Problem 24 - Chemistry and Chemical Reactivity

Question

(a) Write equations for the half-reactions that occur at the cathode and the anode when an aqueous solution of KCl is electrolyzed. Which chemical species is oxidized, and which chemical species is reduced in this reaction? (b) Predict the products formed when an aqueous solution of Csl is electrolyzed.

Step-by-Step Solution

Answer

(a) \( Cl^- \) is oxidized to \( Cl_2 \), \( H_2O \) is reduced to \( H_2 \). (b) CsI electrolysis produces \( H_2 \) and \( I_2 \).

Step 1: Identify Electrolysis Reaction for KCl

In an aqueous solution of KCl, electrolysis involves the decomposition into its elements. The possible ions present are \( K^+ \), \( Cl^- \), \( H_2O \), \( H^+ \), and \( OH^- \). At the electrodes, we consider the reaction of these ions.

Step 2: Determine Anode Reaction for KCl

At the anode (positive electrode), oxidation occurs. The \( Cl^- \) ions are oxidized to form chlorine gas: \[ 2Cl^- \rightarrow Cl_2 + 2e^- \]. Chlorine gas \( Cl_2 \) is formed at the anode.

Step 3: Determine Cathode Reaction for KCl

At the cathode (negative electrode), reduction occurs. Between \( H_2O \) and \( K^+ \), water is reduced to hydrogen gas because it has a lower reduction potential compared to potassium ions: \[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]. Hydrogen gas \( H_2 \) is formed at the cathode.

Step 4: Identify Oxidation and Reduction in KCl Electrolysis

Chloride ions \( Cl^- \) are oxidized to \( Cl_2 \), and water \( H_2O \) is reduced to \( H_2 \). Thus, \( Cl^- \) is oxidized, and \( H_2O \) is reduced.

Step 5: Identify Electrolysis Reaction for CsI

In an aqueous solution of CsI, like in the previous step, we consider the possible ions: \( Cs^+ \), \( I^- \), \( H_2O \), \( H^+ \), and \( OH^- \). Again, we predict the reactions at the cathode and anode.

Step 6: Determine Anode Reaction for CsI

At the anode, \( I^- \) is oxidized to iodine \( I_2 \) as iodine has a lower oxidation potential than water: \[ 2I^- \rightarrow I_2 + 2e^- \]. Iodine gas \( I_2 \) is formed at the anode.

Step 7: Determine Cathode Reaction for CsI

At the cathode, water is reduced to hydrogen gas as it has a lower reduction potential compared to cesium: \[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]. Hydrogen gas \( H_2 \) is produced at the cathode.

Step 8: Predict Products of CsI Electrolysis

The products of the electrolysis of aqueous CsI are hydrogen gas \( H_2 \) at the cathode and iodine \( I_2 \) at the anode.

Key Concepts

Oxidation

Oxidation is a key process in electrolysis where a chemical species loses electrons. This occurs at the anode, the positive electrode in an electrolytic system. For example, during the electrolysis of an aqueous solution of KCl, chloride ions (Cl^-) undergo oxidation to form chlorine gas (Cl_2). The reaction can be written as:
\[ 2Cl^- \rightarrow Cl_2 + 2e^- \]

• The chloride ions (Cl^-) lose two electrons.
• These electrons travel through the external circuit to the cathode.
• This process creates chlorine gas.

Understanding oxidation involves recognizing which species loses electrons and produces a gas at the anode.

Reduction

Reduction is the gain of electrons by a chemical species and occurs at the cathode, which is the negative electrode. During the electrolysis of KCl, water (\(H_2O\)) is reduced instead of potassium ions (K^+) because water has a greater tendency to gain electrons. This reduction is represented as:
\[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]

• Water molecules gain electrons.
• This results in the production of hydrogen gas (H_2) and hydroxide ions (OH^-).

The choice of reduction at the cathode is influenced by the reduction potentials of the possible ions present in the solution.

Half-Reactions

Half-reactions are individual reactions at each electrode in an electrolytic cell that illustrate either oxidation or reduction. Each half-reaction shows the transfer of electrons and the change in oxidation state of involved species.
- **Anode Half-Reaction:** Represents oxidation. \[ 2Cl^- \rightarrow Cl_2 + 2e^- \] - **Cathode Half-Reaction:** Represents reduction. \[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]These half-reactions help us understand the movement of electrons and the changes in the chemical species. They are crucial for predicting the products of electrolysis.

Aqueous Solution

An aqueous solution is a solution where water is the solvent. In electrolysis, ions in the solution participate in electrode reactions. For example, in the electrolysis of both KCl and CsI:

• Cations ( K^+ and Cs^+ ) and anions ( Cl^- , I^- ) are dissolved in water.
• Water itself can participate in electrode reactions because it can be oxidized or reduced.
• The presence of water often leads to the formation of hydrogen and oxygen gases, depending on the reaction conditions.

Being aware of the ions present in the aqueous solution is essential for predicting the behavior of the system during electrolysis.

Electrode Reactions

Electrode reactions are the oxidation and reduction processes occurring at the anode and cathode of an electrolytic cell. They determine the overall chemical changes and products in electrolysis.
- **At the Anode:** Where oxidation happens, such as converting Cl^- to Cl_2 or I^- to I_2 .
- **At the Cathode:** Where reduction occurs, like transforming H_2O into H_2 .
Understanding electrode reactions helps in predicting what substances will be formed at each electrode. These reactions are determined by the potential differences and the specific ions available in the aqueous solution.