Problem 42

Question

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) \(\mathrm{Cu}(\mathrm{s}) \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}\)

Step-by-Step Solution

Verified
Answer
The reaction at the anode is \( 2 \text{Cl}^- \longrightarrow \text{Cl}_2(\text{g}) + 2 \text{e}^- \) (option a).
1Step 1: Identify the Electrolysis Process
Electrolysis of molten \( \text{CuCl}_2 \) involves breaking it into \( \text{Cu}^{2+} \) and \( \text{Cl}^- \) ions, allowing for reactions at the electrodes.
2Step 2: Determine the Reactions at Anode and Cathode
During electrolysis, the anode involves oxidation (loss of electrons), and the cathode involves reduction (gain of electrons). At the anode, \( \text{Cl}^- \) ions are oxidized to chlorine gas. At the cathode, \( \text{Cu}^{2+} \) ions gain electrons to form solid copper.
3Step 3: Analyze Each Option
Option (a) describes \( 2 \text{Cl}^- \) losing electrons to form \( \text{Cl}_2 \), which represents oxidation at the anode. Option (b) shows \( \text{Cl}_2 \) gaining electrons, which is incorrect for an anode process as it represents reduction. Option (c) describes reduction that occurs at the cathode, not the anode. Option (d) depicts copper oxidation, but not relevant for \( \text{Cl}^- \) ions at an anode.
4Step 4: Select the Correct Reaction
The correct reaction occurring at the anode should involve oxidation of chloride ions to chlorine gas. Hence, option (a) \( 2 \text{Cl}^- \longrightarrow \text{Cl}_2(\text{g}) + 2 \text{e}^- \) correctly represents the anode reaction.

Key Concepts

Anode ReactionChlorine GasOxidation Process
Anode Reaction
In an electrolysis process, the electrodes are essential sites for chemical reactions. The anode is the electrode where oxidation occurs. During oxidation, a species loses electrons.
The reaction at this electrode is termed the 'anode reaction'. It is important to remember:
  • Oxidation happens at the anode, meaning electron loss.
  • The anode appears as the positive electrode in electrolytic cells.
  • Substances at the anode donate electrons that travel through the circuit towards the cathode.

In the exercise with molten \( \mathrm{CuCl}_2 \), the anode reaction involves chloride ions (\( \mathrm{Cl}^- \)).These ions lose electrons to form chlorine gas (\( \mathrm{Cl}_2 \)); this is a classic example of an oxidation process taking place at the anode.
Chlorine Gas
Chlorine gas, represented as \( \mathrm{Cl}_2 \), is a diatomic molecule often encountered during electrolysis. It forms under specific conditions when chloride ions are oxidized.
  • During electrolysis of \( \mathrm{CuCl}_2 \), chloride ions are turned into chlorine gas.
  • This process occurs at the anode where oxidation - loss of electrons - takes place.
  • Chlorine gas has a pungent odor and is toxic, necessitating careful handling.

In many industrial applications, chlorine gas produced through electrolysis is essential. It's used in producing bleaching agents, disinfectants, and PVC. Understanding the generation of chlorine gas aids in comprehending the broader applications of electrolysis.
Oxidation Process
The oxidation process is fundamentally about the loss of electrons. It plays a critical role in electrochemical reactions, particularly at the anode in electrolysis.
Oxidation can be remembered with the mnemonic 'OIL' - Oxidation Is Loss (of electrons).
When analyzing electrolysis of molten \( \mathrm{CuCl}_2 \):
  • The chloride ions undergo oxidation.
  • Each \( \mathrm{Cl}^- \) ion loses an electron, subsequently forming chlorine gas \( \mathrm{Cl}_2 \).
  • This reaction results in two electrons being donated to the circuit.

The oxidation process not only helps in forming chlorine gas but also in understanding the flow of electrons. Mastering this concept is crucial for a comprehensive understanding of electrochemical reactions.