Problem 30 - A Complete Resource Book in Chemistry for JEE Main

Question

Electrolysis of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) will give (a) \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) (b) \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{H}_{2}\) (d) \(\mathrm{SO}_{2}\) and \(\mathrm{H}_{2}\)

Step-by-Step Solution

Answer

(b) \( \mathrm{H}_{2} \) and \( \mathrm{O}_{2} \).

Step 1: Understand Electrolysis of \\ \( \mathrm{H}_{2} \mathrm{SO}_{4} \)

When dilute sulfuric acid is subjected to electrolysis, acid dissociates into ions: \( \mathrm{H}^{+} \) and \( \mathrm{SO}_{4}^{2-} \). The water in the solution also dissociates into \( \mathrm{H}^{+} \) and \( \mathrm{OH}^{-} \).

Step 2: Identify the Reactions at Electrodes

At the cathode, \( \mathrm{H}^+ \) ions gain electrons to form hydrogen gas (\( \mathrm{H}_{2} \)), according to the reaction: \[ 2\mathrm{H}^{+} (aq) + 2e^{-} \rightarrow \mathrm{H}_{2} (g) \]At the anode, \( \mathrm{OH}^{-} \) ions lose electrons to form oxygen gas (\( \mathrm{O}_{2} \)), according to the reaction: \[ 4\mathrm{OH}^{-} (aq) \rightarrow \mathrm{O}_{2} (g) + 2\mathrm{H}_{2}\mathrm{O} (l) + 4e^{-} \]

Step 3: Determine the Overall Reaction

Combining the cathode and anode reactions, we get the overall balanced chemical equation:\[ 2\mathrm{H}_{2}\mathrm{O} (l) \rightarrow 2\mathrm{H}_{2} (g) + \mathrm{O}_{2} (g) \]

Step 4: Select the Correct Option

According to the electrolysis reaction, the products formed are \( \mathrm{H}_2 \) and \( \mathrm{O}_2 \). Therefore, the correct option is (b) \( \mathrm{H}_{2} \) and \( \mathrm{O}_{2} \).

Key Concepts

Dilute Sulfuric Acid

Dilute sulfuric acid, denoted as \( \mathrm{H}_2\mathrm{SO}_4 \), plays a central role in the process of electrolysis. It is an aqueous solution where the sulfuric acid is only partially dissociated into ions. These ions include hydrogen ions (\( \mathrm{H}^+ \)) and sulfate ions (\( \mathrm{SO}_4^{2-} \)). The presence of water in the mixture further dissociates into \( \mathrm{H}^+ \) and hydroxide ions (\( \mathrm{OH}^- \)). This abundance of ions facilitates the ability of the solution to conduct electricity, which is a crucial requirement for electrolysis.
This understanding of how dilute sulfuric acid dissociates into ions helps us grasp why it is efficiently used in this process for splitting water into hydrogen and oxygen.

Electrode Reactions

During electrolysis, different reactions occur at the two electrodes involved - the cathode and the anode. These reactions are specifically referred to as electrode reactions. At the cathode, which is negatively charged, hydrogen ions (\( \mathrm{H}^+ \)) gain electrons (a process known as reduction) to form hydrogen gas:

• Reaction: \( 2\mathrm{H}^{+} (aq) + 2e^{-} \rightarrow \mathrm{H}_{2} (g) \).

On the other side, the anode is positively charged. Here, hydroxide ions (\( \mathrm{OH}^- \)) lose electrons, which is referred to as oxidation, to form oxygen gas. The reaction is as follows:

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

Understanding these reactions helps us see how specific ions are transformed into gaseous products during electrolysis.

Hydrogen Production

Hydrogen gas production is one of the primary outcomes of the electrolysis of dilute sulfuric acid. This process occurs at the cathode, where the \( \mathrm{H}^+ \) ions from both the sulfuric acid and water accept electrons to become hydrogen gas (\( \mathrm{H}_2 \)).
The reaction can be summarized as follows: At the cathode, the reaction \( 2\mathrm{H}^{+} (aq) + 2e^{-} \rightarrow \mathrm{H}_{2} (g) \) indicates that two hydrogen ions require electrons to form one molecule of hydrogen gas.
This electrolytic method of producing hydrogen is significant due to the clean nature of the chemical reaction, which results in a gas widely used in various industrial applications including energy generation and as a fuel source.

Oxygen Production

Oxygen is the other principal gas produced during the electrolysis of dilute sulfuric acid. This happens at the anode, where hydroxide ions (\( \mathrm{OH}^- \)) from the dissociation of water release electrons and form oxygen gas.
The anode reaction is demonstrated by: \( 4\mathrm{OH}^- (aq) \rightarrow \mathrm{O}_2 (g) + 2\mathrm{H}_2\mathrm{O} (l) + 4e^- \). This shows that four hydroxide ions are needed to produce one molecule of oxygen gas, along with water molecules and electrons as by-products.
The production of oxygen during electrolysis is not only important for industrial applications but also exemplifies a method for obtaining this essential element in a pure form, potentially for use in medical and environmental technologies.