Question

A current of $40$ microampere is passed through silver nitrate solution for $16$ minutes using platinum electrodes. $50$% of the cathode is occupied by a single atom thick silver layer. Calculate the total surface area of the cathode is one silver atom occupies $5.5\times {10}^{-16}c{m}^2$ surface area.

Answer 1

Mass of silver deposited may be calculated according to Faraday's firs law of electrolysis.
$W=\frac{ItE}{96500}$
$=\frac{40\times {10}^{-6}\times 60\times 16\times 108}{96500}$
$-42.976\times {10}^{-6}g$
Total number of deposited ${}^{\prime }A{g}^{\prime }$ atoms
$=\frac{42.976\times {10}^{-6}}{108}\times 6.023\times {10}^{23}$
$=2.3967\times {10}^{17}atoms$
Surface occupied by deposited silver
$=$ Number of silver atoms $\times$ Area occupied by a single atom
$=2.3967\times {10}^{17}\times 5.5\times {10}^{-16}$
$=131.818c{m}^2$
Since, deposited silver occupies $50$% of total area of cathode, hence, total surface area of cathode $=2\times 131.818$
$=263.636c{m}^2$