For electroplating,1.5 A current is passed f | vedclass

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(A) Given: Current $I = 1.5 \ A$,Time $t = 250 \ s$,Volume $V = 250 \ mL = 0.25 \ L$,Initial Molarity $M_i = 0.15 \ M$.Initial moles of $MSO_4 = M_i \

Vedclass · Accepted Answer

$0.14$

Vedclass · Answer

(A) Given: Current $I = 1.5 \ A$,Time $t = 250 \ s$,Volume $V = 250 \ mL = 0.25 \ L$,Initial Molarity $M_i = 0.15 \ M$.Initial moles of $MSO_4 = M_i 	imes V = 0.15 	imes 0.25 = 0.0375 \ mol$.Charge passed $Q = I 	imes t = 1.5 	imes 250 = 375 \ C$.Effective charge used for electrolysis $Q_{eff} = 375 	imes 0.85 = 318.75 \ C$.For $MSO_4 ightarrow M^{2+} + SO_4^{2-}$,the reaction is $M^{2+} + 2e^- ightarrow M(s)$. Thus,$n = 2$.Moles of $M^{2+}$ deposited $= \frac{Q_{eff}}{n 	imes F} = \frac{318.75}{2 	imes 96500} \approx 0.00165 \ mol$.Moles of $MSO_4$ remaining $= 0.0375 - 0.00165 = 0.03585 \ mol$.Final Molarity $= \frac{0.03585 \ mol}{0.25 \ L} = 0.1434 \ M$.The value is closest to $0.14 \ M$.

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