Question:
What is the value of 'x' in order to balance the following redox reaction by ion electron method?
$$ x\text{H}_2\text{O}_2 + \text{ClO}_4^- \rightarrow x\text{O}_2 + \text{ClO}_2^- + 2\text{H}_2\text{O} $$
What is the value of 'x' in order to balance the following redox reaction by ion electron method?
$$ x\text{H}_2\text{O}_2 + \text{ClO}_4^- \rightarrow x\text{O}_2 + \text{ClO}_2^- + 2\text{H}_2\text{O} $$
$$ x\text{H}_2\text{O}_2 + \text{ClO}_4^- \rightarrow x\text{O}_2 + \text{ClO}_2^- + 2\text{H}_2\text{O} $$
Show Hint
A quick shortcut is balancing by atom counting: look closely at the hydrogen count. The right side has $2\text{H}_2\text{O}$ ($4$ hydrogens total). To balance this on the left, $\text{H}_2\text{O}_2$ must have a coefficient of $x = 2$, which automatically balances the oxygen atom pool.
Updated On: Jun 4, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The problem requires balancing a redox reaction to find the stoichiometric coefficient $x$ accompanying the hydrogen peroxide ($\text{H}_2\text{O}_2$) and oxygen gas ($\text{O}_2$).
Step 2: Detailed Explanation:
Let us balance the equation using the half-reaction method in an acidic/neutral framework:
1.
Oxidation Half-Reaction ($\text{H}_2\text{O}_2 \rightarrow \text{O}_2$): Oxygen changes its oxidation state from $-1$ in peroxide to $0$ in molecular oxygen. Balancing atoms and charges:
$$ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^- \quad \text{--- (Equation 1)} $$ 2.
Reduction Half-Reaction ($\text{ClO}_4^- \rightarrow \text{ClO}_2^-$): Chlorine drops from an oxidation state of $+7$ to $+3$ (a gain of $4\text{e}^-$). Balancing oxygens with water and hydrogens with $\text{H}^+$:
$$ \text{ClO}_4^- + 4\text{H}^+ + 4\text{e}^- \rightarrow \text{ClO}_2^- + 2\text{H}_2\text{O} \quad \text{--- (Equation 2)} $$ 3.
Equalizing Electron Flow: Multiply Equation 1 by 2 so that it releases $4\text{e}^-$, balancing Equation 2:
$$ 2\text{H}_2\text{O}_2 \rightarrow 2\text{O}_2 + 4\text{H}^+ + 4\text{e}^- $$ 4.
Combining the Half-Reactions: Adding the balanced expressions together cancels out the $4\text{e}^-$ and $4\text{H}^+$ ions perfectly:
$$ 2\text{H}_2\text{O}_2 + \text{ClO}_4^- \rightarrow 2\text{O}_2 + \text{ClO}_2^- + 2\text{H}_2\text{O} $$ Comparing this balanced result to the question's format yields $x = 2$.
Step 3: Final Answer: The correct coefficient value is 2, matching option (D).
The problem requires balancing a redox reaction to find the stoichiometric coefficient $x$ accompanying the hydrogen peroxide ($\text{H}_2\text{O}_2$) and oxygen gas ($\text{O}_2$).
Step 2: Detailed Explanation:
Let us balance the equation using the half-reaction method in an acidic/neutral framework:
1.
Oxidation Half-Reaction ($\text{H}_2\text{O}_2 \rightarrow \text{O}_2$): Oxygen changes its oxidation state from $-1$ in peroxide to $0$ in molecular oxygen. Balancing atoms and charges:
$$ \text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^- \quad \text{--- (Equation 1)} $$ 2.
Reduction Half-Reaction ($\text{ClO}_4^- \rightarrow \text{ClO}_2^-$): Chlorine drops from an oxidation state of $+7$ to $+3$ (a gain of $4\text{e}^-$). Balancing oxygens with water and hydrogens with $\text{H}^+$:
$$ \text{ClO}_4^- + 4\text{H}^+ + 4\text{e}^- \rightarrow \text{ClO}_2^- + 2\text{H}_2\text{O} \quad \text{--- (Equation 2)} $$ 3.
Equalizing Electron Flow: Multiply Equation 1 by 2 so that it releases $4\text{e}^-$, balancing Equation 2:
$$ 2\text{H}_2\text{O}_2 \rightarrow 2\text{O}_2 + 4\text{H}^+ + 4\text{e}^- $$ 4.
Combining the Half-Reactions: Adding the balanced expressions together cancels out the $4\text{e}^-$ and $4\text{H}^+$ ions perfectly:
$$ 2\text{H}_2\text{O}_2 + \text{ClO}_4^- \rightarrow 2\text{O}_2 + \text{ClO}_2^- + 2\text{H}_2\text{O} $$ Comparing this balanced result to the question's format yields $x = 2$.
Step 3: Final Answer: The correct coefficient value is 2, matching option (D).
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