/*! 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 71 During a period of discharge of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 71

During a period of discharge of a lead-acid battery, \(402 \mathrm{~g}\) of \(\mathrm{Pb}\) from the anode is converted into \(\mathrm{PbSO}_{4}(s) .\) What mass of \(\mathrm{PbO}_{2}(s)\) is reduced at the cathode during this same period?

Short Answer

Expert verified
Approximately \(464\ \mathrm{g}\) of \(\mathrm{PbO}_{2}\) is reduced at the cathode during the period of discharge of the lead-acid battery.

Step by step solution

01

Calculate the number of moles of \(\mathrm{Pb}\)

Given mass of \(\mathrm{Pb} = 402\ \mathrm{g}\) and molar mass of \(\mathrm{Pb} = 207.2\ \mathrm{g\:mol^{-1}}\). Dividing the mass by its molar mass gives us the number of moles: Moles of \(\mathrm{Pb} = \frac{402}{207.2} \approx 1.94\ \mathrm{mol}\)
02

Use stoichiometric ratios to determine moles of \(\mathrm{PbO}_{2}\)

From the balanced chemical equation above, we can see that 1 mole of \(\mathrm{Pb}\) reacts with 1 mole of \(\mathrm{PbO}_{2}\). Therefore, the number of moles of \(\mathrm{PbO}_{2}\) that got reduced will also be the same as the number of moles of \(\mathrm{Pb}\). Moles of \(\mathrm{PbO}_{2} = 1.94\ \mathrm{mol}\)
03

Determine the mass of reduced \(\mathrm{PbO}_{2}\)

Finally, we need to convert the number of moles of \(\mathrm{PbO}_{2}\) to mass. We will use the molar mass, which can be calculated as follows: Molar mass of \(\mathrm{PbO}_{2} = 207.2\ (\mathrm{Pb}) + (16\times 2)\ (\mathrm{O}) = 239.2\ \mathrm{g\:mol^{-1}}\) Now we can determine the mass using the number of moles: Mass of \(\mathrm{PbO}_{2} = 1.94 \times 239.2 \approx 464\ \mathrm{g}\) So, during the period of discharge of the lead-acid battery, approximately \(464\ \mathrm{g}\) of \(\mathrm{PbO}_{2}\) is reduced at the cathode.

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

A voltaic cell is constructed that uses the following halfcell reactions: $$ \begin{gathered} \mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{gathered} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=3.5 \mathrm{M}\). (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 \(\left[\mathrm{Cu}^{+}\right]\) was equal to \(0.15 \mathrm{M}\), at what concentration of \(\mathrm{I}^{-}\) would the cell have zero potential?

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a \(\mathrm{NaCl}\) solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

For each of the following reactions, write a baFor each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\), and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}{ }^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)lanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\), and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}{ }^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \mathrm{AgCl}(s)+\mathrm{e}^{-}--\rightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two cell compartments have \(\left[\mathrm{Cl}^{-} \mathrm{J}=0.0150 \mathrm{M}\right.\) 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.

A voltaic cell utilizes the following reaction and operates at \(298 \mathrm{~K}\) : $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s)-\rightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 M\), \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ce}^{4+}\right]=0.10 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=1.75 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=2.5 \mathrm{M} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

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.