Problem 113 - A Complete Resource Book in Chemistry for JEE Main

Question

A dilute aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is electrolyzed using platinum electrodes. The product at the anode and cathode are (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

(a) \( \mathrm{O}_2, \mathrm{H}_2 \)

Step 1: Understanding the Electrolysis Process

Electrolysis involves passing an electric current through an electrolyte solution to cause a chemical reaction at the electrodes. In this problem, we have an aqueous solution of \( \mathrm{Na}_{2}\mathrm{SO}_{4} \), and we need to identify the products at the anode and the cathode.

Step 2: Identifying the Ions in Solution

When \( \mathrm{Na}_{2}\mathrm{SO}_{4} \) dissolves in water, it dissociates into \( \mathrm{Na}^+ \) and \( \mathrm{SO}_{4}^{2-} \) ions. Additionally, water itself can dissociate into \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions.

Step 3: Determining the Reaction at the Cathode

At the cathode, reduction takes place. Between \( \mathrm{H}^+ \) and \( \mathrm{Na}^+ \) ions, \( \mathrm{H}^+ \) ions are reduced more easily to produce hydrogen gas. The reduction reaction is: \[ 2\mathrm{H}^+ (aq) + 2e^- \rightarrow \mathrm{H}_2 (g) \]

Step 4: Determining the Reaction at the Anode

At the anode, oxidation occurs. Here, \( \mathrm{O}H^- \) produced from water is oxidized more readily than \( \mathrm{SO}_{4}^{2-} \), leading to oxygen gas. The oxidation reaction is: \[ 4\mathrm{OH}^- (aq) \rightarrow \mathrm{O}_2 (g) + 2\mathrm{H}_2\mathrm{O} (l) + 4e^- \]

Step 5: Conclusion on Anode and Cathode Products

From the reactions, hydrogen gas \( \mathrm{H}_2 \) is produced at the cathode and oxygen gas \( \mathrm{O}_2 \) at the anode.

Key Concepts

Reduction and Oxidation Reactions

In electrolysis, two fundamental reactions occur: reduction and oxidation. These are often remembered through the mnemonic "OIL RIG," meaning "Oxidation Is Loss and Reduction Is Gain." This refers to the transfer of electrons. During electrolysis, different ions in a solution undergo these changes.

**Oxidation** takes place at the anode. This is where electrons are lost from ions, transforming into a different form. In our process with sodium sulfate (\( \mathrm{Na}_{2} \mathrm{SO}_{4} \)), it's the hydroxide ions (\( \mathrm{OH}^- \)) that lose electrons to create oxygen gas. The equation representing this is:

\[ 4\mathrm{OH}^- (aq) \rightarrow \mathrm{O}_2 (g) + 2\mathrm{H}_2\mathrm{O} (l) + 4e^- \]

**Reduction** happens at the cathode. This reaction is marked by the gain of electrons. In our case, hydrogen ions (\( \mathrm{H}^+ \)) gain electrons to form hydrogen gas. The equation is:

\[ 2\mathrm{H}^+ (aq) + 2e^- \rightarrow \mathrm{H}_2 (g) \].

Understanding these reactions helps in predicting the products of electrolysis in various solutions.

Cathode and Anode Reactions

In the electrolysis of aqueous solutions, it’s essential to understand the roles of the cathode and anode.

The **cathode** is the negatively charged electrode. At the cathode, the process of reduction occurs due to electron gain. During the electrolysis of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \), positively charged hydrogen ions (\( \mathrm{H}^+ \)) are reduced. This reduction leads to the production of hydrogen gas, indicated by the equation:

\[ 2\mathrm{H}^+ (aq) + 2e^- \rightarrow \mathrm{H}_2 (g) \].

The **anode** is the positively charged electrode. Here, oxidation takes place as electrons are lost. In our solution, the hydroxide ions (\( \mathrm{OH}^- \)) undergo oxidation, leading to the creation of oxygen gas. This reaction is depicted by:

\[ 4\mathrm{OH}^- (aq) \rightarrow \mathrm{O}_2 (g) + 2\mathrm{H}_2\mathrm{O} (l) + 4e^- \].

These reactions at each electrode help determine the resultant products from the electrolysis process, which in this scenario are hydrogen and oxygen gases.

Electrolysis Products Determination

Determining the products of electrolysis involves considering the types of ions present and their potential reactions at the electrodes. In the example of \( \mathrm{Na}_{2} \mathrm{SO}_{4} \), the solution contains \( \mathrm{Na}^+ \), \( \mathrm{SO}_{4}^{2-} \), \( \mathrm{H}^+ \), and \( \mathrm{OH}^- \) ions.

The **cathode reaction**: The most easily reducible ion is typically the one that will be reduced. Between hydrogen ions and sodium ions, \( \mathrm{H}^+ \) ions are reduced more readily, producing hydrogen gas (\( \mathrm{H}_2 \)).

The **anode reaction**: The oxidation potential of ions will guide which reaction will occur. Here, hydroxide ions are oxidized more easily than sulfate ions, resulting in the production of oxygen gas (\( \mathrm{O}_2 \)).

By evaluating these ion behaviors and their respective reactions, we conclude the products of the diluted aqueous \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) electrolysis are hydrogen gas at the cathode and oxygen gas at the anode, matching answer choice (a) from the original exercise.