Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 10

Discuss the spontaneity of an electrochemical reaction in terms of its standard emf \(\left(E_{\text {cell }}^{\circ}\right)\).

Short Answer

Expert verified
The spontaneity of an electrochemical reaction can be determined by its standard emf. If the standard emf \(E_{\text {cell }}^{\circ}\) is positive, the reaction is spontaneous, and if negative, the reaction is non-spontaneous.

Step by step solution

01

Understanding standard emf and spontaneity

Consider the relationship between the standard emf of a cell and the Gibbs free energy change (\(\Delta G\)) for the reaction. The relationship can be given as \(\Delta G^{\circ} = -nFE_{\text {cell }}^{\circ}\), where \(n\) is the number of moles of electrons transferred in the half reactions, \(F\) is the Faraday's constant approx. 96500 C/mol and \(E_{\text {cell }}^{\circ}\) is the standard cell potential.
02

Relating Gibbs free energy change to spontaneity

For a spontaneous reaction, the Gibbs free energy change, \(\Delta G\) must be negative. Thus, if \(\Delta G^{\circ} = -nFE_{\text {cell }}^{\circ}\), for the reaction to be spontaneous, \(-nFE_{\text {cell }}^{\circ}\) must be less than 0, which implies that the standard emf of the cell, \(E_{\text {cell }}^{\circ}\), must be greater than 0.
03

Conclusion

So, if the value of standard emf \(\left(E_{\text {cell }}^{\circ}\right)\) for an electrochemical reaction is positive, the reaction is spontaneous, and if it is negative, the reaction is non-spontaneous.

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

Based on the following standard reduction potentials: $$ \begin{aligned} \mathrm{Fe}^{2+}(a q)+2 e^{-} & \longrightarrow \mathrm{Fe}(s) & & E_{1}^{\circ}=-0.44 \mathrm{~V} \\ \mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & & E_{2}^{\circ}=& 0.77 \mathrm{~V} \end{aligned} $$ calculate the standard reduction potential for the half-reaction $$ \mathrm{Fe}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Fe}(s) \quad E_{3}^{\circ}=? $$

How many moles of electrons are required to produce (a) \(0.84 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at exactly 1 atm and \(25^{\circ} \mathrm{C}\) from aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution; (b) \(1.50 \mathrm{~L}\) of \(\mathrm{Cl}_{2}\) at \(750 \mathrm{mmHg}\) and \(20^{\circ} \mathrm{C}\) from molten \(\mathrm{NaCl} ;\) (c) \(6.0 \mathrm{~g}\) of Sn from molten \(\mathrm{SnCl}_{2} ?\)

In recent years there has been much interest in electric cars. List some advantages and disadvantages of electric cars compared to automobiles with internal combustion engines.

Predict whether \(\mathrm{Fe}^{3+}\) can oxidize \(\mathrm{I}^{-}\) to \(\mathrm{I}_{2}\) under standard-state conditions.

What is the difference between a galvanic cell (such as a Daniell cell) and an electrolytic cell?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

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