Question

What amount of current is required to be passed during a period of 5 minutes in order to deposit 0.6354gm of copper by electrolysis of aqueous cupric sulphatesolution? (Molar wt. of Copper = 63.5gm/mole)

• A. 10.24 A
• B. 6.43 A
• C. 12.68 A
• D. 1930 C

Answer 1

The correct option is B 6.43 A

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 Felectricity is required .
so, to obtain $0.01$ moles of copper, 0.02 F of electricity will be needed.
$Q=I\times t=0.02F=0.02\times 96500C=1930C$ of electricity
Or, $I\times (5\times 60)=1930$
$\Rightarrow I=\frac{193}{30}=6.43A$