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, $I = 5 A$
time, $t = 40 \min $
$= 40 \times 60=2400 \, s $
Amount of electricity passed
$Q=It $
$Q=5 \times 2400 $
$Q=12000 \, C. $
$Zn^{2+}+2e^- \rightarrow 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, $12000\, C$ charge will deposite
$ =\frac {65.39 \times 12000}{2 \times 96500}$
$=4.065 \, g$ of zinc