Question

A current of 1.5 A flows through a copper voltameter. The thickness of copper deposited on the electrode surface of area 50cm in 20 minutes will be $($Density of copper $=9000$ kg/m${}^3$ and e.c.e. of copper $=$ 0.00033 g/C$)$

• A. $2.6\times {10}^{-5}m$
• B. $2.6\times {10}^{-4}m$
• C. $1.3\times {10}^{-5}m$
• D. $1.3\times {10}^{-4}m$

Answer 1

Answer: (C)

$1.3\times {10}^{-5}m$

Given that

$I=1.5$ A

$A=50c{m}^{-2}=50\times {10}^{-2}{m}^{-2}$

$t=20min=20\times 60$

$\rho =9000kg{m}^{-3}=9\times {10}^{3}kg{m}^{-3}$

e.c.e. of copper, $Z=0.00033g{C}^{-1}=3.3\times {10}^{-7}Kg{C}^{-1}$

The mass of deposited copper is

$m=ZIt$

$m=3.3\times {10}^{-7}\times 1.5\times 20\times 60$

$m=5.94\times {10}^{-5}Kg$

Now,

The thickness of copper deposited on the electrode surface is

$t=\frac{m}{A\rho }$

$t=\frac{5.94\times {10}^{-5}}{50\times {10}^{-2}\times 9\times {10}^3}$

$t=\frac{5.94}{450}\times {10}^{-6}$

$t=1.3\times {10}^{-5}m$