Question

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

5.34 g Cu and 13.42 g Br₂ are produced.

Step-by-step solution

Understand the Reaction

Copper(II) bromide, CuBr₂, dissociates into Cu²⁺ and Br⁻ ions in solution. During electrolysis, Cu²⁺ ions are reduced to form copper metal at the cathode, and Br⁻ ions are oxidized to form Br₂ gas at the anode.

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.\(n = \frac{16200}{96485} \approx 0.168\, \text{mol of electrons}\).

Calculate Moles of Copper and Bromine Produced

The reduction of Cu²⁺ to Cu requires 2 moles of electrons per mole of Cu, and the oxidation of Br⁻ to Br₂ requires 2 moles of electrons per mole of Br₂.Moles of Cu produced: \( \frac{0.168}{2} = 0.084\, \text{mol of Cu}\).Moles of Br₂ produced: \( \frac{0.168}{2} = 0.084\, \text{mol of Br}_{2}\).

Convert Moles to Grams

Calculate grams of Cu and Br₂ using their molar masses. Molar mass of Cu is 63.55 g/mol, and molar mass of Br₂ is 159.808 g/mol.Grams of Cu: \(0.084 \times 63.55 = 5.34 \text{ g of Cu}\).Grams of Br₂: \(0.084 \times 159.808 = 13.42 \text{ g of Br}_{2}\).

Key concepts

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^- → Cu- Copper ions gain electrons and deposit onto the cathode as metallic copper.
• Anode Reaction: 2Br^- → 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.