Question

A current of 1.50 A was passed through an electrolytic cell containing AgNO3 solution with inert electrodes. The weight of the silver deposited was 1.50g. How long did the current flow?
(Molar mass of Ag=108gmol^-1 and 1F=96500 Cmol^-1)

Please give correct answer with explanation plzz!!

Pls don't copy paste from Internet..​

Answer 1

14.9 Minutes

Explanation:

Given:

Current = 1.5 Amperes

Weight of the silver deposited = 1.5 grams

Now Solving:

When Ag gains an electron:

Ag  +  e- ------------> Ag (108 gram or 1 mol)

1 mol or 108 gram of silver needs:

= 1 Faraday

= 96500 Coulombs

(As we know 1 Farday is 96500 Coulombs)

Therefore, 1.5 grams of silver needs =  $\frac{96500 * 1.50}{108}$

= 1340.3 Coulombs

Now using the Formula:

$\fbox{Q = I * T}$

Substituting the known values in the formula we get:

1340.3 = 1.5 Amp * Time (in seconds)

Time(in seconds) =  $\frac{1340.3}{1.5}$

= $\frac{13403}{15}$

= $893.5$ seconds

Converting seconds to minutes:

$\fbox{Mins = Seconds / 60}$

$Mins = \frac{893.5}{60}$

$Mins = \frac{8935}{60}$

∴ $14.9$ Minutes is how long the current flowed.

Answer 2

Answer:

14.9. ................... this is the answer