Electrolysis can be used to determine atomic masses. A current of \(0.550\) A deposits \(0.55 \mathrm{~g}\) of a certain metal in 100 minutes. Calculate the atomic mass of the metal if eq. wt. = mole. wt./3 (a) 100 (b) \(45.0\) (c) \(48.25\) (d) \(144.75\)

Short Answer

Expert verified
The atomic mass of the metal is approximately 48.25 g/mol.

Step by step solution

01

Calculate the amount of substance deposited

Use Faraday's law of electrolysis to determine the amount of substance deposited. The law states that the amount of substance released at an electrode during electrolysis is directly proportional to the amount of electricity that passes through the electrolyte. The formula is given by \( m = \frac{MIt}{nF} \), where \( m \) is the mass of substance deposited (0.55 g), \( M \) is the molar mass of the substance, \( I \) is the current (0.550 A), \( t \) is the time in seconds (100 minutes \( \times \) 60 seconds/minute), \( n \) is the number of moles of electrons required to deposit one mole of substance, and \( F \) is Faraday's constant (approximately \( 96485 \) C/mol).
02

Convert time to seconds

First, convert the electrolysis time from minutes to seconds to use in the formula. \( t = 100 \text{ minutes} \times 60 \text{ seconds/minute} = 6000 \text{ seconds} \).
03

Solve for the molar mass (M)

Since we know the equivalent weight is a third of the molar weight (eq. wt. = mole wt./3), and that the equivalent weight corresponds to the molar mass divided by the valence (n), we can derive that the valence is 3. So, \( n = 3 \). Now, rearrange the formula to solve for \( M \): \( M = \frac{m \cdot n \cdot F}{I \cdot t} \).
04

Calculate the molar mass (M)

Substitute the known quantities into the rearranged formula: \( M = \frac{0.55 \text{ g} \cdot 3 \cdot 96485 \text{ C/mol}}{0.550 \text{ A} \cdot 6000 \text{ s}} = \frac{0.55 \cdot 3 \cdot 96485}{0.550 \cdot 6000} \text{ g/mol} = \frac{0.55 \cdot 3 \cdot 96485}{3300} \text{ g/mol} = \) approximately \( 48.25 \text{ g/mol} \).

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 Vaia!

Key Concepts

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

Chemical Calculations
Understanding chemical calculations is an integral part of mastering chemistry, especially when delving into concepts such as electrolysis. These calculations are necessary for predicting the outcome of reactions and understanding the quantitative aspects of chemical processes. When we discuss the context of electrolysis, we must become familiar with Faraday's laws of electrolysis, which provide the quantitative basis to relate the amount of substance deposited on an electrode to the quantity of electric charge passed through the solution.

Let's consider the problem presented. It involves calculating the atomic mass of a metal based on the amount of metal deposited during electrolysis. To approach this problem systematically, we apply chemical calculations using the formula derived from Faraday's law:
\( m = \frac{MIt}{nF} \)
Where each variable represents a specific quantity: mass deposited (m), molar mass (M), electric current (I), time (t), valence number (n), and Faraday's constant (F). By rearranging the equation and plugging in known values, we can solve for the unknown molar mass of the metal.

Performing these calculations correctly is essential for students to understand not only the specific procedure in question but also the broader application of fundamental chemical principles in practical scenarios like electrochemical deposition.
Molar Mass Determination
Molar mass determination is a key concept in chemistry that allows us to understand the relationship between the mass of a substance and the amount of substance present in moles. In an electrolysis experiment, accurately determining the molar mass is crucial because it helps identify the substance being deposited. This process also showcases the importance of stoichiometry and chemical calculations in everyday laboratory practices.

According to the problem we have, the equivalent weight of a substance is indicated to be one-third of the molecular weight (eq. wt. = mole wt./3). This piece of information, combined with Faraday's law, gives us insight into how to calculate the molar mass of the metal. Equivalent weight relates to the molar mass divided by the valence; thus, we can infer that the valence for our substance is three. By utilizing the electrolysis formula and solving for the molar mass (M), students can determine the molar mass of the unknown metal to be approximately 48.25 g/mol.

Understanding how to calculate molar mass directly from experimental data, as in the case of electrochemical deposition, not only reinforces the student's understanding of theoretical concepts but also provides valuable laboratory skills.
Electrochemical Deposition
Electrochemical deposition is a fascinating process where material from an electrolyte is deposited onto an electrode due to the flow of electric current. This process is not only vital for determining chemical properties like atomic mass but also has widespread applications in industries for plating, corrosion protection, and manufacturing of electronic components.

In the context of our problem, electrochemical deposition is used to deposit a known mass of a metal from its ionic solution. By measuring the mass of the deposited metal and knowing the amount of electric current used, we can apply Faraday's law to determine properties such as the metal's atomic or molar mass. This experimentally-driven method allows for the direct verification of theoretical molecular weights, highlighting the synergistic relationship between theoretical concepts and practical applications. Through such hands-on experiments, students can witness firsthand the principles of chemistry in action, cementing their understanding of both the scientific theory and the methodology behind determining chemical properties via electrochemical means.

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

Standard electrode potential of SHE at \(298 \mathrm{~K}\) is : (a) \(0.05 \mathrm{~V}\). (b) \(0.10 \mathrm{~V}\) (c) \(0.50 \mathrm{~V}\) (d) \(0.00 \mathrm{~V}\)

Consider the following equations for a cell reaction $$ \begin{gathered} A+B \rightleftharpoons C+D ; \quad E^{\circ}=x \text { volt, } K_{e q}=K_{1} \\\ 2 A+2 B \rightleftharpoons 2 C+2 D ; E^{\circ}=y \text { volt, } K_{e q}=K_{2} \end{gathered} $$ then : (a) \(x=y, K_{1}=K_{2}\) (b) \(x=2 y, K_{1}=2 K_{2}\) (c) \(x=y, K_{1}^{2}=K_{2}\) (d) \(x^{2}=y, K_{1}^{2}=K_{2}\)

Based on the following information arrange four metals, \(A, B, C\) and \(D\) in order of increasing ability to act as reducing agents : (I) Only \(C\) react with \(1 M \mathrm{HCl}\) to give \(\mathrm{H}_{2}(g)\) (II) When \(A\) is added to solution of the other metal ions, metallic \(D\) is formed but not \(B\) or \(C\) (a) \(D

Consider an electrochemical cell in which the following reaction occurs and predict which changes will decrease the cell voltage : $$ \mathrm{Fe}^{2+}(a q)+\mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Ag}(s)+\mathrm{Fe}^{3+}(a q) $$ (I) decrease the [Ag^ '] (II) increase in \(\left[\mathrm{Fe}^{3+}\right.\) ] (1II) increase the amount of \(\mathrm{Ag}\) (a) I (b) II and III (c) II (d) I and II

A conductance cell was filled with a \(0.02 \mathrm{M} \mathrm{KCl}\) solution which has a specific conductance of \(2.768 \times 10^{-3} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\). If its resistance is \(82.4 \mathrm{ohm}\) at \(25^{\circ} \mathrm{C}\), the cell constant is : (a) \(0.2182 \mathrm{~cm}^{-1}\) (b) \(0.2281 \mathrm{~cm}^{-1}\) (c) \(0.2821 \mathrm{~cm}^{-1}\) (d) \(0.2381 \mathrm{~cm}^{-1}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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