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 at the anode is \(\mathrm{H}_{2} \mathrm{S}_{2}\mathrm{O}_{8}\).

Step 1: Identify the Ions Present

In a 50% aqueous solution of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), the ions present are \(\mathrm{H}^{+}\) from water and acid, \(\mathrm{SO_{4}}^{2-}\), and \(\mathrm{OH}^{-}\) from water itself.

Step 2: Identify Electrode Reactions

At the anode, oxidation occurs. We need to consider the possible oxidation reactions for the anions present: \(\mathrm{OH}^{-}\) and \(\mathrm{SO_{4}}^{2-}\).- \(\mathrm{OH}^{-}\) can oxidize to produce \(\mathrm{O}_{2}\).- \(\mathrm{SO_{4}}^{2-}\) can undergo further oxidation to form \(\mathrm{S_{2}O_{8}}^{2-}\) (peroxodisulfate ion).

Step 3: Determine the Favorable Anodic Reaction

The electrode potential determines which reaction will occur.- The oxidation potential for \(\mathrm{OH}^{-}\) to \(\mathrm{O}_{2}\) is high.- However, under the concentrated condition of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), \(\mathrm{SO_{4}}^{2-}\) is more readily oxidized to \(\mathrm{S_{2}O_{8}}^{2-}\).

Step 4: Conclude the Anodic Product

Given the high concentration of sulfate ion and competition, the reaction favoring the formation of \(\mathrm{H}_{2} \mathrm{S}_{2}\mathrm{O}_{8}\) is more likely. Thus, the main product at the anode is \(\mathrm{H}_{2} \mathrm{S}_{2}\mathrm{O}_{8}\).

Key Concepts

Anode reactions

In electrolysis, the anode is where oxidation occurs. During this process, electrons are removed from ions, converting them to their elemental or a different ionic form. In the context of a 50% aqueous solution of sulfuric acid ( H_2SO_4 ) being electrolyzed, we focus on two potential anode reactions.
- Hydroxide ions ( OH^- ) can be oxidized, producing oxygen gas ( O_2 ), water, and electrons. - Sulfate ions ( SO_4^{2-} ) can be oxidized further to form peroxodisulfate ions ( S_2O_8^{2-} ).
This competition at the anode results in the formation of specific oxidation products depending on the conditions.

Electrode potential

Electrode potential plays a crucial role in determining which anode reaction occurs. It is essentially the ability of a substance to be oxidized or reduced, measured in volts under standard conditions. In electrolysis, the material with the more positive electrode potential will likely undergo oxidation.
- The oxidation reaction of OH^- to O_2 involves a relatively high electrode potential. - The SO_4^{2-} oxidation to S_2O_8^{2-} has a lower electrode potential under concentrated sulfuric acid conditions, making it the favorable reaction. Therefore, due to the specific high sulfate ion concentration, S_2O_8^{2-} formation is preferred over O_2 as the anode product.

Peroxodisulfate ion

The peroxodisulfate ion ( S_2O_8^{2-} ) is formed at the anode during the electrolysis of concentrated sulfuric acid. It is a strong oxidizing agent containing a peroxide bond where two sulfate ions are linked by an extra oxygen. This bond is formed when sulfate ions are oxidized during electrolysis.
Key properties include: - S_2O_8^{2-} ions are more stable in acidic conditions. - It serves as an efficient oxidizing agent in several industrial and laboratory processes. The formation of S_2O_8^{2-} demonstrates how specific electrolytic conditions can alter ionic interactions leading to different products.

Platinum electrodes

In electrolysis, electrodes facilitate the transfer of electrons and are crucial for the reaction's efficiency. Platinum is used as an electrode material due to its excellent conductive properties and chemical inertness. This means it does not react with the electrolyte, ensuring it remains unchanged throughout the process.
Key advantages of platinum electrodes include: - High resistance to corrosion, which is significant in the presence of strong acids like H_2SO_4 . - Capable of conducting electricity efficiently without participating in the chemical reactions. Therefore, platinum electrodes provide a stable environment for consistent and safe electrolysis, contributing to the successful formation of desired products like the peroxodisulfate ion in concentrated sulfuric acid.