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(B) To balance the redox reaction,we look at the change in oxidation states:$Mn^{2+} \rightarrow MnO_2$: $Mn$ changes from $+2$ to $+4$ (oxidation,los

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$x=2$

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(B) To balance the redox reaction,we look at the change in oxidation states:$Mn^{2+} \rightarrow MnO_2$: $Mn$ changes from $+2$ to $+4$ (oxidation,loss of $2e^-$).$ClO_3^- \rightarrow ClO_2$: $Cl$ changes from $+5$ to $+4$ (reduction,gain of $1e^-$).To balance the electrons,we multiply the reduction half-reaction by $2$:$Mn^{2+} + 2 ClO_3^- \rightarrow MnO_2 + 2 ClO_2$.Comparing this to the given equation $Mn^{2+} + x ClO_3^- \rightarrow MnO_2 + x ClO_2$,we find $x=2$.

What is the value of $x$ in order to balance the following redox reaction?
$Mn_{(aq)}^{2+} + x ClO_{3_{(aq)}}^{-} \rightarrow MnO_{2_{(s)}} + x ClO_{2_{(aq)}}$

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What is the value of $x$ in order to balance the following redox reaction?$Mn_{(aq)}^{2+} + x ClO_{3_{(aq)}}^{-} \rightarrow MnO_{2_{(s)}} + x ClO_{2_{(aq)}}$