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

The correct answer is (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\).

Step 1: Understand the Electrolysis Setup

We have a dilute aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) with platinum electrodes. This solution dissociates into \(\mathrm{Na}^{+}, \mathrm{SO}_{4}^{2-}, \mathrm{H}^{+},\) and \(\mathrm{OH}^{-}\) ions.

Step 2: Consider Possible Reactions at the Anode

At the anode, oxidation reactions occur. Possible reactions are: oxidation of \(\mathrm{OH}^{-}\) to \(\mathrm{O}_{2}\) and oxidation of \(\mathrm{SO}_{4}^{2-}\) to \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\). In dilute solutions, \(\mathrm{OH}^{-}\) typically gets oxidized, producing \(\mathrm{O}_{2}\).

Step 3: Determine the Reaction at the Cathode

At the cathode, reduction reactions take place. We can either reduce \(\mathrm{Na}^{+}\) to \(\mathrm{Na}\) or \(\mathrm{H}^{+}\) to \(\mathrm{H}_{2}\). The reduction of \(\mathrm{H}^{+}\) to \(\mathrm{H}_{2}\) is more likely because it has a lower reduction potential compared to the reduction of \(\mathrm{Na}^{+}\).

Step 4: Combine Anode and Cathode Reactions

From Steps 2 and 3, the likely reactions at the anode and cathode in the electrolysis of dilute \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are: Anode: \(2\mathrm{OH}^{-} \rightarrow \mathrm{O}_{2} + 2\mathrm{H}^{+} + 4e^{-}\); Cathode: \(2\mathrm{H}^{+} + 2e^{-} \rightarrow \mathrm{H}_{2}\). Thus, the products are \(\mathrm{O}_{2}\) at the anode and \(\mathrm{H}_{2}\) at the cathode.

Step 5: Choose the Correct Option

Match the predicted outcomes with the options given: The correct products \(\mathrm{O}_{2}\) at the anode and \(\mathrm{H}_{2}\) at the cathode correspond to option (a).

Key Concepts

Platinum Electrodes

Platinum electrodes are often used in electrolysis due to their exceptional chemical stability and conductivity. Unlike other metals, platinum does not react with the substances in the solution upon which it is meant to work.
This characteristic makes platinum ideal for conducting electricity without contaminating the reactions taking place at the electrode surfaces. In electrolysis, conductivity is essential, as it allows electrons to flow freely, facilitating oxidation and reduction reactions.
Moreover, platinum's resistance to corrosion guarantees that it can be used repeatedly without degrading over time, making it a long-lasting choice for electrochemical cells.

Oxidation Reactions

During electrolysis, the anode is the site of oxidation reactions, where substances lose electrons. In the context of a dilute aqueous solution of sodium sulfate egin{equation} ext{Na}_{2} ext{SO}_{4} ightleftharpoons ext{Na}^{+}, ext{SO}_{4}^{2-}, ext{H}^{+}, ext{OH}^{-} ext, egin{equation} several species can potentially be oxidized.
Typically, in such a dilute solution, hydroxide ions ( OH^{-} ) are oxidized instead of sulfate ions ( SO_{4}^{2-} ). The reaction at the anode can thus be represented as:

• egin{equation} 2 ext{OH}^{-} ightarrow ext{O}_{2} + 2 ext{H}^{+} + 4e^{-} ext, egin{equation}

where oxygen gas is released along with hydrogen ions.

Reduction Reactions

At the cathode, reduction reactions occur, which involve gaining electrons. In a dilute aqueous solution of sodium sulfate, the reduction focus is generally between sodium ions ( Na^{+} ) and hydrogen ions ( H^{+} ).
Despite sodium ions being present, the availability of water and the electrochemical conditions favor the reduction of hydrogen ions over sodium ions. Specifically, the reaction tends to proceed as follows:

• egin{equation} 2 ext{H}^{+} + 2e^{-} ightarrow ext{H}_{2} ext, egin{equation}

This results in the formation of hydrogen gas at the cathode instead of metallic sodium, which has a higher reduction potential.

Aqueous Solution Chemistry

Aqueous solution chemistry focuses on reactions occurring in water-based solutions. The electrolysis of ext{Na}_{2} ext{SO}_{4} ext, a common laboratory setup, illustrates key principles.
When ext{Na}_{2} ext{SO}_{4} ext is dissolved in water, it dissociates into sodium ( Na^{+} ) and sulfate ions ( SO_{4}^{2-} ), as well as hydrogen ( H^{+} ) and hydroxide ions ( OH^{-} ).
This dissociation indicates that both the original compound and water itself engage in ionic processes.

• Ion mobility in water is crucial, enabling the transport of charge via ions to electrodes where redox reactions occur.
• The intermingling of water's own dissociation products amplifies this effect.

Overall, understanding how compounds like ext{Na}_{2} ext{SO}_{4} ext behave in an aqueous environment is essential to predicting outcomes in electrolysis.