Problem 160 - A Complete Resource Book in Chemistry for JEE Main

Question

Consider the reaction, \(3 \mathrm{Br}_{2}+6 \mathrm{NaOH} \longrightarrow 5 \mathrm{NaBr}+\) \(\mathrm{NaBrO}_{3}+3 \mathrm{H}_{2} \mathrm{O} .\) If \(\mathrm{E}\) and \(\mathrm{A}\) represent equivalent weight and atomic weight of bromine, then \(5 \mathrm{E}=\mathrm{x} \mathrm{A}\). The value of \(x\) is ____ .

Step-by-Step Solution

Answer

The value of \( x \) is 2.5.

Step 1: Understand the Reaction

The chemical reaction provided is \( 3 \mathrm{Br}_{2} + 6 \mathrm{NaOH} \rightarrow 5 \mathrm{NaBr} + \mathrm{NaBrO}_{3} + 3 \mathrm{H}_{2} \mathrm{O} \). In this reaction, bromine (\( \mathrm{Br}_2 \)) undergoes a redox process. Bromine is both reduced to \( \mathrm{NaBr} \) and oxidized to \( \mathrm{NaBrO}_{3} \).

Step 2: Assign Oxidation Numbers

Determine the oxidation numbers for bromine in the reaction. In \( \mathrm{Br}_2 \), bromine is in the 0 oxidation state. In \( \mathrm{NaBr} \), bromine is in the -1 oxidation state, and in \( \mathrm{NaBrO}_{3} \), bromine is in the +5 oxidation state.

Step 3: Calculate Change in Oxidation State

For the reduction, \( \mathrm{Br}_2 \) goes from 0 to -1, a change of -1 per bromine atom. For oxidation, \( \mathrm{Br}_2 \) goes from 0 to +5, a change of +5 per bromine atom.

Step 4: Account for Moles in Reaction

The reaction shows that 3 moles of \( \mathrm{Br}_2 \) are used. Out of these, 5 bromine atoms are reduced and 2 are oxidized. Thus, 5 moles of \( \mathrm{Br}^- \) ions are produced and for oxidation, 1 mole of \( \mathrm{BrO}_3^- \) ions is produced from 3 atoms of bromine.

Step 5: Define Equivalent Weight

The equivalent weight \( E \) for bromine in this reaction is based on the amount of substance that can provide or react with one mole of electrons. Since 5 moles of \( \mathrm{Br}^- \) are involved, each providing one electron, the equivalent weight in terms of \( xA \) must match the change in moles as related to the electrons.

Step 6: Compute \( x \) in \( 5E = xA \)

Since 5 moles of bromine atoms give -1 charge each (total 5 electrons transferred), while 1 mole of \( \mathrm{Br}_2 \) contains 2 bromine atoms (atomic weight \( A \)), then for bromine, 1 mole of gained or lost electrons corresponds to 2 atoms, thus \( x = \frac{5}{2} = 2.5 \).

Key Concepts

Oxidation States

In the realm of chemistry, understanding oxidation states is key to deciphering redox reactions. Oxidation states, or oxidation numbers, are theoretical charges assigned to atoms in a compound. They help chemists track how electrons are transferred between atoms.For example:

• In its elemental form, like in the molecule \(\mathrm{Br}_2\), bromine has an oxidation state of \(0\).
• In \(\mathrm{NaBr}\), bromine gains an electron becoming \(\mathrm{Br}^-\), resulting in an oxidation state of \(-1\).
• In \(\mathrm{NaBrO}_3\), bromine loses electrons reaching an oxidation state of \(+5\).

By assigning these numbers, you can determine which atoms are oxidized and which are reduced.
Oxidation involves an increase in oxidation state, while reduction involves a decrease. So, bromine is oxidized when it goes from \(0\) to \(+5\) and reduced when it goes from \(0\) to \(-1\).
Understanding these changes helps one to balance redox reactions and calculate electron transfer.

Equivalent Weight

Equivalent weight is a crucial concept in understanding how substances react chemically. It is defined as the mass of a substance that will combine with or displace one mole of hydrogen atoms in a chemical reaction. In the context of bromine in our exercise:

• The equivalent weight \(E\) can be calculated based on electron exchange in the reaction.
• 5 moles of \(\mathrm{Br}^-\) ions result from the reduction, each transferring one electron to the process.
• Thus, the total electron exchange helps calculate \(E\) relative to the atomic weight \(A\) for bromine.

Since each \(\mathrm{Br}\) atom can change the way it interacts with electrons, the equivalence aids in standardizing these interactions, aligning with the atomic weight.
This is essentially what allows chemists to predictably measure and react chemical substances based on their fundamental weights, simplifying many calculations.

Electron Transfer

Electron transfer is the heartbeat of redox reactions. It involves the movement of electrons from one atom or molecule to another. Here's how it plays out in our example:

• In the reaction provided, some bromine atoms (\(\mathrm{Br}_2\)) are oxidized, meaning they lose electrons and transfer them to another substance.
• Meanwhile, other bromine atoms are reduced, meaning they gain electrons.
• Specifically, the reaction shows that 5 bromine atoms are reduced to form \(\mathrm{NaBr}\) with a \(-1\) oxidation state for each bromine.
• Simultaneously, 2 atoms are oxidized to form \(\mathrm{NaBrO}_3\) with a \(+5\) oxidation state.

The total electron gain or loss must balance, meaning all electrons lost by oxidation are gained during reduction.
This balance is fundamental in redox reactions and crucial for stoichiometric calculations, ensuring the reaction complies with conservation laws.