Question

In which of the following reactions of H$_2$O$_2$, dioxygen is evolved? (only = O$_2$ evolved)
I) With HOCl in acidic medium
II) With permanganate in acidic medium
III) With permanganate in basic medium

• A. I& II only
• B. I, II& III
• C. II& III only
• D. I& III only

Hint:

When \( \text{H}_2\text{O}_2 \) acts as a reducing agent, it is often oxidized to \( \text{O}_2 \). Balance the redox reactions in the given medium (acidic or basic) to check if \( \text{O}_2 \) appears as a product.

Answer 1

The Correct Option is B

Step 1: Analyze Reaction I - H$_2$O$_2$ with HOCl in Acidic Medium
\( \text{H}_2\text{O}_2 \) can act as a reducing agent or oxidizing agent. \( \text{HOCl} \) (hypochlorous acid) is an oxidizing agent because chlorine is in the +1 oxidation state and can be reduced to \( \text{Cl}^- \) (–1). Let’s write the possible reaction:
\( \text{HOCl} \) is reduced:
\[ \text{HOCl} + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{Cl}^- + \text{H}_2\text{O} \] \( \text{H}_2\text{O}_2 \) is oxidized to \( \text{O}_2 \):
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^- \] Combine the half-reactions:
\[ \text{H}_2\text{O}_2 + \text{HOCl} \rightarrow \text{O}_2 + \text{Cl}^- + \text{H}_2\text{O} \] Balance atoms:
Left: 3H, 3O, 1Cl
Right: 2H, 3O, 1Cl
Add 1 \( \text{H}^+ \) on the right to balance hydrogen:
\[ \text{H}_2\text{O}_2 + \text{HOCl} \rightarrow \text{O}_2 + \text{Cl}^- + \text{H}_2\text{O} + \text{H}^+ \] This reaction evolves \( \text{O}_2 \), so dioxygen is produced in Reaction I. Step 2: Analyze Reaction II - H$_2$O$_2$ with Permanganate in Acidic Medium
In acidic medium, \( \text{MnO}_4^- \) is a strong oxidizing agent, reduced to \( \text{Mn}^{2+} \):
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \] \( \text{H}_2\text{O}_2 \) is oxidized to \( \text{O}_2 \):
\[ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^- \] Balance electrons (LCM of 5 and 2 is 10):
Multiply the reduction by 2:
\[ 2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \] Multiply the oxidation by 5:
\[ 5\text{H}_2\text{O}_2 \rightarrow 5\text{O}_2 + 10\text{H}^+ + 10\text{e}^- \] Combine:
\[ 2\text{MnO}_4^- + 16\text{H}^+ + 5\text{H}_2\text{O}_2 \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{O}_2 + 10\text{H}^+ \] Simplify:
\[ 2\text{MnO}_4^- + 6\text{H}^+ + 5\text{H}_2\text{O}_2 \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{O}_2 \] This reaction evolves \( \text{O}_2 \), so dioxygen is produced in Reaction II. Step 3: Analyze Reaction III - H$_2$O$_2$ with Permanganate in Basic Medium
In basic medium, \( \text{MnO}_4^- \) is typically reduced to \( \text{MnO}_2 \):
\[ \text{MnO}_4^- + 2\text{H}_2\text{O} + 3\text{e}^- \rightarrow \text{MnO}_2 + 4\text{OH}^- \] \( \text{H}_2\text{O}_2 \) is oxidized to \( \text{O}_2 \):
\[ \text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2\text{e}^- \] Balance electrons (LCM of 3 and 2 is 6):
Multiply the reduction by 2:
\[ 2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6\text{e}^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^- \] Multiply the oxidation by 3:
\[ 3\text{H}_2\text{O}_2 + 6\text{OH}^- \rightarrow 3\text{O}_2 + 6\text{H}_2\text{O} + 6\text{e}^- \] Combine:
\[ 2\text{MnO}_4^- + 4\text{H}_2\text{O} + 3\text{H}_2\text{O}_2 + 6\text{OH}^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^- + 3\text{O}_2 + 6\text{H}_2\text{O} \] Simplify:
\[ 2\text{MnO}_4^- + 3\text{H}_2\text{O}_2 \rightarrow 2\text{MnO}_2 + 2\text{OH}^- + 3\text{O}_2 + 2\text{H}_2\text{O} \] This reaction also evolves \( \text{O}_2 \), so dioxygen is produced in Reaction III. Step 4: Determine Which Reactions Evolve Dioxygen
Reaction I: \( \text{O}_2 \) is evolved.
Reaction II: \( \text{O}_2 \) is evolved.
Reaction III: \( \text{O}_2 \) is evolved.
All three reactions evolve dioxygen (\( \text{O}_2 \)). Step 5: Analyze Options
Option (1): I& II only. Incorrect, as Reaction III also evolves \( \text{O}_2 \).
Option (2): I, II& III. Correct, as all three reactions evolve \( \text{O}_2 \).
Option (3): I& III only. Incorrect, as Reaction II also evolves \( \text{O}_2 \).
Option (4): I& II only. Incorrect (same as Option 1).