/*! 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 150 Given the following two standard... [FREE SOLUTION] | 91影视

91影视

Given the following two standard reduction potentials, $$ \begin{array}{ll} \mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.10 \mathrm{~V} \\ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.50 \mathrm{~V} \end{array} $$ solve for the standard reduction potential of the half-reaction $$ \mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+} $$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

Short Answer

Expert verified
The standard reduction potential of the half-reaction M鲁鈦 + e鈦 鈫 M虏鈦 is \(\mathscr{E}^{\circ}= -0.70 \mathrm{~V}\).

Step by step solution

01

Write the given half-reactions and the Nernst equation

The given half-reactions are: (a) M鲁鈦 + 3e鈦 鈫 M \(\mathscr{E}^{\circ}=-0.10 \mathrm{~V}\) (b) M虏鈦 + 2e鈦 鈫 M \(\mathscr{E}^{\circ}=-0.50 \mathrm{~V}\) We have to find the standard reduction potential of the following half-reaction: (c) M鲁鈦 + e鈦 鈫 M虏鈦 \(\mathscr{E}^{\circ}=?\) The Nernst equation relating 螖G潞 and the standard reduction potentials is given by: 螖G潞 = -nFE潞 Here, 螖G掳 is the standard Gibbs free energy change, n is the number of moles of electrons transferred, F is the Faraday constant, and \(\mathscr{E}^{\circ}\) is the standard reduction potential. Let's apply this equation to each half-reaction.
02

Calculate the Gibbs free energy change for the given half-reactions

The Nernst equation for the given half-reactions (a) and (b) can be written as: 螖G鈦(a) = -3FE鈦(a) 螖G鈦(b) = -2FE鈦(b) Now we substitute the values of standard reduction potentials: 螖G鈦(a) = -3F(-0.10V) 螖G鈦(b) = -2F(-0.50V)
03

Calculate the Gibbs free energy change for the desired half-reaction

Since 螖G掳 is an extensive property, we can now find the standard Gibbs free energy change for the desired half-reaction (c) by combining equations 螖G鈦(a) and 螖G鈦(b): 螖G鈦(c) = 螖G鈦(a) - 螖G鈦(b) 螖G鈦(c) = -3F(-0.10V) - (-2F(-0.50V))
04

Determine the number of electrons transferred in the desired half-reaction

In the desired half-reaction (c) M鲁鈦 鈫 M虏鈦, only one electron is transferred from M鲁鈦 to M虏鈦. Thus, n = 1.
05

Use the Nernst equation to find the standard reduction potential of the desired half-reaction

We know that 螖G鈦(c) = -nFE鈦(c). In our case, as n=1, we have 螖G鈦(c) = -FE鈦(c). Now we can find the standard reduction potential for half-reaction (c) by rearranging the equation: E鈦(c) = - (螖G鈦(c) / F) Substitute 螖G鈦(c) from step 3: E鈦(c) = - (-3F(-0.10V) - (-2F(-0.50V))) / F Simplify and calculate the value of E鈦(c): E鈦(c) = (0.3F - 1.0F) / F E鈦(c) = -0.7 V
06

Write the final answer

The standard reduction potential of the half-reaction M鲁鈦 + e鈦 鈫 M虏鈦 is \(\mathscr{E}^{\circ}= -0.70 \mathrm{~V}\).

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.

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.

Understanding the Nernst Equation
The Nernst equation is a fundamental equation in electrochemistry that describes the relationship between the reduction potential of a half-reaction at any point in time and the standard reduction potential, temperature, concentration of reactants, and products. It's named after Walther Nernst, a German chemist who formulated the relation in the late 19th century. In essence, the Nernst equation allows us to calculate the cell potential under non-standard conditions by adjusting for the concentration of ionic species.

The Nernst equation is a valuable tool when solving for the standard reduction potential of a half-reaction, as it incorporates the very important concept of Gibbs free energy to relate electrical energy with chemical potential. This interplay between electric potential and free energy is central to understanding the spontaneity of redox reactions.

While solving exercises concerning standard reduction potential, students must be attentive to the coefficients which denote the number of electrons transferred in each half-reaction. This becomes crucial when applying the Nernst equation, which integrates not just the potentials but also the stoichiometry of the process. Misinterpreting these coefficients can lead to incorrect calculations of cell potentials.
The Role of Gibbs Free Energy in Electrochemistry
Gibbs free energy, often denoted as \( G \), plays a pivotal role in determining the spontaneity of chemical reactions. It is essentially the energy available to do work during a process at constant temperature and pressure. In electrochemistry, the change in Gibbs free energy \( \Delta G \) can tell us whether a reaction will proceed spontaneously.

When \( \Delta G \) is negative, the reaction is spontaneous and can do work; if positive, the reaction is non-spontaneous, and work must be done to make it occur. The relationship between Gibbs free energy and the standard reduction potential is described quantitatively by \( \Delta G^\circ = -nFE^\circ \), where \( n \) is the number of moles of electrons, \( F \) is Faraday's constant, and \( E^\circ \) is the standard reduction potential.

Understanding this relationship can assist students in not just calculating cell potentials, but also in understanding the energetic implications of electron transfer processes. This concept is also the backbone of the exercise example provided, where Gibbs free energy is manipulated to arrive at the unknown standard reduction potential.
Essential Electrochemistry Concepts
Electrochemistry is an area of chemistry that studies the movement of electrons to generate electrical currents and the accompanying chemical changes. It's a bridge between the macroscopic world we observe and the submicro level where atomic particles, like electrons, play a crucial role. Fundamentally, it deals with how chemicals produce electricity and how electricity leads to chemical changes, which is particularly applicable in batteries, corrosion, and electrolysis.

In the context of our exercise, the critical points relate to the transfer of electrons between species through redox reactions. These reactions are harnessed in batteries to store and produce electrical energy. Redox reactions also form the basis for the development of sensors and the processing of materials.

Comprehending concepts like standard reduction potential, and how they are affected by variables such as concentration and temperature, equips students with tools to predict and quantify the behavior of electrochemical systems. The more comfortable students become with these fundamentals, the more adept they will be at applying this knowledge to real-world applications such as energy storage systems and corrosion prevention strategies.

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

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of \(1.00\) million A, what mass of aluminum can be produced in \(2.00 \mathrm{~h}\) ?

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

Consider a concentration cell similar to the one shown in Exercise 69, except that both electrodes are made of \(\mathrm{Ni}\) and in the left-hand compartment \(\left[\mathrm{Ni}^{2+}\right]=1.0 M .\) Calculate the cell potential at \(25^{\circ} \mathrm{C}\) when the concentration of \(\mathrm{Ni}^{2+}\) in the compartment on the right has each of the following values. a. \(1.0 M\) b. \(2.0 \mathrm{M}\) c. \(0.10 \mathrm{M}\) d. \(4.0 \times 10^{-5} M\) e. Calculate the potential when both solutions are \(2.5 M\) in \(\mathrm{Ni}^{2+}\). For each case, also identify the cathode, anode, and the direction in which electrons flow.

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$ \operatorname{AgI}(s)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s)+\mathrm{I}^{-}(a q) $$ (Hint: Reference the \(K_{\text {sp }}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+} .\) )

Calculate the \(\mathrm{pH}\) of the cathode compartment for the following reaction given \(\mathscr{E}_{\text {cell }}=3.01 \mathrm{~V}\) when \(\left[\mathrm{Cr}^{3+}\right]=0.15 \mathrm{M}\), \(\left[\mathrm{Al}^{3+}\right]=0.30 M\), and \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.55 M\) \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow\) \(2 \mathrm{Al}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.