What will be the value of x and y by balancin | vedclass

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(A) To balance the given redox reaction:$C_2H_5OH + xI_2 + yOH^{-} 	o CHI_3 + HCO_2^- + 5I^{-} + 5H_2O$$1$. Oxidation half-reaction:$C_2H_5OH + 5OH^{

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$4, 6$

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(A) To balance the given redox reaction:$C_2H_5OH + xI_2 + yOH^{-} 	o CHI_3 + HCO_2^- + 5I^{-} + 5H_2O$$1$. Oxidation half-reaction:$C_2H_5OH + 5OH^{-} 	o HCO_2^- + 4H_2O + 4e^-$$2$. Reduction half-reaction:$I_2 + 2e^- 	o 2I^-$To balance electrons,multiply the reduction half-reaction by $2$:$2I_2 + 4e^- 	o 4I^-$Adding the two half-reactions:$C_2H_5OH + 2I_2 + 5OH^{-} 	o CHI_3 + HCO_2^- + 4I^- + 4H_2O$Wait,checking the stoichiometry for $I$ atoms:Reactants: $2I_2$ ($4I$ atoms) + $5I^-$ (from the provided equation structure).Balancing the full equation:$C_2H_5OH + 4I_2 + 6OH^{-} 	o CHI_3 + HCO_2^- + 5I^{-} + 5H_2O$Comparing with the given equation,$x = 4$ and $y = 6$.

What will be the value of $x$ and $y$ by balancing the reaction
$C_2H_5OH + xI_2 + yOH^{-} \to CHI_3 + HCO_2^- + 5I^{-} + 5H_2O$

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What will be the value of $x$ and $y$ by balancing the reaction$C_2H_5OH + xI_2 + yOH^{-} \to CHI_3 + HCO_2^- + 5I^{-} + 5H_2O$