In an alkaline medium,ClO_2 oxidizes H_2O_2 t | vedclass

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Question

(C) The reduction half-reaction is: $ClO_2 + 2H_2O + 5e^- 	o Cl^- + 4OH^-$The oxidation half-reaction is: $H_2O_2 + 2OH^- 	o O_2 + 2H_2O + 2e^-$To b

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$2.5$

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(C) The reduction half-reaction is: $ClO_2 + 2H_2O + 5e^- 	o Cl^- + 4OH^-$The oxidation half-reaction is: $H_2O_2 + 2OH^- 	o O_2 + 2H_2O + 2e^-$To balance the electrons,multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:$2ClO_2 + 4H_2O + 10e^- 	o 2Cl^- + 8OH^-$$5H_2O_2 + 10OH^- 	o 5O_2 + 10H_2O + 10e^-$Adding these gives the balanced equation:$2ClO_2 + 5H_2O_2 + 2OH^- 	o 2Cl^- + 5O_2 + 6H_2O$From the stoichiometry,$2$ moles of $ClO_2$ react with $5$ moles of $H_2O_2$.Therefore,$1$ mole of $ClO_2$ reacts with $2.5$ moles of $H_2O_2$.

In an alkaline medium,$ClO_2$ oxidizes $H_2O_2$ to $O_2$ and is reduced to $Cl^-$. How many moles of $H_2O_2$ will be oxidized by $1$ mole of $ClO_2$?

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