Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 7

The hydrogen electrode is dipped in a solution of \(\mathrm{pH}=3\) at \(25^{\circ} \mathrm{C}\). The potential of the cell would be (a) \(0.177 \mathrm{~V}\) (b) \(0.087 \mathrm{~V}\) (c) \(-0.177 \mathrm{~V}\) (d) \(0.059 \mathrm{~V}\)

Short Answer

Expert verified
The potential of the hydrogen electrode dipped in the solution with a pH of 3 at 25°C is approximately -0.177 V. So the correct option is (c) -0.177 V.

Step by step solution

01

Understand the Nernst Equation

The Nernst equation is a formula that calculates the electric potential of a half-cell or full cell reaction in electrochemistry. Written for hydrogen it is \(E=E_{0}-\frac{2.303 RT}{nF} pH\). Here \(E\) is the electrode potential, \(E_{0}\) is the standard reduction potential of a half-cell reduction reaction (for hydrogen, \(E_0 = 0V\)), \(R\) is the ideal gas constant (use \(R=0.08V\), as we will be using voltage, temperature in Kelvin and mole), \(T\) is the temperature in kelvin, \(n\) is the number of moles of electrons per mole of the substance (for hydrogen, \(n = 1\)), \(F\) is the Faraday constant (\(F = 96485 C mol^{-1}\)), and \(pH\) is the acidity of the solution.
02

Plug the given values into the Nernst equation

Plug the given values into the formula, \(E=E_{0}-\frac{2.303 RT}{nF}pH\). As this is a hydrogen electrode, \(E_{0}\) equals 0. The temperature is given as \(25^{\circ} \mathrm{C}\), but for the Nernst equation we need to convert this to Kelvin by adding 273, so \(T = 298K\). The number of moles of electrons, \(n\), for hydrogen is 1. The standard gas constant, \(R\), is 0.08 and the Faraday constant, \(F\), is 96485. Lastly, the given pH of the solution is 3. Now you can plug in these values: \(E = 0 - \frac{2.303*0.08*298}{1*96485}*3\).
03

Calculate the potential of the cell

Perform the calculation in the equation to find \(E\). It should result in a cell potential of approximately -0.177 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

In the electrochemical cell, flow of electrons is from (a) cathode to anode in the solution (b) cathode to anode through external supply (c) cathode to anode through internal supply (d) anode to cathode through internal supply

For a cell reaction involving \(2 e^{-}\) change, the standard emf of the cell is found to be \(0.295\) \(\mathrm{V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction will be (a) \(2.95 \times 10^{2}\) (b) 10 (c) \(1 \times 10^{10}\) (d) \(1 \times 10^{-10}\)

Three elements \(A, B\) and \(C\) have reduction potentials \(-1.5,-0.05\), and \(+1.50\). Then the correct order of their reducing power is (a) \(A>B>C\) (b) \(B>A>C\) (c) \(C>B>A\) (d) \(B>C>A\)

Saturated solution of \(\mathrm{KNO}_{3}\) is used to make a 'salt bridge' because (a) velocity of \(\mathrm{K}^{+}\) is greater than that of \(\mathrm{NO}^{-}{ }_{3}\) (b) velocity of \(\mathrm{NO}_{3}^{-}{ }_{3}\) is greater than that of \(\mathrm{K}^{+}\) (c) velocities of \(\mathrm{K}^{+}\) and \(\mathrm{NO}^{-}{ }_{3}\) both are nearly the same (d) none of the above

In an electrochemical cell, (a) potential energy decreases (b) kinetic energy decreases (c) potential energy changes into electrical energy (d) chemical energy changes into electrical energy.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

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