Problem 24 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A dilute aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is electrolyzed using platinum electrodes. The products at the anode and cathode are: [1996-1 Mark] (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}, \mathrm{Na}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{H}_{2}\)

Step-by-Step Solution

Answer

The products are \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \), which correspond to option (a).

Step 1: Analyzing the Electrolysis Process

In electrolysis of \( \mathrm{Na}_2 \mathrm{SO}_4 \), water solvent breaks down at the electrodes because \( \mathrm{Na}_2 \mathrm{SO}_4 \) itself does not undergo electrolysis. Consequently, at the anode, water is oxidized to oxygen (\( \mathrm{O}_2 \)), and at the cathode, water is reduced to hydrogen (\( \mathrm{H}_2 \)).

Step 2: Writing the Electrode Reactions

At the anode, water is decomposed by the reaction: \( 2\mathrm{H_2O} \rightarrow \mathrm{O_2(g)} + 4\mathrm{H^+} + 4\mathrm{e^-} \). At the cathode, water is also decomposed: \( 2\mathrm{H_2O} + 2\mathrm{e^-} \rightarrow \mathrm{H_2(g)} + 2\mathrm{OH^-} \). Thus, the products are oxygen at the anode and hydrogen at the cathode.

Step 3: Determining the Correct Option

Based on the reactions outlined in Step 2, oxygen is produced at the anode and hydrogen at the cathode. This corresponds to option (a): \( \mathrm{O}_2, \mathrm{H}_2 \).

Key Concepts

Platinum electrodes

In electrolysis, it is essential to select a suitable material for the electrodes, as this choice can impact the overall reaction and efficiency. Platinum electrodes are a popular choice in many applications due to their excellent properties.
Platinum is a noble metal, which means it is resistant to corrosion and oxidation even at high temperatures, making it very durable during electrolysis processes. This resistance to reactive substances ensures that platinum electrodes have minimal participation in unwanted side reactions, thus, they maintain the purity of the products formed.
Additionally, platinum has a high electrical conductivity, which allows for efficient electron transfer from the power source to the solutions. This ensures that the electrolysis is conducted smoothly, leading to clear and precise outcomes. Its inertness and high conductivity make platinum electrodes a reliable component in electrolysis setups, especially when the goal is to accurately produce certain gases or compounds.

Aqueous solution

An aqueous solution is one where water serves as the solvent. In the context of electrolysis, the solution of interest is typically a dissolved ionic compound in water. The role of water here is quite significant. Water molecules are polar, meaning they can help dissociate ionic compounds into their respective ions.
When an aqueous solution, such as the one containing sodium sulfate (\(\mathrm{Na}_2 \mathrm{SO}_4\)), is subjected to electrolysis, the water solvent itself can participate in the electrochemical reactions. This is particularly true if the salt does not easily undergo electrolysis, such as sodium sulfate. In such cases, the ions from water, rather than those from the salt, will be involved in the reactions at the electrodes.
With water's participation, hydrogen and oxygen gases can be produced, demonstrating the importance of understanding the behavior of water in an electrolytic process. It also emphasizes the need to examine both the solute and solvent components for a complete grasp of the process.

Anode and cathode reactions

In electrolysis, reactions occur at two electrodes termed as the anode and the cathode. Understanding these reactions is crucial in predicting the products of the electrolysis process. The anode is the positive electrode where oxidation occurs, while the cathode is the negative electrode where reduction takes place.

• Anode Reaction: At the anode, oxidation happens, meaning electrons are liberated. In the case of the electrolysis of an aqueous solution of sodium sulfate, water is oxidized, resulting in the liberation of oxygen gas. The reaction can be represented as: \(2\mathrm{H_2O} \rightarrow \mathrm{O_2(g)} + 4\mathrm{H^+} + 4\mathrm{e^-}\).
• Cathode Reaction: At the cathode, the electrons are consumed in a reduction reaction. Here, water molecules are reduced to produce hydrogen gas and hydroxide ions. This can be depicted using the equation: \(2\mathrm{H_2O} + 2\mathrm{e^-} \rightarrow \mathrm{H_2(g)} + 2\mathrm{OH^-}\).

These reactions are foundational in understanding electrolysis, as they dictate the nature and type of substances formed at each electrode, which in this exercise, results in the production of oxygen and hydrogen gases.