Problem 98 - A Complete Resource Book in Chemistry for JEE Main

Question

In electrolysis of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what is liberated at anode? (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{SO}_{4}^{2-}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{O}_{2}\)

Step-by-Step Solution

Answer

Oxygen ( \(\mathrm{O}_{2}\) ) is liberated at the anode.

Step 1: Understanding Electrolysis

In electrolysis, electrical energy is used to drive a non-spontaneous chemical reaction. Electrolysis of dilute \( ext{H}_2 ext{SO}_4\) involves passing an electric current through the solution, leading to the decomposition of the constituent ions at the electrodes.

Step 2: Identifying Anode and Cathode

The anode is the positive electrode where oxidation occurs, and the cathode is the negative electrode where reduction occurs. In electrolyte solutions, cations move towards the cathode, and anions move towards the anode.

Step 3: Reactions at the Anode

In the electrolysis of dilute \(\text{H}_2\text{SO}_4\), the major species present are \(\text{H}^+\), \(\text{OH}^-\), and \(\text{SO}_4^{2-}\). Water can also act as a source of ions through a reaction: \(2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2\). At the anode, \(\text{OH}^-\) ions are discharged over \(\text{SO}_4^{2-}\).

Step 4: Oxygen Liberation

The discharge of \(\text{OH}^-\) ions at the anode results in the formation of \(\text{O}_2\) according to the reaction: \((4\text{OH}^-) \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-\). The evolution of oxygen gas is observed at the anode.

Key Concepts

Dilute sulfuric acid

Dilute sulfuric acid, a common laboratory reagent, plays a pivotal role in the electrolysis process. Comprised primarily of water (H₂O) and sulfuric acid (H₂SO₄) in lower concentrations, this solution provides both hydrogen ions (H⁺) and sulfate ions (SO₄²⁻) for the electrolysis to occur.

The presence of water is equally significant as it supplies hydroxide ions (OH⁻) upon dissociation. These ions are crucial for the reactions at the electrodes. In the electrolyte solution, water can ionize slightly, creating a small number of H⁺ and OH⁻ ions, thus participating in the overall reaction.

• Composition: Consists mainly of H₂SO₄ dissolved in water.
• Conductivity: The dilution allows the solution to conduct electricity due to the presence of free ions.
• Electrolyte Role: Sulfuric acid provides sulphate and hydrogen ions necessary for redox reactions at the electrodes.

Thus, dilute sulfuric acid effectively facilitates the flow of electricity, enabling the electrolysis process to proceed.

Electrode reactions

In the electrolysis of dilute sulfuric acid, electrode reactions are essential components that drive the process. Electrolytic cells comprise two electrodes: an anode and a cathode, submerged in the electrolyte solution. These electrodes are typically inert materials like platinum that do not interfere with the reactions.

• Anode: This is the positively charged electrode. Here, oxidation occurs, which involves the loss of electrons from species in the solution.
• Cathode: The negatively charged electrode where reduction, or gain of electrons, takes place.

The reactions at these electrodes determine which products are formed during electrolysis.

At the anode, the hydroxide ions (OH⁻) from water get discharged more readily than the sulphate ions (SO₄²⁻), leading to the production of oxygen gas. Meanwhile, at the cathode, hydrogen ions (H⁺) gain electrons to form hydrogen gas (H₂). Overall, the reactions show a beautiful interplay of oxidation and reduction leading to gas liberation at both electrodes.

Oxygen liberation

Oxygen liberation is a fascinating outcome of the electrolysis of dilute sulfuric acid, primarily occurring at the anode. When the electric current is applied, hydroxide ions (OH⁻) present in the solution undergo oxidation. This process leads to the release of oxygen gas.

The detailed reaction occurring at the anode is given by:\[4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^-\] Here, every four hydroxide ions release one molecule of oxygen, forming water in the process.

• Oxidation Process: At the anode, OH⁻ ions lose electrons resulting in oxygen gas.
• Visual Observation: Bubbles of oxygen gas are visible at the anode, a direct indicator of the electrolysis taking place.
• Gas Production: This reaction is a clear example of a non-spontaneous process facilitated by the energy input from an electrical source.

The ability to liberate oxygen through electrolysis is not only a critical concept in chemistry but also an underpinning technology for various industrial processes.