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(B) The reduction reaction is: $Fe^{3+} + e^- ightarrow Fe^{2+}$.Number of moles of $Fe_2(SO_4)_3 = 	ext{Molarity} 	imes 	ext{Volume in Liters} =

Vedclass · Accepted Answer

$772$

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(B) The reduction reaction is: $Fe^{3+} + e^- ightarrow Fe^{2+}$.Number of moles of $Fe_2(SO_4)_3 = 	ext{Molarity} 	imes 	ext{Volume in Liters} = 0.1 	imes 0.080 = 0.008 \ mol$.Since $1 \ mol$ of $Fe_2(SO_4)_3$ contains $2 \ mol$ of $Fe^{3+}$,total moles of $Fe^{3+} = 2 	imes 0.008 = 0.016 \ mol$.To reduce $0.016 \ mol$ of $Fe^{3+}$ to $Fe^{2+}$,we need $0.016 \ mol$ of electrons.Using Faraday's law: $Q = n 	imes F = 0.016 	imes 96500 \ C$.Given current $I = 2 \ A$,time $t = \frac{Q}{I} = \frac{0.016 	imes 96500}{2} = 0.008 	imes 96500 = 772 \ s$.

How long would it take to reduce completely $80 \ mL$ of $0.1 \ M$ $Fe_2(SO_4)_3$ to $FeSO_4$ by passing a current of $2 \ A$?

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