In the reaction X^- + XO_3^- + H^+ longrighta | vedclass

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(B) The given reaction is a comproportionation reaction where $X$ is oxidized from $-1$ to $0$ and reduced from $+5$ to $0$.Step $1$: Write the half-r

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$5 : 1$

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(B) The given reaction is a comproportionation reaction where $X$ is oxidized from $-1$ to $0$ and reduced from $+5$ to $0$.Step $1$: Write the half-reactions:Oxidation: $X^- \longrightarrow X_2 + e^-$Reduction: $XO_3^- + 6H^+ + 5e^- \longrightarrow X_2 + 3H_2O$Step $2$: Balance the electrons by multiplying the oxidation half-reaction by $5$:$5X^- \longrightarrow 2.5X_2 + 5e^-$Step $3$: Add the two half-reactions:$5X^- + XO_3^- + 6H^+ \longrightarrow 3X_2 + 3H_2O$From the balanced equation,the molar ratio of $X^-$ to $XO_3^-$ is $5 : 1$.

In the reaction $X^- + XO_3^- + H^+ \longrightarrow X_2 + H_2O$,the molar ratio in which $X^-$ and $XO_3^-$ react is:

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