Question

If the density of copper is $8.94g/{cm}^3$, the number of faradays required to plate an area $(10cm\times 10cm)$ of thickness of ${10}^{-2}$ cm using ${CuSO}_4$ solution as electrolyte is:

• A. $0.1F$
• B. $0.28F$
• C. $0.4F$
• D. $0.5F$

Answer 1

The correct option is B $0.28F$

Vol. of plate $=10\times 10\times \frac{1}{100}=1c{m}^3$

$\therefore$ density $=\frac{mass}{volume}$

$\therefore$ mass of $Cu$ to be deposited $=8.94g$

${Cu}^{2+}+2e^{-}\to Cu$

moles of $Cu=\frac{8.94}{63.5}$

Faraday required $=\frac{2\times 8.94}{63.5}=0.28F$