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 \(\text{Cl}\) changes from 0 to \(+5\) when \(\text{Cl}_2\) is passed through hot \(\text{NaOH}\), so the answer is (c).
1Step 1: Identify the Reaction
When chlorine gas \(\mathrm{Cl}_{2}\) is passed through hot \(\mathrm{NaOH}\), a disproportionation reaction occurs. The reaction is: \[\mathrm{3Cl}_{2} + 6\mathrm{NaOH} \rightarrow \mathrm{5NaCl} + \mathrm{NaClO}_{3} + 3\mathrm{H}_{2}\mathrm{O}\] In this reaction, chlorine is both oxidized and reduced.
2Step 2: Determine initial Oxidation State of Chlorine
In \(\mathrm{Cl}_{2}\), the oxidation state of chlorine is 0 because it is a diatomic molecule of the same element and is electrically neutral.
3Step 3: Determine Reduced Form of Chlorine
Chlorine is reduced to chloride ions \(\mathrm{Cl}^{-}\) in \(\mathrm{NaCl}\). Here, the oxidation state of chlorine is \(-1\).
4Step 4: Determine Oxidized Form of Chlorine
Chlorine is oxidized to chlorate ions in \(\mathrm{NaClO}_{3}\). In this compound, the oxidation state of chlorine is \(+5\).
5Step 5: Conclude the Change in Oxidation State
The initial oxidation state of chlorine is 0. It changes to \(-1\) when reduced, and changes to \(+5\) when oxidized. Hence, chlorine undergoes a change in oxidation state from 0 to \(+5\).
Key Concepts
Understanding Oxidation StatesThe Dynamics of Chlorine ReactionsIntroduction to Redox Reactions
Understanding Oxidation States
Oxidation states, also referred to as oxidation numbers, are a concept used in chemistry to describe the degree of oxidation of an atom in a compound. They help track electron transfer in reactions. An atom's oxidation state is generally determined using the rules based on electronegativity and standard oxidation numbers. For chlorine in its diatomic form, each chlorine atom has an oxidation state of 0 because it shares electrons equally between the atoms. Note that in compounds with other elements, chlorine usually takes an oxidation state that reflects its tendency to gain one electron to form a chloride ion (\(\text{Cl}^{-}\)), which has an oxidation state of \(-1\).
- The oxidation state can increase (indicating loss of electrons or oxidation),
- or it can decrease (indicating gain of electrons or reduction).
The Dynamics of Chlorine Reactions
Chlorine reactions cover a fascinating area of chemistry, especially when chlorine interacts with alkalis like sodium hydroxide (\(\text{NaOH}\)). The reaction between chlorine gas (\(\text{Cl}_2\)) and hot \(\text{NaOH}\) results in a disproportionation reaction. In this type of reaction, the same element undergoes simultaneous oxidation and reduction.
- Chlorine is reduced to \(\text{NaCl}\), where it is in the \(-1\) oxidation state.
- It is also oxidized to \(\text{NaClO}_3\), where the oxidation state rises to \(+5\).
Introduction to Redox Reactions
Redox reactions are fundamental to understanding chemical changes, as they involve the transfer of electrons between substances. The term "redox" is a combination of "reduction" and "oxidation," the two processes that occur simultaneously in these reactions.
- Oxidation refers to the loss of electrons, often resulting in an increase in oxidation state.
- Reduction describes the gain of electrons, leading to a decrease in oxidation state.
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
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