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(B) To balance the redox reaction,we determine the change in oxidation states:$1$. Oxidation: $O$ in $H_2O_2$ (oxidation state $-1$) is oxidized to $O

Vedclass · Accepted Answer

$x= 2, y =5$

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(B) To balance the redox reaction,we determine the change in oxidation states:$1$. Oxidation: $O$ in $H_2O_2$ (oxidation state $-1$) is oxidized to $O_2$ (oxidation state $0$). The increase in oxidation state per oxygen atom is $1$. Since there are two oxygen atoms in $H_2O_2$,the total increase is $2$.$2$. Reduction: $Cl$ in $ClO_2$ (oxidation state $+4$) is reduced to $Cl^-$ (oxidation state $-1$). The decrease in oxidation state is $5$.$3$. To balance the electrons,we multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$.$4$. This gives $x = 2$ and $y = 5$ in the balanced equation: $5H_2O_2 + 2ClO_2 + 2OH^- 	o 2Cl^- + 5O_2 + 6H_2O$.

Find the values of $x$ and $y$ for the following balanced equation:
$5H_2O_2 + xClO_2 + 2OH^- \to xCl^- + yO_2 + 6H_2O$

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Find the values of $x$ and $y$ for the following balanced equation:$5H_2O_2 + xClO_2 + 2OH^- \to xCl^- + yO_2 + 6H_2O$