/*! 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 98 A common shorthand way to repres... [FREE SOLUTION] | 91影视

91影视

A common shorthand way to represent a voltaic cell is 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 Fe \(\left|\mathrm{Fe}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag}_{\mathrm{g}}\) sketch the cell. (b) Write the half-reactions and overall cell reaction represented by \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{H}^{+}\right| \mathrm{H}_{3}\) s sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: $$ \begin{aligned} \mathrm{ClO}_{3}^{-}(a q)+3 \mathrm{Cu}(s)+6 \mathrm{H}^{+}(a q) & \longrightarrow \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Pt is used as an inert electrode in contact with the \(\mathrm{ClO}_{3}^{-}\)and Cl. Sketch the cell.

Short Answer

Expert verified
The shorthand notations for the voltaic cells are as follows: a) \(Fe | Fe^{2+} || Ag^+ | Ag\) b) \(Zn | Zn^{2+} || H^{+} | H_{2}\) c) \(Pt | ClO_3^-, Cl^-, H^+ || Cu^{2+} | Cu\)

Step by step solution

01

1. Identify Anode and Cathode

In this case, Fe is the anode (where oxidation occurs) and Ag is the cathode (where reduction occurs).
02

2. Write Half-Reactions

Anode half-reaction (oxidation): \(Fe \rightarrow Fe^{2+} + 2 e^-\) Cathode half-reaction (reduction): \(Ag^{+} + e^- \rightarrow Ag\)
03

3. Write Overall Cell Reaction

To get the overall cell reaction, balance the electrons and add the half-reactions: \(Fe + 2Ag^+ \rightarrow Fe^{2+} + 2Ag\)
04

4. Represent Voltaic Cell Using Shorthand Notation

The shorthand notation for this cell is: \[Fe | Fe^{2+} || Ag^+ | Ag\] #Part B: Zn/Zn虏鈦 and H鈦/H鈧 System#
05

1. Identify Anode and Cathode

In this case, Zn is the anode (oxidation) and H鈦/H鈧 is the cathode (reduction).
06

2. Write Half-Reactions

Anode half-reaction (oxidation): \(Zn \rightarrow Zn^{2+} + 2 e^-\) Cathode half-reaction (reduction): \(2H^{+} + 2 e^- \rightarrow H_{2}\)
07

3. Write Overall Cell Reaction

Balance the electrons and add the half-reactions: \(Zn + 2H^{+} \rightarrow Zn^{2+} + H_{2}\)
08

4. Represent Voltaic Cell Using Shorthand Notation

The shorthand notation for this cell is: \[Zn | Zn^{2+} || H^{+} | H_{2}\] #Part C: ClO鈧冣伝/Cl鈦 and Cu/Cu虏鈦 System#
09

1. Identify Anode and Cathode

In this case, ClO鈧冣伝 is the anode (oxidation) and Cu is the cathode (reduction).
10

2. Write Half-Reactions

Anode half-reaction (oxidation): \(2ClO_3^- + 12 H^+ + 10 e^- \rightarrow 2Cl^- + 6 H_2O\) Cathode half-reaction (reduction): \(3 Cu^{2+} + 6 e^- \rightarrow 3 Cu\)
11

3. Write Overall Cell Reaction

Balance the electrons and add the half-reactions: \(2 ClO_3^- + 3 Cu + 12 H^+ \rightarrow 2Cl^- + 3 Cu^{2+} + 6 H_2O\)
12

4. Represent Voltaic Cell Using Shorthand Notation

The shorthand notation for this cell is: \[Pt | ClO_3^-, Cl^-, H^+ || Cu^{2+} | Cu\] The sketch of the cell should show the two compartments with the platinum electrode in contact with the ClO鈧冣伝 and Cl鈦 in the anode half-cell and the copper electrode in contact with the Cu虏鈦 in the cathode half-cell. A salt bridge or porous barrier should be shown connecting the two compartments.

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.

Half-reactions
In electrochemistry, half-reactions are crucial for understanding how voltaic cells work. A half-reaction is a representation of either the oxidation or reduction process separately in a redox reaction. Each half-reaction shows how electrons are either gained or lost. In the voltaic cell, the sum of these half-reactions gives the overall cell reaction.
The anode is where oxidation occurs, and electrons are released. For example, in the provided exercise, the half-reaction for iron is:
  • Anode reaction: \( Fe \rightarrow Fe^{2+} + 2e^- \)
Here, iron (Fe) loses electrons to form iron ions (\( Fe^{2+} \)).
The cathode is where reduction occurs, and electrons are gained. It involves a half-reaction such as:
  • Cathode reaction: \( Ag^{+} + e^- \rightarrow Ag \)
This indicates that silver ions (\( Ag^{+} \)) gain electrons to form metallic silver (Ag).
By balancing the electrons lost and gained in these half-reactions, we can derive the overall reaction occurring in the cell.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the interplay between electrical energy and chemical reactions. Voltaic, or galvanic, cells are devices that convert chemical energy into electrical energy through redox reactions. These cells consist of two half-cells connected by a salt bridge or a porous partition.
A salt bridge serves to maintain electrical neutrality within the internal circuit, preventing the solutions in each half-cell from becoming excessively charged. Each half-cell is prepared with specific electrodes immersed in electrolyte solutions. In a voltaic cell:
  • The anode is the electrode where oxidation occurs, and it has a negative charge, as it releases electrons.
  • The cathode is the electrode where reduction occurs, and it has a positive charge, as it gains electrons.
The potential difference between the electrodes, known as the cell potential or electromotive force (EMF), drives the flow of electrons from the anode to the cathode through an external circuit, producing electricity. Understanding these basic principles of electrochemistry helps explain how devices like batteries and fuel cells work.
Redox reactions
Redox, or reduction-oxidation reactions, are chemical processes in which electrons are transferred between two substances. These reactions occur in each half-cell of a voltaic cell, allowing it to generate electrical energy.
In a redox reaction:
  • Oxidation is defined as the loss of electrons by a substance. For instance, in the half-reaction: \( Zn \rightarrow Zn^{2+} + 2e^- \), zinc loses electrons, undergoing oxidation.
  • Reduction involves the gain of electrons by a substance. For example, in the half-reaction: \( 2H^{+} + 2e^- \rightarrow H_2 \), hydrogen ions gain electrons, resulting in reduction.
Combining the oxidation and reduction reactions allows us to write the net ionic equation for the electrochemical cell. This demonstrates how the flow of electrons between the anode and cathode generates electrical energy.
Understanding the redox chemistry in voltaic cells is essential for designing and optimizing electrochemical devices used in a wide range of technological applications, from energy storage to chemical synthesis.

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

During the discharge of an alkaline battery, \(4.50 \mathrm{~g}\) of \(\mathrm{Zn}\) is consumed at the anode of the battery. (a) What mass of \(\mathrm{MnO}_{2}\) is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from \(\mathrm{Zn}\) to \(\mathrm{MnO}_{2}\) ?

(a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: \begin{tabular}{lr} \hline Reduction Half-Reaction & \(E^{+}(\mathrm{V})\) \\ \hline\(A^{+}(a q)+\mathrm{c}^{-} \longrightarrow \mathrm{A}(\mathrm{s})\) & \(1.33\) \\ \(\mathrm{~B}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{B}(s)\) & \(0.87\) \\ \(\mathrm{C}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{C}^{2+}(a q)\) & \(-0.12\) \\ \(\mathrm{D}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{D}(s)\) & \(-1.59\) \\ \hline \end{tabular} (a) Which substance is the strongest oxidixing agent? Which is weakest? (b) Which substance is the strongest reducing agent? Which is weakest? (c) Which substance(s) can oxidize \(C^{24}\) ? [Sections 20.4 and 20.5]

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)

A plumber's handbook states that you should not connect a copper pipe directly to a steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use instead an insulating fitting to connect them. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

See all solutions

Recommended explanations on Chemistry Textbooks

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.