A current of 1.5, A flows through a copper vo | vedclass

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(C) The mass of copper deposited is given by Faraday's law of electrolysis: $m = Z i t$.Also,mass can be expressed as $m = 	ext{Density} (ho) 	ime

Vedclass · Accepted Answer

$1.3 	imes 10^{-5}\, m$

Vedclass · Answer

(C) The mass of copper deposited is given by Faraday's law of electrolysis: $m = Z i t$.Also,mass can be expressed as $m = 	ext{Density} (ho) 	imes 	ext{Volume} (V) = ho 	imes A 	imes x$,where $A$ is the area and $x$ is the thickness.Equating the two expressions: $ho A x = Z i t$.Therefore,the thickness $x = \frac{Z i t}{A ho}$.Given values:$Z = 0.00033\, g/C = 0.00033 	imes 10^{-3}\, kg/C = 3.3 	imes 10^{-7}\, kg/C$$i = 1.5\, A$$t = 20\, min = 20 	imes 60 = 1200\, s$$A = 50\, cm^2 = 50 	imes 10^{-4}\, m^2$$ho = 9000\, kg/m^3$Substituting the values:$x = \frac{0.00033 	imes 10^{-3} 	imes 1.5 	imes 1200}{50 	imes 10^{-4} 	imes 9000}$$x = \frac{0.00033 	imes 1.5 	imes 1.2}{50 	imes 9} = \frac{0.000594}{450} = 1.32 	imes 10^{-6}\, m$ (Wait,re-calculating: $x = \frac{3.3 	imes 10^{-7} 	imes 1.5 	imes 1200}{50 	imes 10^{-4} 	imes 9000} = \frac{5.94 	imes 10^{-4}}{45} = 1.32 	imes 10^{-5}\, m$).Thus,the thickness is approximately $1.3 	imes 10^{-5}\, m$.

$A$ current of $1.5\, A$ flows through a copper voltameter. The thickness of copper deposited on the electrode surface of area $50\, cm^2$ in $20\, min$ will be (Density of copper $= 9000\, kg/m^3$ and $E.C.E.$ of copper $= 0.00033\, g/C$).

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