Question

Observe the following reaction: \[ 2KClO_3 (s) \xrightarrow{\Delta} 2KCl (s) + 3O_2 (g) \] In this reaction: 

• A. Cl is oxidized and O is reduced
• B. Cl is reduced and O is oxidized
• C. K is oxidized and O is reduced
• D. K is reduced and Cl is also reduced

Hint:

Oxidation refers to an increase in oxidation state, while reduction refers to a decrease. In this reaction, oxygen is oxidized, and chlorine is reduced, demonstrating a classic redox process.

Answer 1

The Correct Option is B

Step 1: Assign Oxidation Numbers 
In potassium chlorate (\( KClO_3 \)): - Potassium (\( K \)) has an oxidation state of +1. - Oxygen (\( O \)) in oxoanions typically has an oxidation state of -2. - Let the oxidation state of chlorine (\( Cl \)) be \( x \). Using the sum rule for a neutral compound: \[ (+1) + x + 3(-2) = 0 \] \[ x - 6 + 1 = 0 \] \[ x = +5 \] Thus, in \( KClO_3 \), \( Cl \) has an oxidation state of \( +5 \). In potassium chloride (\( KCl \)): - \( K \) is still \( +1 \). - \( Cl \) must be \( -1 \) to balance the charge. Thus, the oxidation state of \( Cl \) changes from \( +5 \) in \( KClO_3 \) to \( -1 \) in \( KCl \), meaning \( Cl \) is reduced. 

Step 2: Identify Oxidation of Oxygen 
In \( KClO_3 \), oxygen is in the \(-2\) oxidation state. In \( O_2 \) gas, oxygen exists in its elemental form, which has an oxidation state of \( 0 \). Since oxygen's oxidation state increases from \(-2\) to \( 0 \), oxygen is oxidized. 

Step 3: Verify the Correct Answer 
- Chlorine is reduced (from \( +5 \) to \( -1 \)). - Oxygen is oxidized (from \( -2 \) to \( 0 \)). Thus, the correct answer is Option (2): Cl is reduced and O is oxidized.