Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 23

Calculate \(E^{\circ}\) for the following cells: (a) \(\mathrm{Mn}\left|\mathrm{Mn}^{2+} \| \mathrm{H}^{+}\right| \mathrm{H}_{2} \mid \mathrm{Pt}\) (b) \(\mathrm{Au}\left|\mathrm{AuCl}_{4}^{-} \| \mathrm{Co}^{3+}, \mathrm{Co}^{2+}\right| \mathrm{Pt}\) (c) \(\mathrm{Pt}\left|\mathrm{S}^{2-}\right| \mathrm{S} \| \mathrm{NO}_{3}^{-}|\mathrm{NO}| \mathrm{Pt} \quad\) (basic medium)

Short Answer

Expert verified
In summary, the standard cell potentials for the given electrochemical cells are: Cell (a): \(E^{\circ}_\text{cell(a)} = 1.18\,\mathrm{V}\) Cell (b): \(E^{\circ}_\text{cell(b)} = 1.21\,\mathrm{V}\) Cell (c): \(E^{\circ}_\text{cell(c)} = 1.456\,\mathrm{V}\)

Step by step solution

01

Identify the half-cell reactions

The given cell describes the following half-cell reactions: 1. \(\mathrm{Mn} \rightarrow \mathrm{Mn}^{2+} + 2e^-\) (Oxidation half-cell) 2. \(\mathrm{H}^{+} + e^- \rightarrow \frac{1}{2} \mathrm{H}_{2}\) (Reduction half-cell)
02

Find the standard reduction potentials

The standard reduction potentials can be found in a standard reduction potential table: 1. \(\mathrm{Mn}^{2+} + 2e^- \rightarrow \mathrm{Mn}\): \(E^{\circ} = -1.18\,\mathrm{V}\) 2. \(\mathrm{H}^{+} + e^- \rightarrow \frac{1}{2} \mathrm{H}_{2}\): \(E^{\circ} = 0\,\mathrm{V}\)
03

Calculate the standard cell potential

Now we can calculate the standard cell potential by subtracting the standard reduction potentials of the oxidation reaction from the reduction reaction: \(E^{\circ}_\text{cell(a)} = E^{\circ}_\text{reduction} - E^{\circ}_\text{oxidation} = 0\,\mathrm{V} - (-1.18\,\mathrm{V}) = \boxed{1.18\,\mathrm{V}}\) _Cell (b):_ Au|AuCl₄⁻||Co³⁺, Co²⁺|Pt
04

Identify the half-cell reactions

The given cell describes the following half-cell reactions: 1. \(\mathrm{AuCl}_{4}^{-} + e^- \rightarrow \mathrm{Au} + 4\mathrm{Cl}^-\) (Reduction half-cell) 2. \(\mathrm{Co}^{3+} + e^- \rightarrow \mathrm{Co}^{2+}\) (Reduction half-cell)
05

Find the standard reduction potentials

The standard reduction potentials can be found in a standard reduction potential table: 1. \(\mathrm{AuCl}_{4}^{-} + e^- \rightarrow \mathrm{Au} + 4\mathrm{Cl}^-\): \(E^{\circ} = 0.93\,\mathrm{V}\) 2. \(\mathrm{Co}^{3+} + e^- \rightarrow \mathrm{Co}^{2+}\): \(E^{\circ} = -0.28\,\mathrm{V}\)
06

Calculate the standard cell potential

Now we can calculate the standard cell potential by subtracting the standard reduction potentials of the oxidation reaction from the reduction reaction: \(E^{\circ}_\text{cell(b)} = E^{\circ}_\text{reduction} - E^{\circ}_\text{oxidation} = 0.93\,\mathrm{V} - (-0.28\,\mathrm{V}) = \boxed{1.21\,\mathrm{V}}\) _Cell (c):_ Pt|S²⁻|S||NO₃⁻|NO|Pt (basic medium)
07

Identify the half-cell reactions

The given cell describes the following half-cell reactions: 1. \(\mathrm{S}^{2-} \rightarrow \mathrm{S} + 2e^-\) (Oxidation half-cell) 2. \(\mathrm{NO}_{3}^{-} + 3e^- + 3\mathrm{OH}^- \rightarrow \mathrm{NO} + 2\mathrm{H}_{2}\,\mathrm{O}\) (Reduction half-cell, balanced in basic medium)
08

Find the standard reduction potentials

The standard reduction potentials can be found in a standard reduction potential table: 1. \(\mathrm{S}^{2-} \rightarrow \mathrm{S} + 2e^-\): \(E^{\circ} = -0.48\,\mathrm{V}\) 2. \(\mathrm{NO}_{3}^{-} + 3e^- + 3\mathrm{OH}^- \rightarrow \mathrm{NO} + 2\mathrm{H}_{2}\,\mathrm{O}\): \(E^{\circ} = 0.976\,\mathrm{V}\)
09

Calculate the standard cell potential

Now we can calculate the standard cell potential by subtracting the standard reduction potentials of the oxidation reaction from the reduction reaction: \(E^{\circ}_\text{cell(c)} = E^{\circ}_\text{reduction} - E^{\circ}_\text{oxidation} = 0.976\,\mathrm{V} - (-0.48\,\mathrm{V}) = \boxed{1.456\,\mathrm{V}}\)

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

What is \(E^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the following reaction? $$ \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \longrightarrow \mathrm{BaSO}_{4}(s) $$ \(K_{s p}\) for \(\mathrm{BaSO}_{4}\) is \(1.1 \times 10^{-10}\)

An alloy made up of tin and copper is prepared by simultaneously electroplating the two metals from a solution containing \(\mathrm{Sn}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\). If \(20.0 \%\) of the total current is used to plate tin, while \(80.0 \%\) is used to plate copper, what is the percent composition of the alloy?

In a nickel-cadmium battery (Nicad), cadmium is oxidized to \(\mathrm{Cd}(\mathrm{OH})_{2}\) at the anode, while \(\mathrm{Ni}_{2} \mathrm{O}_{3}\) is reduced to \(\mathrm{Ni}(\mathrm{OH})_{2}\) at the cathode. A portable CD player uses \(0.175\) amp of current. How many grams of \(\mathrm{Cd}\) and \(\mathrm{Ni}_{2} \mathrm{O}_{3}\) are consumed when the CD player is used for an hour and a half?

Write the equation for the reaction, if any, that occurs when each of the following experiments is performed under standard conditions. (a) Sulfur is added to mercury. (b) Manganese dioxide in acidic solution is added to liquid mercury. (c) Aluminum metal is added to a solution of potassium ions.

Consider a salt bridge cell in which the anode is a manganese rod immersed in an aqueous solution of manganese(II) sulfate. The cathode is a chromium strip immersed in an aqueous solution of chromium(III) sulfate. Sketch a diagram of the cell, indicating the flow of the current throughout. Write the half- equations for the electrode reactions, the overall equation, and the abbreviated notation for the cell.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

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