Question

How many atoms of calcium will be deposited from a solution of $CaC{l}_2$ by a current of $5mA$ flowing for $60s$?

• A. $4.68\times 10^{18}$
• B. $4.68\times 10^{15}$
• C. $4.68\times 10^{12}$
• D. $4.68\times 10^9$

Answer 1

Answer: (A)

Given, $i=25mA=0.0025A$
$t=60s$
$Q=$ it
$=0.0025\times 60$
$=1.5C$
number of electrons in $1.5C$
$=\frac{Q\times \text{ Avogadro number }}{96500}$
$=\frac{1.5\times 6.023\times 10^{23}}{96500}$
$=9.36\times 10^{18}$
$Ca\to C{a}^{2+}+2e^{-}$
$2e^{-}$ are required to deposit $1Ca$ atom.
$\therefore$ number of $Ca$ atoms deposited
$=\frac{\text{ no. of electrons }}{2}$
$=\frac{9.36\times 10^{18}}{2}$
$=4.68\times 10^{18}$