Question

$Zn$ Amalgam is prepared by the electrolysis of aqueous $ZnC{l}_2$ using $Hg$ cathode (9 gm). How much current should be passed through $ZnC{l}_2$ solution for 1000 seconds to prepare a $Zn$ amalgam with 25% $Zn$ by weight ?$(Zn$atomic weight $=65.4)$

• A. 5.6 A
• B. 7.2 A
• C. 8.85 A
• D. 11.2 A

Answer 1

Answer: (C)

8.85 A

First calculate the quantity of zinc required.

Let $x$ be the total mass of amalgam.

$\text{Mass of mercury}=0.75x=9g⟹x=12g$

$\text{Mass of zinc}=0.25x=0.25\times 12g=3g$

Quantity of electricity passed$=Q\left(C\right)=I\left(A\right)\times t\left(s\right)=I\left(A\right)\times 1000s$

The atomic weight of zinc is 65.4 $\frac{g}{mol}$.

Moles of electrons passed $=\frac{Q\left(C\right)}{96500\frac{C}{mol{e}^{-}}}=\frac{I\left(A\right)\times 1000s}{96500\frac{C}{mol{e}^{-}}}$

The weight of zinc deposited,

$3g=\text{molar mass of zinc}\times \text{mole ratio}\times \text{moles of electrons passed}$

$3g=65.4\frac{g}{mol}\times \frac{1molzinc}{2mol{e}^{-}}\times \frac{I\left(A\right)\times 1000s}{96500\frac{C}{mol{e}^{-}}}$

$I\left(A\right)=8.85A$