For the reaction I^{-} + ClO_3^{-} + H_2SO_4 | vedclass

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(C) The oxidation half-reaction is: $2I^{-} 	o I_2 + 2e^{-}$ (multiplied by $3$ to balance electrons: $6I^{-} 	o 3I_2 + 6e^{-}$). The reduction half

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Stoichiometric coefficient of $HSO_4^{-}$ is $6$

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(C) The oxidation half-reaction is: $2I^{-} 	o I_2 + 2e^{-}$ (multiplied by $3$ to balance electrons: $6I^{-} 	o 3I_2 + 6e^{-}$). The reduction half-reaction is: $ClO_3^{-} + 6H^{+} + 6e^{-} 	o Cl^{-} + 3H_2O$. Adding these,we get: $6I^{-} + ClO_3^{-} + 6H^{+} 	o 3I_2 + Cl^{-} + 3H_2O$. Since $H_2SO_4$ provides $H^{+}$,we add $6HSO_4^{-}$ to both sides: $6I^{-} + ClO_3^{-} + 6H_2SO_4 	o 3I_2 + Cl^{-} + 3H_2O + 6HSO_4^{-}$. Comparing the coefficients,the stoichiometric coefficient of $HSO_4^{-}$ is $6$.

For the reaction $I^{-} + ClO_3^{-} + H_2SO_4 \to Cl^{-} + HSO_4^{-} + I_2$ (no free $H^{+}$ is present),select the correct statement for the balanced equation (coefficients in smallest whole number ratio).

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