Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 41

Sketch the galvanic cells based on the following half- reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{V}$ $\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{V}$ b. $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{V}$ $\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \quad \mathscr{E}^{\circ}=1.60 \mathrm{V}$

Short Answer

Expert verified
For the first cell: 1. The reduction half-reaction is \(\mathrm{Cl}_{2} + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-}, \mathscr{E}^{\circ}=1.36\,\text{V}\) and the oxidation half-reaction is \(2\mathrm{Br}^{-}\rightarrow\mathrm{Br}_{2}+2\mathrm{e}^{-}, \mathscr{E}^{\circ}=1.09\,\text{V}\). 2. The overall balanced equation is \(2\mathrm{Br}^{-}+\mathrm{Cl}_{2}\rightarrow\mathrm{Br}_{2}+2\mathrm{Cl}^{-}\). 3. The overall cell potential, \(\mathscr{E}^{\circ}_{\text{cell}}\), is \(0.27\,\text{V}\). For the second cell: 1. The reduction half-reaction is \(\mathrm{MnO}_{4}^{-}+8\mathrm{H}^{+}+5\mathrm{e}^{-}\rightarrow\mathrm{Mn}^{2+}+4\mathrm{H}_{2}\mathrm{O}, \mathscr{E}^{\circ}=1.51\,\text{V}\) and the oxidation half-reaction is \(\mathrm{IO}_{4}^{-}\rightarrow\mathrm{IO}_{3}^{-}+2\mathrm{e}^{-}, \mathscr{E}^{\circ}=1.60\,\text{V}\). 2. The overall balanced equation is \(10\,\mathrm{MnO}_{4}^{-}+16\,\mathrm{H}^{+}+5\,\mathrm{IO}_{4}^{-}\rightarrow10\,\mathrm{Mn}^{2+}+20\,\mathrm{H}_{2}\,\mathrm{O}+5\,\mathrm{IO}_{3}^{-}\). 3. The overall cell potential, \(\mathscr{E}^{\circ}_{\text{cell}}\), is \(-0.09\,\text{V}\).

Step by step solution

01

Identify the reduction half-reaction

Chlorine has a higher standard potential, so chlorination occurs. A higher potential indicates a stronger oxidizing agent, which will cause a reduction of Cl2: \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{V}\)
02

Identify the oxidation half-reaction

Bromine has a lower standard potential, so bromination occurs. A lower potential indicates a less strong oxidizing agent, which will cause an oxidation of Br-: \(2 \mathrm{Br}^{-} \rightarrow \mathrm{Br}_{2} + 2\mathrm{e}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{V}\) Now, we need to set up our galvanic cell such that the reduction (chlorination) occurs at the cathode and the oxidation (bromination) occurs at the anode.
03

Sketch the galvanic cell

Draw two separate half-cells. One half-cell contains the anode where bromination occurs, and the other half-cell contains the cathode where chlorination occurs. Electron flow is from anode (-) to cathode (+). Through the salt bridge, negative ion migration occurs from the cathode half-cell (Cl- ions) to the anode half-cell (Br- ions), while positive ion migration occurs from the anode half-cell (Br2 ions) to the cathode half-cell (Cl2 ions).
04

Write the overall balanced equation

Add the oxidation half-reaction and the reduction half-reaction together: \(2 \mathrm{Br}^{-} +\mathrm{Cl}_{2} \rightarrow \mathrm{Br}_{2} + 2 \mathrm{Cl}^{-}\)
05

Calculate the overall cell potential

Subtract the oxidation potential from the reduction potential: \(\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{reduction} - \mathscr{E}^{\circ}_{oxidation} = 1.36\,\mathrm{V} - 1.09\,\mathrm{V} = 0.27\,\mathrm{V}\) b. Once again, we will compare the standard potentials of the two half-reactions to determine which will undergo oxidation and which will undergo reduction. In this case, the MnO4- half-reaction has a lower potential than the IO4- half-reaction, so the MnO4- half-reaction will undergo reduction and the IO4- half-reaction will undergo oxidation.
06

Identify the reduction half-reaction

Since the MnO4- potential is smaller, the reduction half-reaction will be: $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{V}$
07

Identify the oxidation half-reaction

The oxidation half-reaction will be: $\mathrm{IO}_{4}^{-} \rightarrow \mathrm{IO}_{3}^{-}+ 2\mathrm{e}^{-} \quad \mathscr{E}^{\circ}=1.60 \mathrm{V}$ Now we need to set up our galvanic cell such that the reduction (MnO4-) occurs at the cathode and the oxidation (IO4-) occurs at the anode.
08

Sketch the galvanic cell

Draw two separate half-cells. One half-cell contains the anode where oxidation of IO4- occurs, and the other half-cell contains the cathode where reduction of MnO4- occurs. Electron flow is from anode (-) to cathode (+). Through the salt bridge, positive ion migration occurs from the anode half-cell to the cathode half-cell, while negative ion migration occurs from the cathode half-cell to the anode half-cell.
09

Balance the overall equation

Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5 so that the electron transfer is equal in both directions: $2(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O})$ \(5(\mathrm{IO}_{4}^{-} \rightarrow \mathrm{IO}_{3}^{-}+ 2\mathrm{e}^{-})\) Now, add both reactions together: \(10\,\mathrm{MnO}_{4}^{-} + 16\,\mathrm{H}^{+} + 5\,\mathrm{IO}_{4}^{-} \rightarrow 10\,\mathrm{Mn}^{2+} +20\,\mathrm{H}_{2}\,\mathrm{O} + 5\,\mathrm{IO}_{3}^{-}\)
10

Calculate the overall cell potential

Subtract the oxidation potential from the reduction potential: \(\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{reduction} - \mathscr{E}^{\circ}_{oxidation} = 1.51\,\mathrm{V} - 1.60\,\mathrm{V} = -0.09\,\mathrm{V}\)

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 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.)

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at $25^{\circ} \mathrm{C}\( is 0.195 \)\mathrm{V}\( . What is \)\left[\mathrm{Cu}^{2+}\right] ?$ (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.

Consider the following reduction potentials: $$\begin{array}{ll}{\mathrm{Co}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Co}} & {\mathscr{E}^{\circ}=1.26 \mathrm{V}} \\ {\mathrm{Co}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}} & {\mathscr{E}^{\circ}=-0.28 \mathrm{V}}\end{array}$$ a. When cobalt metal dissolves in 1.0\(M\) nitric acid, will \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}^{2+}\) be the primary product (assuming standard conditions)? b. Is it possible to change the concentration of \(\mathrm{HNO}_{3}\) to get a different result in part a? Concentrated \(\mathrm{HNO}_{3}\) is about 16 \(M\) .

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

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