Observe the following reaction:2 KClO_{3(s)} | vedclass

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(B) Assign oxidation states to each element:In $KClO_3$,the oxidation state of $K$ is $+1$,$Cl$ is $+5$,and $O$ is $-2$.In $KCl$,the oxidation state o

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$Cl$ is reduced and $O$ is oxidized

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(B) Assign oxidation states to each element:In $KClO_3$,the oxidation state of $K$ is $+1$,$Cl$ is $+5$,and $O$ is $-2$.In $KCl$,the oxidation state of $K$ is $+1$ and $Cl$ is $-1$.In $O_2$,the oxidation state of $O$ is $0$.Comparing the oxidation states:$Cl$ changes from $+5$ to $-1$ (decrease in oxidation state,so $Cl$ is reduced).$O$ changes from $-2$ to $0$ (increase in oxidation state,so $O$ is oxidized).Therefore,$Cl$ is reduced and $O$ is oxidized.

Observe the following reaction:
$2 KClO_{3(s)} \xrightarrow{\Delta} 2 KCl_{(s)} + 3 O_{2(g)}$
In this reaction:

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Observe the following reaction:$2 KClO_{3(s)} \xrightarrow{\Delta} 2 KCl_{(s)} + 3 O_{2(g)}$In this reaction: