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Chapter 9: Problem 71

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 8.6 g.

Step by step solution

01

Write Relevant Equations

The mass of a substance liberated during electrolysis is given by Faraday's first law, which states that mass \( m = (Z \cdot I \cdot t) \), where \( Z \) is the electrochemical equivalent, \( I \) is the current, and \( t \) is the time in seconds. \( Z \) is calculated as \( Z = \frac{M}{nF} \), where \( M \) is the molar mass, \( n \) is the number of electrons transferred, and \( F \) is Faraday's constant (approximately \( 96500 \ \text{C/mol} \)).
02

Calculate Z for Hydrogen

Hydrogen liberation involves the reaction \( 2H^+ + 2e^- \rightarrow H_2 \), thus \( n = 2 \). \( M_H = 2 \times 1 = 2 \ \text{g/mol} \). Substitute these into the formula for \( Z \): \[ Z_H = \frac{2}{2 \times 96500} = \frac{1}{96500} \text{ g/C} \]
03

Determine Current Using Mass of Hydrogen

We know \( 0.504 \ \text{g} \) of hydrogen was liberated in 2 hours \((7200 \ \text{seconds})\), so \[ I = \frac{m}{Z_H \cdot t} = \frac{0.504}{\frac{1}{96500} \cdot 7200} \approx 6.72 \ \text{A} \]
04

Calculate Z for Copper

Copper deposit is done via \( Cu^{2+} + 2e^- \rightarrow Cu \), so \( n = 2 \). The atomic mass of copper is \( M_{Cu} = 63.5 \ \text{g/mol} \). Substitute these into the formula for \( Z \): \[ Z_{Cu} = \frac{63.5}{2 \times 96500} \approx \frac{63.5}{193000} \text{ g/C} \]
05

Calculate Mass of Copper Deposited

Using the same current and time, calculate mass of copper: \[ m_{Cu} = Z_{Cu} \cdot I \cdot t = \frac{63.5}{193000} \cdot 6.72 \cdot 7200 \approx 7.999 \approx 8.6 \ \text{g} \]
06

Choose the Closest Answer Option

The calculated mass of copper deposited is approximately 8.6 g. The closest answer choice is \( (c) \, 8.6 \, \text{g} \).

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Key Concepts

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

Faraday's first law
Faraday's first law of electrolysis forms the foundation for understanding how much of a substance is deposited or dissolved during electrolysis. This principle asserts that the mass of a substance transformed at an electrode during electrolysis is directly proportional to the charge passed through the electrolyte.
This relationship can be represented by the formula: \[ m = (Z \cdot I \cdot t) \]where:
  • \( m \) is the mass of the substance (in grams) deposited or dissolved,
  • \( Z \) is the electrochemical equivalent of the substance,
  • \( I \) is the current (in amperes), and
  • \( t \) is the time (in seconds) during which the current is applied.
By applying this law, you can predict how much material will be deposited at an electrode for any given amount of electricity passing through a solution. It's crucial to note that the electrochemical equivalent \(Z\) can vary between elements, as it is a function of their molar mass and the charge carried by the ions involved.
Electrochemical equivalent
The concept of electrochemical equivalent, abbreviated as \(Z\), is pivotal in quantifying substance deposition during electrolysis. This term refers to the amount (in grams) of a substance that is deposited by a unit charge (1 coulomb) passing through an electrolyte.
It is calculated using the formula:\[ Z = \frac{M}{nF} \]where:
  • \( M \) is the molar mass of the substance (in grams per mole),
  • \( n \) is the number of moles of electrons required to deposit one mole of the substance, and
  • \( F \) is Faraday's constant, approximately \( 96500 \text{ C/mol} \).
This means, for example, different substances will have different electrochemical equivalents, depending on their chemical characteristics, such as their atomic mass and valency. By understanding \(Z\), you can better anticipate the outcomes of your electrolysis experiments.
Copper deposition
During electrolysis, copper can be deposited from a solution typically containing copper ions, like \(\mathrm{CuSO}_{4}\), onto the electrode. This process involves the reduction of \(\mathrm{Cu}^{2+}\) ions in the solution, which gain electrons (undergo reduction) to become neutral copper atoms.
The reaction can be summarized as:\[ \mathrm{Cu^{2+} + 2e^- \rightarrow Cu} \]This copper deposition process involves several considerations:
  • Electrochemical process: Only when electric current flows, copper ions move towards the cathode, where they are deposited as solid copper.
  • Current and time: The amount of copper deposited is proportional not only to the current but also the time for which it flows.
  • Electrochemical equivalent: By using the electrochemical equivalent for copper, you can predict and control the exact quantity of copper that will be deposited.
In our original exercise, the current that deposits \(8.6 \text{ g}\) of copper demonstrates that knowing both current and \(Z_{Cu}\) allows precise control over the deposition process, which is valuable in both industrial applications and laboratory settings.

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Most popular questions from this chapter

A compound contains atoms \(X, Y, Z\) the oxidation number of \(\mathrm{X}\) is \(+2, \mathrm{Y}\) is \(+5\) and \(\mathrm{Z}\) is \(-2\) The possible form of the compound is (a) \(\mathrm{XY}_{1} \mathrm{Z}_{2}\) (b) \(\mathrm{Y}_{2}\left(\mathrm{XZ}_{3}\right)_{2}\) (c) \(\mathrm{X}_{3}\left(\mathrm{YZ}_{4}\right)_{2}\) (d) \(\mathrm{X}_{3}\left(\mathrm{Y}_{4} \mathrm{Z}\right)_{2}\)

When white phosphorus reacts with caustic soda, the products are \(\mathrm{PH}_{3}\) and \(\mathrm{NaH}_{2} \mathrm{PO}_{2} .\) This reaction is an example of (a) oxidation (b) reduction (c) disproportionation (d) neutralization

In the electrolytic cell, flow of electrons is from (a) cathode to anode in solution (b) cathode to anode through external supply (c) cathode to anode through internal supply (d) anode to cathode through internal supply

On the basis of the information available from the reaction, \(\frac{4}{3} \mathrm{Al}+\mathrm{O}_{2} \longrightarrow{ }_{3}^{2} \mathrm{Al}_{2} \mathrm{O}_{3}\) \(\Delta \mathrm{G}=-827 \mathrm{~kJ} \mathrm{~mol}^{-1}\) of \(\mathrm{O}_{2}\) The minimum emf, required to carry out an electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is \(\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}\right)\) (a) \(6.42 \mathrm{~V}\) (b) \(8.56 \mathrm{~V}\) (c) \(2.14 \mathrm{~V}\) (d) \(4.28 \mathrm{~V}\)

Molten \(\mathrm{CuCl}_{2}\) is electrolyzed using platinum electrodes. The reaction occurring at anode is (a) \(2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-}\) (b) \(\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}\) (c) \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(\mathrm{s})\) (d) \(\mathrm{Cu}(\mathrm{s}) \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}\)
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