Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 34

What is the emf of a cell consisting of a \(\mathrm{Pb}^{2+} / \mathrm{Pb}\) half-cell and a \(\mathrm{Pt} / \mathrm{H}^{+} / \mathrm{H}_{2}\) half-cell if \(\left[\mathrm{Pb}^{2+}\right]=0.10 \mathrm{M}\), \(\left[\mathrm{H}^{+}\right]=0.050 \mathrm{M},\) and \(P_{\mathrm{H}}=2.0 \mathrm{~atm} ?\)

Short Answer

Expert verified
The EMF of the cell is approximately -0.03V.

Step by step solution

01

Identify the Half-Cell Reactions and Standard Reduction Potentials

First, you need to identify each half-cell´s reaction and its standard reduction redox potential (E0). The Pb2+ /Pb half-cell, where Pb2+ is reduced to Pb, has a standard reduction potential \(E^0_{Pb2+/Pb}= -0.13V\). The Pt/ H+/ H2 half-cell is the hydrogen half-cell often used as a reference electrode; its standard reduction potential is always 0V.
02

Write the Cell Reaction

The overall cell reaction is: \(Pb^{2+} + H_{2} -> Pb + 2H^{+}\).
03

Applying the Nernst Equation

We apply the Nernst Equation: \(E_cell= E^0 - \frac{RT}{nF}lnQ\). However, for the condition of the problem, temperature T is assumed to be standard condition (25C & equivalent to 298K), R (gas constant) is \(8.31 JK^{-1}mol^{-1}\), n is the number of the electrons transferred in the reaction (which equals 2) and F (Faraday constant) is \(96500 C/mol \) . Then, this equation can be simplified as \(E_cell= E^0 - \frac{0.0591V}{n}logQ\).
04

Locate the Values

The E0 for cell can be calculated as: \(E^0_{cell} =E^0_{cathode}-E^0_{anode}\) = \( E^0_{Pb2+/Pb} - E^0_{ H+/ H2}\) = \( -0.13 - 0 = -0.13 V\). Q, the reaction quotient, is \([H^{+}]^2 / [Pb^{2+}]\) = \((0.050)^2 / 0.10 = 0.025\).
05

Calculate the EMF

Substitute the values into the equation from Step 3, we get: \(E_cell= -0.13V - \frac{0.0591V}{2}log(0.025) = - 0.03 V \).

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

Comment on whether \(\mathrm{F}_{2}\) will become a stronger oxidizing agent with increasing \(\mathrm{H}^{+}\) concentration.

The \(E^{\circ}\) value of one cell reaction is positive and that of another cell reaction is negative. Which cell reaction will proceed toward the formation of more products at equilibrium?

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

The half-reaction at an electrode is $$\mathrm{Mg}^{2+}(\text { molten })+2 e^{-} \longrightarrow \mathrm{Mg}(s)$$ Calculate the number of grams of magnesium that can be produced by supplying \(1.00 \mathrm{~F}\) to the electrode.

Consider a galvanic cell consisting of a magnesium electrode in contact with \(1.0 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and a cadmium electrode in contact with \(1.0 \mathrm{M}\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate \(E^{\circ}\) for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

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