/*! 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 14 A certain current liberates \(0.... [FREE SOLUTION] | 91Ó°ÊÓ

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A certain current liberates \(0.504 \mathrm{~g}\) of hydrogen in 2 hour. The amount (or mass) of copper deposited at the cathode during the electrolysis of \(\mathrm{CuSO}_{4}\) (aq) solution by the same current flowing for the same time is (atomic masses \(\mathrm{H}=1.0, \mathrm{Cu}=63.5\) ) a. \(16.0 \mathrm{~g}\) b. \(32.0 \mathrm{~g}\) c. \(8.6 \mathrm{~g}\) d. \(1.60 \mathrm{~g}\)

Short Answer

Expert verified
The mass of copper deposited is 16.0 g.

Step by step solution

01

Understand the Concept

The problem involves the electrolysis process, where the amount of substance deposited is proportional to the charge passed through the solution. This is described by Faraday’s laws of electrolysis.
02

Calculate Moles of Hydrogen Liberated

Given that 0.504 g of hydrogen is liberated, use the molar mass of hydrogen to calculate the moles. The molar mass of hydrogen is 2 g/mol (since diatomic Hâ‚‚ is considered). Thus, the moles of hydrogen liberated are \( \frac{0.504}{2} = 0.252 \) moles.
03

Determine the Charge Used

According to Faraday's first law, the charge passed is equal to the moles of electrons transferred times Faraday's constant (96000 C/mol). Since 2 electrons are needed to liberate one molecule of Hâ‚‚, the total moles of electrons for hydrogen are \( 0.252 \times 2 = 0.504 \) moles. Thus, the charge used is \(0.504 \times 96000 = 48384\) C.
04

Calculate Moles of Copper Deposited

Use Faraday’s second law, which states that the same charge passing through the electrolysis setup will result in the deposition of different equivalent weights of different substances. Copper requires 2 electrons to deposit one atom, so the moles of copper deposited will be the same as the moles of electrons used for hydrogen, 0.252 moles.
05

Calculate Mass of Copper Deposited

Finally, use the moles of copper and its atomic mass to find the mass deposited. The molar mass of copper is 63.5 g/mol. Thus, mass of copper = \(0.252 \times 63.5 = 15.998 \approx 16.0\) g.

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Key Concepts

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

Understanding the Electrolysis Process
Electrolysis is a fascinating process where electrical energy is used to drive a non-spontaneous chemical reaction. At its core, it involves passing an electric current through a solution or molten compound to cause a chemical change. In the process of electrolysis, cations move towards the cathode to gain electrons, while anions move towards the anode to lose electrons. Faraday's laws of electrolysis explain how these reactions are quantitatively related.
Faraday's first law states that the amount of substance altered at an electrode is proportional to the total charge passed through the electrolyte. Essentially, the more electricity that flows through the system, the more significant the chemical changes.
  • The amount of substance deposited or liberated at an electrode is directly proportional to the electric charge.
  • Charge (Coulombs) = Current (Amps) × Time (Seconds)
This law provides the basis for calculating the mass of a metal deposited during electrolysis, as shown when copper is deposited from a copper sulfate solution.
Moles Calculation in Electrolysis
Calculating moles during electrolysis involves using the mass of substances and their molar masses. This is a critical step for any quantitative electrochemical analysis. Here's how you can calculate moles:
  • Identify the chemical substance involved, such as hydrogen or copper.
  • Use the formula: Moles = \( \frac{\text{Given Mass}}{\text{Molar Mass}} \)
In our specific problem, hydrogen is involved, and we begin by determining the moles of hydrogen. Hydrogen exists as diatomic \( H_2 \), with a molar mass of 2 g/mol. By dividing the mass of hydrogen (0.504 g) by its molar mass, we get 0.252 moles.
This calculation is essential because, in electrolysis, knowing the number of moles allows us to predict the behavior of other chemical species in the same reaction. Similar calculations are carried out for copper, ensuring that each step follows the principle that like charges assist in determining similar amounts of chemical species.
Understanding Atomic Mass
Atomic mass, also known as atomic weight, is crucial for chemical calculations, especially those involving moles and stoichiometry. Each element on the periodic table has a unique atomic mass that helps in these calculations.
The atomic mass of an element, like copper or hydrogen, indicates its average mass per atom, usually expressed in atomic mass units (amu). For calculations in chemical reactions:
  • Copper has an atomic mass of 63.5, meaning every mole of copper weighs 63.5 grams.
  • Hydrogen, when considering its diatomic molecule \( H_2 \), has a molar mass of 2 grams per mole.
Understanding these values assists us in determining how much of a substance is consumed or produced in a reaction. In this electrolysis problem, knowing the atomic mass of copper was necessary to find out how much copper is deposited when 0.252 moles are considered. These calculations are the backbone of working with chemical reactions up to a quantitative level.
Cathode Deposition in Electrolysis
In the electrolysis process, the cathode is where reduction, or gain of electrons, occurs. Metals like copper are deposited at the cathode because they are cations, positively charged ions that are attracted to the negative charge of the cathode.
When electricity is applied to a solution like \( \text{CuSO}_4 \), copper ions \( \text{Cu}^{2+} \) move towards the cathode, where they gain electrons to become neutral copper atoms. Key aspects of cathode deposition include:
  • Reduction Reaction: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)
  • One mole of copper requires two moles of electrons to deposit as a solid metal.
Calculating how much copper is deposited is akin to knowing the moles of electrons transferred, as viewed with Faraday's second law. By understanding cathode deposition, students can predict the outcomes of electrolysis experiments and calculate the necessary conditions for metal recovery efficiently.

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Most popular questions from this chapter

In the reaction, \(\mathrm{As}_{2} \mathrm{~S}_{5}+\mathrm{xHNO}_{3} \rightarrow 5 \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{yNO}_{2}+2 \mathrm{H}_{3} \mathrm{AsO}_{4}+\) \(12 \mathrm{H}_{2} \mathrm{O}\) The values of \(\mathrm{x}\) and \(\mathrm{y}\) are a. 30,30 b. 40,40 c. 10,10 d. 20,20

In the following question two statements (Assertion) A and Reason (R) are given. Mark a. if \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. if \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of A; c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. A is false but \(\mathrm{R}\) is true, e. A and \(R\) both are false. (A): Any cell having \(\mathrm{E}^{\text {o }}\) cell \(=5.91\). Volt, will have equilibrium constant \(10^{3}\) at \(298 \mathrm{~K}\). \((\mathbf{R}):\) As at \(298 \mathrm{~K}, \mathrm{Keq}\) \(=10\left(\mathrm{E}^{\circ}\right.\) cell \(\left./ 0.0591\right)\).

Which is/are true for standard electrode potentials? a. \(\mathrm{E}^{\circ}\) for oxidation is the negative of \(\mathrm{E}^{\circ}\) for reduction. b. Cell constituents are in their standard states. c. The potential for the standard hydrogen electrode is chosen to be \(+1.00 \mathrm{~V}\). d. The half reactions are written as reductions.

The standard reduction potentials of \(\mathrm{Zn}\) and Ag in water are \(\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \leftrightarrow \mathrm{Zn}\); \(\left(\mathrm{E}^{\circ}=-0.76 \mathrm{~V}\right)\) and \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \leftrightarrow \mathrm{Ag}\) \(\left(\mathrm{E}^{\circ}=+0.80 \mathrm{~V}\right)\) at \(298 \mathrm{~K}\). Which of the following reaction is/are not feasible? a. \(\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{Ag}(\mathrm{s})\) b. \(\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) \rightarrow 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})\) c. \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq})\) d. \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}\) (s)

Ammonia is always is added in this reaction. Which of the following must be incorrect? a. \(\mathrm{NH}_{3}\) combines with \(\mathrm{Ag}^{+}\)to form a complex b. \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\)is a stronger oxidizing reagent than \(\mathrm{Ag}^{+}\). c. In absence of \(\mathrm{NH}_{3}\) silver salt of gluconic acid is formed d. \(\mathrm{NH}_{3}\) has affected the standard reduction potential of glucose/gluconic acid electrode.

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