Question

A $5$ A current is passed through a solution of zinc sulphate for $40\min$. The amount of zinc deposited at the cathode is

• A. 40.65 g
• B. 0.4065 g
• C. 4.065 g
• D. 65.04 g

Answer 1

Answer: (C)

Current, $I=5A$
time, $t=40\min$
$=40\times 60=2400s$
Amount of electricity passed
$Q=It$
$Q=5\times 2400$
$Q=12000C.$
$Z{n}^{2+}+2e^{-}\to Zn$
$n=2e^{-}$
From Faraday first law
$W=ZIt$
$Z=$ equivalent mass
[$65.39=$ mass of zinc]
$=\frac{\text{Mass}}{e^{-}\times F}$
$=\frac{65.39}{2\times 96500}g$ of zinc
therefore, $12000C$ charge will deposite
$=\frac{65.39\times 12000}{2\times 96500}$
$=4.065g$ of zinc