How much time (in sec) is required to coat a | vedclass

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(D) Volume of silver layer $= 	ext{Area} 	imes 	ext{Thickness} = 80 \, cm^2 	imes (0.005 	imes 10^{-1} \, cm) = 0.04 \, cm^3$.
Mass of silver $=

Vedclass · Accepted Answer

$125.09$

Vedclass · Answer

(D) Volume of silver layer $= 	ext{Area} 	imes 	ext{Thickness} = 80 \, cm^2 	imes (0.005 	imes 10^{-1} \, cm) = 0.04 \, cm^3$.
Mass of silver $= 	ext{Density} 	imes 	ext{Volume} = 10.5 \, g \, cm^{-3} 	imes 0.04 \, cm^3 = 0.42 \, g$.
Using Faraday's law of electrolysis: $w = \frac{E 	imes I 	imes t}{96500}$,where $E$ (equivalent weight of $Ag$) $= 108 \, g \, mol^{-1}$.
$0.42 = \frac{108 	imes 3 	imes t}{96500}$.
$t = \frac{0.42 	imes 96500}{108 	imes 3} = 125.09 \, sec$.

How much time (in $sec$) is required to coat a metal surface of $80 \, cm^2$ with a $0.005 \, mm$ thick layer of silver from a silver nitrate solution using a current of $3 \, A$? (Density of silver $= 10.5 \, g \, cm^{-3}$)

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