Potassium chlorate is prepared by the electro | vedclass

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Question

(A) The balanced half-reaction is: $6 OH^{-} + Cl^{-} \rightarrow ClO_{3}^{-} + 3 H_{2}O + 6 e^{-}$.Number of moles of $KClO_{3}$ produced = $\frac{10

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) The balanced half-reaction is: $6 OH^{-} + Cl^{-} ightarrow ClO_{3}^{-} + 3 H_{2}O + 6 e^{-}$.Number of moles of $KClO_{3}$ produced = $\frac{10 \ g}{122 \ g \ mol^{-1}} = 0.08197 \ mol$.From the reaction,$1 \ mol$ of $KClO_{3}$ requires $6 \ mol$ of electrons. Therefore,total charge $Q$ required = $n 	imes z 	imes F = 0.08197 	imes 6 	imes 96500 \ C = 47462.3 \ C$.Given that only $60\%$ of the current is utilized,the effective current $I_{eff} = 2 \ A 	imes 0.60 = 1.2 \ A$.Using the formula $Q = I_{eff} 	imes t$,we have $t = \frac{Q}{I_{eff}} = \frac{47462.3 \ C}{1.2 \ A} = 39551.9 \ s$.Converting to hours: $t = \frac{39551.9}{3600} \approx 10.98 \ hr$.Rounding to the nearest hour,we get $11 \ hr$.

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