Question

How long a current of 2 ampere should be applied through a
solution of silver nitrate solution to obtain 1.08 g Silver?

a) 965 second
b) 9650 second
c) 482.5 second
d) 4825 second​

Answer 1

Given:

Amount of current is 2 amperes. Amount of Silver deposited is 1.08 grams.

To find:

Time for which the current has to be applied to the silver nitrate solution .

Calculation:

$\boxed{\sf{{Ag}^{+}+{e}^{-}\longrightarrow Ag}}$

From the equation , we can easily understand that 1 Faraday of electricity is required to produce 1 mole of Ag (i.e. 108 gm of Ag)

So, amount of electricity required to produce 1.08 grams of Ag :

$= \dfrac{1.08}{108}$

$= {10}^{ - 2} \: F$

Let time taken be t :

$\therefore \: i = \dfrac{q}{t}$

$= > \: t = \dfrac{q}{i}$

$= > \: t = \dfrac{ {10}^{ - 2} \: F}{2 \: amp}$

$= > \: t = \dfrac{ {10}^{ - 2} \times 96500}{2 \: amp}$

$= > \: t = \dfrac{ 965}{2 \:}$

$= > \: t = 482.5 \: sec$

So, final answer is:

$\boxed{ \bold{ \large{\: t = 482.5 \: sec}}}$