Problem 28 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Balance the following equations. (i) \(\mathrm{Cu}_{2} \mathrm{O}+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \rightarrow \mathrm{Cu}^{2+}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) (ii) \(\left.\mathrm{K}_{4} \mathrm{Fe}(\mathrm{CN})_{6}\right]+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\) \(\quad \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{FeSO}_{4}+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}+\mathrm{CO}\) (iii) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{I}_{2}+\mathrm{OH}^{-} \rightarrow \mathrm{CHI}_{3}+\mathrm{HCO}_{3}^{-}+\mathrm{I}^{-}+\mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

(i) \(\mathrm{Cu}_2\mathrm{O} + 2\mathrm{NO}_3^- + 8\mathrm{H}^+ \rightarrow 2\mathrm{Cu}^{2+} + 2\mathrm{NO} + 4\mathrm{H}_2\mathrm{O}\); (ii) \(\mathrm{K}_4\mathrm{Fe}(\mathrm{CN})_6 + 6\mathrm{H}_2\mathrm{SO}_4 + 6\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{K}_2\mathrm{SO}_4 + \mathrm{FeSO}_4 + 3(\mathrm{NH}_4)_2\mathrm{SO}_4 + 6\mathrm{CO}\); (iii) \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 4\mathrm{I}_2 + 6\mathrm{OH}^- \rightarrow \mathrm{CHI}_3 + \mathrm{HCO}_3^- + 5\mathrm{I}^- + 5\mathrm{H}_2\mathrm{O}\).

Step 1: Identify Each Reaction

The chemical equations given are not balanced, which means that the number of each type of atom is not the same on both sides of the equation. Our goal is to ensure that the atom count for each type is balanced between reactants and products.

Step 2: Balance Equation (i)

For the reaction \(\mathrm{Cu}_{2}\mathrm{O}+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \rightarrow \mathrm{Cu}^{2+}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\), first balance copper (Cu), followed by nitrogen (N), oxygen (O), and hydrogen (H). This process might need to be repeated a few times, adjusting coefficients to ensure the balance across all the involved atoms: - Balance Cu: Start with \(\mathrm{Cu}_{2}\mathrm{O}\), so you need 2 Cu atoms on the products side: \(2\ \mathrm{Cu}^{2+}\).- Balance O: Modify H2O to balance oxygen.- Balance H: Adjust \(\mathrm{H}^{+}\).- Balance N: Adjust \(\mathrm{NO}_{3}^{-}\) so that nitrogen atoms are equal.The final balanced equation is: \[\mathrm{Cu}_2\mathrm{O} + 2\mathrm{NO}_3^- + 8\mathrm{H}^+ \rightarrow 2\mathrm{Cu}^{2+} + 2\mathrm{NO} + 4\mathrm{H}_2\mathrm{O}\]

Step 3: Balance Equation (ii)

For the reaction \(\left.\mathrm{K}_{4} \mathrm{Fe}(\mathrm{CN})_{6}\right]+\mathrm{H}_{2}\mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{K}_{2}\mathrm{SO}_{4}+\mathrm{FeSO}_{4}+\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}+\mathrm{CO}\), balance in the following sequence:- Balance K: Start with K4Fe(CN)6 requiring two on the product side (\(2\ \mathrm{K}_{2}\mathrm{SO}_4\)).- Balance Fe: Adjust \(\mathrm{FeSO}_4\).- Balance C: From CO and CN part.- Balance N: From \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}\).- Balance all others by adjusting \(\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{H}_{2}\mathrm{SO}_{4}\).The balanced equation is:\[\mathrm{K}_4\mathrm{Fe}(\mathrm{CN})_6 + 6\mathrm{H}_2\mathrm{SO}_4 + 6\mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{K}_2\mathrm{SO}_4 + \mathrm{FeSO}_4 + 3\left(\mathrm{NH}_4\right)_{2}\mathrm{SO}_4 + 6\mathrm{CO}\]

Step 4: Balance Equation (iii)

For the reaction \(\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}+\mathrm{I}_{2}+\mathrm{OH}^{-} \rightarrow \mathrm{CHI}_{3}+\mathrm{HCO}_{3}^{-}+\mathrm{I}^{-}+\mathrm{H}_{2} \mathrm{O}\), balance the following:- Balance I: Start with \(\mathrm{I}_2\), requiring \(\mathrm{I}^-\) and \(\mathrm{CHI}_3\) on the products side.- Balance C: Balance according to the carbon in CHI₃, HCO₃⁻.- Balance O and H: Adjust with H2O and OH−. The balanced equation is:\[\mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 4\mathrm{I}_2 + 6\mathrm{OH}^- \rightarrow \mathrm{CHI}_3 + \mathrm{HCO}_3^- + 5\mathrm{I}^- + 5\mathrm{H}_2\mathrm{O}\]

Step 5: Confirm Balanced Equations

Check each balanced equation to ensure that the number of each atom on the reactants side matches the number on the products side. No additional adjustments should be necessary as each atom type should have an equal count on both sides.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry. It deals with the quantitative aspects of chemical reactions. In simpler terms, it helps us understand how much of each substance is involved in a chemical reaction. This includes how much reactant is required and how much product is produced.

When balancing chemical equations, stoichiometry is vital. It ensures that the number of atoms for each element is the same on both sides of the equation. This follows the law of conservation of mass, which states that matter cannot be created or destroyed.

To balance an equation, you need to identify the coefficients that make the atom count equal on both sides. This often involves:

• Counting the number of each type of atom in the reactants and products.
• Adjusting the coefficients (the numbers before the chemical formulas) to balance the atoms.
• Re-checking to confirm that every atom is balanced after adjusting the coefficients.

In summary, stoichiometry provides the tools to quantify the materials in a chemical reaction accurately.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are types of chemical reactions that involve the transfer of electrons between two species. These are common in many everyday processes, such as rusting iron or metabolizing food for energy.

In a redox reaction, one element loses electrons (oxidation) while another gains electrons (reduction). It’s crucial to properly balance these reactions, ensuring that the electron transfer is accounted for completely.

• Oxidization involves an increase in oxidation state and loss of electrons.
• Reduction involves a decrease in oxidation state and gain of electrons.

For a balanced redox reaction, both mass and charge need to be balanced. This often requires breaking down the equation into half-reactions that separately represent oxidation and reduction processes. Each half-reaction is balanced individually and then combined to provide the overall balanced redox equation.

Understanding redox reactions can help us comprehend processes in both inorganic and organic chemistry, making it a significant concept in the study of chemical transformations.

Chemical Reactions

Chemical reactions are transformative processes where the initial substances (reactants) are converted into new substances (products). Understanding chemical reactions is key to grasping the dynamics of chemistry.

Chemical reactions involve breaking old bonds and forming new ones. Each type of reaction, whether it be synthesis, decomposition, or single displacement, follows specific patterns that allow predictability. Balancing these reactions is essential for accurately representing the process.

• Synthesis reactions combine elements or simpler compounds to create a more complex compound.
• Decomposition reactions break down a compound into simpler elements or compounds.
• Single displacement reactions involve an element replacing another in a compound.

By balancing a chemical reaction, we ensure adherence to the law of conservation of mass. This is crucial not just for theoretical understanding but also for practical applications such as calculating yields or adjusting industrial chemical processes.

In conclusion, understanding the basics of chemical reactions and how to balance them allows for deeper exploration into the world of chemistry.