Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 78

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is 1.62 \(\mathrm{V}\) . Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Short Answer

Expert verified
The potential of the electrochemical cell at 25°C is 1.40 V. The concentration of Al³⁺ in the unknown solution is approximately 5.95 × 10⁻³ M.

Step by step solution

01

Identify relevant formulas

We will use the Nernst equation to find the potential of the electrochemical cell: \(E = E^\circ - \frac{0.0592}{n} \log Q\) Where \(E\) is the cell potential, \(E^\circ\) is the standard cell potential, \(n\) is the number of moles of electrons transferred in the redox reaction, and \(Q\) is the reaction quotient. In this problem, we are given the concentrations of Ni²⁺ and Al³⁺, and we need to find the standard cell potentials for the nickel and aluminum half-reactions. We can find this information in a standard reduction potential table.
02

Find standard cell potentials

Using a standard reduction potential table, we find that: \(E_{Ni^{2+}/Ni}^\circ = -0.25 \, V\) \(E_{Al^{3+}/Al}^\circ = -1.66 \, V\) As the cell potential is calculated as the difference between the potentials of each half-reaction, the standard cell potential will be: \(E_{cell}^\circ = E_{Ni^{2+}/Ni}^\circ - E_{Al^{3+}/Al}^\circ\)
03

Calculate the standard cell potential

Using the standard cell potentials found in step 2: \(E_{cell}^\circ = (-0.25) - (-1.66) = 1.41 \, V\)
04

Calculate the reaction quotient

The reaction quotient \(Q\) for this cell is given by: \(Q = \frac{[\mathrm{Ni}^{2+}]}{[\mathrm{Al}^{3+}]}\) Using the given concentrations of both ions: \(Q = \frac{1.0}{7.2 \times 10^{-3}} = 138.89\)
05

Calculate the cell potential

Now we can use the Nernst equation to find the cell potential at 25°C: \(E = E^\circ - \frac{0.0592}{n} \log Q\) As aluminum loses 3 electrons during the redox reaction, \(n = 3\). Substitute the values obtained so far: \(E = 1.41 - \frac{0.0592}{3} \log 138.89 = 1.40 \, V\) So the potential of the cell at 25°C is 1.40 V. Now for part b:
06

Set up the Nernst equation with the unknown concentration of Al³⁺

We are given the measured cell potential and asked to find the unknown concentration of Al³⁺ in the solution. We will use the Nernst equation with unknown \(Q\): \(E = E^\circ - \frac{0.0592}{n} \log Q\) Here, \(E\) is given as 1.62 V, and \(Q = \frac{[\mathrm{Ni}^{2+}]}{[\mathrm{Al}^{3+}]}\).
07

Solve for the unknown concentration of Al³⁺

Rearrange the Nernst equation to solve for \(Q\): \(\log Q = \frac{n}{0.0592} (E^\circ - E) = \frac{3}{0.0592} (1.41 - 1.62)\) Calculate \(Q\): \(Q = 10^{-3.773} = 1.68 \times 10^{-4}\) Now use \(Q\) to find the unknown concentration of Al³⁺: \([\mathrm{Al}^{3+}] = \frac{1.0}{Q} = \frac{1.0}{1.68 \times 10^{-4}} = 5.95 \times 10^{-3} M\) The concentration of Al³⁺ in the unknown solution is approximately 5.95 × 10⁻³ M.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

An electrochemical cell consists of a silver metal electrode immersed in a solution with \(\left[\mathrm{Ag}^{+}\right]=1.00 M\) separated by a porous disk from a compartment with a copper metal electrode immersed in a solution of 10.00\(M \mathrm{NH}_{3}\) that also contains $2.4 \times 10^{-3} \mathrm{M} \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .$ The equilibrium between \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_{3}\) is: $$\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq}) \qquad K=1.0 \times 10^{13}$$ and the two cell half-reactions are: $$\begin{array}{rl}{\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}} & {\mathscr{E}^{\circ}=0.80 \mathrm{V}} \\ {\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}} & {\mathscr{E}^{\circ}=0.34 \mathrm{V}}\end{array}$$ Assuming \(\mathrm{Ag}^{+}\) is reduced, what is the cell potential at \(25^{\circ} \mathrm{C} ?\)

Assign oxidation numbers to all the atoms in each of the following: a. \(\mathrm{HNO}_{3} \quad\) g. \(\mathrm{PbSO}_{4}\) b. \(\mathrm{CuCl}_{2} \quad\) h. \(\mathrm{PbO}_{2}\) c. \(\mathrm{O}_{2} \quad\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2} \quad\) j. \(\mathrm{CO}_{2}\) e. $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \quad \mathrm{k} .\left(\mathrm{NH}_{0}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}$ f. \(\mathrm{Ag} \quad\) l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\ \hline {\mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La}} & {-2.37} \\\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {-0.44}\end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} M\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} M,\) what is the expected cell potential?

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C} :\) $$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 M)\right|\left|\mathrm{Cu}^{2+}(2.50 M)\right| \mathrm{Cu}$$ The mass of each electrode is 200. g. a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after 10.0 A of current has flowed for 10.0 h. (Assume each half-cell contains 1.00 L of solution.) c. Calculate the mass of each electrode after 10.0 h. d. How long can this battery deliver a current of 10.0 A before it goes dead?

Which of the following statement(s) is(are) true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow$ Al and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}\) , aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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.

Sign-up for free