Problem 44 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

State with balanced equations what happens when : (i) Sodium iodate is added to a solution of sodium bisulphite. [1990 - 1 Marks] (ii) Sodium chlorate reacts with sulphur dioxide in dilute sulphuric acid medium. [1989-1 Mark] (iii) Sodium bromate reacts with fluorine in presence of alkali. [1989-1 Mark] (iv) Iodate ion reacts with bisulphite ion to liberate iodine. [1988-1 Mark] (v) Sodium iodate is treated with sodium bisulphite solution. [1982-1 Mark]

Step-by-Step Solution

Answer

(i) \( 2\text{NaIO}_3 + 5\text{NaHSO}_3 \rightarrow 2\text{NaI} + 5\text{NaHSO}_4 + \text{H}_2\text{O} \) (ii) \( 2\text{NaClO}_3 + \text{SO}_2 + \text{H}_2\text{SO}_4 \rightarrow 2\text{ClO}_2 + 2\text{NaHSO}_4 \) (iii) \( 2\text{NaBrO}_3 + 4\text{NaOH} + 2\text{F}_2 \rightarrow 2\text{NaBr} + 2\text{NaF} + 5\text{O}_2 + 2\text{H}_2\text{O} \) (iv) \( 2\text{IO}_3^- + 5\text{HSO}_3^- \rightarrow \text{I}_2 + 5\text{SO}_4^{2-} + 3\text{H}_2\text{O} \) (v) Same as (i).

Step 1: Understanding the Reaction for (i)

When sodium iodate (NaIO₃) is added to a solution of sodium bisulphite (NaHSO₃), iodate ions (IO₃⁻) react with bisulphite ions (HSO₃⁻) to produce iodide ions (I⁻), sulfate ions (SO₄²⁻), and water (H₂O). This is a redox reaction where iodate is reduced to iodide and bisulphite is oxidized to sulfate.

Step 2: Balanced Equation for Reaction (i)

The balanced chemical equation for this reaction is:\[ 2 ext{NaIO}_3 + 5 ext{NaHSO}_3 ightarrow 2 ext{NaI} + 5 ext{NaHSO}_4 + ext{H}_2 ext{O} \]

Step 3: Understanding Reaction for (ii)

Sodium chlorate (NaClO₃) reacts with sulfur dioxide (SO₂) in the presence of dilute sulfuric acid to produce chlorine dioxide (ClO₂) and water. This is also a redox reaction where chlorate is reduced and sulfur dioxide is oxidized.

Step 4: Balanced Equation for Reaction (ii)

The balanced chemical equation is:\[ 2 ext{NaClO}_3 + ext{SO}_2 + ext{H}_2 ext{SO}_4 ightarrow 2 ext{ClO}_2 + 2 ext{NaHSO}_4 \]

Step 5: Understanding Reaction for (iii)

Sodium bromate (NaBrO₃) reacts with fluorine (F₂) in the presence of alkali to form sodium fluoride (NaF), oxygen (O₂), and sodium bromide (NaBr). This is another redox reaction where bromate is reduced to bromide and fluorine is reduced to fluoride.

Step 6: Balanced Equation for Reaction (iii)

The balanced chemical equation is:\[ 2 ext{NaBrO}_3 + 4 ext{NaOH} + 2 ext{F}_2 ightarrow 2 ext{NaBr} + 2 ext{NaF} + 5 ext{O}_2 + 2 ext{H}_2 ext{O} \]

Step 7: Understanding the Reaction for (iv)

The iodate ion (IO₃⁻) reacts with bisulphite ion (HSO₃⁻) to form iodine (I₂), sulfate ion (SO₄²⁻), and water. Iodate is reduced and bisulphite is oxidized.

Step 8: Balanced Equation for Reaction (iv)

The balanced chemical equation is:\[ 2 ext{IO}_3^- + 5 ext{HSO}_3^- ightarrow ext{I}_2 + 5 ext{SO}_4^{2-} + 3 ext{H}_2 ext{O} \]

Step 9: Understanding Reaction for (v)

This is a repetition of the first reaction (i), where sodium iodate is treated with sodium bisulphite solution, thus having the same balanced equation.

Step 10: Balanced Equation for Reaction (v)

Since it's identical to (i), the balanced equation remains:\[ 2 ext{NaIO}_3 + 5 ext{NaHSO}_3 ightarrow 2 ext{NaI} + 5 ext{NaHSO}_4 + ext{H}_2 ext{O} \]

Key Concepts

Balanced Equations

In chemistry, balanced equations are vital to describe the quantitative relationships between reactants and products in a chemical reaction. They ensure that the number of each type of atom is the same on both sides of the equation. For a reaction to be balanced, every element must have an equal number of atoms before and after the reaction.
This is essential because of the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. By using coefficients, we adjust the amounts of reactants and products to satisfy this law. For example, the equation: \[ 2 \text{NaIO}_3 + 5 \text{NaHSO}_3 \rightarrow 2 \text{NaI} + 5 \text{NaHSO}_4 + \text{H}_2\text{O} \] shows the balanced reaction of sodium iodate with sodium bisulphite, ensuring the mass and the number of atoms are consistent on both sides.

Iodate Reaction

The iodate reaction focuses on the behavior of iodate ions (IO₃⁻) in redox processes. Iodate ions can participate in redox reactions by accepting electrons, thereby getting reduced. In the context of our exercise, when sodium iodate is mixed with sodium bisulphite, the iodate ions are reduced to iodide ions.
This is clear from the chemical equation where iodate gains electrons to form iodide. The reduction of iodate is balanced by the oxidation of bisulphite ions to sulfate ions (SO₄²⁻). Such reactions are crucial in various analytical methods involving iodine, as they allow us to understand and predict the behavior of iodine compounds under different conditions.

Sodium Bisulphite

Sodium bisulphite (NaHSO₃) is often used in chemistry as a reducing agent. In the reactions discussed, it interacts with iodate to undergo oxidation. This involves the sodium bisulphite giving away electrons and converting into sulfate ions.
Sodium bisulphite finds applications in chemical labs as well as industrial processes due to its ability to act as a mild reducing agent. Its ability to donate electrons is what makes it both versatile and effective in various redox reactions, such as the one with sodium iodate.

• Reactant in water treatments
• Preservative in food industries
• Prominent role in redox reactions

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry that involves ensuring atoms on each side of the equation are equal. It is a representation of a chemical reaction in which the symbols and formulas of the reactants are placed on the left and products on the right. As per the conservation of matter, the same number of each type of atom should exist on both sides after a reaction.
The process typically involves the following steps:

• Identify the reactants and products.
• Count the number of atoms of each element in reactants and products.
• Add coefficients to balance the atoms on both sides.
• Ensure all coefficients are the simplest whole numbers.

Understanding and practicing balancing chemical equations are crucial for accurately predicting the outcomes of chemical reactions and for overall mastery of chemical processes.