Problem 27
Question
If \(\mathrm{Cl}_{2}\) is passed through hot \(\mathrm{NaOH}\), oxidation number of \(\mathrm{Cl}\) changes from (a) \(-1\) to 0 (b) 0 to \(-1\) (c) 0 to \(+5\) (d) 0 to \(+1\)
Step-by-Step Solution
Verified Answer
The oxidation number of Cl changes from 0 to +5 in hot NaOH.
1Step 1: Understand the Reaction
When \(\mathrm{Cl}_{2}\) is passed through hot \(\mathrm{NaOH}\), it undergoes a disproportionation reaction where part of the chlorine gets reduced and another part gets oxidized. The overall reaction is \[\mathrm{3Cl_{2} + 6NaOH \rightarrow 5NaCl + NaClO_{3} + 3H_{2}O}\,\] where chlorine is both reduced and oxidized.
2Step 2: Assign and Identify Oxidation States
In \(\mathrm{Cl}_{2}\), chlorine has an oxidation state of 0. During the reaction, chlorine in \(\mathrm{NaCl}\) is reduced to \(-1\), and chlorine in \(\mathrm{NaClO_{3}}\) is oxidized to \(+5\).
Key Concepts
Disproportionation ReactionOxidation StatesChlorine Chemistry
Disproportionation Reaction
A disproportionation reaction is a unique type of redox reaction. In these reactions, a single element undergoes both oxidation and reduction. Quite fascinatingly, this involves the same element taking on dual roles in the chemical transformation. This is different from a typical redox reaction where there is a clear exchange of electrons between different elements.
In our exercise example, the element chlorine undergoes disproportionation. When chlorine gas \( \mathrm{Cl}_{2} \) interacts with sodium hydroxide \( \mathrm{NaOH} \), it results in chlorine being both oxidized and reduced. This means chlorine atoms gain electrons in some places while losing them in others.
In our exercise example, the element chlorine undergoes disproportionation. When chlorine gas \( \mathrm{Cl}_{2} \) interacts with sodium hydroxide \( \mathrm{NaOH} \), it results in chlorine being both oxidized and reduced. This means chlorine atoms gain electrons in some places while losing them in others.
- Reduction: \( \mathrm{Cl} \) in \( \mathrm{NaCl} \) changes its oxidation state from 0 to -1.
- Oxidation: \( \mathrm{Cl} \) in \( \mathrm{NaClO_3} \) changes its oxidation state from 0 to +5.
Oxidation States
The concept of oxidation states is essential in understanding redox reactions. An oxidation state is like a bookkeeping number associated with electrons in a compound's atoms, reflecting how many electrons are lost or gained. These states are determined based on a series of rules focused on electronegativity and elemental priority in compounds.
In the context of our example, chlorine undergoes changes in its oxidation state during the reaction with sodium hydroxide. Initially, in chlorine gas \( \mathrm{Cl}_2 \), both chlorine atoms are uncharged with an oxidation state of 0. This serves as the reference point in identifying changes as the reaction proceeds:
In the context of our example, chlorine undergoes changes in its oxidation state during the reaction with sodium hydroxide. Initially, in chlorine gas \( \mathrm{Cl}_2 \), both chlorine atoms are uncharged with an oxidation state of 0. This serves as the reference point in identifying changes as the reaction proceeds:
- Sodium chloride (\( \mathrm{NaCl} \)) sees chlorine's oxidation state change to -1. This indicates a gain of one electron, or reduction.
- Sodium chlorate (\( \mathrm{NaClO_3} \)) involves chlorine being oxidized to an oxidation state of +5, meaning a loss of five electrons.
Chlorine Chemistry
Chlorine is an incredibly versatile chemical element, playing roles in many reactions. Its reactivity comes from its ability to easily gain or lose electrons, which is why it often participates in redox reactions.
Chlorine exists mostly as diatomic molecules (\( \mathrm{Cl}_2 \)) and can take several oxidation states. This variability lets it interact with a vast range of other elements and compounds. In the reaction with sodium hydroxide, we see chlorine's ability to form both \( \mathrm{NaCl} \) and \( \mathrm{NaClO_3} \), demonstrating its adaptability.
Chlorine exists mostly as diatomic molecules (\( \mathrm{Cl}_2 \)) and can take several oxidation states. This variability lets it interact with a vast range of other elements and compounds. In the reaction with sodium hydroxide, we see chlorine's ability to form both \( \mathrm{NaCl} \) and \( \mathrm{NaClO_3} \), demonstrating its adaptability.
- Sodium chloride (\( \mathrm{NaCl} \)) is commonly known as table salt, representing chlorine in a more reduced form.
- Sodium chlorate (\( \mathrm{NaClO_3} \)) features chlorine in a highly oxidized form often used in bleaching and as an herbicide.
Other exercises in this chapter
Problem 25
Oxidation numbers of carbon in \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}, \mathrm{CH}_{4}\) and diamond respectively are (a) \(+3,4\) and \(+4\) (b) \(+3,-
View solution Problem 26
Which of the following does not represent redox reactions? (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{CrO}_{4}^{2-}+\ma
View solution Problem 28
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{I}_{2}\) Equiv
View solution Problem 29
The cell constant of a solution, whose specific conductance and observed conductance are same, is equal to (a) 1 (b) 0 (c) 10 (d) 100
View solution