Question

___ A of current is required to be passed for a period of 5 minutes in order to deposit 0.6354gm of copper by the process of electrolysis of aqueous cupric sulphate solution. (Molar wt. of Copper = 63.5gm/mole)

Answer 1

Moles of copper to be obtained $=\frac{0.6354}{63.54}=0.01$ moles
$C{u}^{++}+2e^{-}\to Cu$
In order to obtain 1 mole of copper , 2 F of current is required .
so, to obtain $0.01$ moles of copper, 0.02 F of current is needed.
$Q=I\times t=0.02F=0.02\times 96500C=1930C$ of current.
Or, $I\times (5\times 60)=1930$
$\Rightarrow I=\frac{193}{30}=6.43A$