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Chapter 19: Problem 49

Calculate the amounts of \(\mathrm{Cu}\) and \(\mathrm{Br}_{2}\) produced in \(1.0 \mathrm{~h}\) at inert electrodes in a solution of \(\mathrm{CuBr}_{2}\) by a current of \(4.50 \mathrm{~A}\).

Short Answer

Expert verified
5.34 g Cu and 13.42 g Br鈧� are produced.

Step by step solution

01

Understand the Reaction

Copper(II) bromide, CuBr鈧�, dissociates into Cu虏鈦� and Br鈦� ions in solution. During electrolysis, Cu虏鈦� ions are reduced to form copper metal at the cathode, and Br鈦� ions are oxidized to form Br鈧� gas at the anode.
02

Calculate Moles of Electrons Transferred

Using the formula \(Q = I \times t\), calculate the total charge transferred. Given: \(I = 4.50\, \text{A}\) and \(t = 1.0\, \text{h} = 3600\, \text{s}\).Thus, \(Q = 4.50\, \text{A} \times 3600\, \text{s} = 16200\, \text{C}\).The number of moles of electrons transferred is given by \(n = \frac{Q}{F}\), where \(F = 96485\, \text{C/mol}\) is Faraday's constant.\(n = \frac{16200}{96485} \approx 0.168\, \text{mol of electrons}\).
03

Calculate Moles of Copper and Bromine Produced

The reduction of Cu虏鈦� to Cu requires 2 moles of electrons per mole of Cu, and the oxidation of Br鈦� to Br鈧� requires 2 moles of electrons per mole of Br鈧�.Moles of Cu produced: \( \frac{0.168}{2} = 0.084\, \text{mol of Cu}\).Moles of Br鈧� produced: \( \frac{0.168}{2} = 0.084\, \text{mol of Br}_{2}\).
04

Convert Moles to Grams

Calculate grams of Cu and Br鈧� using their molar masses. Molar mass of Cu is 63.55 g/mol, and molar mass of Br鈧� is 159.808 g/mol.Grams of Cu: \(0.084 \times 63.55 = 5.34 \text{ g of Cu}\).Grams of Br鈧�: \(0.084 \times 159.808 = 13.42 \text{ g of Br}_{2}\).

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes where substances undergo chemical changes to form new substances. In the context of electrolysis, these reactions take place within an electrolyte solution. When a direct electrical current is applied, chemical species at the electrodes either gain or lose electrons.
  • Redox Reactions: Electrolysis involves redox reactions, which are reactions where oxidation and reduction occur simultaneously. - At the cathode, reduction occurs: Copper ions (\( \text{Cu}^{2+} \)) gain electrons to form metallic copper. - At the anode, oxidation happens: Bromide ions (\( \text{Br}^{-} \)) lose electrons to form bromine gas (\( \text{Br}_2 \)).
  • Electrolytes: These are substances that dissociate into ions in solution and conduct electricity. For copper bromide, it dissociates into copper (\( \text{Cu}^{2+} \)) and bromide (\( \text{Br}^{-} \)) ions.
Understanding the flow of electrons and the movement of ions helps explain how substances change their form through chemical reactions during electrolysis.
Faraday's Law
Faraday's laws of electrolysis form the bridge between electrical energy and chemical change. Michael Faraday, a pioneering scientist, detailed these quantitative relationships, which are vital in predicting electrolytic outcomes.
  • First Law: The mass of a substance altered at an electrode during electrolysis is proportional to the amount of electrical charge passed through the circuit.- Mathematically, it translates to \( m = kQ \), where \( m \) is the mass of the substance, \( k \) is a constant, and \( Q \) is the electric charge.
  • Second Law: For the same quantity of electricity, the amount of substance altered is directly proportional to its equivalent weight.- This means different substances require different amounts of charge to undergo change, based on their valency and atomic mass.
Applying these laws helps calculate important entities like moles of electrons, providing a bridge to determine the final amounts of products in technical electrolysis applications.
Copper Bromide Electrolysis
Electrolysis of copper bromide (\( \text{CuBr}_2 \)) is an example of a classic redox reaction at electrodes. For students, understanding this process involves splitting the reaction into half-reactions and applying Faraday's laws to quantify the results.
  • Dissociation: Copper bromide in solution dissociates into copper (\( \text{Cu}^{2+} \)) and bromide (\( \text{Br}^{-} \)) ions.
  • Cathode Reaction: Cu2+ + 2e- 鈫� Cu- Copper ions gain electrons and deposit onto the cathode as metallic copper.
  • Anode Reaction: 2Br- 鈫� Br2 + 2e-- Bromide ions lose electrons to form bromine gas at the anode.
  • Mass Calculation: By calculating the number of moles of electrons (using charge and Faraday's constant), students can determine the moles and subsequently the grams of copper and bromine produced.
The exercise emphasizes using equipment like ammeters and understanding well-established principles to convert abstract equations into real-world quantities.

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Most popular questions from this chapter

In recent years, there has been much interest in electric cars. List some advantages and disadvantages of electric cars compared to automobiles with internal combustion engines.

Define the following terms: anode, cathode, cell voltage, electromotive force, standard reduction potential.

As discussed in Section \(19.5,\) the potential of \(\mathrm{a}\) concentration cell diminishes as the cell operates and the concentrations in the two compartments approach each other. When the concentrations in both compartments are the same, the cell ceases to operate. At this stage, is it possible to generate a cell potential by adjusting a parameter other than concentration? Explain.

Balance the following redox equations by the half-reaction method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution) (e) \(\mathrm{Mn}^{2+}+\mathrm{BiO}_{3}^{-} \longrightarrow \mathrm{Bi}^{3+}+\mathrm{MnO}_{4}^{-}\) (in acidic solution)

The cathode reaction in the Leclanch茅 cell is given by: $$ 2 \mathrm{MnO}_{2}(s)+\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{ZnMn}_{2} \mathrm{O}_{4}(s) $$ If a Leclanch茅 cell produces a current of \(0.0050 \mathrm{~A}\), calculate how many hours this current supply will last if there is initially \(4.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) present in the cell. Assume that there is an excess of \(\mathrm{Zn}^{2+}\) ions.The cathode reaction in the Leclanch茅 cell is given by: $$2 \mathrm{MnO}_{2}(s)+\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{ZnMn}_{2} \mathrm{O}_{4}(s)$$ If a Leclanch茅 cell produces a current of \(0.0050 \mathrm{~A}\), calculate how many hours this current supply will last if there is initially \(4.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) present in the cell. Assume that there is an excess of \(\mathrm{Zn}^{2+}\) ions.
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