Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 94

Electrolysis of an alkaline earth metal chloride using a current of \(5.00 \mathrm{~A}\) for \(748 \mathrm{~s}\) deposits \(0.471 \mathrm{~g}\) of metal at the cathode. What is the identity of the alkaline earth metal chloride?

Short Answer

Expert verified
The identity of the alkaline earth metal chloride is magnesium chloride (MgCl2). This is determined by using Faraday's laws of electrolysis and calculating the molar mass of the deposited metal, which is approximately 24.32 g/mol, close to the molar mass of magnesium (24.305 g/mol).

Step by step solution

01

Gather information from the problem statement

We know the following information: 1. Current, I = 5.00 A 2. Time, t = 748 s 3. Mass of metal deposited, m = 0.471 g
02

Calculate the charge passed during the electrolysis

We can use the formula \(Q = It\) to find the charge passed during the electrolysis. So, \(Q = 5.00 \mathrm{~A} \times 748 \mathrm{~s} = 3740 \mathrm{~C}\)
03

Calculate the equivalent weight of the metal

According to Faraday's second law of electrolysis, the weight of a substance deposited during electrolysis is proportional to its equivalent weight, that is: \(W = Ze\) Where: - W is the weight of the substance deposited, - Z is the electrochemical equivalent (ECE), - e is the charge passed. We can calculate the electrochemical equivalent (ECE) using the formula: \(Z = \frac{W}{e}\) Substituting the given values, we get: \(Z = \frac{0.471 \mathrm{~g}}{3740 \mathrm{~C}} = 1.259 \times 10^{-4} \mathrm{\frac{g}{C}}\)
04

Calculate the molar mass of the metal

Since the alkaline earth metals have a valency of 2, the molar mass of the metal is twice the equivalent weight. So, Molar mass of metal = \(2 \times Z \times Faraday's \,\, constant\), where Faraday's constant (F) is \(96485 \mathrm{\frac{C}{mol}}\). Molar mass of metal = \(2 \times 1.259 \times 10^{-4} \mathrm{\frac{g}{C}} \times 96485 \mathrm{\frac{C}{mol}}\) Molar mass of metal \( = 24.32 \mathrm{~g/mol}\)
05

Identify the alkaline earth metal

The molar mass of the metal is approximately 24.32 g/mol. This value is very close to the molar mass of magnesium (24.305 g/mol). So, the identity of the alkaline earth metal chloride is magnesium chloride (MgCl2).

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

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of \(1.00\) million A, what mass of aluminum can be produced in \(2.00 \mathrm{~h} ?\)

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{~kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

For the following half-reaction, \(\mathscr{E}^{\circ}=-2.07 \mathrm{~V}\) : $$\mathrm{AlF}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}+6 \mathrm{~F}^{-}$$ Using data from Table \(18.1\), calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{Al}^{3+}(a q)+6 \mathrm{~F}^{-}(a q) \rightleftharpoons \mathrm{AlF}_{6}^{3-}(a q) \quad K=?$$

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)\) b. \(\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)\)

A factory wants to produce \(1.00 \times 10^{3} \mathrm{~kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{~h}\) to accomplish this?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear 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.

Sign-up for free