/*! 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 146 In which of the following aqueou... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 146

In which of the following aqueous solutions during electrolysis, \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) are liberated? (a) \(\mathrm{CuCl}_{2}\) (aq) (b) \(\mathrm{KCl}(\mathrm{aq})\) (c) \(\mathrm{MgCl}_{2}(\mathrm{aq})\) (d) \(\mathrm{NaCl}(\mathrm{aq})\)

Short Answer

Expert verified
Solutions (b), (c), and (d) liberate \\( ext{H}_2 \\) and \\( ext{Cl}_2 \\); (a) does not.

Step by step solution

01

Identify Ions in Solution

For electrolysis, the dissolved ions in the aqueous solutions are identified. - In option (a) CuCl2 (aq), the ions present are Cu虏鈦 and Cl鈦. - In option (b) KCl (aq), the ions present are K鈦 and Cl鈦. - In option (c) MgCl2 (aq), the ions present are Mg虏鈦 and Cl鈦. - In option (d) NaCl (aq), the ions present are Na鈦 and Cl鈦.
02

Identify Possible Electrolysis Products

Consider what occurs at each electrode: - At the cathode, reduction occurs. Hydrolysis of water can liberate H鈧 ( ext{H鈦簘 or ext{H鈧侽} 鈬 ext{H}_2). - At the anode, oxidation occurs. Halide ions can give Cl鈧 ( ext{Cl鈦粆 鈬 ext{Cl鈧倉). For H鈧 to be liberated, absence of metal reduction (Cu, K, Mg, Na) at the cathode is required.
03

Cathode Reaction Analysis

Determine which metal ions, if any, are more readily discharged at the cathode compared to H鈦. - Cu虏鈦 is lower than H鈦 in the electrochemical series, so will be reduced first. Thus, ext{CuCl鈧倉 will not liberate H鈧. - K鈦, Mg虏鈦, and Na鈦 are higher than H鈦, so H鈧 is liberated in their solutions.
04

Anode Reaction Analysis

At the anode, the Cl鈦 ion will be oxidized to Cl鈧 over the oxidation of H鈧侽 as Cl鈦 has a lower discharge tendency than OH鈦. This applies to each solution with Cl鈦 present: all can produce Cl鈧.
05

Determine the Correct Solutions

Based on cathode and anode reactions, - KCl, MgCl鈧, and NaCl solutions liberate H鈧 and Cl鈧 as their ions lead to these reactions. - CuCl鈧 does not meet the requirement as it liberates Cu at the cathode instead.

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.

Electrochemical series
The electrochemical series is a list of metal ions arranged according to their ability to gain electrons, which is known as their reduction potential. This series helps to predict which ions will be reduced at the cathode during the process of electrolysis. When deciding which reaction occurs at the electrodes in an electrolyte solution, the electrochemical series serves as a helpful guide. For instance:
  • Ions with higher reduction potentials (lower in the series) are more readily reduced than those with lower reduction potentials (higher in the series).
  • In the context of the given exercise, electrolytes like Cu虏鈦 are lower in the series than H鈦, meaning Cu虏鈦 will be reduced at the cathode instead of H鈦.
Understanding the position of each ion helps determine the course of reactions in electrolysis.
Cathode reaction
The cathode in an electrolysis setup is where reduction reactions occur. During electrolysis, different ions compete to receive electrons and get reduced. The choice of ions reduced at the cathode is largely influenced by the electrochemical series.In aqueous solutions involving chloride salts such as \(\mathrm{KCl}, \mathrm{MgCl}_{2}, \mathrm{NaCl}\), the following can occur:
  • H鈧 is liberated at the cathode since ions such as K鈦, Mg虏鈦, and Na鈦 are positioned higher than H鈦 in the electrochemical series.
  • Thus, instead of the metal ions, it is the hydrogen cations (H鈦) or water molecules that are reduced to hydrogen gas (H鈧).
This results in the liberation of hydrogen gas in these electrolyte solutions.
Anode reaction
At the anode, oxidation reactions take place. For chloride-containing solutions, Cl鈦 ions are oxidized. The ease of this oxidation can be attributed to the position of chloride ions concerning hydroxide ions (OH鈦). In electrolysis:
  • Cl鈦 ions are oxidized to produce chlorine gas (Cl鈧) at the anode.
  • This reaction is more favorable than the oxidation of hydroxide ions due to the discharge potential being lower for chlorine ions.
For any solution containing Cl鈦 ions, the expected outcome is the production of chlorine gas at the anode.
Reduction and oxidation processes
A key principle in electrolysis is understanding the reduction and oxidation processes at the electrodes. Reduction refers to the gain of electrons and occurs at the cathode, while oxidation refers to the loss of electrons and takes place at the anode.The processes can be summarized as:
  • Reduction: An example is \(\mathrm{H^+ + e^- \rightarrow H_2}\), where hydrogen ions gain electrons to form hydrogen gas (H鈧).
  • Oxidation: For example, chloride ions undergo \(\mathrm{2Cl^- \rightarrow Cl_{2} + 2e^-}\), losing electrons to form chlorine gas (Cl鈧).
Understanding these processes helps to predict the course of electrolysis reactions, making it possible to determine the gaseous products that will be liberated at the electrodes.

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

The \(E\left(\mathrm{M}^{3+} / \mathrm{M}^{2+}\right)\) values for \(\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}\) and \(\mathrm{Co}\) are \(-0.41,+1.57,+0.77\) and \(+1.97 \mathrm{~V}\) respectively. For which one of these metals the change in oxidation state form \(+2\) to \(+3\) is easiest? (a) \(\mathrm{Cr}\) (b) \(\mathrm{Mn}\) (c) \(\mathrm{Fe}\) (d) \(\mathrm{Co}\)

The equivalent conductance of \(\mathrm{Ba}^{2+}\) and \(\mathrm{Cl}^{-}\)are respectively 127 and \(76 \mathrm{ohm}^{-1} \mathrm{~cm}^{2}\) equiv \(^{-1}\) at infinite dilution. The equivalent conductance (in ohm \(^{-1} \mathrm{~cm}^{2}\) equiv \(^{-1}\) ) of \(\mathrm{BaCl}_{2}\) at infinite dilution will be (a) \(139.5\) (b) 203 (c) 279 (d) \(101.5\)

If \(E^{\circ}\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)=-0.441 \mathrm{~V}\) and \(E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.771 \mathrm{~V}\) the standard emf of the reaction \(\mathrm{Fe}+2 \mathrm{Fe}^{3+} \longrightarrow 3 \mathrm{Fe}^{2+}\) will be (a) \(1.212 \mathrm{~V}\) (b) \(0.111 \mathrm{~V}\) (c) \(0.330 \mathrm{~V}\) (d) \(1.653 \mathrm{~V}\)

The limiting molar conductivities \(\Lambda^{\circ}\) for \(\mathrm{NaCl}, \mathrm{KBr}\) and \(\mathrm{KCl}\) are 126,152 and \(150 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) respectively. The \(\Lambda^{\circ}\) for \(\mathrm{NaBr}\) is (a) \(278 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (b) \(178 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (c) \(128 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (d) \(306 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)

When a quantity of electricity is passed through \(\mathrm{CuSO}_{4}\) solution, \(0.16 \mathrm{~g}\) of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of \(\mathrm{H}_{2}\) liberated at STP will be (At. \(\mathrm{wt}\) of \(\mathrm{Cu}=64\) ) (a) \(4.0 \mathrm{~cm}^{3}\) (b) \(56 \mathrm{~cm}^{3}\) (c) \(604 \mathrm{~cm}^{3}\) (d) \(8.0 \mathrm{~cm}^{3}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Inorganic 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.