Problem 95 - A Complete Resource Book in Chemistry for JEE Main

Question

\(2 \mathrm{MnO}_{4}+\mathrm{aH}^{+}+\mathrm{bH}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{Mn}^{+2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{dO}_{2}\) What are the values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) and \(\mathrm{d} ?\) (a) \(4,5,4,5\) (b) \(6,5,8,5\) (c) \(3,2,4,8\) (d) \(2,5,4,6\)

Step-by-Step Solution

Answer

The correct answer is option (b) \(6,5,8,5\).

Step 1: Balance Manganese Atoms

The reaction involves the compound \( \text{MnO}_4^- \) changing to \( \text{Mn}^{2+} \). There are 2 manganese atoms on the reactant side and 2 manganese atoms on the product side, thus the manganese atoms are already balanced.

Step 2: Balance Oxygen Atoms

Each \( \text{MnO}_4^- \) has 4 oxygen atoms, and there are 2 such species, making a total of 8 oxygen atoms. These need to be accounted for on the right side. The product \( \text{H}_2\text{O} \) contributes to the oxygen count, along with the \( \text{O}_2 \). To balance 8 oxygens: let’s assume \( b = 5 \times \text{H}_2\text{O}_2 \) contributes 5 oxygens, and \( d = 5 \times \text{O}_2 \) gives 5 oxygens. This results in 10 oxygens on the right side from both sources combined, providing room for the needed hydrogen water balance.

Step 3: Balance Hydrogen Atoms

Since we have \( b = 5 \) water molecules on the left side contributing 10 hydrogen atoms (as each water molecule has 2 H atoms), and also \( \text{H}_2\text{O} \) molecules on the right, we must determine \( \text{a} \) incorrectly while ensuring the hydrogen balances both within the formula and the acid condition. Total hydrogens must end to the 7 required from outputs, thus, \( a = 6 \) balancing.

Step 4: Check Charges

For validating balance, reactant side includes overall: \( 2 \times \text{MnO}_4^-\) gives \(-2 \text{charges}\) plus \( a \) \( \text{H}^+ \) ions. The product, 2 \( \text{Mn}^{2+} \) ions (combined charge of +4), and with neutral water, ensure that \( \text{a} = 6 \) hydrogens maintain the charge balance across the chemical reaction

Key Concepts

Stoichiometry

Stoichiometry is the method of using balanced chemical equations to calculate the quantities of reactants and products involved in a chemical reaction. It serves as a fundamental concept in chemistry because it allows us to predict the outcomes and amounts of substances consumed and produced in chemical reactions.

When balancing a chemical equation, you aim to balance both the atoms and the charges between the reactants and the products. In the equation provided, balancing starts with elements like Manganese, then proceeds to Oxygen and Hydrogen. This process ensures the conservation of mass and charge across a reaction.

• Start with atoms that appear in fewer compounds.
• Balance elements that appear in only one compound on each side of the equation.
• Finally, consider the charge balance, especially in ionic equations.

Understanding stoichiometry is crucial for performing any calculations related to chemical quantities and reactions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. One species undergoes reduction by gaining electrons while the other is oxidized by losing electrons.

In our exercise, the reaction involves permanganate ions ( 2 ext{MnO}_{4}^- ) which are reduced and participate in a redox process. Understanding redox reactions is key for deciphering many chemical processes, such as biological reactions and industrial applications.

• Reduction refers to the gain of electrons or a decrease in oxidation state.
• Oxidation refers to the loss of electrons or an increase in oxidation state.
• These reactions can be identified by changes in oxidation numbers.

Recognizing the interplay of redox components aids in balancing the overall equation through stoichiometric coefficients.

Oxidation and Reduction

Oxidation and Reduction are specifically important in understanding the electron transfer process within redox reactions. They are two sides of the same coin but affect the elements in different ways.

Oxidation involves the loss of electrons, which results in an increase in oxidation state. For example, manganese is reduced from +7 in permanganate to +2 in the solution, undergoing electron gain.
Reduction is the opposite; it is the gain of electrons, resulting in a decrease in oxidation state. These changes are crucial for determining which molecules or ions are the oxidizing and reducing agents in a chemical reaction.

• Oxidizing agents accept electrons and are reduced.
• Reducing agents donate electrons and are oxidized.
• Observe the changes in oxidation states to identify the agent roles.

Deciphering these factors is essential for figuring out the balanced coefficients required in the equation.

Chemical Reactions

A chemical reaction involves the transformation of reactants into products through the breaking and forming of bonds. This rearrangement involves energy changes and the laws of chemistry, such as the Law of Conservation of Mass and Energy.

In balancing chemical equations, these principles dictate the method by which we ensure both sides of the equation are equal. In the exercise, we see an example of a balanced chemical reaction, where the number of atoms for each element is the same on both sides. This balance shows complete transformation into new products.

• Identify all reactants and products involved in the equation.
• Track changes in atom numbers and charges to keep mass conserved.
• In some reactions, catalysts may be involved. However, they do not appear in the equation as reactants or products.

Grasping the underlying process of chemical reactions is pivotal for predicting and facilitating reactions both in lab settings and in practical applications.