/*! 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 32 (a) Write the half-reaction that... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 32

(a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (b) The platinum electrode in a standard hydrogen electrode is specially prepared to have a large surface area. Why is this important? (c) Sketch a standard hydrogen electrode.

Short Answer

Expert verified
(a) The half-reaction that occurs at the hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell is: \(H_2(g) \rightarrow 2H^+(aq) + 2e^-\) (b) The large surface area for the platinum electrode allows more hydrogen gas molecules to come into contact with the electrode surface, leading to better electron transfer and greater overall efficiency of the electrochemical cell. (c) A standard hydrogen electrode consists of a glass tube with a small opening, a platinum electrode with a large surface area inserted into the tube, an attached external wire, a bulb-shaped container below the electrode containing the acidic aqueous solution and hydrogen gas, and a thin, flat sheet of platinum partially submerged in the solution, allowing hydrogen gas to come into contact with the electrode.

Step by step solution

01

Determine the oxidation half-reaction

When the hydrogen electrode serves as the anode of a voltaic cell, it undergoes oxidation. Oxidation is a process in which an element loses electrons. Since the hydrogen electrode is in an acidic aqueous solution, hydrogen ions (H鈦�) are available in the solution. In the oxidation process, hydrogen gas (H鈧�) is converted into hydrogen ions (H鈦�) by losing two electrons. The oxidation half-reaction for this process is as follows: \(H_2(g) \rightarrow 2H^+(aq) + 2e^-\) #b) Importance of a large surface area for the platinum electrode#
02

Explain the role of surface area

The platinum electrode in a standard hydrogen electrode is specially prepared with a large surface area, which is critical for its function. A large surface area for the platinum electrode allows more hydrogen gas molecules to come into contact with the electrode surface. As a result, more redox reactions can occur at the electrode surface, leading to better electron transfer and greater overall efficiency of the electrochemical cell. #c) Sketch a standard hydrogen electrode#
03

Sketch the components

To sketch a standard hydrogen electrode, follow these guidelines: 1. Draw a glass tube with a small opening at one end. 2. Insert the platinum electrode inside the glass tube, with the top covered with a protective cap. The electrode should have a porous surface to represent the large surface area. 3. Attach a wire to the platinum electrode to connect it to the external circuit. 4. Below the electrode, draw a bulb-shaped container, which is the reservoir for the acidic aqueous solution (1M H鈧僌鈦� ions) and some hydrogen gas (H鈧�). 5. Draw a thin, flat sheet of platinum attached to the platinum electrode, partially submerged in the solution. This sheet allows the passage of hydrogen gas to come into contact with the electrode. 6. Add labels for each component: platinum electrode, hydrochloric acid solution, hydrogen gas, and the external wire connection. Use this sketch as a visual aid when discussing the components and function of a standard hydrogen electrode.

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.

Half-reaction
In electrochemistry, understanding half-reactions is crucial. They represent the processes of reduction and oxidation that occur within a cell. For the hydrogen electrode acting as an anode in a voltaic cell, oxidation is the key process. During oxidation, the hydrogen gas, represented by the formula \(H_2(g)\), loses electrons. This transformation is written as the half-reaction: \(H_2(g) \rightarrow 2H^+(aq) + 2e^-\). In this process, the hydrogen gas is split into its ions and electrons:
  • Hydrogen Gas Conversion: It converts from a gaseous state to dissolve into the surrounding solution as ions.
  • Electron Loss: The half-reaction depicts the release of electrons, demonstrating the oxidation process.
Understanding this reaction is essential, as it highlights the movement of electrons, a foundation of electrochemical processes.
Hydrogen electrode
The standard hydrogen electrode (SHE) serves as a reference point for measuring electrode potentials. It's pivotal in electrochemistry due to its standardization:
  • Role: It acts as a reference electrode with a defined potential of 0 volts under standard conditions.
  • Components: It comprises a platinum electrode immersed in a solution containing 1M hydrogen ions, often from hydrochloric acid, and is exposed to hydrogen gas at a pressure of 1 atm.
  • Functionality: The hydrogen electrode can function as an anode or cathode, depending on the cell setup.
It is vital for establishing a baseline for redox potential, helping to calculate the standard reduction potentials of other electrodes relative to the hydrogen electrode.
Anode
An anode is a critical component in electrochemical cells, including voltaic cells. Specifically, in a hydrogen electrode system:
  • Function: The anode is where oxidation occurs, marked by the loss of electrons from hydrogen gas when used in a voltaic cell.
  • Reaction at the Anode: For a hydrogen electrode, the reaction is \(H_2(g) \rightarrow 2H^+(aq) + 2e^-\), which explicitly shows the process of oxidation.
  • Electron Flow: Electrons generated at the anode flow through an external circuit towards the cathode, driving the electrical circuit.
In this context, understanding the role of the anode helps in grasping how different electrodes participate in redox reactions, leading to electric power generation.
Platinum electrode
The use of a platinum electrode in a hydrogen electrode setup is deliberate and strategic:
  • Material Choice: Platinum is chosen for its inert nature and excellent conductivity, ensuring no interference with reactions.
  • Surface Area: Increasing the surface area by making it porous improves reaction efficiency by providing more contact points for hydrogen gas molecules, enhancing the rate of redox reactions.
  • Role in SHE: As part of the standard hydrogen electrode, the platinum electrode is vital for systematic electron transfer.
An increased surface area boosts the performance of the electrode significantly, which is crucial in achieving a steady and reliable redox process in electrochemical measurements.
Voltaic cell
A voltaic cell, also known as a galvanic cell, is a type of electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions:
  • Basic Structure: Consists of two half-cells, each containing an electrode and an electrolyte solution. The hydrogen electrode setup can be used as one half-cell.
  • Redox Reaction: In a voltaic cell, oxidation occurs at the anode and reduction at the cathode, generating electricity via electron flow in the external circuit.
  • Applications: Commonly used in batteries and various electronic devices.
Understanding voltaic cells involves recognizing how electron transfer between electrodes powers devices. The hydrogen electrode acting as part of a voltaic cell demonstrates the fundamental principles of electron movement and energy conversion.

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

A voltaic cell utilizes the following reaction: $$2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q)$$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}\), \(P_{\mathrm{H}_{2}}=0.95 \mathrm{~atm},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M},\) and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

A voltaic cell is constructed with two \(\mathrm{Zn}^{2+}-\mathrm{Zn}\) electrodes. The two half-cells have \(\left[\mathrm{Zn}^{2+}\right]=1.8 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=\) \(1.00 \times 10^{-2} M\), respectively. (a) Which electrode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Zn}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) have an atypical oxidation state.) (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow\) $$ \mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q) \text { (acidic solution) } $$ (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) $$ \mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q) \text { (acidic solution) } $$ (d) \(\mathrm{BrO}_{3}^{-}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{N}_{2}(g)\) (acidic solution) (e) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution)

(a) A \(\mathrm{Cr}^{3+}(a q)\) solution is electrolyzed, using a current of 7.60 A. What mass of \(\mathrm{Cr}(s)\) is plated out after 2.00 days? (b) What amperage is required to plate out \(0.250 \mathrm{~mol}\) Cr from a \(\mathrm{Cr}^{3+}\) solution in a period of \(8.00 \mathrm{~h} ?\)

(a) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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.