How long a current of 3 , A has to be passed | vedclass

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(D) Step $1$: Calculate the volume of the silver layer.$Volume = Area 	imes Thickness = 80 \, cm^2 	imes (0.005 \, mm 	imes 0.1 \, cm/mm) = 80 	im

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$125.09$

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(D) Step $1$: Calculate the volume of the silver layer.$Volume = Area 	imes Thickness = 80 \, cm^2 	imes (0.005 \, mm 	imes 0.1 \, cm/mm) = 80 	imes 0.0005 \, cm^3 = 0.04 \, cm^3$.Step $2$: Calculate the mass of silver deposited.$Mass = Volume 	imes Density = 0.04 \, cm^3 	imes 10.5 \, g/cm^3 = 0.42 \, g$.Step $3$: Use Faraday's law of electrolysis: $W = \frac{M 	imes I 	imes t}{n 	imes F}$.For $Ag^+ + e^- ightarrow Ag$,$n = 1$ and $M = 107.87 \, g/mol$.$0.42 = \frac{107.87 	imes 3 	imes t}{1 	imes 96500}$.$t = \frac{0.42 	imes 96500}{107.87 	imes 3} \approx 125.09 \, s$.

How long a current of $3 \, A$ has to be passed through a solution of $AgNO_3$ to coat a metal surface of $80 \, cm^2$ with $0.005 \, mm$ thick layer? Density of $Ag$ is $10.5 \, g/cm^3$. (Answer in $sec$)

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