Problem 56 - A Complete Resource Book in Chemistry for JEE Main

Question

The product obtained at anode when a \(50 \%\) aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is electrolyzed using platinum electrodes is (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{O}_{2}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) (d) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\)

Step-by-Step Solution

Answer

The product formed at the anode is \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8\) (option d).

Step 1: Understanding the Electrolysis Process

In the electrolysis of a solution, positive ions (cations) move towards the cathode (negative electrode) and negative ions (anions) move towards the anode (positive electrode). Oxygen typically forms at the anode from the oxidation of water or other species with oxygen in their structure.

Step 2: Identify Anions in Solution

In a 50% aqueous solution of \(\mathrm{H}_2\mathrm{SO}_4\), sulfuric acid ionizes to form \(\mathrm{H}^+\) and \(\mathrm{SO}_4^{2-}\) ions. However, in an aqueous solution, we also have water molecules, which can produce \(\mathrm{O}_2\) at the anode through the oxidation process.

Step 3: Analyze Anode Reaction Possibilities

The possible anode reactions are the oxidation of water to form \(\mathrm{O}_2\) and the formation of peroxydisulfuric acid \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8\) when \(\mathrm{SO}_4^{2-}\) ions oxidize. The formation of \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8\) is favored in this specific solution.

Step 4: Determine the Predominant Anode Reaction

Given the strong acid condition and presence of \(\mathrm{SO}_4^{2-}\), forming \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8\) at the anode through the reaction: \(2\, \mathrm{SO}_4^{2-} + 2\, e^- \rightarrow \mathrm{S}_2\mathrm{O}_8^{2-}\) is likely. This reacts with hydrogen ions forming \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8\).

Step 5: Conclusion of the Anode Product Formation

Considering the ion reactions, \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8\) is the more likely product at the anode—a result of electrolyte concentration and reaction tendency in acidic media.

Key Concepts

Anode reactions

In electrolysis, anode reactions are crucial for determining the products formed at the anode. Anodes are positive electrodes where oxidation reactions occur. During electrolysis of aqueous solutions, anions migrate towards the anode.
A common example involves water and ions that contain oxygen, such as sulfate ions. At the anode, different reactions can happen depending on the ions available and their ability to be oxidized. Oxidation is a process where a substance loses electrons. This results in the formation of new species, often releasing oxygen if water is involved. Understanding what occurs at the anode is essential for predicting the correct products in electrolysis processes. Electrode materials and reaction conditions, such as concentration and temperature, can influence the specific reactions that occur at the anode.

Sulfuric acid electrolysis

Electrolyzing sulfuric acid is an interesting process because it involves both water and sulfuric acid. A 50% aqueous solution of sulfuric acid dissociates into \(\mathrm{H}^+\) and \(\mathrm{SO}_4^{2-}\) ions, making the solution highly conductive. This setup leads to a competition at the anode between different potential reactions. Water can be oxidized to release oxygen gas, but in strong acid solutions like sulfuric acid, other reactions may take precedence under the right conditions. For sulfuric acid electrolysis, the reaction at the anode is generally less straightforward due to competing processes involving the sulfate ions. Therefore, understanding the nature of sulfuric acid electrolysis is key for predicting the correct outcome.

Peroxydisulfuric acid formation

In the context of sulfuric acid electrolysis, the formation of peroxydisulfuric acid is particularly noteworthy. When \(\mathrm{SO}_4^{2-}\) ions are oxidized, a reaction occurs leading to the formation of peroxydisulfuric acid, \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8\). The chemical equation for the formation of \(\mathrm{S}_2\mathrm{O}_8^{2-}\) is:

• The reaction: \( 2\, \mathrm{SO}_4^{2-} + 2\, e^- \rightarrow\ \mathrm{S}_2\mathrm{O}_8^{2-}\)
• Occurs in the presence of hydrogen ions to form \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_8\).

This formation is favored under strong acidic conditions and high concentrations of sulfuric acid. Such conditions enable the sulfate ions to exhibit a higher oxidation state. Understanding this process reveals the intricacies involved in electrolysis reactions under different conditions, catering to the notion of optimizing conditions to control the chemical output of electrolysis.