Question

The time period to coat a metal surface of area $80{cm}^2$ with $5\times {10}^{-3}cm$ thick layer of silver(density $=1.05g{cm}^{-3}$) with the passage of $3A$ current through a silver nitrate solution.

• A. $115sec$
• B. $125sec$
• C. $135sec$
• D. $145sec$

Answer 1

Answer: (B)

$125sec$

Mass of silver required $=1.05\times 80\times 5\times {10}^{-3}=0.42g$

By Faraday's law, $w=\frac{M}{96500\times n_f}\times i\times t$

where $M=$ Molecular weight, $n_f=n-factor$

In this case, $n_f=1,M=108g,w=0.42g$

$\therefore 0.42=\frac{108}{96500\times 1}\times 3\times t$

$\therefore t=125sec.$