If nearly 105C charge liberates 1 g equivalent of aluminium then the amount oflaluminium (equivalent weight= 9) deposited through electrolysis in 20 min by a current of 50 A will be

Question

If nearly 105C charge liberates 1 g equivalent of aluminium then the amount oflaluminium (equivalent weight= 9) deposited through electrolysis in 20 min by a current of 50 A will be (A) 0.6 g (B) 0.09 g (C) 5.4 g (D) 10.8 g

Tardigrade Admin · Accepted Answer

According to Faraday's law of electrolysis, the amount of mass deposited,

m = z

It where

z =

electrochemical equivalent of substance,

I =

current pass through it and

t =

time.
Here,

z = \frac{9}{1 0 ^{5}}

thus,

m = \frac{9}{1 0 ^{5}} \times 50 \times (20 \times 60) = 5.4 g m

Q. If nearly 105C charge liberates 1 g equivalent of aluminium then the amount oflaluminium (equivalent weight= 9) deposited through electrolysis in 20 min by a current of 50 A will be

0.6 g

0.09 g

5.4 g

10.8 g

According to Faraday's law of electrolysis, the amount of mass deposited, m=z It where z= electrochemical equivalent of substance, I= current pass through it and t= time. Here, z=1059​ thus, m=1059​×50×(20×60)=5.4gm

Q. If nearly $1 0^{5} C$ charge liberates 1 g equivalent of aluminium then the amount oflaluminium (equivalent weight= 9) deposited through electrolysis in 20 min by a current of 50 A will be

According to Faraday's law of electrolysis, the amount of mass deposited,
$m = z$ It where $z =$ electrochemical equivalent of substance,
$I =$ current pass through it and $t =$ time.
Here, $z = \frac{9}{1 0 ^{5}}$
thus, $m = \frac{9}{1 0 ^{5}} \times 50 \times (20 \times 60) = 5.4 g m$