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)

4.065 g

Answer 2

Current, I = 5 A
time, t = 40 min = 40 $\times 60=2400 \, s$
Amount of electricity passed
\hspace15mm Q=It
\hspace15mm Q=5 \times 2400
\hspace15mm Q=12000 \, C.
\hspace12mm Zn^{2+}+2e^- \rightarrow Zn
\hspace15mm n=2e^-
From Faraday first law
\hspace12mm W=ZIt
\hspace12mm Z=equivalent \, mass
[65.39 = mass of zinc]
\hspace12mm = \frac {Mass}{e^- \times F}
\hspace12mm = \frac {65.39}{2 \times 96500} g \, of \, zinc
therefore, 12000 C charge will deposite
$\, \, \, \, \, =\frac {65.39 \times 12000}{2 \times 96500}=4.065 \, g \, of \, zinc$