Question

A plate of area $10$$c{m}^2$ is to be electroplated with Cu (density $9gm/cc$) to a thickness of $10$ $\mu m$ on both sides using a $12V$ cell. Calculate the energy spent by the cell in this process. (electric chemical equivalent of Cu is $3\times {10}^{-17}$ kg/C)

• A. $1.8$ $\times$ ${10}^{-13}$ $J$
• B. $3.6$ $\times$ ${10}^{-13}$ $J$
• C. $5.4$ $\times$ ${10}^{-13}$ $J$
• D. $7.2$ $\times$ ${10}^{-13}$ $J$

Answer 1

Answer: (D)

$7.2$ $\times$ ${10}^{-13}$ $J$

Area of the plate $A=10c{m}^2={10}^{-3}{m}^2$

Total thickness of Cu deposited $t=2\times 10\mu m=2\times {10}^{-5}m$

Density of Cu $\rho =9gm/cc=9000kg/m^3$

Thus volume of Cu deposited $V=At$

$\therefore$ $V={10}^{-3}\times 2\times {10}^{-5}=2\times {10}^{-8}{m}^3$

Mass of Cu deposited $m=\rho V$

$\therefore$ $m=9000\times 2\times {10}^{-8}=1.8\times {10}^{-4}kg$

Charge flowing during the process $q=\frac{1.8\times {10}^{-4}}{3\times {10}^{-17}}=6\times {10}^{12}$ C

Thus energy spent by the cell $E=q{V}_{cell}$

$\therefore$ $E=6\times {10}^{12}\times 12$

$⟹$ $E=7.2\times {10}^{13}J$