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(D) The given reaction is: $x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$Step $1$: Write the half-reactions.Reduction half-reaction: $C

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$2$

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(D) The given reaction is: $x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$Step $1$: Write the half-reactions.Reduction half-reaction: $ClO_4^{-} \rightarrow ClO_2^{-}$Balancing oxygen and charge: $ClO_4^{-} + 4 H^{+} + 4 e^{-} \rightarrow ClO_2^{-} + 2 H_2O$Step $2$: Oxidation half-reaction.$H_2O_2 \rightarrow O_2 + 2 H^{+} + 2 e^{-}$Step $3$: Equalize the number of electrons.Multiply the oxidation half-reaction by $2$ to match the $4 e^{-}$ in the reduction half-reaction:$2 H_2O_2 \rightarrow 2 O_2 + 4 H^{+} + 4 e^{-}$Step $4$: Add the two half-reactions.$(ClO_4^{-} + 4 H^{+} + 4 e^{-}) + (2 H_2O_2) \rightarrow (ClO_2^{-} + 2 H_2O) + (2 O_2 + 4 H^{+} + 4 e^{-})$Simplifying,we get:$2 H_2O_2 + ClO_4^{-} \rightarrow 2 O_2 + ClO_2^{-} + 2 H_2O$Comparing this with the given equation $x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$,we find $x = 2$.

What is the value of $x$ in order to balance the following redox reaction by the ion-electron method?
$x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$

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What is the value of $x$ in order to balance the following redox reaction by the ion-electron method?$x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$