Total surface area of a cathode is 0.05,m^2 a | vedclass

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(B) Mass deposited $(m)$ is given by the product of density $(\rho)$, area $(A)$, and thickness $(x)$: $m = \rho A x$ ... $(i)$According to Faraday's

Vedclass · Accepted Answer

$2.4\,\mu m$

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(B) Mass deposited $(m)$ is given by the product of density $(ho)$, area $(A)$, and thickness $(x)$: $m = ho A x$ ... $(i)$According to Faraday's first law of electrolysis, $m = ZIt$ ... $(ii)$Equating $(i)$ and $(ii)$, we get $ho A x = ZIt$, which implies $x = \frac{ZIt}{ho A}$.Given: $Z = 3.04 	imes 10^{-4}\,g/C$, $I = 1\,A$, $t = 1\,hour = 3600\,s$, $ho = 9\,g/cm^3 = 9000\,kg/m^3 = 9 	imes 10^6\,g/m^3$, and $A = 0.05\,m^2$.Substituting the values in $SI$ units (converting $Z$ to $kg/C$): $Z = 3.04 	imes 10^{-4} 	imes 10^{-3}\,kg/C = 3.04 	imes 10^{-7}\,kg/C$.$x = \frac{3.04 	imes 10^{-7} 	imes 1 	imes 3600}{9000 	imes 0.05} = \frac{1.0944 	imes 10^{-3}}{450} = 2.432 	imes 10^{-6}\,m \approx 2.4\,\mu m$.

Total surface area of a cathode is $0.05\,m^2$ and $1\,A$ current passes through it for $1\,hour$. The thickness of nickel deposited on the cathode is (Given that density of nickel $= 9\,g/cm^3$ and its $E.C.E. = 3.04 \times 10^{-4}\,g/C$)

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