Question

If the density of copper is $8.94g/c{m}^3$, the quantity of elctricity required to plate an area $100C{M}^2$ of the thickness of ${10}^{-2}$ cm using $CuS{O}_4$ solution as electrolyte is:

Answer 1

1) Vol. of Cu to be plated $=\frac{100\times 1}{10}=10cc$

wt. of cu required $=8.94\times 10=89.4$

Moles of $Cu=\frac{89.4}{63.5}$

$C{u}^{2+}+2e^{-}⟶Cu$

Moles of electron $=89.4\times 10\times 2$

amount. of electricity$=\frac{89.4\times 10\times 2}{96500}$ Farads

Or

2) Volume $\left(V\right)$ of $Cu$ $=Area\times thickness=\frac{10\times 10\times 1}{102}=1$.

$d=\frac{m}{V}$

$m=d\times V=8.94\times 1=8.94$.

Hence, Actual mass of $Cu$ to be plated is $8.94g$.

3) Now, $C{u}^{2+}+2e^{-}⟶Cu$

So, $2F⟶63.5gCu$

$2\times 96500C⟶63.5g$ of $Cu$

$PC⟶8.94gCu$

$P=2\times 96500\times \frac{8.94}{96500}=27172.$