{mi}\) ) is \(300 \mathrm{~atm} ?\) (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

At \(900^{\circ} \mathrm{C}\) titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chloride. (a) Write a balanced equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant?

Complete and balance the following equations, and identify the oxidizing and reducing agents. Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), have an atypical oxidation state. (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)-\cdots\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \rightarrow-\rightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q)\) (acidic solution) (d) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(a q)\) (acidic solution) (e) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution)

The capacity of batteries such as thetypical AA alkaline battery is expressed in units of milliamp-hours (mAh). An "AA" alkaline battery yields a nominal capacity of \(2850 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of mAh? (b) The starting voltage of a fresh alkaline battery is \(1.55 \mathrm{~V}\). The voltage decreases during discharge and is \(0.80 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.