Problem 57 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction, \(3 \mathrm{Br}_{2}+6 \mathrm{CO}_{3}^{2-}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow 5 \mathrm{Br}+\mathrm{BrO}_{3}^{-}+6 \mathrm{HCO}_{3}\) (a) bromine is oxidized and carbonate is reduced (b) bromine is reduced and water is oxidized (c) bromine is neither reduced nor oxidized (d) bromine is both reduced and oxidized

Step-by-Step Solution

Answer

(d) bromine is both reduced and oxidized

Step 1: Understanding Redox Reactions

In a redox reaction, oxidation and reduction occur simultaneously. Something is oxidized when it loses electrons, and something is reduced when it gains electrons.

Step 2: Determining the Oxidation States

Assign oxidation states to the elements involved in the reaction. Bromine in \( \mathrm{Br}_2 \) is 0, in \( \mathrm{Br} \) is -1, and in \( \mathrm{BrO}_3^- \) is +5. Carbon in \( \mathrm{CO}_3^{2-} \) and \( \mathrm{HCO}_3^- \) maintains a +4 oxidation state throughout the reaction.

Step 3: Identifying Changes in Oxidation States

Check the changes in oxidation states for elements to identify what's reduced and what's oxidized. Bromine switches from 0 in \( \mathrm{Br}_2 \) to -1 in \( \mathrm{Br} \) (a reduction) and from 0 to +5 in \( \mathrm{BrO}_3^- \) (an oxidation). The carbon atoms in carbonate and bicarbonate retain the same oxidation states.

Step 4: Conclusion Based on Oxidation and Reduction

Since bromine is both reduced (0 to -1) and oxidized (0 to +5), the option is (d) bromine is both reduced and oxidized.

Key Concepts

Oxidation States

Oxidation states, also known as oxidation numbers, help us keep track of where the electrons are moving in a chemical reaction. It's like a bookkeeping method for electrons. In chemistry, especially in redox reactions, tracking these changes is crucial.

Each element in a compound is assigned an oxidation state, representing the number of electrons an atom gains, loses, or appears to use in bonding. For instance, bromine in the molecular form \( \mathrm{Br}_2 \) has an oxidation state of 0. Why? Because each bromine atom shares electrons equally with its pair, not gaining or losing any.

When atoms form ions or participate in bonds with varying electronegativities, their oxidation states reflect the effective number of electrons transferred. For example, in \( \mathrm{BrO}_3^- \), the bromine has an oxidation state of +5, showing it has "lost" five electrons compared to its elemental form. Calculating oxidation states involves several rules, such as:

• Atoms in their elemental form have an oxidation state of 0.
• For monoatomic ions, the oxidation state equals the ion's charge.
• The sum of oxidation states for all atoms in a neutral compound must be zero.
• The sum in a polyatomic ion must equal the ion’s charge.

Understanding these rules helps us determine which elements are oxidized and which are reduced in a reaction.

Bromine Oxidation

Bromine plays a central role in various redox reactions due to its ability to change oxidation states. In the given reaction, bromine undergoes both oxidation and reduction. This phenomenon is known as disproportionation and happens when an element is both oxidized and reduced in the same reaction.

Breaking it down:

• Initially, bromine is in the 0 oxidation state in \( \mathrm{Br}_2 \).


• It is reduced to -1 when forming \( \mathrm{Br}^- \), by gaining an electron.


• It is oxidized to +5 to form \( \mathrm{BrO}_3^- \), by losing electrons.

This dual behavior is interesting because it showcases bromine's versatility in redox reactions. By identifying how its oxidation state changes, we can see that bromine ends up both donating and accepting electrons. This dual change is useful in various chemical processes and indicates bromine’s role as a "mediator" that can stabilize multiple oxidation states, aiding in balancing the reaction.

Electron Transfer

Electron transfer is the heart of redox reactions, dictating the course of oxidation and reduction processes. Every redox reaction entails an electron moving from one atom or molecule to another, this movement defines which substance acts as the oxidizing agent and which is the reducing agent.

In the reaction you're examining, electron transfer is evident by the change in bromine's oxidation state:

• Bromine molecules \( \mathrm{Br}_2 \), originally having no charge, split, with some atoms gaining electrons to become \( \mathrm{Br}^- \). They are reduced, implying they are the oxidizing agents.


• Conversely, some bromine atoms lose electrons to form \( \mathrm{BrO}_3^- \), implying these bromine atoms are oxidized and thus act as reducing agents.

When you think about electron transfer, it's like a game of hot potato with electrons being passed around. The one who gives away the "potato" is oxidized (loses electrons), while the one who receives it is reduced (gains electrons). Understanding this dance helps explain the balancing act of charges in a redox reaction and is key for predicting the behavior of different substances in a reaction.