/*! 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 73 A solution containing both \(\ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 73

A solution containing both \(\mathrm{Ag}^{+}\) and \(\mathrm{Cu}^{2+}\) ions is subjected to electrolysis. (a) Which metal should plate out first? (b) Plating out is finished after a current of \(0.75 \mathrm{A}\) is passed through the solution for 2.50 hours. If the total mass of metal is \(3.50 \mathrm{g},\) what is the mass percent of silver in the product?

Short Answer

Expert verified
The metal that plates out first is silver. The mass percent of silver in the product is 100%.

Step by step solution

01

Determine the metal that plates out first

The metal with a higher reduction potential (or less negative) will be reduced first. Looking at the standard reduction potentials, Ag+ has a higher reduction potential (+0.80V) compared to Cu2+ (+0.34V), and thus Ag+ will plate out first.
02

Calculate the total charge passed

The total charge passed in the electrolysis process can be calculated using the formula Q=It, where I is the current and t is the time. Here, \(I=0.75 \mathrm{A}\) and \(t=2.5 \mathrm{hr} = 9000 \mathrm{s}\). Substituting these into the formula gives \(Q=0.75 \mathrm{A} \times 9000 \mathrm{s} = 6750 \mathrm{C}\)
03

Calculate the moles of electrons

Using Faraday's law, we can calculate the moles of electrons transferred. According to this law, one mole of electrons carries a charge of \(1 \mathrm{F}=96485 \mathrm{C}\). Therefore, the moles of electrons transferred are \( \frac{6750 \mathrm{C}}{96485 \mathrm{C/mol}} \approx 0.07 \mathrm{mol}\) of electrons.
04

Calculate the moles and mass of Ag and Cu plated

From the stoichiometry of the half reactions, the reduction of Ag+ to Ag requires 1 mole of electrons for 1 mole of Ag, while the reduction of Cu2+ to Cu requires 2 moles of electrons for 1 mole of Cu. However, as Ag plates out first, all electrons will reduce Ag+ until it is completely plated out. Hence, approximately 0.07 moles of Ag are plated out. Using the molar mass of silver (107.87 g/mol), the mass of Ag plated out is \(0.07 \mathrm{mol} \times 107.87 \mathrm{g/mol} \approx 7.55 \mathrm{g}\). However, as the total mass of metal is given to be 3.50 g, this means that all electrons could not have reduced Ag and thus, all 3.50 g is silver.
05

Calculate the mass percent of silver

The mass percent of silver can be calculated as \( \frac{mass \, of \, Ag}{total \, mass} \times 100\). Thus, \( \frac{3.50 \, g}{3.50 \, g} \times 100 = 100\%\)

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.

Reduction Potentials
Reduction potential is a measure of the tendency of a chemical species to gain electrons and be reduced. This concept is essential in electrochemistry, especially during electrolysis processes. In an electrolytic cell, ions with higher reduction potentials are more likely to gain electrons and form solid metals on the electrode.In the case of our example with \( \mathrm{Ag}^+ \) and \( \mathrm{Cu}^{2+} \), we refer to the standard reduction potentials:
  • Silver ion (\( \mathrm{Ag}^+ \)) has a reduction potential of +0.80V.
  • Copper ion (\( \mathrm{Cu}^{2+} \)) has a reduction potential of +0.34V.
Since silver has a less negative reduction potential, it will be reduced first in the electrolysis process. This means that \( \mathrm{Ag}^+ \) will plate out onto the electrode before \( \mathrm{Cu}^{2+} \) starts to deposit.
Faraday's Law
Faraday's laws of electrolysis are key principles that relate the amount of substance altered at an electrode during electrolysis to the quantity of electricity transferred. The first law states that the amount of chemical change is proportional to the amount of electricity used.The formula to express Faraday's Law is:\[ Q = It \]where:
  • \( Q \) is the total charge in coulombs.
  • \( I \) is the current in amperes.
  • \( t \) is the time in seconds.
In our exercise, when a current of 0.75 A is passed for 2.5 hours, we convert this time to seconds, resulting in 9000 seconds. By substituting into the equation, we calculate that a total charge of 6750 coulombs has been passed.
Silver Plating
Silver plating is the process of depositing a thin layer of silver onto the surface of another material. In our case, during the electrolysis of a solution containing silver ions, these ions migrate to the electrode and gain electrons to form a solid layer of silver.Through electrolysis, each silver ion \( \mathrm{Ag}^+ \) accepts an electron to become a silver atom. This process is energetically favorable due to silver's higher reduction potential. In the exercise, we begin with 0.07 moles of electrons transferring to the silver ions. Based on stoichiometry, each mole of electrons reduces one mole of silver ions. Therefore, 0.07 moles of electrons can plate out 0.07 moles of silver, provided they are the primary species being reduced.
Stoichiometry
Stoichiometry is a branch of chemistry dealing with the relative quantities of reactants and products in chemical reactions. It is vital in computing the amounts of substances involved in electrochemical processes, like the electrolysis seen in our exercise.In electrolysis involving ions like \( \mathrm{Ag}^+ \) and \( \mathrm{Cu}^{2+} \), stoichiometry assists in determining how many moles of electrons are needed for reduction:
  • 1 mole of \( \mathrm{Ag}^+ \) requires 1 mole of electrons.
  • 1 mole of \( \mathrm{Cu}^{2+} \) requires 2 moles of electrons.
From our calculation of 0.07 moles of electrons, and considering silver's need for only 1 mole of electrons per mole of ions, the stoichiometry confirms that these electrons could reduce 0.07 moles of silver ions. As the total weight of metal is 3.50 g in the given scenario, the calculated amount corresponds to silver, leading to a mass of 3.50 g of silver. Hence, silver constitutes 100% of the total mass of metal plated, according to the stoichiometric calculation.

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

Describe how you might construct batteries with each of the following voltages: (a) \(0.10 \mathrm{V} ;\) (b) \(2.5 \mathrm{V} ;\) (c) \(10.0 \mathrm{V}\). Be as specific as you can about the electrodes and solution concentrations you would use, and indicate whether the battery would consist of a single cell or two or more cells connected in series.

The quantity of electric charge that will deposit \(4.5 \mathrm{g}\) Al at a cathode will also produce the following volume at STP of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}^{+}(\) aq) at a cathode: (a) \(44.8 \mathrm{L} ;\) (b) \(22.4 \mathrm{L} ;\) (c) \(11.2 \mathrm{L} ;\) (d) \(5.6 \mathrm{L}\).

Show that for nonstandard conditions the temperature variation of a cell potential is $$E\left(T_{1}\right)-E\left(T_{2}\right)=\left(T_{1}-T_{2}\right) \frac{\left(\Delta S^{\circ}-R \ln Q\right)}{z F}$$ where \(E\left(T_{1}\right)\) and \(E\left(T_{2}\right)\) are the cell potentials at \(T_{1}\) and \(T_{2},\) respectively. We have assumed that the value of \(Q\) is maintained at a constant value. For the nonstandard cell below, the potential drops from \(0.394 \mathrm{V}\) at \(50.0^{\circ} \mathrm{C}\) to \(0.370 \mathrm{V}\) at \(25.0^{\circ} \mathrm{C} .\) Calculate \(Q\) \(\Delta H^{\circ},\) and \(\Delta S^{\circ}\) for the reaction, and calculate \(K\) for the two temperatures. $$\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$$ Choose concentrations of the species involved in the cell reaction that give the value of \(Q\) that you have calculated, and then determine the equilibrium concentrations of the species at \(50.0^{\circ} \mathrm{C}\)

Write an equation to represent the oxidation of \(\mathrm{Cl}^{-}(\mathrm{aq})\) to \(\mathrm{Cl}_{2}(\mathrm{g})\) by \(\mathrm{PbO}_{2}(\mathrm{s})\) in an acidic solution. Will this reaction occur spontaneously in the forward direction if all other reactants and products are in their standard states and (a) \(\left[\mathrm{H}^{+}\right]=6.0 \mathrm{M} ;\) (b) \(\left[\mathrm{H}^{+}\right]=1.2 \mathrm{M}\) (c) \(\mathrm{pH}=4.25 ?\) Explain.

Dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) in acidic solution is a good oxidizing agent. Which of the following oxidations can be accomplished with dichromate ion in acidic solution? Explain. (a) \(\operatorname{sn}^{2+}\left(\text { aq) to } \operatorname{Sn}^{4+}(\text { aq })\right.\) (b) \(\mathrm{I}_{2}(\mathrm{s})\) to \(\mathrm{IO}_{3}^{-}(\mathrm{aq})\) (c) \(\mathrm{Mn}^{2+}(\mathrm{aq})\) to \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

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.