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(C) The given reaction is a disproportionation reaction where $Cl_2$ is both oxidized and reduced.$Cl_2^0 	o ClO_3^- (+5)$ (Oxidation,loss of $5$ ele

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$x = 3, y = 5$

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(C) The given reaction is a disproportionation reaction where $Cl_2$ is both oxidized and reduced.$Cl_2^0 	o ClO_3^- (+5)$ (Oxidation,loss of $5$ electrons per $Cl$ atom)$Cl_2^0 	o Cl^- (-1)$ (Reduction,gain of $1$ electron per $Cl$ atom)To balance the electrons,we multiply the reduction half-reaction by $5$ and the oxidation half-reaction by $1$:$1Cl_2 + 5Cl_2 + 12OH^- 	o 2ClO_3^- + 10Cl^- + 6H_2O$Simplifying the equation by dividing by $2$:$3Cl_2 + 6OH^- 	o ClO_3^- + 5Cl^- + 3H_2O$Comparing this with the given equation $xCl_2 + 6OH^- 	o ClO_3^- + yCl^- + 3H_2O$,we get $x = 3$ and $y = 5$.

Find the values of $x$ and $y$ for the following balanced equation:
$xCl_2 + 6OH^- \to ClO_3^- + yCl^- + 3H_2O$

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Find the values of $x$ and $y$ for the following balanced equation:$xCl_2 + 6OH^- \to ClO_3^- + yCl^- + 3H_2O$