Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 9: Problem 98

In electrolysis of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what is liberated at anode? (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{SO}_{4}^{2-}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{O}_{2}\)

Short Answer

Expert verified
Oxygen (2) is liberated at the anode.

Step by step solution

01

Understanding Electrolysis

Electrolysis involves passing an electrical current through a solution to cause a chemical reaction. In this case, we are looking at the electrolysis of dilute sulfuric acid, which contains water and ions such as hydrogen ions (H⁺) and sulfate ions (SO₄²⁻).
02

Identifying Anode Reaction

During electrolysis, different reactions occur at the cathode and the anode. At the anode, oxidation takes place. In the case of dilute H₂SO₄, the most likely reaction is the oxidation of water molecules to produce oxygen gas and hydrogen ions: 2H₂O → O₂ + 4H⁺ + 4e⁻.
03

Evaluating Other Options

The other ions in the solution such as hydrogen ions (H⁺) and sulfate ions (SO₄²⁻) either do not participate in the anode reaction or are not oxidized under these conditions. For example, sulfate ions are quite stable and do not get oxidized easily.
04

Selecting the Correct Answer

Based on the oxidation reaction at the anode, oxygen gas (O₂) is liberated. Therefore, the correct answer is oxygen (O₂), which corresponds to option (d).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Anode Reaction
An important part of understanding electrolysis is knowing what happens at the anode. In electrolysis, the anode is the positive electrode where oxidation reactions occur. These reactions involve the loss of electrons by molecules or ions. Because of this, the anode is a site where electrons go to be released. In the specific case of the electrolysis of dilute sulfuric acid (\(\text{H}_2\text{SO}_4\)), understanding the anode reaction helps us predict the products formed.
When the electrical current is passed through the sulfuric acid solution, water molecules become the key players at the anode. They undergo oxidation, meaning they lose electrons. This process leads to the formation of oxygen gas (\(\text{O}_2\)) as a primary product, along with additional hydrogen ions. This is because unlike other ions like sulfate (\(\text{SO}_4^{2-}\)), water oxidizes much more easily, making oxygen gas production the most favorable anode reaction.
Oxidation
Oxidation is a fundamental concept in many chemical processes, including electrolysis. It involves the loss of electrons when a molecule, atom, or ion reacts. During the electrolysis of sulfuric acid, this process is crucial in determining what gets liberated at the anode.
Let's consider the oxidation of water molecules in this context. As electrolysis occurs, water (\(\text{H}_2\text{O}\)) is oxidized. This oxidation process is represented by the chemical equation:
  • \[2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-\]
This equation tells us that each pair of water molecules loses four electrons and produces an oxygen molecule along with hydrogen ions. Oxidation thus not only helps form new substances but also plays a critical role in balancing the overall reaction during electrolysis.
This knowledge is key because any change in electron transfer impacts the type of gases or ions released during the process.
Electrolysis of Sulfuric Acid
The electrolysis of sulfuric acid is a classic example used to illustrate key electrochemical principles. When sulfuric acid is diluted and subjected to an electric current, it facilitates the separation of its components, illustrating the process of electrolysis in action.
Sulfuric acid, when diluted, breaks down into hydrogen ions (\(\text{H}^+\)) and sulfate ions (\(\text{SO}_4^{2-}\)). During electrolysis, however, it is important to consider the presence of water molecules as they have a crucial role. At the anode of the electrolysis cell, water molecules undergo oxidation to produce oxygen gas and hydrogen ions, as previously discussed.
  • At the cathode, a different reduction process occurs, where hydrogen ions gain electrons to form hydrogen gas (\(\text{H}_2\)).
While sulfate ions remain largely inactive during this process, as they are stable and do not readily undergo reduction or oxidation, understanding why these ions don't participate offers insight into ion activity in electrolysis.
This combination of reactions results in oxygen being released at the anode and hydrogen being released at the cathode, making the electrolysis of sulfuric acid a textbook example of redox reactions during electrochemical processes.

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

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration \(0.1 \mathrm{M}\) is \(100 \Omega\). The conductivity of this solution is \(1.29 \mathrm{~S} \mathrm{~m}^{-1}\). Resistance of the same cell when filled with \(0.2 \mathrm{M}\) of the same solution is \(520 \Omega\). The molar conductivity of \(0.02 \mathrm{M}\) solution of the electrolyte will be (a) \(124 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) (b) \(1240 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) (c) \(1.24 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\) (d) \(12.4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)

The specific conductance of \(0.1 \mathrm{~N} \mathrm{KCl}\) solution at \(23^{\circ} \mathrm{C}\) is \(0.012 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\). The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be (a) \(0.142 \mathrm{~cm}^{-1}\) (b) \(0.616 \mathrm{~cm}^{-1}\) (c) \(6.16 \mathrm{~cm}^{-1}\) (d) \(616 \mathrm{~cm}^{-1}\)

The standard reduction potentials of \(\mathrm{Ag}, \mathrm{Cu}, \mathrm{Co}\) and \(\mathrm{Zn}\) are \(0.799,0.337,-0.277\) and \(-0.762 \mathrm{~V}\) respectively. Which of the following cells will have maximum cell emf? (a) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{IM}) \| \mathrm{Cu}^{2+}(1 \mathrm{M})\right| \mathrm{Cu}\) (b) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{lM}) \| \mathrm{Ag}^{+}(\mathrm{lM})\right| \mathrm{Ag}\) (c) \(\mathrm{Cu}\left|\mathrm{Cu}^{2+}(\mathrm{lM}) \| \mathrm{Ag}^{+}(\mathrm{IM})\right| \mathrm{Ag}\) (d) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{lM}) \| \mathrm{Co}^{2+}(\mathrm{IM})\right| \mathrm{Co}\)

Time required to deposit 1 millimol of aluminium metal by the passage of \(9.65\) ampere of current through aqueous solution of aluminium ion, is (a) \(30 \mathrm{~s}\) (b) \(10 \mathrm{~s}\) (c) \(30,000 \mathrm{~s}\) (d) \(10,000 \mathrm{~s}\)

For the electrolysis of \(\mathrm{CuSO}_{4}\) solution which is/are correct? (a) Cathode reaction: \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\) using \(\mathrm{Cu}\) electrode (b) Anode reaction: \(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2}++2 \mathrm{e}^{-}\)using \(\mathrm{Cu}\) electrode (c) Cathode reaction: \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) using Pt electrode (d) Anode reaction: \(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}\) using \(\mathrm{Pt}\) electrode
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

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