Q51 / 180 · COMEDK 2023 (Evening Shift) · Chemistry
In the given Redox equation, identify the stoichiometric coefficients $\mathrm{w}, \mathrm{x}, \mathrm{y}$ and $\mathrm{z}$.
$\mathrm{ClO}_3^{-1}+\mathrm{w} \mathrm{Cl}^{-1}+\mathrm{xH}^{+} \rightarrow \mathrm{y} \mathrm{H}_2 \mathrm{O}+\mathrm{zCl}_2$
- A$w=6 \quad x=5 \quad y=2.5 \quad z=3$
- Correct answerB$w=5 \quad x=6 \quad y=3 \quad z=3$
- C$w=7 \quad x=6 \quad y=3 \quad z=4$
- D$w=3 \quad x=6 \quad y=3 \quad z=2$
Solution
The given redox reaction is $\mathrm{ClO}_3^{-} + \mathrm{w} \mathrm{Cl}^{-} + \mathrm{x} \mathrm{H}^{+} \rightarrow \mathrm{y} \mathrm{H}_2 \mathrm{O} + \mathrm{z} \mathrm{Cl}_2$.
First, identify the oxidation states of chlorine. In $\mathrm{ClO}_3^{-}$, the oxidation state of Cl is $+5$. In $\mathrm{Cl}^{-}$, it is $-1$. In $\mathrm{Cl}_2$, it is $0$.
The reduction half-reaction is $\mathrm{ClO}_3^{-} + 6 \mathrm{H}^{+} + 5 \mathrm{e}^{-} \rightarrow \dfrac{1}{2} \mathrm{Cl}_2 + 3 \mathrm{H}_2 \mathrm{O}$.
The oxidation half-reaction is $\mathrm{Cl}^{-} \rightarrow \dfrac{1}{2} \mathrm{Cl}_2 + 1 \mathrm{e}^{-}$.
To balance the electrons, multiply the oxidation half-reaction by 5:
$5 \mathrm{Cl}^{-} \rightarrow \dfrac{5}{2} \mathrm{Cl}_2 + 5 \mathrm{e}^{-}$.
Adding the two half-reactions:
$\mathrm{ClO}_3^{-} + 5 \mathrm{Cl}^{-} + 6 \mathrm{H}^{+} \rightarrow \dfrac{1}{2} \mathrm{Cl}_2 + \dfrac{5}{2} \mathrm{Cl}_2 + 3 \mathrm{H}_2 \mathrm{O}$.
Simplifying the equation:
$\mathrm{ClO}_3^{-} + 5 \mathrm{Cl}^{-} + 6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Cl}_2 + 3 \mathrm{H}_2 \mathrm{O}$.
Comparing this with the given equation $\mathrm{ClO}_3^{-} + \mathrm{w} \mathrm{Cl}^{-} + \mathrm{x} \mathrm{H}^{+} \rightarrow \mathrm{y} \mathrm{H}_2 \mathrm{O} + \mathrm{z} \mathrm{Cl}_2$, we get $\mathrm{w} = 5$, $\mathrm{x} = 6$, $\mathrm{y} = 3$, and $\mathrm{z} = 3$.
Answer: $w=5 \quad x=6 \quad y=3 \quad z=3$
First, identify the oxidation states of chlorine. In $\mathrm{ClO}_3^{-}$, the oxidation state of Cl is $+5$. In $\mathrm{Cl}^{-}$, it is $-1$. In $\mathrm{Cl}_2$, it is $0$.
The reduction half-reaction is $\mathrm{ClO}_3^{-} + 6 \mathrm{H}^{+} + 5 \mathrm{e}^{-} \rightarrow \dfrac{1}{2} \mathrm{Cl}_2 + 3 \mathrm{H}_2 \mathrm{O}$.
The oxidation half-reaction is $\mathrm{Cl}^{-} \rightarrow \dfrac{1}{2} \mathrm{Cl}_2 + 1 \mathrm{e}^{-}$.
To balance the electrons, multiply the oxidation half-reaction by 5:
$5 \mathrm{Cl}^{-} \rightarrow \dfrac{5}{2} \mathrm{Cl}_2 + 5 \mathrm{e}^{-}$.
Adding the two half-reactions:
$\mathrm{ClO}_3^{-} + 5 \mathrm{Cl}^{-} + 6 \mathrm{H}^{+} \rightarrow \dfrac{1}{2} \mathrm{Cl}_2 + \dfrac{5}{2} \mathrm{Cl}_2 + 3 \mathrm{H}_2 \mathrm{O}$.
Simplifying the equation:
$\mathrm{ClO}_3^{-} + 5 \mathrm{Cl}^{-} + 6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Cl}_2 + 3 \mathrm{H}_2 \mathrm{O}$.
Comparing this with the given equation $\mathrm{ClO}_3^{-} + \mathrm{w} \mathrm{Cl}^{-} + \mathrm{x} \mathrm{H}^{+} \rightarrow \mathrm{y} \mathrm{H}_2 \mathrm{O} + \mathrm{z} \mathrm{Cl}_2$, we get $\mathrm{w} = 5$, $\mathrm{x} = 6$, $\mathrm{y} = 3$, and $\mathrm{z} = 3$.
Answer: $w=5 \quad x=6 \quad y=3 \quad z=3$
Hence, the correct answer is option: 'B'.