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

Atomic masses can be determined by electrolysis. In one hour, a current of \(0.600\) A deposits \(2.42 \mathrm{~g}\) of a certain metal, \(\mathrm{M}\), which is present in solution as \(\mathrm{M}^{+}\) ions. What is the atomic mass of the metal?

Short Answer

Expert verified
Answer: The atomic mass of the metal is approximately 108 g/mol.

Step by step solution

01

Determine the electric charge passed

The first step is to calculate the total electric charge passed during electrolysis. Electric charge (Q) can be calculated using the formula: $$Q = It$$ where \(I\) is the current and \(t\) is the time (in seconds). The given current, \(I\), is 0.600 A and the time, \(t\), is 1 hour (which is equivalent to 3600 seconds). To find the charge, we just need to multiply these values: $$Q = (0.600 A)(3600 s) = 2160 C$$
02

Calculate the moles of the metal ions deposited

The second step is to calculate the moles of metal ions deposited during the electrolysis process. We will use Faraday's law of electrolysis to determine this: $$\text{moles of ions} = \frac{\text{charge}}{\text{Faraday's constant} \times \text{number of electrons per ion}}$$ We are given that the metal's ions have a charge of +1 (as \(\mathrm{M}^{+}\) ions), so the number of electrons per ion would also be 1. Faraday's constant is approximately \(9.65 \times 10^{4} \mathrm{C/mol}\). Now we can calculate the moles of ions deposited: $$\text{moles of ions} = \frac{2160 C}{(9.65 \times 10^{4} \mathrm{C/mol}) \times 1} \approx 0.0224 \mathrm{moles}$$
03

Calculate the atomic mass of the metal

Finally, we will determine the atomic mass of the metal using the relationship between moles, mass, and atomic mass: $$\text{atomic mass} = \frac{\text{mass}}{\text{moles}}$$ We are given that the mass of the metal deposited is 2.42 g. Now we can find the atomic mass of the metal: $$\text{atomic mass} = \frac{2.42 g}{0.0224 \mathrm{moles}} \approx 108 \mathrm{g/mol}$$ So the atomic mass of the metal is approximately 108 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.

Electrolysis
Electrolysis is a fascinating process used to drive chemical reactions using an electric current. During electrolysis, an electrical current passes through a solution or molten substance to cause a non-spontaneous chemical change. This process is widely used in various industrial applications, such as metal plating, extracting metals from ores, and even in water splitting to produce hydrogen and oxygen.
In our exercise, electrolysis is used to determine the atomic mass of a metal. The metal ions in the solution deposit onto an electrode, and by measuring this deposition, we indirectly gather information about the atomic mass of the metal. Electrolysis relies on the movement of ions within the solution, allowing metals like copper, zinc, and silver to deposit in their metallic form upon the electrode as they gain electrons to form the neutral metal atoms.
Faraday's Law of Electrolysis
Faraday's Law of Electrolysis forms the foundation of quantifying electrolysis. It allows us to connect the amount of substance deposited at an electrode to the amount of electric charge passed through the solution. This law states that the amount of substance deposited at each electrode is directly proportional to the total electric charge passed through the electrolyte.
  • The constant of proportionality used in Faraday's Law is Faraday's constant, approximately \( 96,500 \ C/mol \), representing the electric charge per mole of electrons.
  • For our exercise, Faraday's Law helps us calculate the number of moles of metal ions deposited during the electrolysis process. Given the charge on the ions and the current passed through the system, we can derive the moles of metal deposited with high precision.
Understanding Faraday's Law is crucial because it bridges chemical knowledge with electrical concepts, showing how the two fields cooperate to achieve meaningful results in electrolysis.
Atomic Mass Calculation
Atomic mass calculation is a crucial step in determining the identity of unknown substances. The atomic mass is a number that reflects the average mass of atoms of an element, measured in atomic mass units (amu).
In the context of our problem, once we have calculated the moles of the metal using the amount of charge passed and Faraday's Law, we can determine its atomic mass through the relationship:
  • Atomic Mass = \( \frac{\text{Mass of Metal Deposited}}{\text{Moles of Metal}} \)
This formula helps us find the atomic mass by dividing the mass of the deposited metal we have by the moles calculated. The result gives insight into the metal's characteristics and aids in identifying the metal, rounding out our electrolysis experiment nicely.
Electric Charge
Electric charge is a fundamental property of matter that explains how particles interact through electromagnetic forces. It is measured in coulombs (\(C\)) and can be positive or negative, which affects how particles attract or repel one another.
For electrolysis, electric charge is crucial because the entire process is driven by passing a current through the electrolyte, which involves moving electric charges. We calculate the electric charge (\(Q\)) using the formula:
  • \( Q = It \)
Here, \(I\) is the current in amperes, and \(t\) is time in seconds. In our example, by calculating \(Q\), we understand how much total electric charge has passed through the circuit, which is essential for using Faraday's Law to determine the number of moles of metal ions deposited. Knowing how to calculate electric charge allows us to quantify the results of our electrolysis experiments.

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

Consider a voltaic cell in which the following reaction occurs. $$ \mathrm{Zn}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Sn}(s) $$ (a) Calculate \(E^{\circ}\) for the cell. (b) When the cell operates, what happens to the concentration of \(\mathrm{Zn}^{2+}\) ? The concentration of \(\mathrm{Sn}^{2+}\) ? (c) When the cell voltage drops to zero, what is the ratio of the concentration of \(\mathrm{Zn}^{2+}\) to that of \(\mathrm{Sn}^{2+}\) ? (d) If the concentration of both cations is \(1.0 \mathrm{M}\) originally, what are the concentrations when the voltage drops to zero?

Consider the following reaction at \(25^{\circ} \mathrm{C}\). $$ \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{Br}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Br}_{2}(l) $$ If \(\left[\mathrm{H}^{+}\right]\) is adjusted by adding a buffer that is \(0.100 M\) in sodium acetate and \(0.100 \mathrm{M}\) in acetic acid, the pressure of oxygen gas is \(1.00 \mathrm{~atm}\), and the bromide concentration is \(0.100 \mathrm{M}\), what is the calculated cell voltage? ( \(K_{\mathrm{a}}\) acetic acid \(\left.=1.8 \times 10^{-5} .\right)\).

Draw a diagram for a salt bridge cell for each of the following reactions. Label the anode and cathode, and indicate the direction of current flow throughout the circuit. (a) \(\mathrm{Zn}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cd}(s)\) (b) \(2 \mathrm{AuCl}_{4}^{-}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{Au}(s)+8 \mathrm{Cl}^{-}(a q)+3 \mathrm{Cu}^{2+}(a q)\) (c) \(\mathrm{Fe}(s)+\mathrm{Cu}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{Cu}(s)+\mathrm{Fe}(\mathrm{OH})_{2}(s)\)

For the following half-reactions, answer the questions below: $$ \begin{array}{cc} \mathrm{Ce}^{4+}(a q)+e^{-} \longrightarrow \mathrm{Ce}^{3+}(a q) & E^{\circ}=+1.61 \mathrm{~V} \\ \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=+0.80 \mathrm{~V} \\ \mathrm{Sn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Sn}(s) & E^{\circ}=-0.14 \mathrm{~V} \\ \mathrm{Ni}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Ni}(s) & E^{\circ}=-0.24 \mathrm{~V} \\ \mathrm{Al}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s) & E^{o}=-1.68 \mathrm{~V} \end{array} $$ (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will \(\mathrm{Sn}(s)\) reduce \(\mathrm{Ag}^{+}(\mathrm{aq})\) to \(\mathrm{Ag}(s) ?\) (f) Will \(\mathrm{Hg}(l)\) reduce \(\mathrm{Sn}^{2+}(a q)\) to \(\mathrm{Sn}(s) ?\) (g) Which ion(s) can be reduced by \(\operatorname{Sn}(s)\) ? (h) Which metal(s) can be oxidized by \(\mathrm{Ag}^{+}(a q)\) ?

Calculate \(E^{\circ}\) for the following cells: (a) \(\mathrm{Pb}\left|\mathrm{PbSO}_{4} \| \mathrm{Pb}^{2+}\right| \mathrm{Pb}\) (b) \(\mathrm{Pt}\left|\mathrm{Cl}_{2}\right| \mathrm{ClO}_{3}^{-} \| \mathrm{O}_{2}\left|\mathrm{H}_{2} \mathrm{O}\right| \mathrm{Pt}\) (c) \(\mathrm{Pt}\left|\mathrm{OH}^{-}\right| \mathrm{O}_{2} \| \mathrm{ClO}_{3}^{-}, \mathrm{Cl}^{-} \mid \mathrm{Pt} \quad\) (basic medium)

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