Problem 6 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider the reaction: $$ \begin{aligned} &\mathrm{H}_{2} \mathrm{SO}_{3}(\mathrm{aq})+\mathrm{Sn}^{4+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ &\quad \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}(\mathrm{aq})+3 \mathrm{H}^{+}(\mathrm{aq}) \end{aligned} $$ Which of the following statements is correct? (a) \(\mathrm{Sn}^{4+}\) is the oxidizing agent because it undergoes oxidation (b) \(\mathrm{Sn}^{4+}\) is the reducing agent because it undergoes oxidation (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) is the reducing agent because it undergoes oxidation (d) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) is the reducing agent because it undergoes reduction

Step-by-Step Solution

Answer

(d): \( \text{H}_2\text{SO}_3 \) is the reducing agent because it undergoes oxidation.

Step 1: Identify the agents in the reaction

To solve this, first identify the oxidation states of the elements involved in the reaction. The species that lose electrons get oxidized and act as reducing agents, while species that gain electrons get reduced and act as oxidizing agents.

Step 2: Determine the oxidation states

Estimate the oxidation states: \( \text{Sn}^{4+} \) has an oxidation state of +4, \( \text{Sn}^{2+} \) has +2. In \( \text{H}_2\text{SO}_3 \), hydrogen is +1, and assuming oxygen is -2, calculate sulfur's oxidation state which is +4. \( \text{HSO}_4^{-} \) gives sulfur an oxidation state of +6. \( \text{H}^+ \) is +1.

Step 3: Write the half-reactions

Write the half-reactions to show oxidation and reduction. \( \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \) is the reduction half-reaction because electrons are gained. \( \text{H}_2\text{SO}_3 \rightarrow \text{HSO}_4^- + 2H^+ + 2e^- \) is the oxidation half-reaction since electrons are lost.

Step 4: Analyze the agents

Since \( \text{Sn}^{4+} \) is reduced from +4 to +2, it acts as the oxidizing agent. \( \text{H}_2\text{SO}_3 \), which is oxidized as it increases sulfur's oxidation state from +4 to +6, acts as the reducing agent.

Step 5: Select the correct statement

Based on the analysis, the correct statement is (d): \( \text{H}_2\text{SO}_3 \) is the reducing agent because it undergoes oxidation.

Key Concepts

Oxidation States

Understanding oxidation states is crucial for analyzing redox reactions. Oxidation states, also known as oxidation numbers, offer a way to keep track of electrons in chemical reactions. They represent the number of electrons an atom gains, loses, or shares when it forms chemical bonds.
To determine oxidation states, follow these rules:

• The oxidation state of a free element is always zero. For example, O₂ and Sn in their natural states both have oxidation states of 0.
• In simple ions, the oxidation state is equal to the charge of the ion. Sn⁴⁺ has an oxidation state of +4 and Sn²⁺ has +2.
• Hydrogens in compounds usually have an oxidation state of +1.
• Oxygens are generally assigned an oxidation state of -2.
• For compounds, the sum of the oxidation states for all atoms must equal the overall charge of the compound.

For example, in H₂SO₃, sulfur's oxidation state can be calculated by considering the oxidation state of hydrogen (+1) and oxygen (-2). Such calculations help identify which atoms are oxidized or reduced.

Oxidizing Agents

An oxidizing agent, or oxidant, is a substance that facilitates the oxidation of another substance while it undergoes reduction itself.
In redox reactions, when a substance is reduced, it gains electrons. This gain stabilizes the oxidizing agent and allows it to take electrons from the reducing agent.
Taking a closer look at our reaction:

• The oxidizing agent in the equation of Sn⁴⁺ being reduced to Sn²⁺ is Sn⁴⁺ itself. As it gains 2 electrons to become Sn²⁺, it causes oxidation of another substance.

Remember, identifying the oxidizing agent requires observing the change in oxidation states: the species that reduces (gains electrons) serves as the oxidizing agent in the process.

Reducing Agents

Reducing agents play a central role in redox reactions by donating electrons to another substance.
These agents are substances that are themselves oxidized, meaning they lose electrons, become more positively charged, or increase their oxidation number.
In our example:

• The reducing agent is H₂SO₃, where sulfur increases its oxidation state from +4 to +6, indicating it has donated electrons.

As the oxidation state of sulfur in H₂SO₃ increases, it releases electrons. This makes H₂SO₃ the reducing agent, supplying necessary electrons for the reduction of Sn⁴⁺ into Sn²⁺.

Half-Reactions

Half-reactions break down redox reactions into two parts: oxidation and reduction processes. They ensure the conservation of mass and charge in a chemical reaction by accounting for the transfer of electrons.
The two half-reactions must balance not only mass but also with respect to charge.

• Oxidation Half-Reaction: This shows the transfer of electrons away from the reducing agent. For instance, H₂SO₃ is oxidized to HSO₄^-, releasing electrons in the process.
• Reduction Half-Reaction: This involves the transfer of electrons to the oxidizing agent. An example is the conversion of Sn⁴⁺ to Sn²⁺, a process that involves gaining 2 electrons.

Both half-reactions are crucial for understanding complete redox processes, as they illustrate the electron shifts responsible for the changes in chemical oxidation states.