/*! 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 79 Silver dissolves in sulfuric aci... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 79

Silver dissolves in sulfuric acid to form silver sulfate and \(\mathrm{H}_{2},\) but gold does not dissolve in sulfuric acid to form gold sulfate. Which of the two metals is the better reducing agent?

Short Answer

Expert verified
Answer: Silver is the better reducing agent when reacting with sulfuric acid because it dissolves in sulfuric acid and forms silver sulfate, indicating that it's more willing to lose electrons and reduce other substances. Gold, on the other hand, does not dissolve in sulfuric acid, suggesting it is less willing to lose electrons in the reaction.

Step by step solution

01

Write down the balanced chemical equation for the reaction of silver with sulfuric acid

The balanced chemical equation for silver reacting with sulfuric acid is: $$2\mathrm{Ag(s)} + 2\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \rightarrow \mathrm{Ag}_{2}\mathrm{SO}_{4}(s) + 2\mathrm{H}_{2}(g)$$
02

Write down the balanced chemical equation for the reaction of gold with sulfuric acid (if it were to occur)

For comparison purposes, we can also write down the balanced chemical equation for the reaction of gold with sulfuric acid, even though it doesn't actually occur: $$2\mathrm{Au(s)} + 3\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \rightarrow \mathrm{Au}_{2}(\mathrm{SO}_{4})_{3}(s) + \mathrm{3H}_{2}(g)$$
03

Examine the redox behavior of silver and gold during the reactions

A reducing agent is a substance that donates (loses) electrons to another substance in a redox reaction. Let's examine the redox behavior of silver and gold in their respective reactions by identifying the oxidation states of each element before and after the reaction: For silver: - In elemental form, silver has an oxidation state of 0. - In silver sulfate (Ag2SO4), silver has an oxidation state of +1. For gold: - In elemental form, gold has an oxidation state of 0. - In gold sulfate (Au2(SO4)3), gold would have an oxidation state of +3 (if the reaction happened).
04

Determine which metal is the better reducing agent

Since a reducing agent loses electrons in the redox reaction, we can look at how the oxidation state of each metal changes to assess their reducing ability: - Silver goes from an oxidation state of 0 to +1, meaning it loses one electron per atom, reacting with sulfuric acid. - Gold would go from an oxidation state of 0 to +3 if it were to react with sulfuric acid, meaning it would lose three electrons per atom. Since gold doesn't dissolve in sulfuric acid, this indicates that gold is not willing to lose three electrons in the reaction. Therefore, silver, which dissolves in sulfuric acid and forms silver sulfate, is the better reducing agent because it's more willing to lose electrons and reduce other substances.

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

How many milliliters of \(0.100 M\) HNO \(_{3}\) are needed to neutralize the following solutions? a. \(45.0 \mathrm{mL}\) of \(0.667 M \mathrm{KOH}\) b. \(58.5 \mathrm{mL}\) of \(0.0100 M \mathrm{Al}(\mathrm{OH})_{3}\) c. \(34.7 \mathrm{mL}\) of \(0.775 M \mathrm{NaOH}\)

A solution of table salt is a good conductor of electricity, but a solution containing an equal molar concentration of table sugar is not. Why?

The stalactites and stalagmites in most caves are made of calcium carbonate (see Figure 8.10 ). In the Lower Kane Cave in Wyoming, however, they are made of gypsum (calcium sulfate). The presence of \(\mathrm{CaSO}_{4}\) is explained by the following sequence of reactions: $$ \begin{array}{c} \mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \\ \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(g) \end{array} $$ a. Which (if either) of these reactions is a redox reaction? b. Write a net ionic equation for the reaction of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) with \(\mathrm{CaCO}_{3}\) c. How would the net ionic equation be different if the reaction were written as follows? $$ \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$

To determine the concentration of \(\mathrm{SO}_{4}^{2-}\) ion in a sample of groundwater, \(100.0 \mathrm{mL}\) of the sample is titrated with \(0.0250 M \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2},\) forming insoluble \(\mathrm{BaSO}_{4} .\) If \(3.19 \mathrm{mL}\) of the \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution is required to reach the end point of the titration, what is the molarity of the \(\mathrm{SO}_{4}^{2-} ?\)

A puddle of coastal seawater, caught in a depression formed by some coastal rocks at high tide, begins to evaporate on a hot summer day as the tide goes out. If the volume of the puddle decreases to \(23 \%\) of its initial volume, what is the concentration of \(\mathrm{Na}^{+}\) after evaporation if it was \(0.449 \mathrm{M}\) initially?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

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.