/*! 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 94 Explain why most useful galvanic... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 94

Explain why most useful galvanic cells give voltages of no more than 1.5 to \(2.5 \mathrm{~V}\). What are the prospects for developing practical galvanic cells with voltages of \(5 \mathrm{~V}\) or more?

Short Answer

Expert verified
The voltage of typical galvanic cells usually falls within the 1.5 to 2.5 Volts range due to the standard reduction potentials of the substances used, and their internal resistance. Though it's theoretically possible to create a cell with greater than 5 Volts, it's impractical due to the potential instability of the cell, decompositions of the electrolyte, and cost factor of the substances with high reduction potentials.

Step by step solution

01

Understanding Galvanic Cells

Firstly, it's important to understand what a galvanic cell is. A galvanic cell, or voltaic cell, is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell.
02

Reason for 1.5 to 2.5 Voltage for Galvanic Cells

The voltage of a galvanic cell is directly related to the difference in reduction potential between the two half-cells. The voltage is approximately equivalent to the difference in the standard reduction potentials, reported in volts, of the two half-cells. The absolute values of standard reduction potentials usually do not exceed 3 volts hence, after considering losses such as internal resistance, achievable voltages are usually in the range of 1.5 to 2.5 volts.
03

Feasibility of Galvanic Cells with 5V or more

Although possible in theory, in practice, creating galvanic cells with a 5V or higher potential is challenging due to several reasons. It could lead to decomposition of the electrolyte which could potentially make the cell unstable and dangerous. Also, most substances with high reduction potentials are not readily available or are prohibitively expensive, making such cells impractical for common use.

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!

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

Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is insoluble in water. This property has been used to determine the amount of \(\mathrm{Ca}^{2+}\) ions in blood. The calcium oxalate isolated from blood is dissolved in acid and titrated against a standardized \(\mathrm{KMnO}_{4}\) solution, as described in Problem \(19.66 .\) In one test it is found that the calcium oxalate isolated from a 10.0 -mL sample of blood requires \(24.2 \mathrm{~mL}\) of \(9.56 \times 10^{-4} \mathrm{M} \mathrm{KMnO}_{4}\) for titration. Calculate the number of milligrams of calcium per milliliter of blood.

To remove the tarnish \(\left(\mathrm{Ag}_{2} \mathrm{~S}\right)\) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about \(80^{\circ} \mathrm{C}\). After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding \(\mathrm{NaCl}\) instead of \(\mathrm{NaHCO}_{3}\) would also work because both compounds are strong electrolytes. What is the added advantage of using \(\mathrm{NaHCO}_{3}\) ? (Hint: Consider the \(\mathrm{pH}\) of the solution.) (c) What is the purpose of heating the solution? (d) Some commercial tarnish removers containing a fluid (or paste) that is a dilute \(\mathrm{HCl}\) solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disadvantages of using this procedure compared to the one described here.

How many faradays of electricity are required to produce (a) \(0.84 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at exactly \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) from aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution; (b) \(1.50 \mathrm{~L}\) of \(\mathrm{Cl}_{2}\) at \(750 \mathrm{mmHg}\) and \(20^{\circ} \mathrm{C}\) from molten \(\mathrm{NaCl}\); (c) \(6.0 \mathrm{~g}\) of Sn from molten \(\mathrm{SnCl}_{2} ?\)

Balance the following redox equations by the halfreaction method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution)

Calculate the emf of the following concentration cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(0.080 \mathrm{M}) \| \mathrm{Cu}^{2+}(1.2 \mathrm{M})\right| \mathrm{Cu}(s) $$
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

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.