Problem 117 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{O}_{2}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}^{2}, \mathrm{SO}_{2}\) (d) \(\mathrm{H}_{2}, \mathrm{O}_{2}\)

Step-by-Step Solution

Answer

The products are (d) \( \text{H}_2, \text{O}_2 \).

Step 1: Identify the Ions Present

In a solution of sodium sulfate (\( \text{Na}_2\text{SO}_4 \)), the ions present are \( \text{Na}^+ \), \( \text{SO}_4^{2-} \), \( \text{H}^+ \), and \( \text{OH}^- \) from water dissociation.

Step 2: Identify Possible Reactions at the Cathode

At the cathode, the possible reduction reactions are: \( \text{2H}^+ + 2e^- \rightarrow \text{H}_2 \) (producing hydrogen gas) or \( \text{Na}^+ + e^- \rightarrow \text{Na} \).However, sodium is not deposited from aqueous solution, so \( \text{H}_2 \) is formed.

Step 3: Identify Possible Reactions at the Anode

At the anode, the possible oxidation reactions are: \( \text{4OH}^- \rightarrow \text{2H}_2\text{O} + \text{O}_2 + 4e^- \) (producing oxygen gas) or \(\text{SO}_4^{2-} + 2e^- \rightarrow \text{SO}_2 \). The oxidation of water is more favorable, forming \( \text{O}_2 \).

Step 4: Determine the Products

Based on the favorable reactions, \( \text{H}_2 \) is produced at the cathode and \( \text{O}_2 \) is produced at the anode. Therefore, the products are \( \text{H}_2, \text{O}_2 \).

Step 5: Match with the Given Options

Compare our results with the provided options: (a) \( \text{O}_2, \text{H}_2 \), (b) \( \text{O}_2, \text{Na} \), (c) \( \text{O}_2^{2}, \text{SO}_2 \), (d) \( \text{H}_2, \text{O}_2 \). The correct pairing of products is choice (d): \( \text{H}_2, \text{O}_2 \).

Key Concepts

Sodium Sulfate Solution

When sodium sulfate (\( \text{Na}_2\text{SO}_4 \)) is dissolved in water, it dissociates into its constituent ions: sodium ions (\( \text{Na}^+ \)) and sulfate ions (\( \text{SO}_4^{2-} \)).
Besides these, water itself splits into hydrogen ions (\( \text{H}^+ \)) and hydroxide ions (\( \text{OH}^- \)).
This mixture of ions is part of an aqueous solution, often used in electrolysis for various reactions.
In electrolysis, the presence of these ions allows for the conduction of electricity through the solution, facilitating chemical changes at the electrodes.

• The key point is that sodium sodium sulfate doesn’t itself dissociate into gases or metals during electrolysis with inert electrodes.
• Instead, it acts as a medium to transfer charge via the movement of ions.

Understanding the nature of the solution is important in predicting what occurs during the electrolysis process.

Cathode Reaction

In electrolysis, the cathode is the electrode where reduction (gain of electrons) occurs.
For a sodium sulfate solution, at the cathode, the most likely reaction is the reduction of hydrogen ions to form hydrogen gas:

• The reaction is: \( \text{2H}^+ + 2e^- \rightarrow \text{H}_2 \)

Sodium ions are present, but they do not form metallic sodium because water is the solvent and sodium metal is unstable in water.
Thus, hydrogen ions derived from water are reduced instead.
The production of hydrogen gas at the cathode is a classic example of preferential discharge, where the more easily reducible ion wins in a competitive environment.

Anode Reaction

The anode is the site of oxidation (loss of electrons) during electrolysis.
In a sodium sulfate solution using inert electrodes, water oxidation is the more favorable reaction at the anode.

• The relevant reaction is: \( \text{4OH}^- \rightarrow \text{2H}_2\text{O} + \text{O}_2 + 4e^- \)

Although sulfate ions are available, the oxidation of water to form oxygen gas is energetically favored.
This results in the release of \( \text{O}_2 \) at the anode.

• It's crucial to note that inert electrodes do not partake in the reaction; instead, they serve as surfaces for these reactions to happen.

An understanding of the preferred reactions is essential to correctly predict the products of electrolysis.

Water Electrolysis

Water electrolysis is the process of using electricity to decompose water into hydrogen and oxygen gases.
In the context of sodium sulfate solution electrolysis, water provides \( \text{H}^+ \) and \( \text{OH}^- \) ions that further aid the electrolysis process.

• At the cathode, hydrogen ions are reduced to form \( \text{H}_2 \).
• At the anode, hydroxide ions are oxidized to form \( \text{O}_2 \).

This decomposition highlights how essential water is in the formation of gases during electrolysis.
It's a vital supportive mechanism when working with aqueous solutions in electrolysis, as it leads to the formation of well-known gaseous products that are easy to identify.

Inert Electrodes

Inert electrodes are materials that do not react with the electrolyte or the products formed during electrolysis.
They simply provide a surface for oxidation and reduction reactions.

• Common inert materials include platinum and graphite.

Using inert electrodes ensures that the electrode material does not interfere with or alter the chemical reactions in any way.
During the electrolysis of sodium sulfate solution, the inert nature preserves the focus on the solution itself, leading to gas formation at both the cathode and anode:

• At cathode: formation of hydrogen gas (\( \text{H}_2 \))
• At anode: formation of oxygen gas (\( \text{O}_2 \))

Understanding the role of inert electrodes helps one comprehend how clean and uncontaminated product gases are generated in these setups.