Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 113

The value of \(E_{\text {cell }}^{\circ}\) for the reaction \(\mathrm{Zn}(\mathrm{s})+\) \(\mathrm{Pb}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Pb}(\mathrm{s})\) is \(0.66 \mathrm{V} .\) This means that for the reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{Pb}^{2+}(0.01 \mathrm{M})\) \(\rightarrow \mathrm{Zn}^{2+}(0.10 \mathrm{M})+\mathrm{Pb}(\mathrm{s}), E_{\text {cell }}\) equals \((\mathrm{a}) 0.72 \mathrm{V}\) (b) \(0.69 \mathrm{V} ;\) (c) \(0.66 \mathrm{V} ;\) (d) \(0.63 \mathrm{V}\)

Short Answer

Expert verified
The correct answer is (d) 0.63 V.

Step by step solution

01

Understanding the Nernst Equation

The Nernst equation allows us to calculate the potential of a cell under non-standard conditions (not at equilibrium). It relates the cell potential at any point in time to the standard cell potential, temperature, reaction quotient, and the number of electrons transferred in the redox reaction. The formula usually takes two forms: \n1. \(E_{cell}=E_{cell}^{o}-\frac{0.0592}{n}logQ\) at 25°C \n2. \(E_{cell}=E_{cell}^{o}-\frac{(RT)}{nF}lnQ\) at other temperatures. Here, \(R\) is the gas constant (8.314 J/K mol), \(F\) is Faraday's constant (96485 C/mol), \(T\) is the temperature (298 K), and \(n\) is the moles of electrons transferred in the redox reaction. Hence, for first part, we'll use the equation at 25°C.
02

Identifying the Values

We have \(E_{cell}^{o}=0.66 V\), \(Q=\frac{[Zn^{2+}]}{[Pb^{2+}]}=\frac{0.10 M}{0.01 M}=10\), and \(n=2\) electrons are exchanged in the reaction. In the Nernst equation, we plug these values in.
03

Solving the Equation

Insert the given values into the formula: \(E_{cell}=0.66 V- \frac{0.0592}{2}log10 = 0.66 V - 0.03 V = 0.63 V \) Thus, we've calculated the cell potential under the given conditions.

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

Of the following statements concerning electrochemical cells, the correct ones are: (a) The cathode is the negative electrode in both voltaic and electrolytic cells. (b) The function of a salt bridge is to permit the migration of electrons between the half-cell compartments of an electrochemical cell. (c) The anode is the negative electrode in a voltaic cell. (d) Electrons leave the cell from either the cathode or the anode, depending on what electrodes are used. (e) Reduction occurs at the cathode in both voltaic and electrolytic cells. (f) If electric current is drawn from a voltaic cell long enough, the cell becomes an electrolytic cell. (g) The cell reaction is an oxidationreduction reaction.

For the reduction half-reaction \(\mathrm{Hg}_{2}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}\) \(\longrightarrow 2 \mathrm{Hg}(1), E^{\circ}=0.797 \mathrm{V} .\) Will \(\mathrm{Hg}(\mathrm{l})\) react with and dissolve in HCl(aq)? in HNO3(aq)? Explain.

Only a tiny fraction of the diffusible ions move across a cell membrane in establishing a Nernst potential (see Focus On 20: Membrane Potentials), so there is no detectable concentration change. Consider a typical cell with a volume of \(10^{-8} \mathrm{cm}^{3},\) a surface area \((A)\) of \(10^{-6} \mathrm{cm}^{2},\) and a membrane thickness \((l)\) of \(10^{-6} \mathrm{cm}\) Suppose that \(\left[\mathrm{K}^{+}\right]=155 \mathrm{mM}\) inside the cell and \(\left[\mathrm{K}^{+}\right]=4 \mathrm{mM}\) outside the cell and that the observed Nernst potential across the cell wall is \(0.085 \mathrm{V}\). The membrane acts as a charge-storing device called a capacitor, with a capacitance, \(C,\) given by $$C=\frac{\varepsilon_{0} \varepsilon A}{l}$$ where \(\varepsilon_{0}\) is the dielectric constant of a vacuum and the product \(\varepsilon_{0} \varepsilon\) is the dielectric constant of the membrane, having a typical value of \(3 \times 8.854 \times 10^{-12}\) \(\mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}\) for a biological membrane. The SI unit of capacitance is the firad, \(1 \mathrm{F}=1\) coulomb per volt \(=1 \mathrm{CV}^{-1}=1 \times \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-1}\) (a) Determine the capacitance of the membrane for the typical cell described. (b) What is the net charge required to maintain the observed membrane potential? (c) How many \(\mathrm{K}^{+}\) ions must flow through the cell membrane to produce the membrane potential? (d) How many \(\mathrm{K}^{+}\) ions are in the typical cell? (e) Show that the fraction of the intracellular \(K^{+}\) ions transferred through the cell membrane to produce the membrane potential is so small that it does not change \(\left[\mathrm{K}^{+}\right]\) within the cell.

For the voltaic cell, $$\begin{array}{l} \mathrm{Ag}(\mathrm{s}) | \mathrm{Ag}^{+}(0.015 \mathrm{M}) \| \mathrm{Fe}^{3+}(0.055 \mathrm{M}) \\ \quad \mathrm{Fe}^{2+}(0.045 \mathrm{M}) | \mathrm{Pt}(\mathrm{s}) \end{array}$$ (a) what is \(E_{\text {cell initially? }}\) (b) As the cell operates, will \(E_{\text {cell increase }}\) decrease, or remain constant with time? Explain. (c) What will be \(E_{\text {cell }}\) when \(\left[\mathrm{Ag}^{+}\right]\) has increased to \(0.020 \mathrm{M} ?\) (d) What will be \(\left[\mathrm{Ag}^{+}\right]\) when \(E_{\text {cell }}=0.010 \mathrm{V} ?\) (e) What are the ion concentrations when \(E_{\text {cell }}=0 ?\)

For the reaction \(2 \mathrm{Cu}^{+}(\mathrm{aq})+\mathrm{Sn}^{4+}(\mathrm{aq}) \longrightarrow\) \(2 \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq}), E_{\mathrm{cell}}^{\circ}=-0.0050 \mathrm{V}\) (a) can a solution be prepared at \(298 \mathrm{K}\) that is \(0.500 \mathrm{M}\) in each of the four ions? (b) If not, in which direction will a reaction occur?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

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