Problem 155 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the reaction of \(\mathrm{Br}_{2}\) with KOH as shown below and match the following: \(\mathrm{Br}_{2}+\mathrm{KOH} \rightarrow \mathrm{KBr}+\mathrm{KBrO}_{\mathrm{a}}+\mathrm{H}_{2} \mathrm{O}\) $$ \begin{array}{ll} \text { Column-I (Compound) } & \text { Column-II (n-Facto r) } \\ \hline \text { (a) } \mathrm{KOH} & \text { p. } 1 \\ \text { (b) } \mathrm{KBr} & \text { q. } 5 \\ \text { (c) } \mathrm{KBrO}_{3} & \text { r. } 5 / 3 \\ \text { (d) } \mathrm{Br}_{2} & \text { s. } 5 / 6 \\ & \text { t. } 2 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

(a) KOH matches with p. 1, (b) KBr matches with t. 2, (c) KBrO₃ matches with q. 5, (d) Br₂ matches with s. 5/6.

Step 1: Identify the Oxidation States

Determine the oxidation states of bromine in the reactants and products. In \( \mathrm{Br}_{2} \), bromine is in the 0 oxidation state. In \( \mathrm{KBr} \), bromine is in the -1 oxidation state, and in \( \mathrm{KBrO}_a \), the common oxidation state of bromate ions varies, but typically corresponds to +5 for \( \mathrm{KBrO}_3 \) (where \( a = 3 \)).

Step 2: Write the Half-Reactions

The given reaction involves both reduction and oxidation of bromine. The reduction half-reaction is \( \mathrm{Br}_{2} \rightarrow \mathrm{2Br}^{-} \), resulting in the formation of \( \mathrm{KBr} \). The oxidation half-reaction is \( \mathrm{Br}_{2} \rightarrow \mathrm{2BrO}_{3}^{-} \), where each bromine atom increases its oxidation state from 0 to +5.

Step 3: Calculate 'n-Factor' for Each Species

The 'n-factor' is the change in oxidation state per molecule. For each compound: - \( \mathrm{KOH} \): Not directly involved in redox, so n-factor = 1.- \( \mathrm{KBr} \): Changes from \( \mathrm{Br}_{2} \) to \( \mathrm{Br}^{-} \), n-factor = 2 (for two KBr per Br2).- \( \mathrm{KBrO}_{3} \): Each \( \mathrm{Br} \) changes from 0 to +5, n-factor = 5 (for one molecule with three oxygens, i.e., 5 per equation).- \( \mathrm{Br}_{2} \): Combines both changes, calculate n-factor in terms of least common multiple, resulting in \(5/6\) as average.

Step 4: Match Compounds with n-Factors

Using the calculated n-factors: - \( \mathrm{KBr} \) matches with \( t. 2 \) since its n-factor is 2.- \( \mathrm{KBrO}_{3} \) matches with \( q. 5 \) since its n-factor is essentially 5 in full.- \( \mathrm{Br}_{2} \) matches with \( s. 5/6 \) (average oxidative change for overall reaction).- \( \mathrm{KOH} \) matches with \( p. 1 \) since it does not change in oxidation state, so by default acts as per stoichiometry.

Key Concepts

Understanding Oxidation States

Oxidation states are a useful way of keeping track of electrons in chemical reactions, especially those involving redox processes. The oxidation state indicates the degree of oxidation or reduction of an atom in a compound. By convention, the oxidation state of a free element is always 0. For instance, in the molecule \( \mathrm{Br}_2 \), bromine has an oxidation state of 0.
In compounds, the oxidation state is determined by hypothetical charges assigned based on specific rules such as electronegativity. For example, in \( \mathrm{KBr} \), bromine carries an oxidation state of \(-1\), while in \( \mathrm{KBrO}_3 \), bromine typically has an oxidation state of \(+5\). By understanding these states, we can analyze how electrons are transferred in reactions, deducing which atoms are oxidized and which are reduced.
Mastering the calculation of oxidation states is crucial, as changes in these states reveal the electron transfer occurring during redox reactions.

Decoding Half-Reactions

Redox reactions involve both reduction and oxidation processes happening simultaneously, and can be split into what are known as half-reactions. This decomposition helps in tracking the flow of electrons, making it easier to balance the reaction.
In our exercise, bromine in \( \mathrm{Br}_2 \) can both gain and lose electrons, depending on its final compound. The reduction half-reaction involves \( \mathrm{Br}_2 \rightarrow 2\mathrm{Br}^{-} \) which illustrates the gain of electrons, thereby reducing bromine from an oxidation state of 0 to \(-1\).
The oxidation half-reaction, \( \mathrm{Br}_2 \rightarrow 2 \mathrm{BrO}_3^{-} \), shows the loss of electrons, where bromine's oxidation state increases from 0 to \(+5\). Splitting the main reaction into these half-reactions clarifies how each bromine atom behaves in the process, offering a clearer depiction of redox changes.
Understanding half-reactions is a fundamental technique in tackling redox equations, enabling simple balancing of the electron flow.

The Importance of n-Factor Calculation

The 'n-factor' in a redox reaction represents the number of electron exchanges per molecule, determining the extent of change in oxidation state. This factor is essential for understanding the stoichiometry of redox reactions.
In our reaction, we calculate the n-factor for each compound:

• For \( \mathrm{KOH} \), as it's not directly involved in electron exchange, the n-factor is 1.
• In \( \mathrm{KBr} \), the transition of bromine from 0 to \(-1\) implies an n-factor of 2.
• For \( \mathrm{KBrO}_3 \), each bromine moves from 0 to \(+5\), achieving an n-factor of 5.

As for \( \mathrm{Br}_2 \), the overall n-factor calculation combines its roles in both reduction and oxidation, rendering an average n-factor of \( \frac{5}{6} \).
The n-factor is invaluable for understanding the proportional balance needed in redox reactions, which is critical for reaction analysis and adjustment.

Conceiving Oxidation and Reduction

Oxidation and reduction are twin processes occurring jointly in redox reactions, characterized by the transfer of electrons. In simple terms:

• Oxidation involves the loss of electrons, usually resulting in an increase in oxidation state.
• Reduction involves the gain of electrons, typically resulting in a decrease in oxidation state.

In our given reaction, bromine in \( \mathrm{Br}_2 \) undergoes both processes. It is reduced to \( \mathrm{Br}^{-} \) in \( \mathrm{KBr} \), and oxidized to \( \mathrm{Br}^{+5} \) in \( \mathrm{KBrO}_3 \).
These processes demonstrate the dynamic interchange of electron density among atoms, pivotal in various chemical and biological processes. Recognizing and balancing oxidation and reduction reactions is fundamental in chemistry, particularly for learning how to balance redox reactions.

Insight into Bromine Reactions

Bromine, as an element, commonly participates in redox reactions given its unique chemistry. These reactions allow bromine to exhibit multiple oxidation states, lending it versatility in forming diverse compounds.
In our example, bromine reacts with \( \mathrm{KOH} \) in two ways: it reduces to form \( \mathrm{KBr} \), and oxidizes to become \( \mathrm{KBrO}_3 \). This showcases its capacity to engage in both steps of redox processes within a single reaction framework.
The study of bromine reactions is significant due to its applicability in many chemical industries, particularly in the formation of brominated chemical products and waste treatment processes.
Understanding how bromine and similar elements participate in redox reactions is crucial for mastering inorganic chemistry, illustrating the broader implications of electron transfer in creating various compounds.